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Pre-requisite : Before going through this chapter, youshould be thorough with the basic concepts of thechapter explained in X NCERT.

CONSERVATION OF ELECTRIC CHARGE

Whenever two bodies are charged by rubbing, one getspositively charged and the other gets negativelycharged. The net charge on the two bodies, however,remains zero�the same as that before rubbing. In other

words, charge is conserved. It can neither be creatednor be destroyed. The only thing that happens onrubbing is that charged particles (electrons) gettransferred from one body to the other.

In some phenomena, charged particles are created.But even then the conservation of charge holds. Forexample, a free neutron converts itself into an electronand the proton taken together is also zero. So, there isno change in the conversion of a neutron to an electronand a proton.

COULOMB�S LAW

Charles Augustine de Coulomb studied the interactionforces of charged particles in detail in 1784. He used atorsion balance. On the basis of his experiments heestablished Coulomb�s law. According to this law the

magnitude of the electric force between two pointcharges is directly proportional to the product of themagnitude of the two charges and inverselyproportional to the square of the distance betweenthem and acts along the straight line joining the twocharges.

In mathematical terms, the force that each of the twopoint charges q1 and q2 at a distance r apart exerts onthe other can be expressed as�

F = 221

r

qqk

This force is repulsive for like charges and attractivefor unlike charges.

Where k is a constant of proportionality. k = 04

1

,

here 0 is absolute permittivity of free space.The force is directed along the line joining the centresof the two charged particles.For any other medium except air, free space or vacuumcoulomb�s law reduces to

F = 1 22

q q14 r

= Permittivity of the mediumand = 0r

r = relative Permittivity or dielectric constant of the me-dium.Coulomb�s law is based on physical observation and

it is not logically derived from any other concept.

ILLUSTRATIONS

1. Find out the electrostatics force between two pointcharges placed in air (each of +1 C) if they areseparated by 1m .

Sol. Fe = 2

21

r

qkq = 2

9

1

11109 = 9×109 N

Note : From the above result we can say that 1 Ccharge is too large to realize. In nature, charge isusually of the order of C

2. A particle of mass m carrying charge q1 is revolving

around a fixed charge �q2 in a circular path of radius

r. Calculate the period of revolution and its speedalso.

Sol.04

1 2

21

r

qq = mr2 = 2

2

T

mr4'

T2 = 21

220

qq

)mr4(r)4(

or T = 4r 21

0

qq

mr

and also we can say that

20

21

r4

qq

=

rmv2

V = mr4

qq

0

21

PROPERTIES OF ELECTRIC FIELD INTENSITY

(i) It is a vector quantity. Its direction is the sameas the force experienced by positive charge.

(ii) Electric field due to positive charge is alwaysaway from it while due to negative charge alwaystowards it.

(iii) Its S.. unit is Newton/Coulomb.

(iv) Electric force on a charge q placed in a regionof electric field at a point where the electric field

intensity is E

is given by EqF

.

Electric force on point charge is in the samedirection of electric field on positive chargeand in opposite direction on a negative charge.

ELECTRICITY

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(v) It obeys the superposition principle, that is, thefield intensity at a point due to a system of charges isvector sum of the field intensities due to individualpoint charges.

321 EEEE

+ .....

(vi) It is produced by source charges. The electricfield will be a fixed value at a point unless we changethe distribution of source charges.

3. Five point charges, each of value q are placed on fivevertices of a regular hexagon of side L. What is themagnitude of the force on a point charge of value �

q

coulomb placed at the centre of the hexagon?

qq C

qqDE

O

-q

q

BA

F

L

Sol. If there had been a sixth charge +q at the remainingvertex of hexagon force due to all the six charges on �q at O would be zero (as the forces due to individualcharges will balance each other), i.e.,

0FR

Now if f

is the force due to sixth charge and F

due toremaining five charges.

F

+ f

= 0 i.e. F

= � f

or, |F| = |f| = 04

1 2L

qq = 2

2

0 L

q4

1

NetF

= COF

= 2

2

L

q4

1

along CO

4. Calculate the electric field intensity which would bejust sufficient to balance the weight of a particle ofcharge �10 c and mass 10 mg.

Sol. As force on a charge q in an electric field E

is

F

q = q E

So according to given problem

q E

W

A

Fe

|W| |F| q

i.e., |q|E = mg

i.e., E = |q|mg

= 10 N/C., in downward direction.

ELECTROSTATIC EQUILIBRIUM

The position where the resultant force on a chargedparticle becomes zero is called equilibrium position.

(a) Stable Equilibrium :

A charge is initially in equilibrium position and isdisplaced by a small distance. If the charge tries toreturn back to the same equilibrium position then thisequilibrium is called position of stable equilibrium.

(b) Unstable Equilibrium :

If charge is displaced by a small distance from itsequilibrium position and the charge has no tendencyto return to the same equilibrium position. Instead itgoes away from the equilibrium position.

(c) Neutral Equilibrium :

If charge is displaced by a small distance and it is stillin equilibrium condition then it is called neutralequilibrium.

5. Two equal positive point charges 'Q' are fixed at pointsB(a, 0) and A(�a, 0). Another test charge q

0 is also

placed at O(0, 0). Show that the equilibrium at 'O' is(i) stable for displacement along X-axis.(ii) unstable for displacement along Y-axis.

Sol. (i)

Initially AOF

+ BOF

= 0 |F| AO

= |F| BO

= 2

0

a

KQq

When charge is slightly shifted towards + x axis by asmall distance x, then.

|F| AO

< |F| BO

3PAGE # 3

Therefore the particle will move towards origin (itsoriginal position) hence the equilibrium is stable.(ii) When charge is shifted along y axis

After resolving components net force will be along yaxis so the particle will not return to its originalposition so it is unstable equilibrium. Finally thecharge will move to infinity.

ELECTRIC LINES OF FORCE (ELOF)

The line of force in an electric field is an imaginaryline, the tangent to which at any point on itrepresents the direction of electric field at the givenpoint.

(a) Properties :

(i) Line of force originates out from a positivecharge and terminates on a negative charge. Ifthere is only one positive charge then lines startfrom positive charge and terminate at . If thereis only one negative charge then lines start from and terminates at negative charge.

(ii) The electric intensity at a point is the numberof lines of force streaming through per unit areanormal to the direction of the intensity at that point.The intensity will be more where the density oflines is more.

(iii) Number of lines originating (terminating) from(on) is directly proportional to the magnitude ofthe charge.

(iv) ELOF of resultant electric field can never intersectwith each other.

(v) Electric lines of force produced by staticcharges do not form close loop.

(vi ) E lec tr ic l ines of force end or startperpendicularly on the surface of a conductor.

(vii) Electric l ines of force never enter intoconductors.

6. If number of electric lines of force from charge qare 10 then find out number of electric lines offorce from 2q charge.

Sol. No. of ELOF charge10 q 20 2qSo number of ELOF will be 20.

7. A charge + Q is fixed at a distance of d in front ofan infinite metal plate. Draw the lines of forceindicating the directions clearly.

4PAGE # 4

Sol. There will be induced charge on two surfaces ofconducting plate, so ELOF will start from +Q chargeand terminate at conductor and then will again startfrom other surface of conductor.

ELECTRIC FLUX

Consider some surface in an electric field E

. Let us

select a small area element dS on this surface.

The electric flux of the field over the

area element is given by dE =

ds.E

Direction of dS is normal to the surface. It is along

n�or d

E = EdS cos

or dE = (E cos ) dS

dS E

or dE = E

n dS

where En is the component of electric field in the

direction of dS .

If the electric field is uniform over that area then

E = SE

(a) Physical Meaning :

The electric flux through a surface inside an electricfield represents the total number of electric lines offorce crossing the surface in a direction normal thesurface. It is a property of electric field

(b) Unit :

(i) The SI unit of electric flux is Nm2 C�1 (gauss) or Jm2 C�1.

(ii) Electric flux is a scalar quantity. (It can be positive,negative or zero)

8. The electric field in a region is given by,

jE54

iE53

E 00

with E0 = 2.0 × 103 N/C. Find the

flux of this field through a rectangular surface of area0.2m2 parallel to the Y�Z plane.

Sol. E = SE

=

jE

5

4iE

5

300

. i�2.0 = CmN

2402

ELECTRIC POTENTIAL ENERGY

Consider a charge Q placed at a point P as shown infigure. If another charge q of the same sign is nowbrought from a very far away distance (infinity) to pointO near P, then charge q will experience a force ofrepulsion due to charge Q. If charge q is still pushedtowards P, work is done. This work done is the potentialenergy of the system of these two charges.

Q

P

q

O

r q

From infinity

Thus, the electric potential energy of a system ofcharges is defined as the amount of work done inbringing the various charges from infinite separationto their present positions to form the requiredsystem. It is denoted by U. For the system of twocharges separated by distance r as shown in figure,the electric potential energy is given by :

U = r

kQq

Electric potential energy is the from of energy, thereforeit is measured in joule (J).

SUPER CONDUCTOR AND ITS APPLICATIONS

Prof. K. Onnes in 1911 discovered that certain metalsand alloys at very low temperature lose theirresistance considerably. This phenomenon is knownas super-conductivity. As the temperature decreases,the resistance of the material also decreases, but whenthe temperature reaches a certain critical value (calledcritical temperature or transition temperature), theresistance of the material completelydisappears i.e. it becomes zero. Then the materialbehaves as if it is a super-conductor and there will beflow of electrons without any resistance whatsoever.The critical temperature is different for different material.It has been found that mercury at critical temperature4.2 K, lead at 7.25 K and niobium at critical temperature9.2 K become super-conductor.

Applications of super conductors :

(i) Super conductors are used for making very strongelectromagnets.

(ii) Super conductivity is playing an important role inmaterial science research and high energy particlephysics.(iii) Super conductivity is used to produce very highspeed computers.

(iv) Super conductors are used for the transmission ofelectric power.

5PAGE # 5

CELL

It converts chemical energy into electrical energy.Electrochemical cells are of three types :(a) Primary cell (b) Secondary cell(c) Fuel cell

(a) Primary Cell :

It is an electrochemical cell, which cannot be recharged,but the chemicals have to be replaced after a long use.The reactions taking place in the cell are irreversible.

Eg. : Daniel cell, Lechlanche cell, Dry cell etc.

(b) Secondary Cell :

Electrical energy can be converted into chemical energy

and chemical energy can be converted into electrical

energy in these cells, i.e. secondary cells can be

recharged after use. Chemical reaction taking place in

these cells are reversible.

Example : Lead accumulator, Edison cell (alkali cell)

and iron nickel cell.

(c) Electro Motive Force of a Cell (E.M.F.) :

It is the maximum potential difference between the twoelectrodes of the cell when no current is drawn fromthe cell or cell is in the open circuit.

(d) Potential Difference of a Cell :

It is the difference of potential between two terminalsof the cell when current is drawn from it or the cell is inclosed circuit.

(e) Internal Resistance of a Cell :

It is the resistance offered to the flow of current insidethe cell i.e. internal resistance is the resistance offeredto the flow of current by electrolyte. Internal resistancedecreases with the increase of the area of plates andalso with the decrease of the distance between plates.

Determination of internal resistance of a cell :

Connect a voltmeter to a cell through key K1. Also

connect a resistor R to cell through K2. First put in key

K1. The reading shown by voltmeter gives us the e.m.f.

of the cell since negligible current flows through celldue to high resistance of the voltmeter. Insert key K

2

also so that current flows through resistor R. If r is theinternal resistance of the cell and V is the readingshown by voltmeter, then

I = rR

E

E = I (R + r)

E = IR + IrHere, IR = V the potential differenceSo, E = V + r

r =I

V�E.........(i)

V = IR or I = RV

So for equation (i)

r = VV)R�(E

........(ii)

GROUPING OF CELLS

(a) Cells in Series :

BE ,r1 1 E ,r2 2 E ,r3 3

E rn n

A BE ,req eq

Equivalent EMFE

eq = E

1 + E

2 + ......... + E

n (write EMF�s with polarity)

Equivalent internal resistance req

= r1 + r

2 + r

3 + r

4 + ........ r

n

If n cells each of emf E, are arranged in series and if r isinternal resistance of each cell, then total emf = nE

R

E,r E,r E,r E,r

IUpto n

A B

So current in the circuit, I = nrR

nE

There may by two cases :

(i) If nr << R, then I =RnE

= n × current due to one cell.

So, series combination is advantageous.

(ii) If nr >> R, then I =rE

= current due to one cell.

So, Series combination is not advantageous.

Note : If polarity of m cells is reversed, then equivalente.m.f. = (n�2m) E while the equivalent resistance is

still nr + R, so current in R will be

i = Rnr

E)m2n(

6PAGE # 6

(b) Cells in Parallel :

If m cells each of emf E and internal resistance r beconnected in parallel and if this combination isconnected to an external resistance then the emf ofthe circuit is E.

Internal resistance of the circuit = mr

.

and I = rmRmE

mr

R

E

There may by two cases :

(i) If mR << r, then I =r

mE = m × current due to one cell.

So, Parallel combination is advantageous.

(ii) If mR >> r, then I = RE

= current due to one cell.

So, parallel combination is not advantageous.

If emf and internal resistances of each cell are different,then,

Eeq =n21

nn2211

r1r1r1rErErE

/.....///.....//

for two cells E = 21

1221

rrrErE

(Use emf with polarity)

E1

E2

E3

En

r2

r3

rn

r1

(c) Cells in Multiple Arc :

n = number of rowsm = number of cells in each row.

mn = N (total number of identical cells) :

The combination of cells is equivalent to single cell of

emf = mE and internal resistance = n

mr

Current I =

nmr

R

mE

For maximum current, nR = mr

or R = n

mr = internal resistance of the equivalent

battery.

Imax

= R2

mEr2

nE .

using mn = N in above equation we get number of

rows n = RNr

9. 9 cells, each having the same emf and 3 ohm internalresistance, are used to draw maximum current throughan external resistance of 3 ohm. find the combinationof cells.

Sol. For the condition of maximum current number of rows

n = RNr

so n = 3

39 = 3

so combination will be like 3 rows and 3 cells in eachrow.

BATTERY

Battery is an arrangement that creates a constantpotential difference between its terminals. It is acombination of a number of cells in series.

The impact of battery :

With the discovery of voltaic cell, it was soon realisedthat if one constructs a number of cells and joins thenegative terminal of one with the positive terminal ofthe other and so on, then the emf (which is the potentialdifference between the electrodes in an open circuit)of the combination of cells will be the sum of the emf�sof the individual cells. This observation led to a burst ofscientific activity in 1802. Humphrey Davy, an Englishchemist, made a battery of 60 pairs of zinc and copperplates. The large emf thus produced, was used to gethigh current, which could melt iron and platinum wires.By 1807, he had a battery of almost 300 plates withwhich he was able to decompose chemical salts. Thisled to the discovery of new elements.

7PAGE # 7

By 1808, Davy had assembled 2,000 pairs of plates.With this battery, he created electric arcs andsucceeded in extracting the elements like barium,calcium and magnesium from their compounds. Thus,electricity took a front seat in exploring the nature ofmatter.

ELECTRICAL RESISTANCE

The property of a substance by virtue of which itopposes the flow of electric current through it istermed as electrical resistance. Electrical resistancedepends on the size, geometry, temperature andinternal structure of the conductor.

We known that, vd =

meE

=

meV

I = Anevd

= Anem

eV

I = mVAne2

2Ane

mIV

R =

2Ane

mIV

R = 2Ane

m

R = A

=

RA

= 2ne

m

is called resistivity (it is also called specific

resistance), and =

2ne

m =

1, is cal led

conductivity. Therefore current in conductors isproportional to potential difference applied across

its ends. This is Ohm's Law. Units: 11m

also called siemens m�1.

10. If a copper wire is stretched to make its radiusdecrease by 0.15%, Find the percentage increasein resistance (approximately).

Sol. Due to stretching resistance changes are in the ratio4

2

1

1

2

r

r

R

R

or 4rR

orrr

4RR

%15.04 = 0.60%

EFFECT OF STRETCHING OF A WIRE ON RESISTANCE

In stretching, the density of wire usually does notchange. Therefore

Volume before stretching = Volume after stretching

2211 AA

and2

1

1

2

1

2

AA

RR

If information of lengths before and after stretching

is given, then use 1

2

2

1

AA

2

1

2

1

2

R

R

If information of radius r1 and r

2 is given then use

2

1

1

2

AA

2

2

1

1

2

A

A

R

R

4

2

1

r

r

CONDUCTIVITY :

(a) Reciprocal of resistivity of a conductor is calledits conductivity. It is generally represented by .

(b)

1

(c) Unit : 11 metre.ohm

EFFECT OF TEMPERATURE ON RESISTANCE

AND RESISTIVITY

The resistance of a conductor depends upon thetemperature. As the temperature increases, therandom motion of free electrons also increases. Ifthe number density of charge carrier electronsremains constant as in the case of a conductor, thenthe increase of random motion increases theresistivity. The variation of resistance with temperatureis given by the following relation

20t tt1RR

8PAGE # 8

where Rt and R

0 are the resistance at t0C and 00C

respectively and and are constants. The

constant is very small so its may be assumed

negligible.

t1RR 0t

or tR

RR

0

0t

This constant is called as temperature coefficientof resistance of the substance.

If R0 = 1 ohm, t = 10C, then

0t RR

Thus, the temperature coefficient of resistance isequal to the increase in resistance of a conductorhaving a resistance of one ohm on raising itstemperature by 10C. The temperature coefficient ofresistance may be positive or negative.

From calculations it is found that for most of the

metals the value of is nearly C/273

1 0. Hence

substituting in the above equation

273t

1RR 0t

273

TR

273

t273R 00

where T is the absolute temperature of the conductor.

TR t

Thus, the resistance of a pure metallic wire is directlyproportional to its absolute temperature.

The graph drawn between the resistance Rt and

temperature t is found to be a straight line

Rt

tºC

R0

The resistivity or specific resistance varies withtemperature. This variation is due to change inresistance of a conductor with temperature. Thedependence of the resistivity with temperature isrepresented by the following equation.

t10t

With the rise of temperature the specific resistanceor resistivity of pure metals increases and that ofsemi-conductors and insulators decrease.

The resistivity of alloys increases with the rise oftemperature but less than that of metals.

On applying pressure on pure metals, its resistivitydecreases but on applying tension, the resistivityincreases.

The resistance of alloys such as eureka, manganinetc., increases in smaller amount with the rise intemperature. Their temperature coefficient ofresistance is negligible. On account of their highresistivity and negligible temperature coefficient ofresistance these alloys are used to make wires forresistance boxes, potentiometer, meter bridge etc.,

The resistance of semiconductors, insulators,electrolytes etc., decreases with the rise intemperature. Their temperature coefficients ofresistance are negative.

On increasing the temperature of semi conductors alarge number of electrons get free after breaking theirbonds. These electrons reach the conduction bandfrom valence band. Thus conductivity increases orresistivity decreases with the increase of free electrondensity.

11. A wire has a resistance of 2 ohm at 273 K and aresistance of 2.5 ohm at 373 K. What isthe temperature coefficient of resistance of thematerial?

Sol. 273373225.2

TTR

RR

00

0

K/105.2200

5.0 03

WHEATSTONE BRIDGE

Wheatstone bridge is an arrangement of four resistorsin the shape of a quadrilateral which can be used tomeasure unknown resistance in terms of theremaining three resistances.

The arrangement of Wheatstone bridge is shown infigure below. Out of four resistors, two resistances R1,R2 and R3, R4 are connected in series and are joined inparallel across two points a and c. A battery of emf E isconnected across junctions a and c and a galvanometer(G) between junction b and d. The keys K

1 and K

2 are

used for the flow of current in the various branches ofbridge.

Principle of Wheatstone Bridge :

When key K1 is closed, current i from the battery is

divided at junction a in two parts. A part i1 goes throughR1 and the rest i2 goes through R3. When key K

2 is

closed, galvanometer shows a deflection.

9PAGE # 9

The direction of deflection depends on the value of

potential difference between b and d. When the value

of potential at b and d is same, then no current will flow

through galvanometer. This condition is known as the

condition of balanced bridge or null deflection

condition. This situation can be obtained by choosing

suitable values of the resistances. Thus, in null

deflection state, we have :

Va � V

b = V

a � V

d

or i1 R

1 = i

2 R

3...(i)

Similarly :

Vb � V

c = V

d � V

c

or i1 R

2 = i

2 R

4...(ii)

On dividing equation (i) by (ii), we get

21

11

RiRi

= 42

32

Ri

Rior

4

3

2

1

R

R

RR

...(iii)

Equation (iii) states the condition of balanced bridge.

Thus, in null deflection condition the ratio of

resistances of adjacent arms of the bridge are same.

The resistor of unknown resistance is usually

connected in one of the arm of the bridge. The

resistance of one of the remaining three arms is

adjusted such that the galvanometer shows zero

deflection. If resistance of unknown resistor is R4. Then

R4 = (R3)

1

2

RR

For better accuracy of the bridge one should choose

resistances R1, R2, R3 and R4 of same order.

GALVANOMETER

Galvanometer is a simple device, used to detect thecurrent, to find direction of current and also to comparethe currents.With the help of galvanometer we make twoimportant devises known as Ammeter and voltmeteras discussed below.

(a) Ammeter :

Ammeter is an electrical instrument which measuresthe strength of current in �ampere� in a circuit. Ammeter

is a pivoted coil galvanometer which is alwaysconnected in series in circuit so that total current (to bemeasured) may pass through it. For an ammeter ofgood quality, the resistance of its coil should be verylow so that it may measure the strength of currentaccurately (without affecting the current passingthrough the circuit). The resistance of an ideal ammeteris zero (practically it should be minimum). So, tominimize the effective resistance of an ammeter, a lowvalue resistance (shunt) as per requirement isconnected in parallel to the galvanometer to convert itto ammeter of desired range.In electric circuit, the positive terminal of an ammeteris connected to positive plate and negative terminal isconnected to negative plate of battery.

Desired value of shunt depends on the range(measurable maximum current) of ammeter convertedfrom galvanometer.If pivoted galvanometer of resistance G is to measurecurrent i (as an ammeter) then from figure.

Gi ig ig i

is S

ig G = (i � i

g) S or S = )ii(

Gi

g

g

Where ig is an amount of current required for fulldeflection in galvanometer. By using a low value ofresistance S (shunt) in parallel to the galvanometer(resistance G), the effective resistance of

converted ammeter RA = S)(GGS

becomes very low..

NOTE :

Shunt : If anyhow, the flowing current throughgalvanometer becomes more than its capacity, the coilhas possibility of burning due to heat produced byflowing current. Secondly, its pointer may break up dueto impact with �stop pin� as its proportional deflection

as per amount of flowing current.

In order to minimize these possibilities a low resistancewire (or strip) is connected in parallel with galvanometer,which is known as shunt.

(b) Voltmeter :

It is an electrical instrument which measures thepotential difference in �volt� between two points of

electric circuit. It�s construction is similar as that of

ammeter. The only difference between ammeter andvoltmeter is that ammeter has its negligible(approximately zero) resistance so that it may measurecurrent of circuit passing through it more accuratelygiving the deflection accordingly, while the voltmeterpasses negligible current through itself so that potentialdifference developed due to maximum current passing

10PAGE # 10

through circuit may be measured. Therefore, anappropriate value of high resistance is required to beconnected in series of galvanometer to convert it into avoltmeter of desired range.Voltmeter is connected in parallel to the electric circuit.

If a galvanometer of resistance G is to be convertedinto a voltmeter of range V, then required value of highresistance RH will be

V = ig (R

H + G)

or RH =

g

VI � G

G

i

ig

RH

V

Connecting this value of high resistance in the seriesof galvanometer, it will be converted to a voltmeter ofrange V. After connecting high resistance RH in seriesof galvanometer of resistance G, the effective resistanceof voltmeter becomes RV = (RH + G) very high (high incomparison to G).

Ideal voltmeter has infinite resistance of its own. Whenideal voltmeter is connected parallel to a part of anelectric circuit, it passes zero amount of current throughitself from the circuit so that measurement of potentialdifference across the points of connection may beperfectly accurate.

KIRCHHOFF'S LAWS

(a) Kirchhoff�s Current Law (Junction law) :

This law is based on law of conservation of charge.It states that "The algebraic sum of the currentsmeeting at a point of the circuit is zero" or totalcurrent entering a junction is equal total currentleaving the junction.

in = out.

It is also known as KCL (Kirchhoff's current law).

(b) Kirchhof f �s Voltage Law (Loop law) :

�The algebraic sum of all the potential differences

along a closed loop is zero. IR + EMF =0�. The

closed loop can be traversed in any direction. Whiletraversing a loop if potential increases, put a positivesign in expression and if potential decreases puta negative sign.

V1

V2 + V

3 V

4 = 0. Boxes may contain resistor

or battery or any other element (linear or nonlinear).It is also known as KVL

12. Figure shows, current in a part of electrical circuit, whatwill be the value of current (i) ?

2 A 1 A

1.3 A

i 3 A

2 A

P Q RS

Sol.

2 A 1 A

1.3 A

i 3 A

2 A

P Q RS

i 1i 2

i 3

From KCL, current at junction P, i1 = 2 + 3 = 5 A

From KCL, current at junction Q, i2 = i

1 + 1 = 5 + 1 = 6 A

From KCL, current at junction R, i3 = i

2 � 2 = 6 � 2 = 4 A

From KCL, current at junction S, i = i3 � 1.3 = 4 � 1.3

= 2.7 A

13. In the circuit shown, calculate the value of R in ohmthat will result in no current through the 30 V battery.

Sol. Applying KVL in loop CEFDC50 � iR � 20 i = 0

i = R20

50

20R

50V

10

i

i E

FD iB

A C

Potential drop across R = Potential drop across ABiR = 30

R20

50

.R = 30

R = 30

11PAGE # 11

EXERCISE

Charge and coulomb�s Law :

1. Conductivity of superconductor is :(A) infinite (B) very large(C) very small (D) zero

2. Two charges of +1 C & + 5 C are placed 4 cm apart,the ratio of the force exerted by both charges on eachother will be -(A) 1 : 1 (B) 1 : 5(C) 5 : 1 (D) 25 : 1

3. A body has 80 microcoulomb of charge. Number ofadditional electrons on it will be :(A) 8 x 10�5 (B) 80 x 1015

(C) 5 x 1014 (D) 1.28 x 10�17

4. Which of the following relation is wrong ?(A) Q = It

(B) 1 ampere = Second1Coulomb1

(C) V = Wq

(D) V = qW

5. Two particles having charges q1 and q

2 when kept at a

certain distance, exert force F on each other. If distanceis reduced to half, force between them becomes :

(A) 2F

(B) 2F

(C) 4F (D) 4F

6.254

Coulomb of charge contains.............................

electrons :(A) 1015 (B) 1018

(C) 1020 (D) none of these7. 5 charges each of magnitude 10�5 C and mass 1 kg

are placed (fixed) symmetrically about a movablecentral charges of magnitude 5 × 10�5C and mass 0.5kg as shown. The charges at P

1 is removed. The

acceleration of the central charge is : (KVPY/2009)

P1

P2

P3 P4

P5

O

[Given OP1 = OP

2 = OP

3 = OP

4 = OP

5 1 m ;

041

= 9 ×

109 in SI units](A) 9 m s�2 upwards (B) 9 m s�2 downwards(C) 4.5 m s�2 upwards (D) 4.5 m s�2 downwards

8. 12 positive charges of magnitude q are placed on a

circle of radius R in a manner that they are equally

spaced. A charge +Q is placed at the centre. If one of

the charges q is removed, then the force on Q is :

(KVPY/2010)(A) zero

(B) 20R4

qQ

away from the position of the removed

charge.

(C) 20R4

qQ11

away from the position of the removed

charge.

(D) 20R4

qQ

towards the position of the removed

charge.

9. In a neon discharge tube 2.8 × 1018 Ne+ ions move to

the right per second while 1.2 ×1018 electrons move to

the left per second. Therefore , the current in the

discharge tube is : (IJSO/Stage-I/2011)(A) 0.64 A towards right

(B) 0.256 A towards right

(C) 0.64 A towards left

(D) 0.256 A towards left

Electric filed and Potential :

10. If Q = 2 coloumb and force on it is F = 100 newton,

then the value of field intensity will be:

(A) 100 N/C (B) 50 N/C

(C) 200 N/C (D) 10 N/C

11. In the electric field of charge Q, another charge is

carried from A to B. A to C, A to D and A to E, then work

done will be -

BQ +

A

C D E

centre

(A) minimum along path AB.

(B) minimum along path AD.

(C) minimum along path AE.

(D) zero along all the paths.

12PAGE # 12

12. A negatively charged particle initially at rest is placed in

an electric field that varies from point to point. There

are no other fields. Then : (KVPY/2008)

(A) the particle moves along the electric line of force

passing through it.

(B) the particle moves opposite to the electric line of

force passing through it.

(C) the direction of acceleration of the particle is

tangential to the electric line of force at every instant.

(D) the direction of acceleration of the particle is normal

to the electric line of force at every instant.

13. Two charges +q and �q are placed at a distance b

apart as shown in the figure below. (KVPY/2009)

b/2

b

+q �q

B

AP

C

The electric field at a point P on the perpendicular

bisector as shown as :

(A) along vector A (B) along vector B

(C) along vector C (D) Zero

14. Two charges +Q and _2Q are located at points A and B

on a horizontal line as shown below :

The electric field is zero at a point which is located at a

finite distance : (KVPY/2011)

(A) On the perpendicular bisector of AB

(B) left of A on the line

(C) between A and B on the line

(D) right of B on the line

Resistance :

15. There are three resistance 5, 6and 8connected in

parallel to a battery of 15 V and of negligible resistance.

The potential drop across 6resistance is :

(A) 10 V (B) 15 V

(C) 20 V (D) 8 V

16. In the given circuit, the equivalent resistance between

points A and B will be.

(A) 38

R (B) 4R

(C) 6R (D) 10R

17. Resistance of a conductor of length 75 cm is 3.25 .What will be the length of a similar conductor whoseresistance is 13.25 ?(A) 305.76 cm (B) 503.76 cm(C) 200 cm (D) 610 cm

18. A piece of wire of resistance 4 is bent through 1800

at its mid point and the two halves are twisted together,then resistance is :(A) 1 (B) 2 (C) 5 (D) 8

19. In how many parts (equal) a wire of 100 be cut sothat a resistance of 1 is obtained by connectingthem in parallel ?(A) 10 (B) 5(C) 100 (D) 50

20. If a wire of resistance 1 is stretched to double itslength, then the resistance will become :

(A) 21

(B) 2

(C) 41 (D) 4

21. Two copper wires, one of length 1 m and the other oflength 9 m, are found to have the same resistance.Their diameters are in the ratio :(A) 3 : 1 (B) 1 : 9(C) 9 : 1 (D) 1 : 3

22. Reading of ammeter in ampere for the following circuitis :

(A) 0.8 (B) 1(C) 0.4 (D) 2

13PAGE # 13

23. Two resistors are joined in series, then their equivalentresistance is 90 . When the same resistors arejoined in parallel, the equivalent resistance is 20 .The resistances of the two resistors will be :(A) 70 , 20 (B) 80 , 10 (C) 60 , 30 (D) 50 , 40

24. In the ladder network shown, current through theresistor 3

is 0.25 A. The input voltage �V� is equal

to

(A) 10 V (B) 20 V

(C) 5 V (D) 2

15 V

25. The reading of voltmeter is

(A) 50V (B) 60 V(C) 40V (D) 80 V

26.

2

525

10

1.4A1.4A

A

(A) 0.4 (B) 1(C) 0.6 (D) 1.2

27. Three identical bulbs are connected in parallel with abattery. The current drawn from the battery is 6 A. If one of the bulbs gets fused, what will be the totalcurrent drawn from the battery ?(A) 6A (B) 2A(C) 4A (D)

28. A uniform wire of resistance R is uniformly compressedalong its length, until its radius becomes n times theoriginal radius. Now, the resistance of the wirebecomes :(A) R/n (B) R/n4

(C) R/n2 (D) n R

29. The resistance of a wire of cross-section �a� and length

� � is R ohm. The resistance of another wire of the

same material and of the same length but cross-sec-tion �4a� will be

(A) 4R (B) R4

(C) R16

(D) 16 R

30. In the following circuit the value of total resistance be-tween X and Y in ohm is :

r to rr

r r r

rrrX

Y

(A) (1 + 3 )R (B) ( 3 � 1)R

(C) (D) 50 r

31. Wires A and B are made from the same material. WireA has length 12m and weight 50 g, while wire B is 18 mlong and weighs 40 g. Then the ratio (R

A / R

B) of their

resistances will be : (IAO/Jr./Stage-I/2008)(A) 16 / 45 (B) 4 / 5(C) 8 / 15 (D) 4 / 9

32. In case of the circuit arrangement shown below, the

equivalent resistance between A and B is :

(IAO/Jr./Stage-I/2009)

A B

(A) 10 (B) 2.5 W

(C) 340

W (D) None of the above

33. The net resistance between points P and Q in the

circuit shown in fig. is

(A) R/2 (B) 2R/5

(C) 3R/5 (D) R/3

14PAGE # 14

34. A wire of resistance 10.0 ohm is stretched so as to

increase its length by 20%. Its resistance then would

be : (IAO/Sr/Stage-I/2008)

(A) 10.0 ohm (B) 12.0 ohm

(C) 14.4 ohm (D) 10.2 ohm

35. In the circuit shown below, all the resistances are equal,

each equal to R. The equivalent resistance between

points A and C is : (IAO/Sr./Stage-I/2009)

R R

R

RR

RR

R

A D

CB

(A) R (B) 4R

(C) R /2 (D) none of the above

36. A battery or 10 V and negligible internal resistance is

connected across the diagonally opposite corner of a

cubical network consisting of 12 resistors each of

resistance 1 . The total current 1 in the circuit external

to the network is : (KVPY/2007)

A

10V

(A) 0.83 A (B) 12 A

(C) 1 A (D) 4 A

37. Figure (a) below shows a Wheatstone bridge in whichP, Q, R, S are fixed resistances, G is a galvanometerand B is a battery. For this particular case thegalvanometer shows zero deflection. Now, only thepositions of B and G are interchanged,. as shown infigure (b). The new deflection of the galvanometer.

(KVPY/2010)

(A) is to the left.(B) is to the right.(C) is zero.(D) depends on the values of P, Q, R, S

38. In the circuit arrangement shown, if the point A and B arejoined by a wire the current in this wire will be :

(IJSO/Stage-I/2011)

A

B

24Volt

(A) 1A. (B) 2A.(C) 4A. (D) zero.

39. In the following circuit, each resistor has a resistance

of 15 and the battery has an e.m.f. of 12 V with

negligible internal resistance. (IJSO/Stage-II/2011)

When a resistor of resistance R is connected betweenD & F, no current flows through the galvanometer (notshown in the figure) connected between C & F.Calculate the value of R.(A) 10 (B) 15

(C) 5 (D) 30

40. The circuit given below is for the operation of anindustrial fan. The resistance of the fan is 3 ohms. Theregulator provided with the fan is a fixed resistor and avariable resistor in parallel. (IJSO/Stage-II/2011)

Under what value of the variable resistance givenbelow, Power transferred to the fans will be maximum?The power source of the fan is a dc source with internalresistnace of 6 ohm.(A) 3 (B) 0(C) (D) 6

15PAGE # 15

41. When all the resistances in the circuit are 1each,

then the equivalent resistance across points A & B will

be : (IJSO/Stage-II/2011)

(A) 5/6 (B) 1/2

(C) 2/3 (D) 1/3

42. A cylindrical copper rod has length L and resistance R.

If it is melted and formed into another rod of length 2L.

the resistance will be : (KVPY/2011)(A) R (B) 2R

(C) 4R (D) 8R

43. There are four resistors of 12 ohm each. Which of thefollowing values is/are possible by their combinations

(series and / or parallel) ? (IJSO/Stage-I/2008)(A) 9 ohm (B) 16 ohm

(C) 12 ohm (D) 30 ohm

44. In case of the circuit shown below, which of the followingstatements is/are true ? (IJSO/Stage-I/2009)

�

+ �

�R1 R2 R3

AB

2

1

3

4

(A) R1, R

2 and R

3 are in series.

(B) R2 and R

3 are in series.

(C) R2 and R

3 are in parallel.

(D) The equivalent resistance of the circuit is

R1+

32

32

RRRR

45. A current i reaching at a point in a circuit gets branchedand flows through two resistors R

1 and R

2. Then, the

current through R1 varies as : (IAO/Jr./Stage-I/2007)

(A) R1

(B) R2

(C) (R1+ R

2) (D) 1l (R

1 + R

2)

46. In the circuit shown below, (IAO/Sr./Stage-I/2007)

10V

Y

X

(A) current flowing in the circuit is 200 mA(B) power supplied by the battery is 2 watt(C) current from X to Y is zero(D) potential difference across 10 is equal to zero

47. We are given n resistors, each of resistance R. Theratio of the maximum to minimum resistance that canbe obtained by combining them is : (KVPY/2008)(A) nn (B) n(C) n2 (D) logn

Cell :

48. A cell of emf E is connected across a resistance R.The potential difference between the terminals of thecell is found to be V. The internal resistance of the cellis given as :

(A) R(E � V) (B) R

VE

(C) ER

)VE( (D) R

V)VE(

49. 24 cells, each having the same e.m.f. and 2 ohminternal resistance, are used to draw maximum currentthrough an external resistance of 3 ohm. The cellsshould be connected(A) in series(B) in parallel(C) in 4 rows, each row having 6 cells(D) in 6 rows, each row having 4 cells

50. The cells are joined in parallel to get the maximumcurrent when(A) external resistance is very large as compared tothe total internal resistance(B) internal resistance is very large as compared tothe external resistance(C) internal resistance and external resistance areequal(D) emf of each cell is very large

51. In secondary cells :(A) Chemical changes can be reversed by heatingelectrodes(B) Chemical changes can be reversed by passingelectric current(C) Current is produced by photo chemical reactions(D) None of these

16PAGE # 16

52. Three types of electric cells which provide current are :(A) Button cell, solar cell & secondary cell(B) Solar cell, electrolytic cell, electro chemical cell(C) (A) and (B) both are correct(D) Neither (A) nor (B) is correct

53. In which of the following cells, the potential differencebetween the terminals of a cell exceeds its emf.

(A) a (B) b(C) c (D) d

54. A cell, an ammeter and a voltmeter are all connected inseries. The ammeter reads a current I and thevoltmeter a potential difference V. If a torch bulb isconnected across the voltmeter, then.

(IJSO/Stage-I/2009)(A) both I and V will increase(B) both I and V will decrease(C) I will increase but V will decrease(D) I will decrease but V will increase

55. In the process of electrostatic induction.

(IJSO/Stage-II/2011)(A) a conductor is rubbed with an insulator.

(B) a charge is produced by friction.

(C) negative and positive charges are separated.

(D) electrons are �sprayed� on the object.

56. Consider the circuit below. The bulb will light up if :(KVPY/2009)S1

S2

S3~

(A) S1 S

2 and S

3 are all closed.

(B) S1 is closed but S

2 and S

3 are open.

(C) S1 and S

3 are closed but S

2 is open.

(D) none of these

Electric Energy and Power :

57. An electric iron of heating element of resistance 88 is used at 220 volt for 2 hours. The electric energyspent, in unit, will be :(A) 0.8 (B) 1.1(C) 2.2 (D) 8.8

58. Two identical heater wires are first connected in seriesand then in parallel with a source of electricity. Theratio of heat produced in the two cases is :(A) 2 : 1 (B) 1 : 2(C) 4 : 1 (D) 1 : 4

59. You are given three bulbs 25 W, 40 W and 60 W . Whichof them has the lowest resistance?(A) 25 watt bulb (B) 40 watt bulb(C) 60 watt bulb (D) insufficient data

60. If R1 and R

2 are the filament resistances of a 200 W

bulb and a 100 W bulb respectively designed to operateon the same voltage, then :(A) R

1 = 2 R

2(B) R

2 = 2 R

1

(C) R2 = 4 R

1(D) R

1 = 4 R

2

61. If two bulbs, whose resistance are in the ratio of 1 : 2,are connected in series. The power dissipated in themhas the ratio of :(A) 1 : 1 (B) 1 : 2(C) 2 : 1 (D) 1 : 4

62. When a voltage of 20 volt is applied between the twoends of a coil, 800 cal/s heat is produced. The value ofresistance of the coil is :(1 calorie = 4.2 joule) :(A) 1.2 (B) 1.4 (C) 0.12 (D) 0.14

63. You are given two fuse wires A and B with current rating2.5 A and 6 A respectively. Which of the two wires wouldyou select for use with a 1100 W, 220 V room heater ?(A) A (B) B(C) A and B (D) none of these

64. An electric current of 2.0 A passes through a wire ofresistance 25 . How much heat (in joule) will bedeveloped in 1 minute ?(A) 6 (B) 6000(C) 50 (D) 10

65 . Two bulbs, one of 200W and the other of 100W, areconnected in series with a 100 V battery which has nointernal resistance. Then, (KVPY/2009)

200W 100W

100V

(A) the current passing through the 200W bulb is morethan that through the 100W bulb.(B) the power dissipation in the 200W bulb is morethan that In the 100 W bulb.(C) the voltage drop across the 200W bulb is morethan that across the 100W bulb.(D) the power dissipation In the 100W bulb is morethan that in the 200W bulb.

17PAGE # 17

66. An electric heater consists of a nichrome coil and runs

under 220 V, consuming 1 kW power. Part of its coil

burned out and it was reconnected after cutting off the

burnt portion. The power it will consume now is :

(KVPY/2010)(A) more than 1 kW.

(B) less that 1 kW, but not zero.

(C) 1 kW.

(D) 0 kW.

67. In the following circuit, the 1 resistor dissipates

power P. If the resistor is replaced by 9. the power

dissipated in it is : (KVPY/2011)

(A) P (B) 3P

(C) 9P (D) P/3

68. A neon lamp is connected to a voltage a.c. source. The

voltage is gradually increased from zero volt. It is

observed that the neon flashes at 50 V. The a.c. source

is now replaced by a variable dc source and the

experiment is repeated. The neon bulb will flash at :

(IAO/Sr./Stage-I/2008)(A) 50V (B) 70V

(C) 100V (D) 35V

69. A certain network consists of two ideal and indentical

voltage sources in series and a large number of ideal

resistor. The power consumed in one of the resistor is

4W when either of the two sources is active and other

is replaced by a short circuit. The power consumed by

same resistor when both sources are simultaneously

active would be : (IJSO/Stage-II/2011)(A) 0 or 16W (B) 4W or 8W

(C) 0 or 8W (D) 8W or 16W

Circuit and Other :

70. A galvanometer can be converted into a voltmeter by

connecting

(A) A high resistance in series with the galvanometer

(B) A high resistance in parallel with the galvanometer

(C) a low resistance in series with the galvanometer

(D) a low resistance in parallel with the galvanometer

71. The circuit shown has 3 identical light bulbs A, B, Cand 2 identical batteries E

1, E

2 . When the switch is

open, A and B glow with equal brightness. When theswitch is closed: (KVPY/2007)

A

B

C

S

E1

E2

(A) A and B will maintain their brightness and C will bedimmer than A and B.(B) A and B will become dimmer and C will be brighterthan A and B.(C) A and B will maintain their brightness and C will notglow.(D) A, B and C will be equally bright.

72. A student connects two lamps in the circuit shown.

The emf of the two batteries is different.

(IJSO/Stage-II/2011)

Which of the following statements are correct?

i. When keys 1, 2, 3 and 4 are closed, bulbs

A and B will both glow

ii. When key 2 and 4 are closed bulb A will glow

iii.When 1 and 4 are closed, bulb A will glow

iv.When 2, 3 and 4 are closed, both A and B will

glow(A) only ii (B) only iv

(C) i, ii and iv (D) ii and iii

73. Figure below shows a portion of an electric circuit withthe currents in ampere and their directions. Themagnitude and direction of the current in the portionPQ is : (KVPY/2011)

(A) 0A (B) 3A from P to Q(C) 4A from Q to P (D) 6A from Q to P

PAGE # 18

MOLE CONCEPT

ATOMS

All the matter is made up of atoms. An atom is thesmallest particle of an element that can take part in achemical reaction. Atoms of most of the elementsare very reactive and do not exist in the free state (assingle atom).They exist in combination with the atomsof the same element or another element.Atoms are very, very small in size. The size of an atomis indicated by its radius which is called "atomicradius" (radius of an atom). Atomic radius ismeasured in "nanometres"(nm).1 metre = 109 nanometre or 1nm = 10-9 m.Atoms are so small that we cannot see them underthe most powerful optical microscope.

Note :

Hydrogen atom is the smallest atom of all , having anatomic radius 0.037nm.

(a) Symbols of Elements :

A symbol is a short hand notation of an element whichcan be represented by a sketch or letter etc.Dalton was the first to use symbols to representelements in a short way but Dalton's symbols forelement were difficult to draw and inconvenient touse, so Dalton's symbols are only of historicalimportance. They are not used at all.

It was J.J. Berzelius who proposed the modernsystem of representing en element.The symbol of an element is the "first letter" or the"first letter and another letter" of the English name orthe Latin name of the element.

e.g. The symbol of Hydrogen is H.The symbol of Oxygen is O.There are some elements whose names begin withthe same letter. For example, the names of elementsCarbon, Calcium, Chlorine and Copper all begin withthe letter C. In such cases, one of the elements isgiven a "one letter "symbol but all other elements aregiven a "first letter and another letter" symbol of theEnglish or Latin name of the element. This is to benoted that "another letter" may or may not be the"second letter" of the name. Thus,The symbol of Carbon is C.The symbol of Calcium is Ca.The symbol of Chlorine is Cl.

The symbol of Copper is Cu (from its Latin nameCuprum)

It should be noted that in a "two letter" symbol, thefirst letter is the "capital letter" but the second letter isthe small letter

English Name of the Element

Symbol

Hydrogen HHelium HeLithium LiBoron B

Carbon CNitrogen NOxygen OFluorine F

Neon NeMagnesium MgAluminium Al

Silicon SiPhosphorous P

Sulphur SChlorine ClArgon Ar

Calcium Ca

Symbol Derived from English Names

Symbols Derived from Latin Names

English Name of the Element

SymbolLatin Name of the Element

Sodium Na Natrium

Potassium K Kalium

(b) Significance of The Symbol of an

Element :

(i) Symbol represents name of the element.

(ii) Symbol represents one atom of the element.

(iii) Symbol also represents one mole of the element.That is, symbol also represent 6.023 × 1023 atoms ofthe element.

(iv) Symbol represent a definite mass of the element

i.e. atomic mass of the element.

Example :(i) Symbol H represents hydrogen element.

(ii) Symbol H also represents one atom of hydrogenelement.

(iii) Symbol H also represents one mole of hydrogenatom.(iv) Symbol H also represents one gram hydrogen

atom.

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PAGE # 19

IONS

An ion is a positively or negatively charged atom orgroup of atoms.Every atom contains equal number of electrons(negatively charged) and protons (positively charged).Both charges balance each other, hence atom iselectrically neutral.

(a) Cation :

If an atom has less electrons than a neutral atom,then it gets positively charged and a positivelycharged ion is known as cation.e.g. Sodium ion (Na+), Magnesium ion (Mg2+) etc.A cation bears that much units of positive charge asthere are the number of electrons lost by the neutralatom to form that cation.

e.g. An aluminium atom loses 3 electrons to formaluminium ion, so aluminium ion bears 3 units ofpositive charge and it is represented as Al3+.

(b) Anion :

If an atom has more number of electrons than that ofneutral atom, then it gets negatively charged and anegatively charged ion is known as anion.e.g. Chloride ion (Cl¯), oxide ion (O2-) etc.

An anion bears that much units of negative charge asthere are the number of electrons gained by theneutral atom to form that anion.e.g. A nitrogen atom gains 3 electrons to form nitrideion, so nitride ion bears 3 units of negative chargeand it is represented as N3-.

Note :Size of a cation is always smaller and anion is alwaysgreater than that of the corresponding neutral atom.

(c) Monoatomic ions and polyatomic ions :(i) Monoatomic ions : Those ions which are formedfrom single atoms are called monoatomic ions orsimple ions.e.g. Na+, Mg2+ etc.

(ii) Polyatomic ions : Those ions which are formedfrom group of atoms joined together are calledpolyatomic ions or compound ions.e.g. Ammonium ion (NH

4+) , hydroxide ion (OH�) etc.

which are formed by the joining of two types of atoms,nitrogen and hydrogen in the first case and oxygen andhydrogen in the second.

(d) Valency of ions :The valency of an ion is same as the charge presenton the ion.If an ion has 1 unit of positive charge, its valency is 1and it is known as a monovalent cation. If an ion has2 units of negative charge, its valency is 2 and it isknown as a divalent anion.

LIST OF COMMON ELECTROVALENT POSITIVE RADICALS

LIST OF COMMON ELECTROVALENT NEGATIVE RADICALS

Monovalent Electronegative Bivalent Electronegative

Trivalent Electronegative

Tetravalent Electronegative

1. Fluoride F� 1. Sulphate SO 42- 1. Nitride N3- 1. Carbide C4-

2. Chloride Cl� 2. Sulphite SO 32- 2. Phosphide P3-

3. Bromide Br� 3. Sulphide S2-3. Phosphite PO3

3-

4. Iodide I 4. Thiosulphate S2O32- 4. Phosphate PO4

3-

5. Hydride H� 5. Zincate ZnO22-

6. Hydroxide OH� 6. Oxide O2-

7. Nitrite NO2� 7. Peroxide O2

2-

8.Nitrate NO3� 8. Dichromate Cr2O7

2-

9. Bicarbonate or Hydrogen carbonate HCO3� 9. Carbonate CO3

2-

10. Bisulphite or Hydrogen sulphite HSO3� 10. Silicate SiO 3

2-

11. Bisulphide or Hydrogen sulphide HS�

12. Bisulphate or Hydrogen sulphate HSO4�

13. Acetate CH COO3�

Note :Cation contains less no. of electrons and anion contains more no. of electrons than the no. of protons present inthem.

PAGE # 20

LAWS OF CHEMICAL COMBINATION

The laws of chemical combination are theexperimental laws which led to the idea ofatoms being the smallest unit of matter. The laws ofchemical combination played a significant role in the

development of Dalton�s atomic theory of matter.

There are two important laws of chemical combination.These are:

(i) Law of conservation of mass(ii) Law of constant proportions

(a) Law of Conservation of Mass or Matter :

This law was given by Lavoisier in 1774 . According tothe law of conservation of mass, matter can neitherbe created nor be destroyed in a chemical reaction.

OrThe law of conservation of mass means that in achemical reaction, the total mass of products is equalto the total mass of the reactants. There is no changein mass during a chemical reaction.Suppose we carry out a chemical reaction between Aand B and if the products formed are C and D then,A + B C + D

Suppose 'a' g of A and 'b' g of B react to produce 'c' g ofC and 'd' g of D. Then, according to the law ofconservation of mass, we have,

a + b = c + dExample :When Calcium Carbonate (CaCO

3) is heated, a

chemical reaction takes place to form Calcium Oxide(CaO) and Carbon dioxide (CO

2). It has been found

by experiments that if 100 grams of calcium carbonateis decomposed completely, then 56 grams of CalciumOxide and 44 grams of Carbon dioxide are formed.

Since the total mass of products (100g ) is equal tothe total mass of the reactants (100g), there is nochange in the mass during this chemical reaction.The mass remains same or conserved.

(b) Law of Constant Proportions / Law of

Definite Proportions :

Proust, in 1779, analysed the chemical composition(types of elements present and percentage ofelements present ) of a large number of compoundsand came to the conclusion that the proportion ofeach element in a compound is constant (or fixed).According to the law of constant proportions: Achemical compound always consists of the sameelements combined together in the same proportionby mass.

Note :The chemical composition of a pure substance isnot dependent on the source from which it is obtained.

Example :

Water is a compound of hydrogen and oxygen. It canbe obtained from various sources (like river, sea, welletc.) or even synthesized in the laboratory. Fromwhatever source we may get it, 9 parts by weight ofwater is always found to contain 1 part by weight ofhydrogen and 8 parts by weight of oxygen. Thus, inwater, this proportion of hydrogen and oxygen alwaysremains constant.

Note :The converse of Law of definite proportions that whensame elements combine in the same proportion, thesame compound will be formed, is not always true.

(c) Law of Multiple Proportions :

According to it, when one element combines with theother element to form two or more different compounds,the mass of one element, which combines with aconstant mass of the other, bears a simple ratio toone another.

Simple ratio means the ratio between small naturalnumbers, such as 1 : 1, 1 : 2, 1 : 3e.g.Carbon and oxygen when combine, can form twooxides that are CO (carbon monoxide), CO

2 (carbon

dioxide).In CO,12 g carbon combined with 16 g of oxygen.In CO

2,12 g carbon combined with 32 g of oxygen.

Thus, we can see the mass of oxygen which combinewith a constant mass of carbon (12 g) bear simpleratio of 16 : 32 or 1 : 2

Note :The law of multiple proportion was given by Dalton in1808.

Sample Problem :

1. Carbon is found to form two oxides, which contain42.8% and 27.27% of carbon respectively. Show thatthese figures illustrate the law of multiple proportions.

Sol. % of carbon in first oxide = 42.8% of oxygen in first oxide = 100 - 42.8 = 57.2% of carbon in second oxide = 27.27 % of oxygen in second oxide = 100 - 27.27 = 72.73

For the first oxide -Mass of oxygen in grams that combines with 42.8 gof carbon = 57.2 Mass of oxygen that combines with 1 g of carbon =

1.3442.857.2

g

For the second oxide -Mass of oxygen in grams that combines with 27.27 gof carbon = 72.73 Mass of oxygen that combines with 1 g of carbon =

2.6827.2772.73

g

Ratio between the masses of oxygen that combinewith a fixed mass (1 g) of carbon in the two oxides= 1.34 : 2.68 or 1 : 2 which is a simple ratio. Hence,this illustrates the law of multiple proportion.

PAGE # 21

(d) Law of Reciprocal Proportions :

According to this law the ratio of the weights of twoelement A and B which combine separately with afixed weight of the third element C is either the sameor some simple multiple of the ratio of the weights inwhich A and B combine directly with each other.e.g.

The elements C and O combine separately with thethird element H to form CH

4 and H

2O and they combine

directly with each other to form CO2.

H 24

CH4 H O2

C O16

12

12

CO2 32

In CH4, 12 parts by weight of carbon combine with 4

parts by weight of hydrogen. In H2O, 2 parts by weight

of hydrogen combine with 16 parts by weight ofoxygen. Thus the weight of C and O which combinewith fixed weight of hydrogen (say 4 parts by weight)are 12 and 32 i.e. they are in the ratio 12 : 32 or 3 : 8.Now in CO

2, 12 parts by weight of carbon combine

directly with 32 parts by weight of oxygen i.e. theycombine directly in the ratio 12 : 32 or 3 : 8 which isthe same as the first ratio.

Note :

The law of reciprocal proportion was put forward byRitcher in 1794.

Sample Problem :2. Ammonia contains 82.35% of nitrogen and 17.65%

of hydrogen. Water contains 88.90% of oxygen and11.10% of hydrogen. Nitrogen trioxide contains63.15% of oxygen and 36.85% of nitrogen. Show thatthese data illustrate the law of reciprocal proportions.

Sol

In NH3, 17.65 g of H combine with N = 82.35 g

1g of H combine with N = 17.65

82.35 g = 4.67 g

In H2O, 11.10 g of H combine with O = 88.90 g

1 g H combine with O = 11.10

88.90g = 8.01 g

Ratio of the weights of N and O which combinewith fixed weight (=1g) of H= 4.67 : 8.01 = 1 : 1.72

In N2O

3, ratio of weights of N and O which combine

with each other = 36.85 : 63.15 = 1 : 1.71

Thus the two ratio are the same. Hence it illustratesthe law of reciprocal proportions.

(e) Gay Lussac�s Law of Gaseous Volumes :

Gay Lussac found that there exists a definiterelationship among the volumes of the gaseousreactants and their products. In 1808, he put forwarda generalization known as the Gay Lussac�s Law of

combining volumes. This may be stated as follows :

When gases react together, they always do so in

volumes which bear a simple ratio to one another

and to the volumes of the product, if these are also

gases, provided all measurements of volumes are

done under similar conditions of temperature and

pressure.

e.g.Combination between hydrogen and chlorine to form

hydrogen chloride gas. One volume of hydrogen and

one volume of chlorine always combine to form two

volumes of hydrochloric acid gas.

H2 (g) + Cl

2 (g) 2HCl (g)

1vol. 1 vol. 2 vol.

The ratio between the volume of the reactants and

the product in this reaction is simple, i.e., 1 : 1 : 2.

Hence it illustrates the Law of combining volumes.

(f) Avogadro�s Hypothesis :

This states that equal volumes of all gases under

similar conditions of temperature and pressure

contain equal number of molecules.

This hypothesis has been found to explain elegantly

all the gaseous reactions and is now widely

recognized as a law or a principle known as Avogadro�s

Law or Avogadro�s principle.

The reaction between hydrogen and chlorine can be

explained on the basis of Avogadro�s Law as follows :

Hydrogen + Chlorine Hydrogen chloride gas 1 vol. (By experiment)1 vol. 2 vol.

n molecules. n molecules. 2n molecules.(By Avogadro's Law)

21 molecules. molecules. 1 molecules. (By dividing throughout by 2n)

1 Atom 1 Atom 1 Molecule (Applying Avogadro's hypothesis)

21

It implies that one molecule of hydrogen chloride gas

is made up of 1 atom of hydrogen and 1 atom of

chlorine.

(i) Applications of Avogadro�s hypothesis :

(A) In the calculation of atomicity of elementarygases.

e.g.

2 volumes of hydrogen combine with 1 volume of

oxygen to form two volumes of water vapours.Hydrogen + Oxygen Water vapours2 vol. 1 vol. 2 vol.

PAGE # 22

Applying Avogadro�s hypothesis

Hydrogen + Oxygen Water vapours2 n molecules n molecules 2 n molecules

or 1 molecule 2

1 molecule 1 molecule

Thus1 molecule of water contains 2

1 molecule of

oxygen. But 1 molecule of water contains 1 atom of

oxygen. Hence. 2

1 molecule of oxygen = 1 atom of

oxygen or 1 molecules of oxygen = 2 atoms of oxygeni.e. atomicity of oxygen = 2.

(B) To find the relationship between molecular massand vapour density of a gas.

Vapour density (V.D.) = hydrogenofDensitygas ofDensity

=

epressur and temp. same the at hydrogen of volume same the of Mass

gas the of volume certain a of Mass

If n molecules are present in the given volume of a gasand hydrogen under similar conditions of temperatureand pressure.

V.D. = hydrogen of molecules n of Massgas the of molecules n of Mass

= hydrogen of molecule 1 of Massgas the of molecule 1 of Mass

= hydrogen of mass Moleculargas the of mass Molecular

= 2

mass Molecular

(since molecular mass of hydrogen is 2)Hence, Molecular mass = 2 × Vapour density

ATOMIC MASS UNIT

The atomic mass unit (amu) is equal to one-twelfth(1/12) of the mass of an atom of carbon-12.The massof an atom of carbon-12 isotope was given the atomicmass of 12 units, i.e. 12 amu or 12 u.The atomic masses of all other elements are nowexpressed in atomic mass units.

RELATIVE ATOMIC MASS

The atomic mass of an element is a relative quantityand it is the mass of one atom of the element relativeto one -twelfth (1/12) of the mass of one carbon-12atom. Thus, Relative atomic mass

= atom12Coneofmass

12

1

elementtheofatomoneofMass

[1/12 the mass of one C-12 atom = 1 amu, 1 amu =1.66 × 10�24 g = 1.66 × 10�27 kg.]

Note :One amu is also called one dalton (Da).

GRAM-ATOMIC MASS

The atomic mass of an element expressed in grams

is called the Gram Atomic Mass of the element.

The number of gram -atoms

= elementtheofmassAtomicGram

gramsinelementtheofMass

e.g.

Calculate the gram atoms present in (i) 16g of oxygen

and (ii) 64g of sulphur.

(i) The atomic mass of oxygen = 16.

Gram-Atomic Mass of oxygen (O) = 16 g.

No. of Gram-Atoms = 1616

= 1

(ii) The gram-atoms present in 64 grams of sulphur.

= sulphurofMass AtomicGram64

= 32

64= 2

AtomicNumber Element Symbol

Atomicmass

1 Hydrogen H 12 Helium He 43 Lithium Li 74 Beryllium Be 95 Boron B 116 Carbon C 127 Nitrogen N 148 Oxygen O 169 Fluorine F 1910 Neon Ne 2011 Sodium Na 2312 Magnesium Mg 2413 Aluminium Al 2714 Silicon Si 2815 Phosphorus P 3116 Sulphur S 3217 Chlorine Cl 35.518 Argon Ar 4019 Potassium K 3920 Calcium Ca 40

RELATIVE MOLECULAR MASS

The relative molecular mass of a substance is themass of a molecule of the substance as comparedto one-twelfth of the mass of one carbon -12 atomi.e.,Relative molecular mass

= atom12Coneofmass

12

1

substancetheofmoleculeoneofMass

The molecular mass of a molecule, thus, representsthe number of times it is heavier than 1/12 of themass of an atom of carbon-12 isotope.

PAGE # 23

GRAM MOLECULAR MASS

The molecular mass of a substance expressed in

grams is called the Gram Molecular Mass of the

substance . The number of gram molecules

= ancetsubstheofmassmolecularGram

gramsintancesubstheofMass

e.g.

(i) Molecular mass of hydrogen (H2) = 2u.

Gram Molecular Mass of hydrogen (H2) = 2 g .

(ii) Molecular mass of methane (CH4) = 16u

Gram Molecular Mass of methane (CH4) = 16 g.

e.g. the number of gram molecules present in 64 g of

methane (CH4).

= 4CHofmassmolecularGram

64 =

16

64 = 4.

(a) Calculation of Molecular Mass :

The molecular mass of a substance is the sum of

the atomic masses of its constituent atoms present

in a molecule.

Ex.1 Calculate the molecular mass of water.

(Atomic masses : H = 1u, O = 16u).

Sol. The molecular formula of water is H2O.

Molecular mass of water = ( 2 × atomic mass of H)

+ (1 × atomic mass of O)

= 2 × 1 + 1 × 16 = 18

i.e., molecular mass of water = 18 amu.

Ex.2 Find out the molecular mass of sulphuric acid.

(Atomic mass : H = 1u, O = 16u, S = 32u).

Sol. The molecular formula of sulphuric acid is H2SO

4.

Molecular mass of H2SO

4

= (2 × atomic mass of H) + ( 1 × atomic mass of S)

+ ( 4 × atomic mass of O)

= (2 × 1) + (1× 32) + (4×16) = 2 + 32 + 64 = 98

i.e., Molecular mass of H2SO

4= 98 amu.

FORMULA MASS

The term �formula mass� is used for ionic compounds

and others where discrete molecules do not exist,

e.g., sodium chloride, which is best represented as

(Na+Cl�)n, but for reasons of simplicity as NaCl or

Na+Cl�. Here, formula mass means the sum of the

masses of all the species in the formula.

Thus, the formula mass of sodium chloride = (atomic

mass of sodium) + (atomic mass of chlorine)

= 23 + 35.5

= 58.5 amu

EQUIVALENT MASS

(a) Definition :

Equivalent mass of an element is the mass of theelement which combine with or displaces 1.008 partsby mass of hydrogen or 8 parts by mass of oxygen or35.5 parts by mass of chlorine.

(b) Formulae of Equivalent Masses of differentsubstances :(i) Equivalent mass of an element =

element the ofValency element the of wt.Atomic

(ii) Eq. mass an acid = acid the ofBasicity acid the of wt. Mol.

Basicity is the number of replaceable H+ ions fromone molecule of the acid.

(iii) Eq. Mass of a base = base the ofAcidity base the of wt. Mol.

Acidity is the number of replaceable OH� ions fromone molecule of the base

(iv) Eq. mass of a salt

= metal ofvalency atoms metal of Numbersalt the of wt. Mol.

(v) Eq. mass of an ion = ion the on Charge

ion the of wt.Formula

(vi) Eq. mass of an oxidizing/reducing agent

=

substance the ofmolecule

oneby gained or lost electrons of No.

wt At. or wt.Mol

Equivalent weight of some compounds are given inthe table :

S.No. CompoundEquivalent

weight

1 HCl 36.5

2 H2SO4 49

3 HNO3 63

4 45

5 .2H2O63

6 NaOH 40

7 KOH 56

8 CaCO3 50

9 NaCl 58.5

10 Na2CO3 53

COOH

COOH

PAGE # 24

In Latin, mole means heap or collection or pile. Amole of atoms is a collection of atoms whose totalmass is the number of grams equal to the atomicmass in magnitude. Since an equal number of molesof different elements contain an equal number ofatoms, it becomes convenient to express theamounts of the elements in terms of moles. A molerepresents a definite number of particles, viz, atoms,molecules, ions or electrons. This definite number iscalled the Avogadro Number (now called the Avogadroconstant) which is equal to 6.023 × 1023.

A mole is defined as the amount of a substance thatcontains as many atoms, molecules, ions, electronsor other elementary particles as there are atoms inexactly 12 g of carbon -12 (12C).

(a) Moles of Atoms :

(i) 1 mole atoms of any element occupy a mass whichis equal to the Gram Atomic Mass of that element.

e.g. 1 Mole of oxygen atoms weigh equal to GramAtomic Mass of oxygen, i.e. 16 grams.

(ii) The symbol of an element represents 6.023 x 1023

atoms (1 mole of atoms) of that element.

e.g : Symbol N represents 1 mole of nitrogen atomsand 2N represents 2 moles of nitrogen atoms.

(b) Moles of Molecules :

(i) 1 mole molecules of any substance occupy a masswhich is equal to the Gram Molecular Mass of thatsubstance.

e.g. : 1 mole of water (H2O) molecules weigh equal toGram Molecular Mass of water (H2O), i.e. 18 grams.

(ii) The symbol of a compound represents 6.023 x1023 molecules (1 mole of molecules) of thatcompound.

e.g. : Symbol H2O represents 1 mole of watermolecules and 2 H2O represents 2 moles of watermolecules.

Note :

The symbol H2O does not represent 1 mole of H2molecules and 1 mole of O atoms. Instead, itrepresents 2 moles of hydrogen atoms and 1 moleof oxygen atoms.

Note :The SI unit of the amount of a substance is Mole.

(c) Mole in Terms of Volume :

Volume occupied by 1 Gram Molecular Mass or 1mole of a gas under standard conditions oftemperature and pressure (273 K and 1atm.pressure) is called Gram Molecular Volume. Its valueis 22.4 litres for each gas.Volume of 1 mole = 22.4 litre (at STP)

Note :The term mole was introduced by Ostwald in 1896.

SOME IMPORTANT RELATIONS AND FORMULAE

(i) 1 mole of atoms = Gram Atomic mass = mass of6.023 × 1023 atoms

(ii) 1 mole of molecules = Gram Molecular Mass= 6.023 x 1023 molecules(iii) Number of moles of atoms

= elementofMassAtomicGram

gramsinelementofMass

(iv) Number of moles of molecules

= substanceofMassMolecularGram

gramsinsubstanceofMass

(v) Number of moles of molecules

= AN

N

numberAvogadro

elementofmolecules ofNo.

Ex.3 To calculate the number of moles in 16 grams ofSulphur (Atomic mass of Sulphur = 32 u).

Sol. 1 mole of atoms = Gram Atomic Mass.So, 1 mole of Sulphur atoms = Gram Atomic Mass ofSulphur = 32 grams.Now, 32 grams of Sulphur = 1 mole of SulphurSo, 16 grams of Sulphur= (1/32) x 16 = 0.5 molesThus, 16 grams of Sulphur constitute 0.5 mole ofSulphur.

6.023 × 10

(N ) Atoms

23

A

6.023 × 10

(N ) molecules

23

A

1 Mole

1 gram atomof element

1 gram molecule of substance

1 gram formula mass of substance

In terms of particles

In terms of mass

22.4 litre

In term ofvolume

PROBLEMS BASED ON THE MOLE CONCEPT

Ex.4 Calculate the number of moles in 5.75 g of sodium. (Atomic mass of sodium = 23 u)

Sol. Number of moles

= elementofMassAtomicGram

gramsinelementtheofMass =

23

5.75= 0.25 mole

or,1 mole of sodium atoms = Gram Atomic mass ofsodium = 23g.23 g of sodium = 1 mole of sodium.

5.75 g of sodium = 23

5.75mole of sodium = 0.25 mole

PAGE # 25

Ex.5 What is the mass in grams of a single atom ofchlorine ? (Atomic mass of chlorine = 35.5u)

Sol. Mass of 6.022 × 1023 atoms of Cl = Gram AtomicMass of Cl = 35.5 g.

Mass of 1 atom of Cl =23106.022

g35.5

= 5.9 × 10�23 g.

Ex.6 The density of mercury is 13.6 g cm�3. How manymoles of mercury are there in 1 litre of the metal ?(Atomic mass of Hg = 200 u).

Sol. Mass of mercury (Hg) in grams = Density(g cm�3)× Volume (cm3)= 13.6 g cm�3 × 1000 cm3 = 13600 g.

Number of moles of mercury

= mercuryofMassAtomicGramgramsinmercuryofMass

=200

13600 = 68

Ex.7 The mass of a single atom of an element M is3.15× 10�23 g . What is its atomic mass ? Whatcould the element be ?

Sol. Gram Atomic Mass = mass of 6.022 × 1023 atoms= mass of 1 atom × 6.022 × 1023

= (3.15 × 10�23g) × 6.022 × 1023

= 3.15 × 6.022 g = 18.97 g.

Atomic Mass of the element = 18.97uThus, the element is most likely to be fluorine.

Ex.8 An atom of neon has a mass of 3.35 × 10�23 g.How many atoms of neon are there in 20 g of thegas ?

Sol. Number of atoms

= atom1ofMass

massTotal = 23�103.35

02

= 5.97 × 1023

Ex.9 How many grams of sodium will have the samenumber of atoms as atoms present in 6 g ofmagnesium ?(Atomic masses : Na = 23u ; Mg =24u)

Sol. Number of gram -atom of Mg

= MassAtomicGramgramsinMgofMass

= 246

= 4

1

Gram Atoms of sodium should be = 4

1

1 Gram Atom of sodium = 23 g

4

1 gram atoms of sodium = 23 ×

4

1 = 5.75 g

Ex.10 How many moles of Cr are there in 85g of Cr2S

3 ?

(Atomic masses : Cr = 52 u , S =32 u)Sol. Molecular mass of Cr

2S

3 = 2 × 52 + 3 × 32 = 104

+ 96 = 200 u.200g of Cr

2S

3 contains = 104 g of Cr.

85 g of Cr2S

3 contains =

200

85104 g of Cr = 44.2g

Thus, number of moles of Cr = 52

44.2 = 0.85 .

Ex.11 What mass in grams is represented by

(a) 0.40 mol of CO2,

(b) 3.00 mol of NH3,

(c) 5.14 mol of H5IO

6

(Atomic masses : C=12 u, O=16 u, N=14 u,

H=1 u and I = 127 u)

Sol. Weight in grams = number of moles × molecular

mass.

Hence,

(a) mass of CO2 = 0.40 × 44 = 17.6 g

(b) mass of NH3 = 3.00 × 17 = 51.0 g

(c) mass of H5IO

6 = 5.14 × 228 = 1171.92g

Ex.12 Calculate the volume in litres of 20 g of hydrogen

gas at STP.

Sol. Number of moles of hydrogen

= hydrogenofMassMolecularGram

grams in hydrogenofMass =

220

= 10

Volume of hydrogen = number of moles × Gram

Molecular Volume.

= 10 ×22.4 = 224 litres.

Ex.13 The molecular mass of H2SO

4 is 98 amu.

Calculate the number of moles of each elementin 294 g of H

2SO

4.

Sol. Number of moles of H2SO

4 =

98

294= 3 .

The formula H2SO

4 indicates that 1 molecule of

H2SO

4 contains 2 atoms of H, 1 atom of S and 4

atoms of O. Thus, 1 mole of H2SO

4 will contain 2

moles of H,1 mole of S and 4 moles of O atomsTherefore, in 3 moles of H

2SO

4 :

Number of moles of H = 2 × 3 = 6

Number of moles of S = 1 × 3 = 3

Number of moles of O = 4 × 3 = 12

Ex.14 Find the mass of oxygen contained in 1 kg ofpotassium nitrate (KNO

3).

Sol. Since 1 molecule of KNO3 contains 3 atoms of

oxygen, 1 mol of KNO3 contains 3 moles of

oxygen atoms. Moles of oxygen atoms = 3 × moles of KNO

3

= 3 × 101

1000 = 29.7

(Gram Molecular Mass of KNO3 = 101 g)

Mass of oxygen = Number of moles × Atomic

mass= 29.7 × 16 = 475.2 g

Ex.15 You are asked by your teacher to buy 10 moles ofdistilled water from a shop where small bottleseach containing 20 g of such water are available.How many bottles will you buy ?

Sol. Gram Molecular Mass of water (H2O) = 18 g

10 mol of distilled water = 18 × 10 = 180 g.

Because 20 g distilled water is contained in 1bottle,

180 g of distilled water is contained in =20

180

bottles = 9 bottles.

Number of bottles to be bought = 9

PAGE # 26

Ex.16 6.022 × 1023 molecules of oxygen (O2) is equal tohow many moles ?

Sol. No. of moles =

AN

N

moleculesofno.sAvogadro'

oxygenofmoleculesofNo. =

23

23

106.023

106.023

= 1

PERCENTAGE COMPOSITION

The percentage composition of elements in acompound is calculated from the molecular formulaof the compound.The molecular mass of the compound is calculatedfrom the atomic masses of the various elementspresent in the compound. The percentage by massof each element is then computed with the help of thefollowing relations.Percentage mass of the element in the compound

= massMolecular

elementtheofmassTotal× 100

Ex.17 What is the percentage of calcium in calciumcarbonate (CaCO

3) ?

Sol. Molecular mass of CaCO3 = 40 + 12 + 3 × 16

= 100 amu.Mass of calcium in 1 mol of CaCO

3 = 40g.

Percentage of calcium = 100

10040 = 40 %

Ex.18 What is the percentage of sulphur in sulphuricacid (H

2SO

4) ?

Sol. Molecular mass of H2SO

4 = 1 × 2 + 32 + 16 × 4 = 98

amu.

Percentage of sulphur = 98

10032 = 32.65 %

Ex.19 What are the percentage compositions ofhydrogen and oxygen in water (H

2O) ?

(Atomic masses : H = 1 u, O = 16 u)

Sol. Molecular mass of water, H2O = 2 + 16 = 18 amu.

H2O has two atoms of hydrogen.

So, total mass of hydrogen in H2O = 2 amu.

Percentage of H = 18

1002 = 11.11 %

Similarly,

percentage of oxygen = 18

10016 = 88.88 %

The following steps are involved in determining theempirical formula of a compound :(i) The percentage composition of each element isdivided by its atomic mass. It gives atomic ratio of theelements present in the compound.

(ii) The atomic ratio of each element is divided by theminimum value of atomic ratio as to get the simplestratio of the atoms of elements present in thecompound.

(iii) If the simplest ratio is fractional, then values ofsimplest ratio of each element is multiplied bysmallest integer to get the simplest whole numberfor each of the element.

(iv) To get the empirical formula, symbols of variouselements present are written side by side with theirrespective whole number ratio as a subscript to thelower right hand corner of the symbol.

(v) The molecular formula of a substance may bedetermined from the empirical formula if the molecularmass of the substance is known. The molecularformula is always a simple multiple of empiricalformula and the value of simple multiple (n) isobtained by dividing molecular mass with empiricalformula mass.

n = MassFormulaEmpiricalMassMolecular

Ex-20 A compound of carbon, hydrogen and nitrogencontains these elements in the ratio of 9:1:3.5 respectively.Calculate the empirical formula. If its molecular mass is108, what is the molecular formula ?

Sol.

ElementMassRatio

AtomicMass

Relative Numberof Atoms

Simplest Ratio

Carbon 9 12 0.75×4 = 3

Hydrogen 1 1 1 × 4 = 4

Nitrogen 3.5 14 0.25 ×4 = 1

0.75129

111

0.25143.5

Empirical ratio = C3H

4N

Empirical formula mass = (3 × 12) + (4× 1) + 14 = 54

n = MassFormulaEmpiricalMassMolecular

= 254

108

Thus, molecular formula of the compound

= (Empirical formula)2

= (C3H

4N)

2

= C6H

8N

2

Ex.21 A compound on analysis, was found to have thefollowing composition :(i) Sodium = 14.31%, (ii) Sulphur = 9.97%, (iii) Oxygen= 69.50%, (iv) Hydrogen = 6.22%. Calculate themolecular formula of the compound assuming thatwhole hydrogen in the compound is present as waterof crystallisation. Molecular mass of the compoundis 322.

Sol. Element PercentageAtomic mass

Relative Number of atoms

Simplest ratio

Sodium 14.31 23 0.622

Sulphur 9.97 32 0.311

Hydrogen 6.22 1 6.22

Oxygen 69.50 16 4.34

20.311

0.622

10.311

0.311

200.311

6.22

140.311

4.34

23

31.14

1

22.6

16

50.69

32

97.9

PAGE # 27

The empirical formula = Na2SH

20O

14

Empirical formula mass= (2 × 23) + 32 + (20 × 1) + (14 × 16)

= 322Molecular mass = 322Molecular formula = Na

2SH

20O

14

Whole of the hydrogen is present in the form of waterof crystallisation. Thus, 10 water molecules arepresent in the molecule.So, molecular formula = Na

2SO

4. 10H

2O

CONCENTRATION OF SOLUTIONS

(a) Strength in g/L :

The strength of a solution is defined as the amount ofthe solute in grams present in one litre (or dm3) of thesolution, and hence is expressed in g/litre or g/dm3.

Strength in g/L = litreinsolutionofVolume

graminsoluteofWeight

(b) Molarity :

Molarity of a solution is defined as the number ofmoles of the solute dissolved per litre (or dm3) ofsolution. It is denoted by �M�. Mathematically,

M = litreinsolutiontheofVolume

soluteofmolesofNumber

litreinsolutionofVolume

soluteofMassMoleculargram/GraminsoluteofMass

M can be calculated from the strength as given below :

M = solute of mass Molecular

litre per grams in Strength

If �w� gram of the solute is present in V cm3 of a givensolution , then

M = massMolecular

w ×

V

1000

e.g. a solution of sulphuric acid having 4.9 grams of itdissolved in 500 cm3 of solution will have its molarity,

M = massMolecular

w×

V

1000

M = 98

4.9 ×

500

1000 = 0.1

(c) Formality :

In case of ionic compounds like NaCl, Na2CO

3 etc.,

formality is used in place of molarity. The formality ofa solution is defined as the number of gram formulamasses of the solute dissolved per litre of thesolution. It is represented by the symbol �F�. The term

formula mass is used in place of molecular massbecause ionic compounds exist as ions and not asmolecules. Formula mass is the sum of the atomicmasses of the atoms in the formula of the compound.

litreinsolutionofVolume

soluteofMasslagram/FormuinsoluteofMass

(d) Normality :

Normality of a solution is defined as the number ofgram equivalents of the solute dissolved per litre (dm3)of given solution. It is denoted by �N�.Mathematically,

N = litre in solution the of Volume

solute of sequivalent gram of Number

N = litre in solution the of Volume

solute of weight equivalent/ graminsolute of Weight

N can be calculated from the strength as given below :

N = solute of mass Equivalent

litre per grams in Strength=

E

S

If �w� gram of the solute is present in V cm3 of a givensolution.

N = solute the of mass Equivalentw

× V

1000

e.g. A solution of sulphuric acid having 0.49 gram ofit dissolved in 250 cm3 of solution will have itsnormality,

N = solute the of mass Equivalentw

× V

1000

N = 49

0.49×

250

1000 = 0.04

(Eq. mass of sulphuric acid = 49).

Solution Seminormal

Decinormal

Centinormal

Normality101

1001

21

Some Important Formulae :

(i) Milli equivalent of substance = N × V

where , N normality of solutionV Volume of solution in mL

(ii) If weight of substance is given,

milli equivalent (NV) = E1000w

Where, W Weight of substance in gramE Equivalent weight of substance

(iii) S = N × E

S Strength in g/LN Normality of solutionE Equivalent weight

(iv) Calculation of normality of mixture :

Ex.22 100 ml of 10

NHCl is mixed with 50 ml of

5

NH

2SO

4 .

Find out the normality of the mixture.Sol. Milli equivalent of HCl + milli equivalent H

2SO

4

= milli equivalent of mixtureN

1 V

1 + N

2 V

2 = N

3 V

3 { where, V

3 =V

1 + V

2 )

50

51

100101 N

3 × 150

N3 =

150

20 =

15

2= 0.133

PAGE # 28

Ex.23 100 ml of 10

NHCl is mixed with 25 ml of

5

NNaOH.

Find out the normality of the mixture.

Sol. Milli equivalent of HCl � milli equivalent of NaOH

= milli equivalent of mixtureN

1 V

1 � N

2 V

2 = N

3 V

3 { where, V

3 =V

1 + V

2 )

25

51

�100101

= N3 × 125

N3 =

251

Note :

1 milli equivalent of an acid neutralizes 1 milliequivalent of a base.

(e) Molality :

Molality of a solution is defined as the number ofmoles of the solute dissolved in 1000 grams of thesolvent. It is denoted by �m�.Mathematically,

m = gram in solvent the of Weightsolute the of moles of Number

× 1000 �m� can

be calculated from the strength as given below :

m = solute of mass Molecularsolvent of gram 1000 per Strength

If �w� gram of the solute is dissolved in �W� gram of the

solvent then

m = solute the of mass Mol.

w×

W1000

e.g. A solution of anhydrous sodium carbonate(molecular mass = 106) having 1.325 grams of it,dissolved in 250 gram of water will have its molality -

m = 250

1000106

1.325 = 0.05

Note :

Relationship Between Normality and Molarity of aSolution :Normality of an acid = Molarity × Basicity

Normality of base = Molarity × Acidity

Ex.24 Calculate the molarity and normality of a solutioncontaining 0.5 g of NaOH dissolved in 500 cm3

of solvent.Sol. Weight of NaOH dissolved = 0.5 g

Volume of the solution = 500 cm3

(i) Calculation of molarity :Molecular weight of NaOH = 23 + 16 + 1 = 40

Molarity =litre in solution of Volume

solute of weightmolecularsolute/ of Weight

= 500/1000

0.5/40 = 0.025

(ii) Calculation of normality :Normality

=litre in solution of Volume

solute of weightequivalentsolute/ of Weight

=500/1000

0.5/40 = 0.025

Ex.25 Find the molarity and molality of a 15% solution

of H2SO

4 (density of H

2SO

4 solution = 1.02 g/cm3)

(Atomic mass : H = 1u, O = 16u , S = 32 u)

Sol. 15% solution of H2SO

4 means 15g of H

2SO

4 are

present in 100g of the solution i.e.

Wt. of H2SO

4 dissolved = 15 g

Weight of the solution = 100 g

Density of the solution = 1.02 g/cm3 (Given)

Calculation of molality :

Weight of solution = 100 g

Weight of H2SO

4 = 15 g

Wt. of water (solvent) = 100 � 15 = 85 g

Molecular weight of H2SO

4 = 98

15 g H2SO

4 =

98

15= 0.153 moles

Thus ,85 g of the solvent contain 0.153 moles .

1000 g of the solvent contain= 85

0.153× 1000 = 1.8 mole

Hence ,the molality of H2SO

4 solution = 1.8 m

Calculation of molarity :

15 g of H2SO

4 = 0.153 moles

Vol. of solution = solution ofDensity

solution of Wt.

= 1.02

100 = 98.04 cm3

This 98.04 cm3 of solution contain H2SO

4 = 0.153

moles

1000 cm3 of solution contain H2SO

4

= 98.04

0.153 × 1000 = 1.56 moles

Hence the molarity of H2SO

4 solution = 1.56 M

(f) Mole Fraction :

The ratio between the moles of solute or solvent to

the total moles of solution is called mole fraction.

mole fraction of solute = Nn

n

solutionofMoles

soluteofMoles

= W/Mw/m

w/m

Mole fraction of solvent = Nn

N

solutionofMoles

solventofMoles

=

W/Mw/m

W/M

where,

n number of moles of solute

N number of moles of solvent

m molecular weight of solute

M molecular weight of solvent

w weight of solute

W weight of solvent

PAGE # 29

Ex.26 Find out the mole fraction of solute in 10% (by weight)urea solution.weight of solute (urea) = 10 gweight of solution = 100 gweight of solvent (water) = 100 � 10 = 90g

mole fraction of solute = solutionofMoles

soluteofMoles =

W/Mw/m

w/m

=

18/9060/10

60/10

= 0.032

Note :Sum of mole fraction of solute and solvent is alwaysequal to one.

STOICHIOMETRY

(a) Quantitative Relations in Chemical

Reactions :

Stoichiometry is the calculation of the quantities ofreactants and products involved in a chemicalreaction.

It is based on the chemical equation and on therelationship between mass and moles.N

2(g) + 3H

2(g) 2NH

3(g)

A chemical equation can be interpreted as follows -

1 molecule N2 + 3 molecules H

2 2 molecules

NH3(Molecular interpretation)

1 mol N2 + 3 mol H

2 2 mol NH

3

(Molar interpretation)

28 g N2 + 6 g H

2 34 g NH

3

(Mass interpretation)

1 volume N2 + 3 volume H

2 2 volume NH

3

(Volume interpretation)

Thus, calculations based on chemical equations aredivided into four types -

(i) Calculations based on mole-mole relationship.

(ii) Calculations based on mass-mass relationship.

(iii) Calculations based on mass-volume relationship.

(iv) Calculations based on volume -volumerelationship.

(i) Calculations based on mole-mole relationship :In such calculations, number of moles of reactantsare given and those of products are required.Conversely, if number of moles of products are given,then number of moles of reactants are required.

Ex.27 Oxygen is prepared by catalytic decompositionof potassium chlorate (KClO

3). Decomposition

of potassium chlorate gives potassium chloride(KCl) and oxygen (O

2). How many moles and how

many grams of KClO3 are required to produce

2.4 mole O2.

Sol. Decomposition of KClO3 takes place as,

2KClO3(s) 2KCl(s) + 3O

2(g)

2 mole KClO3 3 mole O

2

3 mole O2 formed by 2 mole KClO

3

2.4 mole O2 will be formed by

4.2

32

mole KClO3 = 1.6 mole KClO

3

Mass of KClO3 = Number of moles × molar mass

= 1.6 × 122.5 = 196 g

(ii) Calculations based on mass-mass relationship:In making necessary calculation, following steps are

followed -

(a) Write down the balanced chemical equation.

(b) Write down theoretical amount of reactants and

products involved in the reaction.

(c) The unknown amount of substance is calculated

using unitary method.

Ex.28 Calculate the mass of CaO that can be prepared

by heating 200 kg of limestone CaCO3 which is

95% pure.

Sol. Amount of pure CaCO3 = 200

10095

= 190 kg

= 190000 g

CaCO3(s) CaO(s) + CO

2(g)

1 mole CaCO3 1 mole CaO

100 g CaCO3 56 g CaO

100 g CaCO3 give 56 g CaO

190000 g CaCO3 will give=

10056

× 190000 g CaO

= 106400 g = 106.4 kg

Ex.29 Chlorine is prepared in the laboratory by treatingmanganese dioxide (MnO

2) with aqueous

hydrochloric acid according to the reaction -

MnO2 + 4HCl MnCl

2 + Cl

2 + 2H

2O

How many grams of HCl will react with 5 g MnO2 ?

Sol. 1 mole MnO2 reacts with 4 mole HCl

or 87 g MnO2 react with 146 g HCl

5 g MnO2 will react with =

87146

× 5 g HCl = 8.39 g HCl

Ex.30 How many grams of oxygen are required to burncompletely 570 g of octane ?

Sol. Balanced equation

2C H + 25O 8 18 2 16CO + 18H O2 2

2 mole2 × 114

25 mole25 × 32

First method : For burning 2 × 114 g of the octane,

oxygen required = 25 × 32 g

For burning 1 g of octane, oxygen required =1142

3225

g

Thus, for burning 570 g of octane, oxygen required

= 1142

3225

× 570 g = 2000 g

PAGE # 30

Mole Method : Number of moles of octane in 570grams

114570

= 5.0

For burning 2.0 moles of octane, oxygen required= 25 mol = 25 × 32 g

For burning 5 moles of octane, oxygen required

= 0.23225

× 5.0 g = 2000 g

Proportion Method : Let x g of oxygen be required forburning 570 g of octane. It is known that 2 × 114 g of

the octane requires 25 × 32 g of oxygen; then, the

proportion.

etanocg1142oxygeng3225

= etanocg570

x

x = 1142

5703225

= 2000 g

Ex.31 How many kilograms of pure H2SO

4 could be

obtained from 1 kg of iron pyrites (FeS2) according to

the following reactions ?4FeS

2 + 11O

2 2Fe

2O

3 + 8SO

2

2SO2 + O

2 2SO

3

SO3 + H

2O H

2SO

4

Sol. Final balanced equation,4FeS + 15O + 8H O2 2 2

2Fe O + 8H SO2 3 2 4

8 mole8 × 98 g

4 mole4 × 120 g

4 × 120 g of FeS2 yield H

2SO

4 = 8 × 98 g

1000 g of FeS2 will yield H

2SO

4 =

1204988

× 1000

= 1633.3 g

(iii) Calculations involving mass-volume relationship :In such calculations masses of reactants are givenand volume of the product is required and vice-versa.1 mole of a gas occupies 22.4 litre volume at STP.Mass of a gas can be related to volume according tothe following gas equation -PV = nRT

PV = mw

RT

Ex-32. What volume of NH3 can be obtained from 26.75 g

of NH4Cl at 27ºC and 1 atmosphere pressure.

Sol. The balanced equation is -

NH Cl(s) 4 NH (g) + HCl(g)3

1 mol1 mol53.5 g

53.5 g NH4Cl give 1 mole NH

3

26.75 g NH4Cl will give

5.531

× 26.75 mole NH3

= 0.5 molePV = nRT1 ×V = 0.5 × 0.0821 × 300

V = 12.315 litre

Ex-33 What quantity of copper (II) oxide will react with2.80 litre of hydrogen at STP ?

Sol. CuO + H 2 Cu + H O2

1 mol79.5 g

1 mol22.4 litre at NTP

22.4 litre of hydrogen at STP reduce CuO = 79.5 g2.80 litre of hydrogen at STP will reduce CuO

= 4.225.79

× 2.80 g = 9.93 g

Ex-34 Calculate the volume of carbon dioxide at STPevolved by strong heating of 20 g calcium carbonate.

Sol. The balanced equation is -

CaCO 3 CaO + CO 2

1 mol = 22.4 litre at STP

1 mol 100 g

100 g of CaCO3 evolve carbon dioxide = 22.4 litre

20 g CaCO3 will evolve carbon dioxide

= 100

4.22 × 20 = 4.48 litre

Ex.35 Calculate the volume of hydrogen liberated at 27ºC

and 760 mm pressure by heating 1.2 g of magnesiumwith excess of hydrochloric acid.

Sol. The balanced equation is

Mg + 2HCl MgCl + H2 2

1 mol 24 g

24 g of Mg liberate hydrogen = 1 mole

1.2 g Mg will liberate hydrogen = 0.05 mole

PV = nRT1 × V = 0.05 × 0.0821 × 300

V = 1.2315 litre

(iv) Calculations based on volume volumerelationship :These calculations are based on two laws :(i) Avogadro�s law (ii) Gay-Lussac�s Law

e.g.N (g) + 3H (g)2 2 2NH (g) (Avogadro's law)3

2 mol 2 × 22.4 L

1 mol1 × 22.4 L

3 mol3 × 22.4 L

(under similar conditions of temperature andpressure, equal moles of gases occupy equalvolumes)N (g) + 3H (g)2 2 2NH (g)3

1 vol 3 vol 2 vol(Gay- Lussac's Law)

(under similar conditions of temperature andpressure, ratio of coefficients by mole is equal to ratioof coefficient by volume).

Ex-36 One litre mixture of CO and CO2 is taken. This is

passed through a tube containing red hot charcoal.The volume now becomes 1.6 litre. The volume aremeasured under the same conditions. Find thecomposition of mixture by volume.

Sol. Let there be x mL CO in the mixture , hence, there willbe (1000 � x) mL CO

2. The reaction of CO

2 with red

hot charcoal may be given as -

CO (g) + C(s)2 2CO(g) 2 vol.2(1000 � x)

1 vol.(1000 �x)

Total volume of the gas becomes = x + 2(1000 � x)

x + 2000 � 2x = 1600

x = 400 mL

volume of CO = 400 mL and volume of CO2 = 600 mL

PAGE # 31

Ex-37 What volume of air containing 21% oxygen by volume

is required to completely burn 1kg of carbon containing

100% combustible substance ?

Sol. Combustion of carbon may be given as,

C(s) + O (g) 2 CO (g)2

1 mol12 g

1 mol32 g

12 g carbon requires 1 mole O2 for complete

combustion

1000 g carbon will require 1000121 mole O

2 for

combustion, i.e. , 83.33 mole O2

Volume of O2 at STP = 83.33 × 22.4 litre

= 1866.66 litre

21 litre O2 is present in 100 litre air

1866.66 litre O2 will be present in

21100

× 1866.66 litre air

= 8888.88 litre or 8.89 × 103 litre

Ex-38 An impure sample of calcium carbonate contains80% pure calcium carbonate 25 g of the impuresample reacted with excess of hydrochloric acid.Calculate the volume of carbon dioxide at STPobtained from this sample.

Sol. 100 g of impure calcium carbonate contains = 80 gpure calcium carbonate25 g of impure calcium carbonate sample will contain

= 10080

× 25 = 20 g pure calcium carbonate

The desired equation is -

CaCO + 2HCl 3 CaCl + CO + H O2 2 2

1 mol100 g

22.4 litre at STP

100 g pure CaCO3 liberate = 22.4 litre CO

2.

20 g pure CaCO3 liberate = 20

1004.22

= 4.48 litre CO2

VOLUMETRIC CALCULATIONS

The quantitative analysis in chemistry is primarily

carried out by two methods, viz, volumetric analysis

and gravimetric analysis.In the first method the mass

of a chemical species is measured by measurement

of volume, whereas in the second method it is deter-

mined by taking the weight.

The strength of a solution in volumetric analysis is

generally expressed in terms of normality, i.e., num-

ber of equivalents per litre but since the volume in the

volumetric analysis is generally taken in millilitres

(mL), the normality is expressed by milliequivalents

per millilitre.

USEFUL FORMULAE FOR

VOLUMETRIC CALCULATIONS

(i) milliequivalents = normality × volume in millilitres.

(ii) At the end point of titration, the two titrants, say 1

and 2, have the same number of milliequivalents,

i.e., N1V

1 = N

2V

2, volume being in mL.

(iii) No. of equivalents = 1000

.e.m.

(iv) No. of equivalents for a gas =

)STPat.eq1of.vol(volumeequivalentSTPatVolume

(v) Strength in grams per litre = normality × equivalent

weight.

(vi) (a) Normality = molarity × factor relating mol. wt.

and eq. wt.

(b) No. of equivalents = no. of moles × factor relat

ing mol. wt. and eq. wt.

Ex.39 Calculate the number of milli equivalent of H2SO

4

present in 10 mL of N/2 H2SO

4 solution.

Sol. Number of m.e. = normality × volume in mL =21

× 10 = 5.

Ex.40 Calculate the number of m.e. and equivalents of

NaOH present in 1 litre of N/10 NaOH solution.

Sol. Number of m.e. = normality × volume in mL

= 101

× 1000 = 100

Number of equivalents = 1000

.e.mof.no =

1000100

= 0.10

Ex.41 Calculate number of m.e. of the acids present in

(i) 100 mL of 0.5 M oxalic acid solution.

(ii) 50 mL of 0.1 M sulphuric acid solution.

Sol. Normality = molarity × basicity of acid

(i) Normality of oxalic acid = 0.5 × 2 = 1 N

m.e. of oxalic acid = normality × vol. in mL = 1 × 100

= 100.

(ii) Normality of sulphuric acid = 0.1 × 2 = 0.2 N

m.e. of sulphuric acid = 0.2 × 50 = 10

Ex.42 A 100 mL solution of KOH contains 10

milliequivalents of KOH. Calculate its strength in nor-

mality and grams/litre.

Sol. Normality = mLinvolume.e.mof.no

= 1.010010

strength of the solution = N/10

Again, strength in grams/litre = normality × eq. wt.

= 56101 = 5.6 gram/litre.

56

156

acidity.wtmolecular

KOHof.wt.eq

PAGE # 32

Ex.43 What is strength in gram/litre of a solution of H2SO

4,

12 cc of which neutralises 15 cc of 10N

NaOH

solution ?

Sol. m.e. of NaOH solution = 101

× 15 = 1.5

m.e. of 12 cc of H2SO

4 = 1.5

normality of H2SO

4 =

125.1

Strength in grams/litre = normality × eq. wt.

= 12

5.1 × 49 grams/litre

= 6.125 grams/litre.

49

298

basicitywt.molecular

SOHofwt.eq. 42

Ex.44 What weight of KMnO4 will be required to prepare

250 mL of its 10N

solution if eq. wt. of KMnO4 is 31.6 ?

Sol. Equivalent weight of KMnO4 = 31.6

Normality of solution (N) = 101

Volume of solution (V) = 250 ml

1000NEV

W ; W = 1000

2506.31101 79.0

406.31 g

Ex.45 100 mL of 0.6 N H2SO

4 and 200 mL of 0.3 N HCl

were mixed together. What will be the normality of theresulting solution ?

Sol. m.e. of H2SO

4 solution = 0.6 × 100 = 60

m.e. of HCl solution = 0.3 × 200 = 60

m.e. of 300 mL (100 + 200) of acidic mixture= 60 + 60 = 120.

Normality of the resulting solution = .voltotal.e.m

= 300120

= 52

N.

Ex.46 A sample of Na2CO

3. H

2O weighing 0.62 g is added

to 100 mL of 0.1 N H2SO

4. Will the resulting solution

be acidic, basic or neutral ?

Sol. Equivalents of Na2CO

3. H

2O =

6262.0

= 0.01

62

2124

OH.CONaof.wt.eq 232

m.e. of Na2CO

3. H

2O = 0.01 × 1000 = 10

m.e. of H2SO

4 = 0.1 × 100 = 10

Since the m.e. of Na2CO

3. H

2O is equal to that of H

2SO

4,

the resulting solution will be neutral.

(a) Introduction :

Volumetric analysis is a method of quantitativeanalysis. It involves the measurement of the volumeof a known solution required to bring about thecompletion of the reaction with a measured volumeof the unknown solution whose concentration orstrength is to be determined. By knowing the volumeof the known solution, the concentration of the solutionunder investigation can be calculated. Volumetricanalysis is also termed as titrimetric analysis.

(b) Important terms used in volumetric

analysis :

(i) Titration : The process of addition of the knownsolution from the burette to the measured volume ofsolution of the substance to be estimated until thereaction between the two is just complete, is termedas titration. Thus, a titration involves two solutions:

(a) Unknown solution and (b) Known solution or stan-dard solution.

(ii) Titrant : The reagent or substance whose solu-tion is employed to estimate the concentration of un-known solution is termed titrant. There are two typesof reagents or titrants:

(A) Primary titrants : These reagents can beaccurately weighed and their solutions are not to bestandardised before use. Oxalic acid, potassiumdichromate, silver nitrate, copper sulphate, ferrousammonium sulphate, sodium thiosulphates etc., arethe examples of primary titrants.

(B) Secondary titrants : These reagents cannotaccurately weighed and their solutions are to bestandardised before use. Sodium hydroxide,potassium hydroxide, hydrochloric acid, sulphuricacid, iodine, potassium permanganate etc. are theexamples of secondary titrants.

(iii) Standard solution : The solution of exactly knownconcentration of the titrant is called the standardsolution.

(iv) Titrate : The solution consisting the substance tobe estimated is termed unknown solution. Thesubstance is termed titrate.

(v) Equivalence point : The point at which the reagent(titrant) and the substance (titrate) under investigationare chemically equivalent is termed equivalence pointor end point.

(vi) Indicator : It is the auxiliary substance used forphysical (visual) detection of the completion of titrationor detection of end point is termed as indicator.Indicators show change in colour or turbidity at thestage of completion of titration.

(c) Concentraion representation of solution

(A) Strength of solution : Grams of solute dissolvedper litre of solution is called strength of solution'

(B) Parts Per Million (ppm) : Grams of solutedissolved per 106 grams of solvent is calledconcentration of solution in the unit of Parts Per Million(ppm). This unit is used to represent hardness ofwater and concentration of very dilute solutions.

(C) Percentage by mass : Grams of solute dissolvedper 100 grams of solution is called percentage bymass.

(D) Percentage by volume : Millilitres of solute per100 mL of solution is called percentage by volume.For example, if 25 mL ethyl alcohol is diluted withwater to make 100 mL solution then the solution thusobtained is 25% ethyl alcohol by volume.

(E) Mass by volume percentage :Grams of solutepresent per 100 mL of solution is called percentagemass by volume.For example, let 25 g glucose is dissolved in water tomake 100 mL solution then the solution is 25% massby volume glucose.

PAGE # 33

(d) Classification of reactions involved in

volumetric analysis

(A) Neutralisation reactions

The reaction in which acids and bases react to formsalt called neutralisation.

e.g., HCI + NaOH NaCI + H2O

H+(acid)

+ OH�

(base) H

2O (feebly ionised)

The titration based on neutralisation is calledacidimetry or alkalimetry.

(B) Oxidation-reduction reactions

The reactions involving simultaneous loss and gainof electrons among the reacting species are calledoxidation reduction or redox reactions, e.g., let usconsider oxidation of ferrous sulphate (Fe2+ ion) bypotassium permanganate (MnO

4� ion) in acidic

medium.

MnO4� + 8H+ + 5e� Mn2+ + 4H

2O

(Gain of electrons or reduction)5 [Fe2+ Fe3+ + e�](Loss of electrons or oxidation)

MnO4� + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H

2O

________________________________________________________________

In the above reaction, MnO4� acts as oxidising agent

and Fe2+ acts as reducing agent.

The titrations involving redox reactions are called redoxtitrations. These titrations are also called accordingto the reagent used in the titration, e.g., iodometric,cerimetric, permanganometric and dichromometrictitrations

(C) Precipitation reactions :A chemical reaction in which cations and anionscombine to form a compound of very low solubility (inthe form of residue or precipitate), is calledprecipitation.BaCl

2 + Na

2SO

4 BaSO

4 + 2NaCl

(white precipitate)

The titrations involving precipitation reactions arecalled precipitation titrations.

(D) Complex formation reactions :

These are ion combination reactions in which asoluble sl ightly dissociated complex ion orcompound is formed.Complex compounds retain their identity in thesolution and have the properties of the constituentions and molecules.e.g. CuSO

4 + 4NH

3 [Cu(NH

3)

4]SO

4

(complex compound)AgNO

3 + 2KCN K[Ag(CN)

2] + KNO

3

(complex compound)

2CuSO4 + K

4[Fe(CN)

6] Cu

2[Fe(CN)

6] + 2K

2SO

4

(complex compound)The titrations involving complex formation reactionsare called complexometric titrations.

The determination of concentration of bases bytitration with a standard acid is called acidimetry andthe determination of concentration of acid by titrationwith a standard base is called alkalimetry.The substances which give different colours withacids and base are called acid base indicators. Theseindicators are used in the visual detection of theequivalence point in acid-base titrations.The acid-base indicators are also called pHindicators because their colour change according tothe pH of the solution.

In the selection of indicator for a titration, followingtwo informations are taken into consideration :

(i) pH range of indicator(ii) pH change near the equivalence point in thetitration.

The indicator whose pH range is included in the pHchange of the solution near the equivalence point, istaken as suitable indicator for the titration.

(i) Strong acid-strong base titration : In the titrationof HCl with NaOH, the equivalence point lies in thepH change of 4�10. Thus, methyl orange, methyl red

and phenolphthalein will be suitable indicators.

(ii) Weak acid-strong base titration : In the titrationof CH

3COOH with NaOH the equivalence point lies

between 7.5 and 10. Hence, phenolphthalein (8.3�10) will be the suitable indicator.

(iii) Weak base-strong acid titration : In the titrationof NH

4OH (weak base) against HCl (strong acid) the

pH at equivalence point is about 6.5 and 4. Thus,methyl orange (3.1�4.4) or methyl red (4.2�6.3) will

be suitable indicators.

(iv) Weak acid-weak base titration : In the titration ofa weak acid (CH

3COOH) with weak base (NH

4OH)

the pH at the equivalence point is about 7, i.e., liesbetween 6.5 and 7.5 but no sharp change in pH isobserved in these titrations. Thus, no simple indica-tor can be employed for the detection of the equiva-lence point.

(v) Titration of a salt of a weak acid and a strongbase with strong acid:

H2CO

3 + 2NaOH Na

2CO

3 + 2H

2O

Weak acid Strong base

Na2CO

3 when titrated with HCl, the following two

stages are involved :Na

2CO

3 + HCl NaHCO

3 + NaCl (First stage)

pH = 8.3, near equivalence pointNaHCO

3 + HCl NaCl + H

2CO

3 (Second stage)

pH = 4, near equivalence point

For first stage, phenolphthalein and for second stage,methyl orange will be the suitable indicator.

PAGE # 34

TITRATION OF MIXTURE OF NaOH, Na2CO3 ANDNaHCO3 BY STRONG ACID LIKE HCl

In this titration the following indicators are mainly used :

(i) Phenolphthalein (weak organic acid) : It showscolour change in the pH range (8 � 10)

(ii) Methyl orange (weak organic base) : It shows

colour change in the pH range (3.1 � 4.4). Due to

lower pH range, it indicates complete neutralisation

of whole of the base.

Let for complete neutralisation of Na2CO

3, NaHCO

3 and NaOH, x,y and z mL of standard HCl are required. The

titration of the mixture may be carried by two methods as summarised below :

Mixture Phenlphthalein Methyl orange Phenolphthalein Methyl orange from from from after first end beginning beginning point

1. NaOH z + (x/2) (x + z) z + x/2 x/2 (for remaining 50% + Na CO2 3

beginning

Na CO )

2. NaOH z + 0 (z + y) z + 0 y (for remaining 100% + NaHCO NaHCO

3. Na CO (x/2) + 0 (x + y) (x/2) + 0 x/2 + y (for remaining 50% + NaHCO of Na CO and 100% NaHCO are indicated)

2 3

3 3

2 3

3 2 3

3

Volume of HClused with

Volume of HCl used

An indicator is a substance which is used to deter-mine the end point in a titration. In acid-base titra-tions organic substances (weak acids or weak bases)are generally used as indicators. They change theircolour within a certain pH range. The colour changeand the pH range of some common indicators aretabulated below:________________________________________Indicator pH range Colour

change________________________________________

Methyl orange 3.2 � 4.5 Orange to red

Methyl red 4.4 � 6.5 Red to yellow

Litmus 5.5 � 7.5 Red to blue

Phenol red 6.8 � 8.4 Yellow to red

Phenolphthalein 8.3 � 10.5 Colourless to pink________________________________________

Theory of acid-base indicators : Two theories havebeen proposed to explain the change of colour ofacid-base indicators with change in pH.

1. Ostwald's theory:According to this theory

(a) The colour change is due to ionisation of the acid-base indicator. The unionised form has differentcolour than the ionised form.

(b) The ionisation of the indicator is largely affected inacids and bases as it is either a weak acid or a weakbase. In case, the indicator is a weak acid, itsionisation is very much low in acids due to commonions while it is fairly ionised in alkalies. Similarly if theindicator is a weak base, its ionisation is large inacids and low in alkalies due to common ions.

Considering two important indicators phenolphtha-lein (a weak acid) and methyl orange (a weak base),Ostwald's theory can be illustrated as follows:

PAGE # 35

Phenolphthalein: It can be represented as HPh. Itionises in solution to a small extent as:

HPh H+ + Ph�

Colourless Pink

Applying law of mass action,

K = ]HPh[

]Ph][H[

The undissociated molecules of phenolphthalein arecolourless while ph� ions are pink in colour. In pres-ence of an acid, the ionisation of HPh is practicallynegligible as the equilibrium shifts to left hand sidedue to high concentration of H+ ions. Thus, the solu-tion would remain colourless. On addition of alkali,hydrogen ions are removed by OH� ions in the form ofwater molecules and the equilibrium shifts to righthand side. Thus, the concentration of ph� ions in-creases in solution and they impart pink colour to thesolution.

Let us derive Hendetson's equation for an indicator

HIn + H2O H

3+O + In�

'Acid form' 'Base form'

Conjugate acid-base pair

KIn =

]HIn[]OH][In[ 3

KIn = Ionization constant of indicator

[H3

+O] = KIn ]In[

]HIn[

pH = �log10

[H3

+O] = �log10

[KIn] � log

10 ]In[

]HIn[

pH = pKIn + log

10

]HIn[]In[

(Henderson's equation for

indicator)At equivalence point ;[In�] = [HIn] and pH = pK

In

Methyl orange : It is a very weak base and can berepresented as MeOH. It is ionised in solution to giveMe+ and OH� ions.

MeOH Me+ +OH�

Orange Red

Applying law of mass action,

K = ]MeOH[

]OH][Me[

In presence of an acid, OH� ions are removed in theform of water molecules and the above equilibriumshifts to right hand side. Thus, sufficient Me+ ions areproduced which impart red colour to the solution. Onaddition of alkali, the concentration of OH� ions in-creases in the solution and the equilibrium shifts toleft hand side, i.e., the ionisation of MeOH is practi-cally negligible. Thus, the solution acquires the colourof unionised methyl orange molecules, i.e. orange.

This theory also explains the reason why phenol-phthalein is not a suitable indicator for titrating a weakbase against strong acid. The OH� ions furnished bya weak base are not sufficient to shift the equilibriumtowards right hand side considerably, i.e., pH is notreached to 8.3. Thus, the solution does not attainpink colour. Similarly, it can be explained why methylorange is not a suitable indicator for the titration ofweak acid with strong base.

SOLUBILITY

The solubility of a solute in a solution is alwaysexpressed with respect to the saturated solution.

(a) Definition :

The maximum amount of the solute which can bedissolved in 100g (0.1kg) of the solvent to form asaturated solution at a given temperature.Suppose w gram of a solute is dissolved in W gramof a solvent to make a saturated solution at a fixedtemperature and pressure. The solubility of the solutewill be given by -

W

w× 100 =

solventtheofMass

solutetheofMass× 100

For example, the solubility of potassium chloride inwater at 20ºC and 1 atm. is 34.7 g per 100g of water.

This means that under normal conditions 100 g ofwater at 20ºC and 1 atm. cannot dissolve more than

34.7g of KCl.

(b) Effect of Temperature and Pressure on

Solubility of a Solids :

The solubility of a substance in liquids generallyincreases with rise in temperature but hardly changeswith the change in pressure. The effect of temperaturedepends upon the heat energy changes whichaccompany the process.

Note :If heat energy is needed or absorbed in the process,it is of endothermic nature. If heat energy is evolvedor released in the process, it is of exothermic nature.

(i) Effect of temperature on endothermic dissolutionprocess : Most of the salts like sodium chloride,potassium chloride, sodium nitrate, ammoniumchloride etc. dissolve in water with the absorption ofheat. In all these salts the solubility increases withrise in temperature. This means that sodium chloridebecomes more soluble in water upon heating.

(ii) Effect of temperature on exothermic dissolutionprocess : Few salts like lithium carbonate, sodiumcarbonate monohydrate, cerium sulphate etc.dissolve in water with the evolution of heat. Thismeans that the process is of exothermic nature. Inthese salts the solubility in water decreases with risein temperature.

PAGE # 36

Note :

1. While expressing the solubility, the solution mustbe saturated but for expressing concentration (masspercent or volume percent), the solution need not tobe saturated in nature.

2. While expressing solubility, mass of solvent isconsidered but for expressing concentration themass or volume of the solution may be taken intoconsideration.

(c) Effect of Temperature on the Solubility

of a Gas

(i) The solubility of a gas in a liquid decreases withthe rise in temperature.

(ii) The solubility of gases in liquids increases onincreasing the pressure and decreases on decreas-ing the pressure.

SAMPLE PROBLEMS

Ex.47 12 grams of potassium sulphate dissolves in75 grams of water at 60ºC. What is the solubility

of potassium sulphate in water at that temperature ?

Sol. Solubility = 100solventofmasssoluteofmass

= 75

12×100 = 16 g

Thus, the solubility of potassium sulphate inwater is 16 g at 60ºC.

Ex.48 4 g of a solute are dissolved in 40 g of water toform a saturated solution at 25ºC. Calculate the

solubility of the solute.

Sol. Solubility = solventofMass

soluteofMass× 100

Mass of solute = 4 gMass of solvent = 40 g

Solubility = 404

× 100 = 10 g

Ex.49 (a) What mass of potassium chloride would beneeded to form a saturated solution in 50 g ofwater at 298 K ? Given that solubility of the salt is46g per 100g at this temperature.

(b) What will happen if this solution is cooled ?

Sol. (a) Mass of potassium chloride in 100 g of waterin saturated solution = 46 gMass of potassium chloride in 50 g of water insaturated solution.

= 10046

× (50g) = 23 g

(b) When the solution is cooled, the solubility ofsalt in water will decrease. This means, that uponcooling, it will start separating from the solutionin crystalline form.

EXERCISE

1. The solubility of K2SO

4 in water is 16 g at 50ºC. The

minimum amount of water required to dissolve 4 g

K2SO

4 is -

(A) 10 g (B) 25 g

(C) 50 g (D) 75 g

2. Molarity of H2SO

4 (density 1.8g/mL) is 18M. The

molality of this solution is -

(A)36 (B) 200

(C) 500 (D) 18

3. 8g of sulphur are burnt to form SO2, which is oxidised

by Cl2 water. The solution is treated with BaCl

2

solution. The amount of BaSO4 precipitated is -

(A) 1.0 mole (B) 0.5 mole

(C) 0.75 mole (D) 0.25 mole

4. In a compound AxB

y -

(A) Mole of A = Mole of B = mole of AxB

y

(B) Eq. of A = Eq. of B = Eq. of AxB

y

(C) X × mole of A = y × mole of B = (x + y) × mole of AxB

y

(D) X × mole of A = y × mole of B

5. The percentage of sodium in a breakfast cereal

labelled as 110 mg of sodium per 100 g of cereal is -

(A) 11% (B) 1.10%

(C) 0.110% (D) 110%

6. Two elements A (at. wt. 75) and B (at. wt. 16) combine

to yield a compound. The % by weight of A in the

compound was found to be 75.08. The empirical

formula of the compound is -

(A) A2B (B) A

2B

3

(C) AB (D) AB2

7. No. of oxalic acid molecules in 100 mL of 0.02 N

oxalic acid are -

(A) 6.023 × 1020 (B) 6.023 × 1021

(C) 6.023 × 1022 (D) 6.023 × 1023

8. Which of the following sample contains the maximum

number of atoms -

(A) 1 mg of C4H

10(B) 1 mg of N

2

(C) 1 mg of Na (D) 1 mL of water

9. The total number of protons, electrons and neutrons

in 12 g of C126 is -

(A) 1.084 × 1025 (B) 6.022 × 1023

(C) 6.022 × 1022 (D) 18

10. 4.4 g of CO2 and 2.24 litre of H

2 at STP are mixed in a

container. The total number of molecules present in

the container will be -

(A) 6.022 × 1023 (B) 1.2044 × 1023

(C) 2 mole (D) 6.023 × 1024

PAGE # 37

11. Which is not a molecular formula ?

(A) C6H

12O

6(B) Ca(NO

3)

2

(C) C2H

4O

2(D) N

2O

12. The hydrated salt, Na2SO

4. nH

2O undergoes 55.9%

loss in weight on heating and becomes anhydrous.

The value of n will be -

(A) 5 (B) 3

(C) 7 (D) 10

13. Which of the following mode of expressing

concentration is independent of temperature -

(A) Molarity (B) Molality

(C) Formality (D) Normality

14. The haemoglobin of the red blood corpuscles of most

of the mammals contains approximately 0.33% of

iron by weight. The molecular weight of haemoglobin

is 67,200. The number of iron atoms in each molecule

of haemoglobin is (Atomic weight of iron = 56) -

(A) 2 (B) 3

(C) 4 (D) 5

15. An oxide of metal have 20% oxygen, the eq. wt. of

metal oxide is -

(A) 32 (B) 40

(C) 48 (D) 52

16. How much water is to be added to dilute 10 mL of 10

N HCl to make it decinormal ?

(A) 990 mL (B) 1010 mL

(C) 100 mL (D) 1000 mL

17. The pair of compounds which cannot exist in solution is -

(A) NaHCO3 and NaOH

(B) Na2SO

3 and NaHCO

3

(C) Na2CO

3 and NaOH

(D) NaHCO3 and NaCl

18. If 250 mL of a solution contains 24.5 g H2SO

4 the

molarity and normality respectively are -

(A) 1 M, 2 N (B) 1M,0.5 N

(C) 0.5 M, 1N (D) 2M, 1N

19. The mole fraction of NaCl, in a solution containing 1

mole of NaCl in 1000 g of water is -

(A) 0.0177 (B) 0.001

(C) 0.5 (D) 0.244

20. 3.0 molal NaOH solution has a density of 1.110 g/

mL. The molarity of the solution is -

(A) 2.9732 (B) 3.05

(C) 3.64 (D) 3.0504

21. How many atoms are contained in a mole of Ca(OH)2 -

(A) 30 × 6.02 × 1023 atoms/mol

(B) 5 × 6.02 × 1023 atoms/mol

(C) 6 × 6.02 × 1023 atoms/mol

(D) None of these

22. Insulin contains 3.4% sulphur. The minimum

molecular weight of insulin is -

(A) 941.176 u (B) 944 u

(C) 945.27 u (D) None of these

23. Number of moles present in 1 m3 of a gas at NTP are -

(A) 44.6 (B) 40.6

(C) 42.6 (D) 48.6

24. Weight of oxygen in Fe2O

3 and FeO is in the simple

ratio of -

(A) 3 : 2 (B) 1 : 2

(C) 2 : 1 (D) 3 : 1

25. 2.76 g of silver carbonate on being strongly heated

yield a residue weighing -

(A) 2.16g (B) 2.48 g

(C) 2.32 g (D) 2.64 g

26. How many gram of KCl would have to be dissolved in

60 g of H2O to give 20% by weight of solution -

(A) 15 g (B) 1.5 g

(C) 11.5 g (D) 31.5 g

27. When the same amount of zinc is treated separately

with excess of H2SO

4 and excess of NaOH, the ratio

of volumes of H2 evolved is -

(A) 1 : 1 (B) 1 : 2

(C) 2 : 1 (D) 9 : 4

28. Amount of oxygen required for combustion of 1 kg of a

mixture of butane and isobutane is -

(A) 1.8 kg (B) 2.7 kg

(C) 4.5 kg (D) 3.58 kg

29. Rakesh needs 1.71 g of sugar (C12

H22

O11

) to sweeten

his tea. What would be the number of carbon atoms

present in his tea ?

(A) 3.6 × 1022 (B) 7.2 × 1021

(C) 0.05 × 1023 (D) 6.6 × 1022

30. The total number of AlF3 molecule in a sample of AlF

3

containing 3.01 × 1023 ions of F� is -

(A) 9.0 × 1024 (B) 3.0 × 1024

(C) 7.5 × 1023 (D)1023

31. The volume occupied by one molecule of water

(density 1 g/cm3) is -

(A) 18 cm3 (B) 22400 cm3

(C) 6.023 × 10�23 (D) 3.0 × 10�23 cm3

PAGE # 38

32. 224 mL of a triatomic gas weigh 1 g at 273 K and 1

atm. The mass of one atom of this gas is -

(A) 8.30 × 10�23 g (B) 2.08 × 10�23 g

(C) 5.53 × 10�23 g (D) 6.24 × 10�23 g

33. The percentage of P2O

5 in diammonium hydrogen

phosphate is -

(A) 77.58 (B) 46.96

(C) 53.78 (D) 23.48

34. The mole fraction of water in 20% (wt. /wt.) aqueous

solution of H2O

2 is -

(A) 68

77(B)

77

68

(C) 80

20(D)

20

80

35. Which of the following has the maximum mass ?

(A) 25 g of Hg

(B) 2 moles of H2O

(C) 2 moles of CO2

(D) 4 g atom of oxygen

36. Total mass of neutrons in 7mg of 14C is -

(A) 3 × 1020 kg (B) 4 × 10�6 kg

(C) 5 × 10�7 kg (D) 4 × 10�7 kg

37. Vapour density of a metal chloride is 66. Its oxide

contains 53% metal. The atomic weight of metal is -

(A) 21 (B) 54

(C) 26.74 (D) 2.086

38. The number of atoms in 4.25 g NH3 is approximately -

(A) 1 × 1023 (B) 1.5 × 1023

(C) 2 × 1023 (D) 6 × 1023

39. The modern atomic weight scale is based on -

(A) C12 (B) O16

(C) H1 (D) C13

40. Amount of oxygen in 32.2g of Na2SO

4. 10H

2O is -

(A) 20.8 g (B) 22.4 g

(C) 2.24 g (D) 2.08 g

41. Which of the followings does not change on dilution ?

(A) Molarity of solution

(B) Molality of solution

(C) Millimoles and milli equivalent of solute

(D) Mole fraction of solute

42. Equal masses of O2, H

2 and CH

4 are taken in a

container. The respective mole ratio of these gases

in container is -

(A) 1 : 16 : 2 (B) 16 : 1 : 2

(C) 1 : 2 : 16 (D) 16 : 2 : 1

43. The number of molecules present in 11.2 litre CO2 at

STP is -

(A) 6.023 × 1032 (B) 6.023 × 1023

(C) 3.011 × 1023 (D) None of these

44. 250 ml of 0.1 N solution of AgNO3 are added to 250

ml of a 0.1 N solution of NaCl. The concentration of

nitrate ion in the resulting solution will be -

(A) 0.1N (B) 1.2 N

(C) 0.01 N (D) 0.05 N

45. Amount of BaSO4 formed on mixing the aqueous

solution of 2.08 g BaCl2 and excess of dilute H

2SO

4 is -

(A) 2.33 g (B) 2.08 g

(C) 1.04 g (D) 1.165 g

46. 2g of NaOH and 4.9 g of H2SO

4 were mixed and

volume is made 1 litre. The normality of the resulting

solution will be -

(A) 1N (B) 0.05 N

(C) 0.5 N (D) 0.1N

47. 1g of a metal carbonate neutralises completely 200

mL of 0.1N HCl. The equivalent weight of metal

carbonate is -

(A) 25 (B) 50

(C) 100 (D) 75

48. 100 mL of 0.5 N NaOH were added to 20 ml of 1N

HCl and 10 mL of 3 N H2SO

4. The solution is -

(A) acidic (B) basic

(C) neutral (D) none of these

49. 1M solution of H2SO

4 is diluted from 1 litre to 5 litres ,

the normality of the resulting solution will be -

(A) 0.2 N (B) 0.1 N

(C) 0.4 N (D) 0.5 N

50. The volume of 7g of N2 at S.T.P. is -

(A) 11.2 L (B) 22.4 L

(C) 5.6 L (D) 6.5 L

51. One mole of calcium phosphide on reaction with

excess of water gives -

(A) three moles of phosphine

(B) one mole phosphoric acid

(C) two moles of phosphine

(D) one mole of P2O

5

52. Mg (OH)2 in the form of milk of magnesia is used to

neutralize excess stomach acid. How many moles of

stomach acid can be neutralized by 1 g of Mg(OH)2 ?

(Molar mass of Mg(OH)2 = 58.33)

(A) 0.0171 (B) 0.0343

(C) 0.686 (D) 1.25

PAGE # 39

53. Calcium carbonate decomposes on heating

according to the following equation -

CaCO3(s) CaO(s) + CO

2(g)

How many moles of CO2 will be obtained by

decomposition of 50 g CaCO3 ?

(A) 23

(B) 25

(C) 21

(D) 1

54. Sulphur trioxide is prepared by the following two

reactions -

S8(s) + 8O

2(g) 8SO

2(g)

2SO2(g) + O

2(g) 2SO

3(g)

How many grams of SO3 are produced from 1 mole

of S8 ?

(A) 1280 (B) 640

(C) 960 (D) 320

55. PH3(g) decomposes on heating to produce

phosphorous and hydrogen. The change in volume

when 100 mL of such gas decomposed is -

(A) + 50 mL (B) + 500 mL

(C) � 50 mL (D) � 500 mL

56. What amount of BaSO4 can be obtained on mixing

0.5 mole BaCl2 with 1 mole of H

2SO

4 ?

(A) 0.5 mol (B) 0.15 mol

(C) 0.1 mol (D) 0.2 mol

57. In the reaction , CrO5 + H

2SO

4 Cr

2(SO

4)3 + H

2O + O

2

one mole of CrO5 will liberate how many moles of O

2 ?

(A) 5/2 (B) 5/4

(C) 9/2 (D) None

58. Calcium carbonate decomposes on heating

according to the equation -

CaCO3(s) CaO(s) + CO

2(g)

At STP the volume of CO2 obtained by thermal

decomposition of 50 g of CaCO3 will be -

(A) 22.4 litre (B) 44 litre

(C) 11.2 litre (D) 1 litre

59. When FeCl3 is ignited in an atmosphere of pure

oxygen, the following reaction takes place-

4FeCl3(s) + 3O

2(g) 2Fe

2O

3(s) + 6Cl

2(g)

If 3 moles of FeCl3 are ignited in the presence of 2

moles of O2 gas, how much of which reagent is

present in excess and therefore, remains unreacted ?

(A) 0.33 mole FeCl3 remains unreacted

(B) 0.67 mole FeCl3 remains unreacted

(C) 0.25 mole O2 remains unreacted

(D) 0.50 mole O2 remains unreacted

60. The volume of CO2 (in litres) liberated at STP when

10 g of 90% pure limestone is heated completely, is-

(A) 22.4 L (B) 2.24 L

(C) 20.16 L (D) 2.016 L

61. A metal oxide has the formula Z2O

3. It can be reduced

by hydrogen to give free metal and water. 0.1596 g of

the metal requires 6 mg of hydrogen for complete

reduction. The atomic mass of the metal is -

(A) 27.9 (B) 159.6

(C) 79.8 (D) 55.8

Question number 62, 63, 64 and 65 are based on the

following information :

Q. Dissolved oxygen in water is determined by using a

redox reaction. Following equations describe the

procedure -

I 2Mn2+(aq) + 4OH�(aq) + O2 (g) 2MnO2(s) + 2H2O( )

II MnO2(s)+2I�(aq)+4H+(aq) Mn2+(aq)+I2(aq) + 2H2O( )

III �232OS2 (aq) + I2(aq) �2

64OS2 (aq) + 2I�(aq)

62. How many moles of �232OS are equivalent to each

mole of O2 ?

(A) 0.5 B) 1

(C) 2 (D) 4

63. What amount of I2 will be liberated from 8 g dissolved

oxygen ?

(A) 127 g (B) 254 g

(C) 504 g (D) 1008 g

64. 3 × 10�3 moles O2 is dissolved per litre of water, then

what will be molarity of I� produced in the given

reaction ?

(A) 3 × 10�3 M (B) 4 × 3 × 10�3 M

(C) 2 × 3 × 10�3 M (D) 3�10321

M

65. 8 mg dissolved oxygen will consume -

(A) 5 × 10�4 mol Mn+2

(B) 2.5 × 10�4 mol Mn2+

(C) 10 mol Mn2+

(D) 2 mol Mn2+

66. 2 g of a base whose eq. wt. is 40 reacts with 3 g of an

acid. The eq. wt. of the acid is-

(A) 40 (B) 60

(C) 10 (D) 80

67. Normality of 1% H2SO

4 solution is nearly -

(A) 2.5 (B) 0.1

(C) 0.2 (D) 1

68. What volume of 0.1 N HNO3 solution can be prepared

from 6.3 g of HNO3 ?

(A) 1 litre (B) 2 litres

(C) 0.5 litre (D) 4 litres

PAGE # 40

69. The volume of water to be added to 200 mL of

seminormal HCl solution to make it decinormal is -

(A) 200 mL (B) 400 mL

(C) 600 mL (D) 800 mL

70. 0.2 g of a sample of H2O

2 required 10 mL of 1N KMnO

4

in a titration in the presence of H2SO

4. Purity of H

2O

2 is-

(A) 25% (B) 85%

(C) 65% (D) 95%

71. Which of the following has the highest normality ?

(A) 1 M H2SO

4(B) 1 M H

3PO

3

(C) 1 M H3PO

4(D) 1 M HNO

3

72. The molarity of 98% H2SO

4(d = 1.8g/mL) by wt. is -

(A) 6 M (B) 18.74 M

(C) 10 M (D) 4 M

73. 0.7 g of Na2CO

3 . xH

2O is dissolved in 100 mL. 20 mL

of which required to neutralize 19.8 mL of 0.1 N HCl.

The value of x is -

(A) 4 (B) 3

(C) 2 (D) 1

74. 0.45 g of an acid of molecular weight 90 was

neutralised by 20 mL of 0.5 N caustic potash. The

basicity of the acid is -

(A) 1 (B) 2

(C) 3 (D) 4

75. 1 litre of 18 molar H2SO

4 has been diluted to 100

litres. The normality of the resulting solution is -

(A) 0.09 N (B) 0.18

(C) 1800 N (D) 0.36

76. 150 ml of 10N

HCl is required to react completely with

1.0 g of a sample of limestone. The percentage purity

of calcium carbonate is -

(A) 75% (B) 50%

(C) 80% (D) 90%

77. 50 ml of 10N

HCl is treated with 70 ml 10N

NaOH.

Resultant solution is neutralized by 100 ml of

sulphuric acid. The normality of H2SO

4 -

(A) N/50 (B) N/25

(C) N/30 (D) N/10

78. 200 mL of 10N

HCl were added to 1 g calcium car-

bonate, what would remain after the reaction ?

(A) CaCO3

(B) HCl

(C) Neither of the two (D) Part of both

79. Equivalent mass of KMnO4, when it is converted to

MnSO4 is -

(A) M/5 (B) M/3

(C) M/6 (D) M/2

80. A 3 N solution of H2SO

4 in water is prepared from

Conc. H2SO

4 (36 N) by diluting [KVPY-PartII-2007]

(A) 20 ml of the conc. H2SO

4 to 240 ml

(B) 10 ml of the conc. H2SO

4 to 240 ml

(C) 1 ml of the conc. H2SO

4 to 36 ml

(D) 20 ml of the conc. H2SO

4 to 36 ml

81. The solubility curve of KNO3 as a function of tempera-

ture is given below [KVPY-Part-II-2007]

0 20 40 60 80 100

0

50

100

150

200

250

Temperature (°C)

Sol

ubili

ty (

g/10

0 m

l wat

er)

The amount of KNO3 that will crystallize when a satu-

rated solution of KNO3 in 100 ml of water is cooled

from 90°C to 30 °C, is

(A) 16 g (B) 100 g

(C) 56 g (D)160 g

82. The volume of 0.5 M aqueous NaOH solution required

to neutralize 10 ml of 2 M aqueous HCl solution is :

[KVPY-Part-I-2008](A) 20ml (B) 40ml

(C) 80ml (D) 120ml

83. 3.01×1023 molecules of elemental Sulphur will react

with 0.5 mole of oxygen gas completely to produce

[KVPY-Part-I-2008](A) 6.02 × 1023 molecules of SO

3

(B) 6.02 × 1023 molecules of SO2

(C) 3.01 × 1023 molecules of SO3

(D) 3.01 x 1023 molecules of SO2

84. The solubility of a gas in a solution is measured in

three cases as shown in the figure given below where

w is the weight of a solid slab placed on the top of the

cylinder lid. The solubility will follow the order :

[KVPY-Part-I-2008]

gas

solution

w

gas

solution

w w

gas

solution

w w w

(A) a > b > c (B) a < b < c

(C) a = b = c (D) a >b < c

PAGE # 41

85. The density of a salt solution is1.13 g cm�3 and itcontains 18% of NaCI by weight. The volume of thesolution containing 36.0 g of the salt will be :

[KVPY-Part-II-2008](A) 200 cm3 (B) 217 cm3

(C) 177 cm3 (D) 157cm3

86. One mole of nitrogen gas on reaction with 3.01 x 1023

molecules of hydrogen gas produces - [KVPY-Part-I-2009]

(A) one mole of ammonia(B) 2.0 x 1023 molecules of ammonia(C) 2 moles of ammonia(D) 3.01 × 1023 molecules of ammonia

87. [KVPY-Part-I-2009]Solubility g/I

20 40 60 80 100

KNO3

KCl

Temperature (ºC)

50

100

150

200

250

Given the solubility curves of KNO3 and KCl, which of

the following statements is not true ?(A) At room temperature the solubility of KNO

3 and

KCI are not equal(B) The solubilities of both KNO

3 and KCI increase

with temperature(C) The solubility of KCI decreases with temperature(D) The solubility of KNO

3 increases much more com-

pared to that of KCl with increase in temperature

88. 10 ml of an aqueous solution containing 222 mg ofcalcium chloride (mol. wt. = 111) is diluted to 100 ml.The concentration of chloride ion in the resulting so-lution is - [KVPY-Part-II-2009](A) 0.02 mol/lit. (B) 0.01 mol/lit.(C) 0.04 mol/lit (D) 2.0 mol/lit.

89. Aluminium reduces manganese dioxide to manga-nese at high temperature. The amount of aluminiumrequired to reduce one gram mole of manganesedioxide is - [KVPY-Part-II-2009](A) 1/2 gram mole (B) 1 gram mole(C) 3/4 gram mole (D) 4/3 gram mole

90. One mole of oxalic acid is equivalent to[IJSO-2009]

(A) 0.5 mole of NaOH (B) 1 mole of NaOH(C) 1.5 mole of NaOH (D) 2 mole of NaOH

42PAGE # 42

(i) Natural numbers :Counting numbers are known as natural numbers.N = { 1, 2, 3, 4, ... }.

(ii) Whole numbers :

All natural numbers together with 0 form the collectionof all whole numbers. W = { 0, 1, 2, 3, 4, ... }.

(iii) Integers :All natural numbers, 0 and negative of natural numbersform the collection of all integers.I or Z = { ..., � 3, � 2, � 1, 0, 1, 2, 3, ... }.

(iv) Rational numbers :These are real numbers which can be expressed in the

form of qp

, where p and q are integers and 0q .

e.g. 2/3, 37/15, -17/19.

All natural numbers, whole numbers and integers arerational.

Rational numbers include all Integers (without anydecimal part to it), terminating fractions ( fractions inwhich the decimal parts are terminating e.g. 0.75,� 0.02 etc.) and also non-terminating but recurringdecimals e.g. 0.666....., � 2.333...., etc.

Fractions :

(a) Common fraction : Fractions whose denominatoris not 10.

(b) Decimal fraction : Fractions whose denominator is10 or any power of 10.

(c) Proper fraction : Numerator < Denominator i.e.5

3.

(d) Improper fraction : Numerator > Denominator i.e.35 .

(e) Mixed fraction : Consists of integral as well as

fractional part i.e.72

3 .

(f) Compound fraction : Fraction whose numerator and

denominator themselves are fractions. i.e.5/72/3

.

Improper fraction can be written in the form of mixedfraction.

(v) Irrational Numbers :

All real number which are not rational are irrationalnumbers. These are non-recurring as well asnon-terminating type of decimal numbers.

For Ex. : 2 , 3 4 , 32 , 32 , 4 7 3 etc.

NUMBER SYSTEM

(vi) Real numbers : Numbers which can represent

actual physical quantities in a meaningful way areknown as real numbers. These can be representedon the number line. Number line is geometrical straightline with arbitrarily defined zero (origin).

(vii) Prime numbers : All natural numbers that have

one and itself only as their factors are called primenumbers i.e. prime numbers are exactly divisible by1 and themselves. e.g. 2, 3, 5, 7, 11, 13, 17, 19, 23,...etc.If P is the set of prime number then P = {2, 3, 5, 7,...}.

(viii) Composite numbers : All natural numbers, whichare not prime are composite numbers. If C is the setof composite number then C = {4, 6, 8, 9, 10, 12,...}.

1 is neither prime nor composite number.

(ix) Co-prime Numbers : If the H.C.F. of the given

numbers (not necessarily prime) is 1 then they areknown as co-prime numbers. e.g. 4, 9 are co-primeas H.C.F. of (4, 9) = 1.

Any two consecutive numbers will always be co-prime.

(x) Even Numbers : All integers which are divisible by 2are called even numbers. Even numbers are denotedby the expression 2n, where n is any integer. So, if E isa set of even numbers, then E = { ..., � 4, �2, 0, 2, 4,...}.

(xi) Odd Numbers : All integers which are notdivisible by 2 are called odd numbers . Oddnumbers are denoted by the general expression2n �1 where n is any integer. If O is a set of odd

numbers, then O = {..., �5, �3, �1, 1, 3, 5,...}.

(xii) Imaginary Numbers : All the numbers whosesquare is negative are called imaginary numbers.

e.g. 3i, -4i, i, ... ; where i = 1- .

(xiii) Complex Numbers : The combined form of realand imaginary numbers is known as complexnumbers. It is denoted by Z = A + iB where A is real part

and B is imaginary part of Z and A, B R.

The set of complex number is the super set of all thesets of numbers.

Squares : When a number is multiplied by itself thenthe product is called the square of that number.

Perfect Square : A natural number is called a perfectsquare if it is the square of any other natural numbere.g. 1, 4, 9,... are the squares of 1, 2, 3,... respectively.

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43PAGE # 43

Ex.1 Find the smallest number by which 300 must bemultiplied so that the product is a perfect square.

Sol. Given number is 300, first we resolve it into primefactors.

1

55

255

753

1502

3002

300 = 2 × 2 × 3 × 5 × 5

Clearly, 3 has no pair. Thus if we multiply it by 3 thenproduct will be a perfect square. Required smallest number is 3 .

Ex.2 Find the smallest number by which 1575 must bedivided so that the quotient becomes a perfect square.

Sol. Given number is 1575, first we write it as the product ofprime factors

1

77

355

1755

5253

15753

1575 = 3 × 3 × 5 × 5 × 7.

Clearly, 7 has no pair, so if we divide it by 7 then quotientbecome a perfect square.

Square roots : The square root of a number x is thatnumber which when multiplied by itself gives x as theproduct. As we say square of 3 is 9, then we can alsosay that square root of 9 is 3.The symbol use to indicate the square root of a number

is � � , i.e. 81 = 9, 225 = 15 ...etc.

We can calculate the square root of positive numbersonly. However the square root of a positive numbermay be a positive or a negative number.

e.g. 25 = + 5 or � 5.

Properties of Square Roots :

(i) If the unit digit of a number is 2, 3, 7 or 8, then it doesnot have a square root in N.

(ii) If a number ends in an odd number of zeros, then itdoes not have a square root in N.

(iii) The square root of an even number is even and

square root of an odd number is odd.e.g. 81 = 9, 256

= 16, 324 = 18 ...etc.

(iv) Negative numbers have no square root in set ofreal numbers.

Ex.3 Find the square root of 3 + 2 .

Sol. Let 23 = p + q

3 + 2 = p + q + 2 pq [By squaring both sides]

p + q = 3 ...(i) [By equating the parts]

2 pq = 2 ...(ii)

4pq = 2 ...(iii) [By squaring both sides ]

(p � q)2 = (p + q)2 � 4 pq

(p � q)2 = 9 � 2

(p � q)2 = 7

p � q = 7 ...(iv)

p + q = 3 [By eqn (i)]

p = 7321

[On adding (i) & (iv)]

q = 7321

[On subtracting (i) & (iv)]

23 = 2

1

7373

Cube : If any number is multiplied by itself three times

then the result is called the cube of that number.

Perfect cube : A natural number is said to be a perfect

cube if it is the cube of any other natural number.

Ex.4 What is the smallest number by which 675 must be

multiplied so that the product is a perfect cube.

Sol. Resolving 675 into prime factor, we get

1

55

255

753

2253

6753

675 = (3 × 3 × 3) × 5 × 5

Grouping the factor in triplets of equal factors, we get

675 = (3 × 3 × 3) × 5 × 5

We find that 3 occurs as a prime factor of 675 thrice but

5 occurs as a prime factor only twice. Thus, if we multiply

675 by 5, 5 will also occur as a prime factor thrice and

the product will be 3 × 3 × 3 × 5 × 5 × 5, which is a

perfect cube.

Hence, we must multiply 675 by 5 so that the product

becomes a perfect cube.

44PAGE # 44

Ex.5 What is the smallest number by which 18522 must bedivided so that the quotient is a perfect cube ?

Sol. Resolving 18522 into prime factors, we get

1

77

497

3437

10293

30873

92613

185222

18522 = 2 × 3 × 3 × 3 × 7 × 7 × 7

Grouping the factors in triplets of equal factors, we get18522 = 2 × (3 × 3 × 3) × (7 × 7 × 7)

Clearly, if we divide 18522 by 2, the quotient would be3 × 3 × 3 × 7 × 7 × 7 = 33 × 73 which is a perfect cube.Therefore, we must divide 18522 by 2 so that thequotient �9261� is a perfect cube.

Any irrational number of the form n a is given a special

name Surd. Where �a� is called radicand, rational. Also

the symbol n is called the radical sign and the index

n is called order of the surd.

n a is read as nth root of �a� and can also be written

as n

1

a .

Identification of Surds :

(i) 3 4 is a surd as radicand is a rational number..

Similar examples : ...,12,7,12,5 543

(ii) 2 + 3 is a surd (as surd + rational number will

give a surd)

Similar examples : ...,13,13,2�3 3

(iii) 34�7 is a surd as 7 � 4 3 is a perfect square

of 3�2 .

Similar examples : ...,549,54�9,347

(iv) 3 3 is a surd as 6613

1

21

3 3333

Similar examples : ...,6,5 4 53 3

(v) These are not a surds :

(A) 3 8 , because 3 33 28 which is a rational

number.

(B) 32 , because 2 + 3 is not a perfect square.

(C) 3 31 , because radicand is an irrational number..

Laws of Surds :

(i) n nnn aa = a

(ii) nnn abba [Here order should be same]

(iii) nnn

ba

ba

(iv) m nnmn m aaa

(v) pn pn aa

or, pn pmn m aa

[Important for changing order of surds]

Ex.6 If x = 1 + 21/3 + 22/3, then find the value of x3 � 3x2 � 3x � 1.

Sol. x = 1 + 21/3 + 22/3

x � 1 = (21/3 + 22/3)(x � 1)3 = ( 21/3 + 22/3)3

x3 � 3x2 + 3x � 1 = (21/3)3 + (22/3)3 + 3. 21/3. 22/3 (21/3 + 22/3)x3 � 3x2 + 3x � 1 = 2 + 22 + 3.21 (x � 1)

x3 � 3x2 + 3x � 1 = 6 + 6 (x � 1)

x3 � 3x2 + 3x � 1 = 6x

x3 � 3x2 � 3x � 1 = 0.

Ex.7 Simplify : 3 225 ba4ba8

Sol. 6 4426 3153 ba4ba8 = 63 ab2ba4 .

Ex.8 Divide : 24 by 3 200

Sol. 66 2

6 3

3 625

216

200

24

200

24

)(

)( .

Comparison of Surds :

It is clear that if x > y > 0 and n > 1 is a (+ve) integer

then n x > .yn e.g. 3 16 > ,3 12 5 36 > 5 25 and so

on.,

Ex.9 Which is greater in each of the following :

(i) 3 6 and 5 8 (ii) 21

and 331

Sol. (i) L.C.M. of 3 and 5 is 15.

1553 53 777666

1553 35 51288

15 51215 7776

5 83 6

(ii) L.C.M. of 2 and 3 is 6.

6

2

6

3

31

and21

= 6691

and81

as 8 < 9 91

81

So, 69

16

8

1 or,

21

> 331

.

45PAGE # 45

Ex.10 Arrange 3 32, and 4 5 in ascending order..

Sol. L.C.M. of 2, 3, 4 is 12.

1262 6 6422 ,

1243 43 8133 ,

and 1234 34 12555

As, 64 < 81 < 125.

121212 1258164

43 532

Conjugate Surds :

R.F. of ba and ba type surds are called

conjugate surds .

Ex.11 (i) 32 � is conjugate of 32 .

(ii) 15 is conjugate of 15 � .

Sometimes conjugate surd and reciprocals are same.

Ex.12 (i) 32 , it�s conjugate is 32 � , its reciprocal is

32 � & vice versa.

(ii) 62�5 , it�s conjugate is 625 , its reciprocal is

625 & vice versa.

Factors : �a� is a factor of �b� if there exists a relationsuch that a × n = b, where �n� is any natural number.

1 is a factor of all numbers as 1 × b = b.

Factor of a number cannot be greater than the number(infact the largest factor will be the number itself). Thusfactors of any number will lie between 1 and the numberitself (both inclusive) and they are limited.

Multiples : �a� is a multiple of �b� if there exists a relationof the type b × n = a. Thus the multiples of 6 are6 × 1 = 6, 6 × 2 = 12, 6 × 3 = 18, 6 × 4 = 24, and so on.

The smallest multiple will be the number itself and thenumber of multiples would be infinite.

NOTE :To understand what multiples are, let�s just take an

example of multiples of 3. The multiples are 3, 6, 9,12,.... so on. We find that every successive multiplesappears as the third number after the previous.So if one wishes to find the number of multiples of 6less than 255, we could arrive at the number through

6255

= 42 (and the remainder 3). The remainder is of

no consequence to us. So in all there are 42 multiples.

If one wishes to find the multiples of 36, find 36

255 = 7

(and the remainder is 3). Hence, there are 7 multiplesof 36.

Factorisation : It is the process of splitting any number

into form where it is expressed only in terms of the

most basic prime factors.

For example, 36 = 22 × 32. 36 is expressed in the

factorised form in terms of its basic prime factors.

Number of factors : For any composite number C,

which can be expressed as C = ap × bq × cr ×....., where

a, b, c ..... are all prime factors and p, q, r are positive

integers, then the number of factors is equal to

(p + 1) × (q + 1) × (r + 1).... e.g. 36 = 22 × 32. So the

factors of 36 = (2 +1) × (2 + 1) = 3 × 3 = 9.

Ex.13 If N = 123 × 34 ×52, find the total number of even

factors of N.

Sol. The factorised form of N is

(22 × 31)3 × 34 × 52 26 × 37 × 52.

Hence, the total number of factors of N is

(6 + 1) (7 + 1) (2 + 1) = 7 × 8 × 3 = 168.

Some of these are odd multiples and some are even.

The odd multiples are formed only with the combination

of 3s and 5s.

So, the total number of odd factors is (7 + 1)(2 + 1) = 24.

Therefore, the number of even factors

= 168 � 24 = 144.

Ex.14 A number N when factorised can be written

N = a4 × b3 × c7. Find the number of perfect squares

which are factors of N (The three prime numbers

a, b, c > 2).

Sol. In order that the perfect square divides N, the powers

of �a� can be 0, 2 or 4, i.e. 3.

Powers of �b� can be 0, 2, i.e. 2. Power of �c� can be 0, 2,

4 or 6, i.e. 4.

Hence, a combination of these powers given 3 × 2 × 4

i.e. 24 numbers.

So, there are 24 perfect squares that divides N.

Ex.15 Directions : (i to iv) Answer the questions based on

the given information.

There are one thousand lockers and one thousand

students in a school. The principal asks the first student

to go to each locker and open it. Then he asks the

second student go to every second locker and close it.

The third student goes to every third locker, and if it is

closed, he opens it, and it is open, he closes it. The

fourth student does it to every fourth locker and so on.

The process is completed with all the thousand

students.

(i) How many lockers are closed at the end of the

process ?

(ii) How many students can go to only one locker ?

(iii) How many lockers are open after 970 students

have done their job ?

(iv) How many student go to locker no. 840 ?

46PAGE # 46

Sol. (i to iv) : Whether the locker is open or not depends onthe number of times it is accessed. If it is accessedodd number of times, then it is open while if it isaccessed even number of times then it is closed.How many times a locker will be accessed dependson the locker no. If it contains odd number of factors,,then it will be open and if it contains even number offactors. Then it will be closed. We know that a perfectsquare contains odd number of factors while a non-perfect square contains even number of factors. Thusthe lockers with perfect square number will be openand the number of these perfect squares from 1 to1000 determines the no. of open lockers.

(i) No. of closed lockers = No. of non-perfect squarenumbers from 1 to 1000 = 1000 � 31 = 969.

(ii) Upto 500 students they can go to two or more thantwo lockers, while the rest 500 can go to only one locker.

(iii) The 31 perfect squares ( the last being 312 = 961)will be open while the lockers from 971 to 1000 is yetto be accessed last time so they all are open. The totalbeing = 31 + 30 = 61

(iv) The no. of students that have gone to locker no.840 is same as the no. of factors of 840.840 = 23 × 3 × 5 × 7.

So, the no. of factors = (3 + 1) (1 + 1) (1 + 1) (1 + 1) = 32.

LCM (least Common Multiple) : The LCM of given

numbers, as the name suggests is the smallestpositive number which is a multiple of each of the givennumbers.

HCF (Highest Common factor) : The HCF of givennumbers, as the name suggests is the largest factorof the given set of numbers.Consider the numbers 12, 20 and 30. The factors andthe multiples are :

FactorsGiven

numbersMultiples

1, 2, 3, 4, 6, 12 12 12, 24, 36, 48, 60, 72, 84, 96, 108, 120....

1, 2, 4, 5, 10, 20 20 20, 40, 60, 80, 100, 120.....

1, 2, 3, 5, 6, 10, 15, 30 30 30, 60, 90, 120....

The common factors are 1 and 2 and the commonmultiples are 60, 120...Thus the highest common factor is 2 and the leastcommon multiple is 60. Meaning of HCF is that theHCF is the largest number that divides all the givennumbers.Also since a number divides its multiple, the meaningof LCM is that it is smallest number which can bedivided by the given numbers.HCF will be lesser than or equal to the least of thenumbers and LCM will be greater than or equal to thegreatest of the numbers.

Ex.16 Find a number greater then 3 which when divided by4, 5, and 6 always leaves the same remainder 3.

Sol. The smallest number which, when divided by 4, 5 and6, leaves the remainder 3 in each case isLCM (4, 5 and 6) + 3 = 60 + 3 = 63.

Ex.17 In a school 437 boys and 342 girls have been dividedinto classes, so that each class has the same numberof students and no class has boys and girls mixed.What is the least number of classes needed?

Sol. We should have the maximum number of students ina class. So we have to find HCF (437, 342) = 19.HCF is also the factor of difference of the number.

Number of classes = 19437

+ 19342

= 23 + 18

= 41 classes.

For any two numbers x and y :x × y = HCF (x, y) × LCM (x, y).

HCF and LCM of fractions :

LCM of fractions = atorsmindenoofHCFnumeratorsofLCM

HCF of fractions = atorsmindenoofLCMnumeratorsofHCF

Make sure the fractions are in the most reducible form.

Ex.18 Find the least number which when divided by 6, 7, 8,9 and 10 leaves remainder 1.

Sol. As the remainder is same Required number = LCM of divisors + Remainder

= LCM (6, 7, 8, 9, 10) +1 = 2520 + 1 = 2521.

Ex.19 Six bells start tolling together and they toll at intervalsof 2, 4, 6, 8, 10, 12 sec. respectively, find.(i) after how much time will all six of them toll together ?(ii) how many times will they toll together in 30 min ?

Sol. The time after which all six bells will toll together mustbe multiple of 2, 4, 6, 8, 10, 12.Therefore, required time = LCM of time intervals.

= LCM (2, 4, 6, 8, 10, 12) = 120 sec.Therefore after 120 s all six bells will toll together.After each 120 s, i.e. 2 min, all bell are tolling together.

Therefore in 30 min they will toll together

1

230

= 16 times1 is added as all the bells are tolling together at thestart also, i.e. 0th second.

Ex.20 LCM of two distinct natural numbers is 211. What istheir HCF ?

Sol. 211 is a prime number. So there is only one pair ofdistinct numbers possible whose LCM is 211,i.e. 1 and 211. HCF of 1 and 211 is 1.

Ex.21 An orchard has 48 apple trees, 60 mango trees and96 banana trees. These have to be arranged in rowssuch that each row has the same number of trees andall are of the same type. Find the minimum number ofsuch rows that can be formed.

Sol. Total number of trees are 204 and each of the treesare exactly divisible by 12. HCF of (48, 60, 96).

12204

= 17 such rows are possible.

47PAGE # 47

Division Algorithm : General representation of result is,

DivisormainderRe

QuotientDivisor

Dividend

Dividend = (Divisor × Quotient ) + Remainder

NOTE :(i) (xn � an) is divisible by (x � a) for all the values of n.

(ii) (xn � an) is divisible by (x + a) and (x � a) for all the

even values of n.(iii) (xn + an) is divisible by (x + a) for all the odd values of n.

Test of Divisibility :

No. Divisiblity Test

2 Unit digit should be 0 or even

3 The sum of digits of no. should be divisible by 3

4 The no formed by last 2 digits of given no. should be divisible by 4.

5 Unit digit should be 0 or 5.

6 No should be divisible by 2 & 3 both

8 The number formed by last 3 digits of given no. should be divisible by 8.

9 Sum of digits of given no. should be divisible by 9

11The difference between sums of the digits at even & at odd places should be zero or multiple of 11.

25 Last 2 digits of the number should be 00, 25, 50 or 75.

Rule for 7 : Double the last digit of given number andsubtract from remaining number the result should bezero or divisible by 7.

Ex.22 Check whether 413 is divisible by 7 or not.Sol. Last digit = 3, remaining number = 41, 41 � (3 x 2) = 35

(divisible by 7). i.e. 413 is divisible by 7.This rule can also be used for number having morethan 3 digits.

Ex.23 Check whether 6545 is divisible by 7 or not.Sol. Last digit = 5, remaining number 654, 654 � (5 x 2)

= 644; 64 � (4 x 2) = 56 divisible by 7. i.e. 6545 is

divisible by 7.

Rule for 13 : Four times the last digit and add toremaining number the result should be divisible by13.

Ex.24 Check whether 234 is divisible by 13 or not .Sol. 234, (4 x 4) + 23 = 39 (divisible by 13), i.e. 234 is divisible

by 13.

Rule for 17 : Five times the last digit of the number andsubtract from previous number the result obtainedshould be either 0 or divisible by 17.

Ex.25 Check whether 357 is divisible by 17 or not.Sol. 357, (7 x 5) � 35 = 0, i.e. 357 is divisible by 17.

Rule for 19 : Double the last digit of given number andadd to remaining number The result obtained shouldbe divisible by 19.

Ex.26 Check whether 589 is divisible by 19 or not.Sol. 589, (9 x 2) + 58 = 76 (divisible by 19), i.e. the number

is divisible by 19.

Ex27. Find the smallest number of six digits which is exactlydivisible by 111.

Sol. Smallest number of 6 digits is which is 100000.On dividing 100000 by 111, we get 100 as remainder.

Number to be added = (111 - 100) = 11.

Hence, required number = 100011.

Ex.28 Find the largest four digit number which whenreduced by 54, is perfectly divisible by all even naturalnumbers less than 20.

Sol. Even natural numbers less than 20 are 2, 4, 6, 8, 12,14, 16, 18.Their LCM = 2 × LCM of first 9 natural numbers

= 2 × 2520 = 5040

This happens to be the largest four-digit numberdivisible by all even natural numbers less than 20. 54was subtracted from our required number to get thisnumber.Hence, (required number � 54) = 5040

required number = 5094.

Ex.29 Ajay multiplied 484 by a certain number to get theresult 3823a. Find the value of �a�.

Sol. 3823a is divisible by 484, and 484 is a factor of 3823a.4 is a factor of 484 and 11 is also a factor of 484.Hence, 3823a is divisible by both 4 and 11.To be divisible by 4, the last two digits have to bedivisible by 4.�a� can take two values 2 and 6.

38232 is not divisible by 11, but 38236 is divisible by11.Hence, 6 is the correct choice.

Ex.30 Which digits should come in place of and $ if thenumber 62684$ is divisible by both 8 and 5 ?

Sol. Since the given number is divisible by 5, so 0 or 5 mustcome in place of $. But, a number ending with 5 innever divisible by 8. So, 0 will replace $.Now, the number formed by the last three digits is 40,which becomes divisible by 8, if is replaced by 4 or 8.Hence, digits in place of and $ are 4 or 8 and 0respectively.

Ex.31 On dividing 15968 by a certain number, the quotientis 89 and the remainder is 37. Find the divisor.

Sol. Divisor = 89

3715968Quotient

mainderReDividend

= 179.

Ex.32 How many numbers between 200 and 600 aredivisible by 4, 5, and 6 ?

Sol. Every such number must be divisible by L.C.M. of 4, 5,6, i.e.60.Such numbers are 240, 300, 360, 420, 480, 540Clearly, there are 6 such numbers.

The method of finding the remainder without actuallyperforming the process of division is termed asremainder theorem.

Remainder should always be positive. For example ifwe divide �22 by 7, generally we get �3 as quotient and

�1 as remainder. But this is wrong because remainder

is never be negative hence the quotient should be �4

and remainder is +6. We can also get remainder 6 byadding �1 to divisor 7 ( 7�1 = 6).

48PAGE # 48

Ex.33 Two numbers, x and y, are such that when divided by6, they leave remainders 4 and 5 respectively. Find theremainder when (x2 + y2) is divided by 6.

Sol. Suppose x = 6k1 + 4 and y = 6k

2 + 5

x2 + y2 = (6k1 + 4)2 + (6k

2 + 5)2

= 36k1

2 + 48k1 + 16 + 36k

22 + 60k

2 + 25

= 36k1

2 + 48k1 + 36k

22 + 60k

2 + 41

Obviously when this is divided by 6, the remainder willbe 5.

Ex.34 A number when divided by 259 leaves a remainder139. What will be the remainder when the samenumber is divided by 37 ?

Sol. Let the number be P.So, P � 139 is divisible by 259.

Let Q be the quotient then, 259

139P = Q

P = 259Q + 139

37P

= 37

139Q259

259 is divisible by 37, When 139 divided by 37, leaves a remainder of 28.

Ex.35 A number being successively divided by 3, 5 and 8leaves remainders 1, 4 and 7 respectively. Find therespective remainders if the order of divisors bereversed.

Sol.

714z81y5

x3

z = (8 × 1 + 7) = 15 ; y (5z + 4) = (5 × 15 + 4) = 79 ;

x = (3y + 1) = (3 × 79 + 1) = 238.

Now, 214536295

2388

Respective remainders are 6, 4, 2.

Ex.36 A number was divided successively in order by 4, 5and 6. The remainders were respectively 2, 3 and 4.Then find out the number.

Sol.

413z62y5

x4

z = (6 × 1 + 4) = 10

y = (5 × z + 3) = (5 × 10 + 53) = 53

x = (4 × y + 2) = (4 × 53 + 2) = 214

Hence, the required number is 214.

Ex.37 In dividing a number by 585, a student employed themethod of short division. He divided the numbersuccessively by 5, 9 and 13 (factors of 585) and got theremainders 4, 8 and 12. If he had divided number by585, then find out the remainder.

Sol.

1218z134y9

x5

Now, 1169 when divided by 585 gives remainder= 584.

To find the remainder of big number

NOTE :

(i) Binomial Expansion :

(a + b)n = an + 1!

nan�1b +

2!

1)n(n an � 2b2 + .... + bn, or

(a � b)n = an � 1!

nan�1b +

2!

1)n(n an� 2b2 � ... + (� 1)nbn.

Hence, first term is pure of a i.e an and last digit is pureof b, i.e. bn.

(ii) Total number of terms in the expansion of (a + b)n is(n + 1).

Ex.38 What is the remainder when 738 is divided by 48.

Sol.48738

=

487

192

=

4849 19

=

48148 19

so by using

binomial expansion, we can say that 18 terms arecompletely divisible by 48 but the last term which is

481 19

is not divisible. So, 191 = 1 is the remainder..

Ex.39 What is the remainder if 725 is divided by 4?Sol. 725 can be written (8�1)25. There are 26 terms in all and

first 25 terms are divisible by 8, hence also by 4. Thelast term is (�1)25. Hence, (8 �1)25 can be written 8X � 1

or 4Y �1 ( where Y = 2X). So, 4Y � 1 divided by 4 leaves

the remainder 3.

Ex.40 What is the remainder if 345 is divided by 8 ?Sol. 345 can be written 922 × 3. 9 can be written as (8 + 1).

Hence, any power of 9 can be written 8N + 1. In otherwords, any power of 9 is 1 more than a multiple of 8.Hence, (8N + 1) × 3 leaves remainder 3 when divided

by 8.

Ex.41 What is the remainder when 161514 is divided by 5 ?

Sol.161514 = (15 �1)odd = 15n + (�1)odd, i.e. a (multiple of 5)

�1. Thus when divided by 5 the remainder will be (�1),

i.e. 4.

Ex.42 What is the remainder when 357 + 27 is divided by28?

Sol. 357 = (33)19

357 + 27 = (27)19 + 27= (28 � 1)19 + 27= 28M + (�1)19 + 27 [Expand by binomial theorem]= 28M � 1 + 27

= 28M + 26When 28M + 26 divided by 28, the remainder is 26.Hence, the required remainder is 26.

49PAGE # 49

Ex.43 What is the remainder when 82361 + 83361 + 84361

+ 85361 + 86361 is divided by 7?Sol. 82361 + 83361 + 84361 + 85361 + 86361 = [(84 � 2)361

+ (84 � 1)361 + + 84361 + (84 + 1)361 + (84 + 2)361]Since, 84 is a multiple of 7, then the remainder will bewhen, (� 2)361 + (�1)361 + 1361 + 2361 is divided by 7.is (� 2)361 + (�1)361 + 1361 + 2361 = 0.So the remainder is zero.

We are having 10 digits in our number systems andsome of them shows special characterstics like they,repeat their unit digit after a cycle, for example 1 repeatits unit digit after every consecutive power. So, itscyclicity is 1 on the other hand digit 2 repeat its unitdigit after every four power, hence the cyclicity of 2 isfour. The cyclicity of digits are as follows :

Digit Cyclicity

0, 1, 5 and 6 1

4 and 9 2

2, 3, 7 and 8 4

So, if we want to find the last digit of 245, divide 45 by 4.The remainder is 1 so the last digit of 245 would besame as the last digit of 21 which is 2.

To Find the Unit Digit in Exponential Expressions :

(i) When there is 2 in unit�s place of any number.

Since, in 21 unit digit is 2, in 22 unit digit is 4, in 23 unitdigit is 8, in 24 unit digit is 6, after that the unit�s digit

repeats. e.g. unit digit(12)12 is equal to the unit digit of,24 i.e.6

Ex.44 In (32)33 unit digit is equal to the unit digit of 21 i.e. 2.

(ii) When there is 3 in unit�s place of any number.

Since, in 31 unit digit is 3, in 32 unit digit is 9, in 33 unitdigit is 7, in 34 unit digit is 1, after that the unit�s digit

repeats.

Ex.45 In (23)13 unit digit be 3

Ex.46 In (43)46 unit digit be 9

(iii) When there is 4 in unit�s place of any number.

Since, in 41 unit digit is 4, in 42 unit digit is 6, after thatthe unit�s digit repeats.

Ex.47 In (34)14 unit digit is 6

Ex.48 In (34)33 unit digit is 4

(iv) When there is 5 in unit�s place of any number.

Since, in 51 unit digit is 5, in 52 unit digit is 5 and so on.

Ex.49 In (25)15 unit digit is 5

(v) When there is 6 in unit�s place of any number.

Since, in 61 unit digit is 6, in 62 unit digit is 6 & so on.

,Ex.50 In (46)13 unit digit is 6.

(vi) When there is 7 in unit�s place of any number.

Since, in 71 unit digit is 7, in 72 unit digit is 9, in 73 unitdigit is 3, in 74 unit digit is 1, after that the unit�s digit

repeats.

Ex.51 In (57)9 unit digit is 7

Ex.52 In (97)99 unit digit is 3

(vii) When there is 8 in unit�s place of any number.

Since, in 81 unit digit is 8, in 82 unit digit is 4, in 83 unitdigit is 2, in 84 unit digit is 6, after that unit�s digit repeats

after a group of 4.

(viii) When there is 9 in unit�s place of any number.

Since, in 91 unit�s digit is 9, in 92 unit�s digit is 1, after

that unit�s digit repeats after a group of 2.

(ix) When there is zero in unit�s place of any number.

There will always be zero in unit�s place.

Ex.53 Find the last digit of(i) 357 (ii) 1359

Sol. (i) The cyclicity of 3 is 4. Hence, 4

57gives the remainder

1. So, the last digit of 357 is same as the last digit of 31,i.e. 3.

(ii) The number of digits in the base will not make adifference to the last digit. It is last digit of the basewhich decides the last digit of the number itself. For

1359, we find 4

59 which gives a remainder 3. So the

last digit of 1359 is same as the last digit of 33, i.e. 7.

Ex.54 Find the last digit of the product 723 x 813.Sol. Both 7 and 8 exhibit a cyclicity of 4. The last digits are

71 = 7 81 = 872 = 9 82 = 473 = 3 83 = 274 = 1 84 = 675 = 7 85 = 8

The cycle would repeat itself for higher powers.723 ends with the same last digit as 73, i.e. 3.813 ends with the same last digit as 81, i.e. 8. Hence,the product of the two numbers would end with thesame last digit as that of 3 × 8, i.e. 4.

Ex.55Find unit�s digit in y = 717 + 734

Sol. 717 + 734 = 71 + 72 = 56, Hence the unit digit is 6

Ex.56What will be the last digit of 766475)73(

Sol. Let 766475)73( = (73)x where x =

766475 = (75)even power

Cyclicity of 3 is 4 To find the last digit we have to find the remainderwhen x is divided by 4.x = (75)even power = (76 � 1)even power , where n is divided by4 so remainder will be 1.

Therefore, the last digit of 766475)73( will be 31 = 3.

Ex.57What will be the unit digit of 556375)87( .

Sol. Let 556375)87( = (87)x where x =

556375 = (75)odd

Cyclicity of 7 is 4. To find the last digit we have to find the remainderwhen x is divided by 4.x = (75)odd power = (76 � 1)odd power

where x is divided by 4 so remainder will be �1 or 3, but

remainder should be positive always

therefore, the last digit of 556375)87( will be 73 = 343.

Hence, the last digit is of 556375)87( is 3.

50PAGE # 50

To Find the Last Two Digits in Exponential

Expressions :

We know that the binomial theorem :

(a + b)n = an + 1!

nan�1b +

2!

1)n(n an � 2b2 + .... + bn.

(i) Last two digits of numbers ending in 1 :

Let's start with some examples.

Ex.58What are the last two digits of 31786 ?

Sol. 31786 = (30 + 1)786 = 30786 + 786 × 30785 × 1

+ 2!

1)(786786 × 30784 × 12 + .... + 1786 .

Note that all the terms excluding last two terms will

end in two or more zeroes. The last two terms are

786 × 30 ×1785 and 1786. Now, the second last term

will end with one zero and the tens digit of the second

last term will be the product of 786 and 3 i.e. 8.

Therefore, the last two digits of the second last term

will be 80. The last digit of the last term is 1. So the

last two digits of 31786 are 81.

Ex.59Find the last two digits of 412789.

Sol. According to the previous example we can calculate

the answer 61 (4 × 9 = 36. Therefore, 6 will be the

tens digit and one will be the units digit).

Ex.60 Find the last two digits of 7156747.

Sol. Last two digits will be 91 (7 × 7 gives 9 and 1 as

units digit)

Ex.61 Find the last two digits of 51456 × 61567.

Sol. The last two digits of 51456 will be 01 and the last

two digits of 61567 will be 21. Therefore, the last two

digits of 51456 × 61567 will be the last two digits of

01 × 21 = 21.

(ii) Last two digits of numbers ending in 3, 7 or 9 :

Ex.62 Find the last two digits of 19266.

Sol. 19266 = (192)133. Now, 192 ends in 61 (192 = 361)

therefore, we need to find the last two digits of (61)133.

Once the number is ending in 1 we can straight away

get the last two digits with the help of the previous

method. The last two digits are 81 (6 × 3 = 18, so

the tenth digit will be 8 and last digit will be 1).

Ex.63 Find the last two digits of 33288.

Sol. 33288 = (334)72. Now 334 ends in 21 (334 = 332 × 332

= 1089 × 1089 = xxxxx21) therefore, we need to find

the last two digits of 2172. By the previous method,

the last two digits of 2172 = 41 (tens digit = 2 × 2 = 4,

unit digit = 1)

Ex.64Find the last two digits of 87474.

Sol. 87474 = 87472 × 872 = (874)118 × 872 = (69 × 69)118 × 69

[The last two digits of 872 are 69]

= 61118 × 69 = 81 × 69 = 89.

(iii) Last two digits of numbers ending in 2, 4, 6 or 8 :There is only one even two-digit number whichalways ends in itself (last two digits) - 76 i.e. 76raised to any power gives the last two digits as 76.Therefore, our purpose is to get 76 as last two digitsfor even numbers. We know that 242 ends in 76 and210 ends in 24. Also, 24 raised to an even poweralways ends with 76 and 24 raised to an odd poweralways ends with 24. Therefore, 2434 will end in 76and 2453 will end in 24.

Ex.65Find the last two digits of 2543.Sol. 2543 = (210)54 × 23 = (24)54 (24 raised to an even

power). 23 = 76 × 8 = 08.

NOTE :Here if you need to multiply 76 with 2n, then you canstraightaway write the last two digits of 2n becausewhen 76 is multiplied with 2n the last two digitsremain the same as the last two digits of 2n.Therefore, the last two digits of 76 ×27 will be the

last two digits of 27 = 28.

Ex.66Find the last two digits of 64236.Sol. 64236 = (26)236 = 21416 = (210)141 × 26 = 24141 (24 raised

to odd power) × 64 = 24 × 64 = 36.

Now those numbers which are not in the form of 2n

can be broken down into the form 2n × odd number.

We can find the last two digits of both the partsseparately.

Ex.67Find the last two digits of 62586.Sol. 62586 = (2 × 31)586 = 2586 × 31586 = (210)58 × 26 × 31586

= 76 × 64 × 81 = 84.

Ex.68Find the last two digits of 54380.Sol. 54380 = (2 × 33)380 = 2380 × 31140 = (210)38 × (34)285 = 76

× 81285 = 76 × 01 = 76.

Ex.69Find the last two digits of 56283.Sol. 56283 = (23 × 7)283 = 2849 × 7283 = (210)84 × 29 × (74)70 × 73

= 76 × 12 × (01)70 × 43 = 16.

Ex.70 Find the last two digits of 78379.Sol. 78379 = (2 × 39)379 = 2379 × 39379 = (210)37 × 29 × (392)189

× 39 = 24 × 12 × 81 × 39 = 92.

Factorial n : Product of n consecutive natural numbers

is known as �factorial n� it is denoted by �n!�.

So, n! = n(n � 1)(n � 2).............321.

e.g. 5! = 5 × 4 × 3 × 2 × 1 = 120.

The value of factorial zero is equal to the value of

factorial one. Hence 0! = 1 = 1!

The approach to finding the highest power of x dividing

y! is

32 x

y

x

yxy

......., where [ ] represents just

the integral part of the answer and ignoring the fractionalpart.

51PAGE # 51

Ex.71 What is the highest power of 2 that divides 20!completely?

Sol. 20! = 1 × 2 × 3 × 4 ×....× 18 × 19 × 20

= 1 × (21) × 3 × (22) × 5 × (21 × 31) × 7 × (23) × ..... so on.

In order to find the highest power of 2 that divides theabove product, we need to find the sum of the powersof all 2 in this expansion. All numbers that the divisibleby 21 will contribute 1 to the exponent of 2 in the product

12

20= 10. Hence, 10 numbers contribute 21 to the

product. Similarly, all numbers that are divisible by 22

will contribute an extra 1 to the exponent of 2 in the

product, i.e 22

20 = 5. Hence, 5 numbers contribute an

extra 1 to exponents. Similarly, there are 2 numbers thatare divisible by 23 and 1 number that is divisible by 24.Hence, the total 1s contributed to the exponent of 2 in20! is the sum of ( 10 + 5 +2 +1) = 18. Hence, group ofall 2s in 20! gives 218 x (N), where N is not divisible by2. If 20! is divided by 2x then maximum value of x is 18.

Ex.72 What is the highest power of 5 that divides ofx = 100! = 100 × 99 × 98 × ...... × 3 × 2 × 1.

Sol. Calculating contributions of the different powers of 5,

we have 15

100= 20, 25

100 = 4.

Hence, the total contributions to the power of 5 is 24, orthe number 100! is divisible by 524.

Ex.73 How many zeros at the end of first 100 multiples of10.

Sol. First 100 multiple of 10 are = 10 × 20 × 30 × ......× 1000

= 10100 (1 × 2 × 3 × .......× 100)

= 10100 × 1024 × N

= 10124 × N

Where N is not divisible by 10So, there are 124 zero at the end of first 100 multiple of10.

Ex.74 What is the highest power of 6 that divides 9!

Sol. By the normal method. 69

= 1 and 26

9 = 0. Thus

answers we get is 1 which is wrong. True there is justone multiple of 6 from 1 to 9 but the product 2 × 3 = 6

and also 4 × 9 = 36, can further be divided by 6. Thus,

when the divisor is a composite number find thehighest power of its prime factors and then proceed. Inthis case, 9! can be divided by 27 and 34 and thus by 64

(In this case we need not have checked power of 2 asit would definitely be greater then that of 3).

Ex.75What is the largest power of 12 that would divide 49! ?Sol. To check the highest power of 12 in 49!, we need to

check the highest powers of 4 and 3 in it.Highest power of 3 in 49! = 22Highest power of 2 in 49! = 46

Highest power of 4 in 49! = 2

46 = 23

Highest power of 12 will be 22. (Since the commonpower between 3 and 4 is 22).

Ex.76 How many zeros will be there at the end of !36!36 ?Sol. Highest power of 5 in 36! is 8.

So there will be 8 zeros at the end of 36!

So at the end of !36!36 , there will be 8 × 36! zeros.

The number system that we work in is called the�decimal system�. This is because there are 10 digitsin the system 0-9. There can be alternative system thatcan be used for arithmetic operations. Some of themost commonly used systems are : binary, octal andhexadecimal.These systems find applications in computing.Binary system has 2 digits : 0, 1.Octal system has 8 digits : 0, 1, 2..., 7.Hexadecimal system has 16 digits : 0, 1, 2,..., 9, A , B,C, D, E, F.After 9, we use the letters to indicate digits. For instance,A has a value 10, B has a value 11, C has a value 12,...so on in all base systems.The counting sequences in each of the systems wouldbe different though they follow the same principle.Conversion : conversion of numbers from (i) decimalsystem to other base system. (ii) other base system todecimal system.

(i) Conversion from base 10 to any other base :

Ex.77 Convert (122)10

to base 8 system.

Sol.

10

718

2158

1228

The number in decimal is consecutively divided by thenumber of the base to which we are converting thedecimal number. Then list down all the remainders inthe reverse sequence to get the number in that base.So, here (122)

10 = (172)

8.

Ex.78 Convert (169)10

in base 7

Sol. 03241697

77

133

Remainder

(169)10 =(331)7

Ex.79 Convert (0.3125)10

to binary equivalent.Sol. Integer

2 0.3125 = 0.625 02 0.625 = 1.25 12 0.25 = 0.50 02 0.50 = 1.00 1Thus

(0.3125)10

= (0.1010)2

52PAGE # 52

Ex.80 Convert (1987.725)10

(........)8

Sol. First convert non-decimal part into base 8.

30

738

0318

32488

19878

(1987)10

= (3703)8

Now we have to convert (0.725)10 (........)

8

Multiply

0.725 × 8 = [5.8] ...5

0.8 × 8 = [6.4]

0.4 × 8 = [3.2]

0.2 × 8 = [1.6]

0.6 × 8 = [4.8]

...6

...3

...1

...4

Keep on accomplishing integral parts after

multiplication with decimal part till decimal part is zero.

(0.725)10

= (0.56314...)8

(1987.725)10

= (3703.56314...)8

(ii) Conversion from any other base to decimal system

Ex.81 Convert (231)8 into decimal system.

Sol. (231)8 , the value of the position of each of the numbers

( as in decimal system) is :

1 = 80 × 1

3 = 81 × 3

2 = 82 × 2

Hence, (231)8 = (80 × 1 + 81 × 3 + 82 × 2)

10

(231)8

= (1 + 24 + 128)10

(231)8

= (153)10

Ex.82 Convert (0.03125)10

to base 16.

Sol. 16 0.3125 = 0.5 0

16 0.5 = 8.0 8

So (0.03125)10

= (0.08)16

Ex.83 Convert (761.56)8 (......)

16

Sol. In such conversion which are standard form

conversions, it is easier to

(761.56)8 (.....)

2 (.....)

16

Converting every digit in base 8 to base 2,

(111110001.10110)2 (1F1.B8)

16

Ex.84 Convert (3C8.08)16

to decimal

Sol. (3C8.08)16

= 3 162 + C 161

+ 8 16 + 0 16�1 + 8 16�2

= 768 + 192 + 8 + 0 + 0.03125 = (968.03125)10

So (3C8.08)16

= (968.03125)10

Ex.85 If a � b = 2,

and 0cc

_____bb

aa

then find the value of a, b and c.

Sol. These problems involve basic number

(i) aa + bb = 11(a + b) (ii) aa, bb are two-digit numbers.

Hence, their sum cannot exceed 198. So, c must be 1.

(iii) Hence, cc0 = 110. This implies a + b = 10 or a = 6

and b = 4.

Such problems are part of a category of problems called

alphanumerices.

Ex.86 If _____9aa

ca

b3a

then find a, b and c if each of them is

distinctly different digit.

Sol. (i) since the first digit of (a 3 b) is written as it is after

subtracting ac carry over from a to 3.

(ii) there must be a carry over from 3 to b, because if no

carry over is there, it means 3 � a = a.

2a = 3

a = 23

which is not possible because a is a digit. For a carry

over 1, 2 � a = a

a = 1

(iii) it means b and c are consecutive digit (2, 3),

(3, 4),.... (8, 9)

Ex.87 If

1 a 4x 3 b

8 c 87 2

5 d 8

s

t

then, find the value of a, b, c, d, s and t, where all of

them are different digits.

Sol. Let us consider 1 a 4 × 3 = s72.

a × 3 results in a number ending in 6.

As 16 and 26 is ruled out, a is 2.

Thus, s = 3, t = 4

Now 1 a 4 × b = 8c8 ; b = 2 or 7

Again 2 is ruled out because in that case, product would

be much less than 800.

b = 7.

Hence, a = 2, b = 7, c = 6, d = 8, s = 3 and t = 4.

53PAGE # 53

1. If the number 357y25x is divisible by both 3 and 5, thenfind the missing digit in the unit�s place and the

thousand place respectively are :(A) 0, 6 (B) 5, 6(C) 5, 4 (D) None of these

2. There are four prime numbers written in ascendingorder. The product of the first three is 385 and that ofthe last three is 1001. The last number is :(A) 11 (B) 13(C) 17 (D) 19

3. Find the square root of 7 � 4 3 .

(A) 3�2 (B) 3�5

(C) 5�2 (D) None of these

4. The number of prime factors of (3 × 5)12 (2 × 7)10 (10)25

is :(A) 47 (B) 60(C) 72 (D) 94

5. How many three-digit numbers would you find, whichwhen divided by 3, 4, 5, 6, 7 leave the remainders 1, 2,3, 4, and 5 respectively ?(A) 4 (B) 3(C) 2 (D) 1

6. Six strings of violin start vibrating simultaneously andthey vibrate at 3, 4, 5, 6,10 and 12 times in a minute,find :i. After how much time will all six of them vibratetogether ?ii. How many times will they vibrate together in 30 min ?(A) 60 sec, 31 times (B) 60 min, 31 times(C) 120 sec, 15 times (D) None of these

7. The HCF of 2 numbers is 11 and their LCM is 693. Iftheir sum is 176, find the numbers.(A) 99,77 (B) 110, 66(C) 88,77 (D) 121, 44

8. If P is a prime number, then the LCM of P and (P + 1) is :(A) P(P +1) (B) (P + 2)P(C) (P + 1)(P � 1) (D) None of these

9. Find out (A + B + C + D) such that AB x CB = DDD, whereAB and CB are two-digit numbers and DDD is a three-digit number.(A) 21 (B) 19(C) 17 (D) 18

10. Three pieces of cakes of weights 21

4 Ibs, 43

6 Ibs and

51

7 Ibs respectively are to be divided into parts of equal

weights. Further, each must be as heavy as possible.If one such part is served to each guest, then what isthe maximum number of guests that could beentertained ?(A) 54 (B) 72(C) 20 (D) 41

11. How many natural numbers between 200 and 400 are

there which are divisible by

i. Both 4 and 5?

ii. 4 or 5 or 8 or 10 ?

(A) 9, 79 (B) 10, 80

(C) 10, 81 (D) None of these

12. 64636261 4444 is divisible by :

(A) 3 (B) 10

(C) 11 (D) 13

13. If x is a whole number, then x2 (x2 � 1) is always divisible

by :

(A) 12 (B) 24

(C) 12 � x (D) Multiple of 12

14. If 653 xy is exactly divisible by 80 then the find the value

of (x + y)

(A) 2 (B) 3

(C) 4 (D) 6

15. Find the unit digit of (795 � 358).

(A) 6 (B) 4

(C) 3 (D) None of these

16. When a number P is divided by 4 it leaves remainder

3. If the twice of the number P is divided by the same

divisor 4 than what will be the remainder ?

(A) 0 (B) 1

(C) 2 (D) 6

17. If (232 +1) is divisible by a certain number then which of

the following is also divisible by that number.

(A) (216 � 1) (B) 216 + 1

(C) 296 + 1 (D) None of these

18. When 1! + 2! + 3! + ... + 25! is divided by 7, what will be

the remainder ?

(A) 0 (B) 5

(C) 1 (D) None of these

19. A number when divided by 342 gives a remainder 47.

When the same number is divided by 19, what would

be the remainder ?

(A) 3 (B) 5

(C) 9 (D) None of these

20. What is the remainder when 9875347 × 7435789

× 5789743 is divided by 4 ?

(A) 1 (B) 2

(C) 3 (D) None of these

21. P is a prime number greater than 5. What is the

remainder when P is divided by 6?

(A) 5 (B) 1

(C) 1 or 5 (D) None of these

22. What is the remainder when 587 is divided by 15?

(A) 0 (B) 5

(C) 10 (D) None of these

54PAGE # 54

23. What is the remainder when 763 is divided by 344?

(A) 1 (B) 343

(C) 6 (D) 338

24. What is the remainder when 3040 is divided by 17?

(A) 1 (B) 16

(C) 13 (D) 4

25. What is the remainder when 7413 � 4113 + 7513 � 4213 is

divided by 66?

(A) 2 (B) 64

(C) 1 (D) 0

26. A number when divided successively by 4 and 5 leaves

remainders 1 and 4 respectively. When it is

successively divided by 5 and 4, then the respective

remainders will be :

(A) 1, 2 (B) 2, 3

(C) 3, 2 (D) 4, 1

27. How many zeros will be there at the end of the product!10!8!6!4!2 !10!8!6!4!2 ?

(A) 10! + 6! (B) 2 (10!)

(C) 10! + 8! + 6! (D) 6! + 8! + 2 (10!)

28. When Sholey screened on the TV there was a

commercial break of 5 min after every 15 min of the

movie. If from the start of the movie to the end of the

movie there was in all 60 min of commercials that was

screened what is the duration the movie ?

(A) 180 min (B) 195 min

(C) 169 min (D) 165 min

29. Which of the following numbers is exactly divisible by

all prime numbers between 1 and 17 ?

(A) 345345 (B) 440440

(C) 510510 (D) 515513

Directions : (32 to 36) Read the following information

carefully and answer the questions given below.

In a big hostel, there are 1,000 rooms. In that hostel

only even numbers are used for room numbers, i.e.

the room numbers are 2, 4, 6, ...., 1998, 2000. All the

rooms have one resident each. One fine morning, the

warden calls all the residents and tells them to go

back to their rooms as well as multiples of their room

numbers. When a guy visits a room and finds the door

open, he closes it, and if the door is closed, he opens

it, All 1,000 guys do this operation. All the doors were

open initially.

30. The last room that is closed is room number ?

(A) 1936 (B) 2000

(C) 1922 (D) None of these

31. The 38th room that is open is room number :

(A) 80 (B) 88

(C) 76 (D) None of these

32. If only 500 guys, i.e. residents of room number 2 to1000 do the task, then the last room that is closed isroom number(A) 2000 (B) 1936(C) 1849 (D) None of these

33. In the case of previous question, how many roomswill be closed in all ?(A) 513 (B) 31(C) 13 (D) 315

34. If you are a lazy person, you would like to stay in a roomwhose number is :(A) more than 500 (B) more than 1000(C) 500 (D) 2000

35. A 4-digit number is formed by repeating a 2-digitnumber such as 2525, 3232 etc. Any number of thisform is exactly divisible by :(A) 7 (B) 11(C) 13(D) Smallest 3-digit prime number

36. If we write all the whole numbers from 200 to 400, thenhow many of these contain the digit 7 only once ?(A) 32 (B) 34(C) 35 (D) 36

37. If (12 + 22 + 32 + .....+ 102) = 385, then the value of(22 + 42 + 62 +...... + 202).(A) 770 (B) 1155(C) 1540 (D) (385 × 385)

38. How many four-digit numbers are there such that thedigits are repeated at least once, i.e. all the digits donot occur only once ?(A) 9000 (B) 4536(C) 4464 (D) None of these

39. Find the total number of prime factors in the expression(4)11 × (7)5 × (11)2.(A) 37 (B) 33(C) 26 (D) 29

40. The largest number which exactly divides the productof any four consecutive natural numbers is :(A) 6 (B) 12(C) 24 (D) 120

41. The largest natural number by which the product ofthree consecutive even natural numbers is alwaysdivisible, is :(A) 6 (B) 24(C) 48 (D) 96

42. The sum of three consecutive odd numbers is alwaysdivisible by :I. 2 II. 3 III. 5 IV. 6(A) Only I (B) Only II(C) Only I and III (D) Only II and IV

43. A number is multiplied by 11 and 11 is added to theproduct. If the resulting number is divisible by 13, thesmallest original number is :(A) 12 (B) 22(C) 26 (D) 53

55PAGE # 55

44. A 3-digit number 4a3 is added to another 3-digitnumber 984 to give the four-digit number 13b7, whichis divisible by 11. Then ,(a + b) is :(A) 10 (B) 11(C) 12 (D) 15

45. A number when divided by the sum of 555 and 445gives two times their difference as quotient and 30 asthe remainder. The number is :(A) 1220 (B) 1250(C) 22030 (D) 220030

46. In a 4-digit number, the sum of the first two digits isequal to that of the last two digits. The sum of the firstand last digits is equal to the third digit. Finally, the sumof the second and fourth digits is twice the sum of theother two digits. What is the third digit of the number ?(A) 5 (B) 8(C) 1 (D) 4

47. Anita had to do a multiplication. Instead of taking 35 asone of the multipliers, she took 53. As a result, theproduct went up by 540. What is the new product ?(A) 1050 (B) 540(C) 1440 (D) 1590

48. Three friends, returning from a movie, stopped to eatat a restaurant. After dinner, they paid their bill andnoticed a bowl of mints at the front counter. Sita took 1/3 of the mints, but returned four because she had amonetary pang of guilt. Fatima then took 1/4 of whatwas left but returned three for similar reasons. Eswarithen took half of the remainder but threw two back intothe bowl. The bowl had only 17 mints left when the raidwas over. How many mints were originally in the bowl?(A) 38 (B) 31(C) 41 (D) 48

49. In a number system, the product of 44 and 11 is 3414.The number 3111 of this system, when converted tothe decimal number system, becomes :(A) 406 (B) 1086(C) 213 (D) 691

50. A set of consecutive positive integers beginning with 1is written on the blackboard. A student came and erasedone number. The average of the remaining numbers

is 177

35 . What was the number erased?

(A) 7 (B) 8(C) 9 (D) None of these

51. Let D be a recurring decimal of the form D = 0. a1 a

2 a

1

a2 a

1 a

2 ....., where digits a

1 and a

2 lie between 0 and 9.

Further, at most one of them is zero. Which of thefollowing numbers necessarily produces an integer,when multiplied by D?(A) 18 (B) 108(C) 198 (D) 288

52. What is the value of the following expression

)120(

1.....

)16(

1

)14(

1

)12(

12222 ?

(A) 199

(B) 1910

(C) 2110

(D) 2111

53. Consider a sequence of seven consecutive integers.The average of the first five integers is n. The averageof all the seven integers is :(A) n(B) n + 1(C) k × n, where k is a function of n

(D) n +

72

54. Let N = 553 + 173 � 723 . N is divisible by :(A) both 7 and 13 (B) both 3 and 13(C) both 17 and 7 (D) both 3 and 17

55. Convert the number 1982 from base 10 to base 12.The results is :(A) 1182 (B) 1912(C) 1192 (D) 1292

56. The LCM of two numbers is 864 and their HCF is 144.If one of the numbers is 288, then the othernumber is :(A) 576 (B) 1296(C) 432 (D) 144

57. The LCM of two numbers is 567 and their HCF is 9. Ifthe difference between the two numbers is 18, find thetwo numbers :(A) 36 and 18 (B) 78 and 60(C) 63 and 81 (D) 52 and 34

58. If the product of n positive real numbers is unity, thentheir sum is necessarily :

(A) a multiple of n (B) equal to n � n1

(C) never less than n (D) a positive integer

59. How many even integers n, where 100 n 200, aredivisible neither by seven nor by nine ?(A) 40 (B) 37(C) 39 (D) 38

60. In a certain examination paper, there are n question. Forj = 1, 2....n, there are 2n-j

students who answered j or

more questions wrongly. If the total number of wronganswers is 4095, then the value of n is :(A) 12 (B) 11(C) 10 (D) 9

61. The number of positive n in the range 12 n 40 suchthat the product (n �1) (n � 2).... 3.2.1 is not divisible by

n is :(A) 5 (B) 7(C) 13 (D) 14

56PAGE # 56

62. The owner of a local jewellery store hired 3 watchmento guard his diamonds, but a thief still got in and stolesome diamonds. On the way out, the thief met each

watchman, one at a time. To each he gave 21

of the

diamonds he had then, and 2 more besides. Heescaped with one diamond. How many did he stealoriginally ?(A) 40 (B) 36(C) 25 (D) none of these

63. A rich merchant had collected many gold coins. He didnot want any body to know about him. One day, his wifeasked, �How many gold coins do we hire?� After

pausing a moment he replied, �Well ! if divide the the

coins into two unequal numbers, then 48 times thedifference between the two numbers equals thedifference between the square of the two numbers.� The wife looked puzzled. Can you help the merchant�s

wife by finding out how many gold coins the merchanthas ?(A) 96 (B) 53(C) 43 (D) 48

64. A child was asked to add first few natural numbers(that is 1 + 2 + 3.....) so long his patience permitted. Ashe stopped, he gave the sum as 575. When the teacherdeclared the result wrong, the child discovered he hadmissed one number in the sequence during addition.The number he missed was :(A) less than 10 (B) 10(C) 15 (D) more than 15

65. 76n � 66n, where n is an integer > 0, is divisible by :(A) 13 (B) 127(C) 559 (D) All of these

66. The value of 2251541082510 is :

(A) 4 (B) 6(C) 8 (D) 10

67 When (55)10

is represented in base 25 then theexpression is :(A) (25)

25(B) (35)

25

(C) (55)25

(D) none of these

68. Arrange the following numbers in ascending order

21

,97

,54

,73

.

(A) 21

,93

,57

,54

(B) 54

,97

,21

,73

(C) 73

,21

,97

,54

(D) 54

,97

,73

,21

69. Which of the following surds is greatest in magnitude :3126 4,25,2,17

(A) 6 17 (B) 12 25

(C) 3 4 (D) 2

70. What is the decimal equivalent of the hexadecimal 25digit number (100.....001)

16 ?

(A) 223 + 1 (B) 224 + 1(C) 292 + 1 (D) 296 + 1

71. The square root of a perfect square containing n digitshas ............ digits.

(A) 2

1n (B) n/2

(C) A or B (D) None

72. If the decimal number 2111 is written in the octal system,then what is its unit place digit ?(A) 0 (B) 1(C) 2 (D) 3

73. The 288th term of the series a, b, b, c, c, c, d, d, d, d, e,e, e, e, e, f, f, f, f, f, f,....is :(A) u (B) v(C) x (D) w

74. If n = 67 then find the unit digit of [3n + 2n ].(A) 1 (B) 10(C) 5 (D) None

75. The number of 2-digit numbers n such that 3divides n � 2 and 5 divides n � 3 is : [KVPY 2007](A) 5 (B) 6(C) 7 (D) 10

76. How many 4-digit numbers are there with the propertythat it is a square and the number obtained byincreasing all its digits by 1 is also a square ?

[KVPY 2007](A) 0 (B) 1(C) 2 (D) 4

77. The number of positive fractions m/n such that1/3 < m/n < 1 and having the property that the fractionremains the same by adding some positive integer tothe numerator and multiplying the denominator by thesame positive integer is : [KVPY 2007](A) 1 (B) 3(C) 6 (D) infinite

78. If a and b are any two real numbers with oppositesigns, which of the following is the greatest ?

[KVPY 2008](A) (a�b)2 (B) (|a| � |b|)2

(C) |a2 � b2| (D) a2 + b2

79. The number (1024)1024 is obtained by raising (16)16

to the power n. What is the value of n ?[KVPY 2008]

(A) 64 (B) 642

(C) 6464 (D) 160

80. The number of integers a such that 1 a 100 andaa is a perfect square is : [KVPY 2008](A) 50 (B) 53(C) 55 (D) 56

81. The sum of 7 consecutive positive integers is equalto the sum of the next five consecutive integers. Whatis the largest among the 12 numbers ? [KVPY 2008](A) 24 (B) 23(C) 22 (D 21

57PAGE # 57

82. Let 0 < a < b < c be 3 distinct digits. The sum of all3-digit numbers formed by using all the 3 digitsonce each is 1554. The value of c is : [KVPY 2008](A) 3 (B) 4(C) 5 (D) 6

83. If the decimal 0.d25d25d25 �� is expressible in the

form n/27, then d + n must be : [KVPY 2008](A) 9 (B) 28(C) 30 (D) 34

84. Let Sn be the sum of all integers k such that

2n < k < 2n+1, for n 1. Then 9 divides Sn if and only if :

[KVPY 2009](A) n is odd(B) n is of the form 3k+1(C) n is even

(D) n is of the form 3k + 2

85. The number of ways in which 1440 can be expressedas a product of two positive integers is : [IAO 2007](A) 17 (B) 18(C) 35 (D) 36

86. Let N = 28 , the sum of all distinct factors of N is : [IAO 2008]

(A) 27 (B) 28(C) 55 (D) 56

87. The units digit of (1 + 9 + 92 + 93 + --------- + 92009) is : [IAO 2009]

(A) 0 (B) 1(C) 9 (D) 3

88. The biggest among the following is : [IAO 2009](A) 21/2 (B) 31/3

(C) 61/6 (D) 81/8

89. When a positive integer x is divided by 47, theremainder is 11. Therefore, when x2 is divided by 47,the remainder will be : [IAO 2009](A) 7 (B) 17(C) 27 (D) 37

90. When the number 72010 is divided by 25, the remainderwill be : [IAO 2009](A) 1 (B) 7(C) 18 (D) 24

91. P, Q and R are three natural numbers such that P andQ are primes and Q divides PR. Then out of thefollowing the correct statement is : [IJSO 2008](A) Q divides R (B) P divides R(C) P divides QR (D) P divides PQ

92. The expression (5a � 3b)3 + (3b � 7c)3 � (5a � 7c)3 isdivisible by : [IJSO 2008](A) (5a + 3b + 7c) (B) (5a � 3b � 7c)

(C) (3b � 7c) (D) (7c � 5a)

93. If a2 + 2b = 7, b2 + 4c = � 7 and c2 + 6a = � 14, then the

value of (a2 + b2 + c2) is : [IJSO 2009](A) 14 (B) 25(C) 36 (D) 47

94. In the familiar decimal number system the base is 10.In another number system using base 4, the countingproceeds as 1, 2, 3, 10, 11, 12, 13, 20, 21 .... Thetwentieth number in this system will be:

[IJSO 2010](A) 40 (B) 320(C) 210 (D) 110

95. If the eight digit number 2575d568 is divisible by 54and 87, the value of the digit �d� is: [IJSO 2011](A) 4. (B) 7.(C) 0. (D) 8.

96. The number of distinct prime divisors of the number

5123 � 2533 � 2593 is : [KVPY 2011]

(A) 4 (B) 5

(C) 6 (D) 7

58PAGE # 58

The logarithm of the number N to the base ' a

' is the

exponent indicating the power to which the base

' a

' must be raised to obtain the number N. This

number is designated as loga

N. Hence:

logaN = x ax = N , a > 0, a 1 & N > 0

Systems of Logarithm :There are two systems of logarithm which are generally

used.

(i) Common logarithm : In this system base is always

taken as 10.

If a = 10, then we write log b rather than log10

b.

(ii) Natural logarithm : In this system the base of the

logarithm is taken as �e�. Where �e� is an irrational

number lying between 2 and 3.

If a = e, we write nb rather than loge b. Here '

e

' is

called as Napiers base & has numerical value equal

to 2.7182.

REMEMBER

log10

2 = 0.3010 ; log10

3 = 0.4771 ;

n 2 = 0.693 ; n 10 = 2.303

REMARK :

The existence and uniqueness of the number

loga

N can be determined with the help of set of

conditions, a > 0 and a 1 and N > 0.

Some Useful Results :

(i) If a > 1 then

(a) loga x < 0 for all x satisfying 0 < x < 1

(b) loga x = 0 for x = 1

(c) loga x > 0 for x > 1

(d) x > y loga x > log

a y i.e. log

ax is an increasing

function.

graph of y = loga x, a > 1

x'

y'

x0 (1,0)

y = log x, a > 1a

y

LOGARITHM

(ii) If 0 < a < 1, then

(a) loga x < 0 for all x > 1

(b) loga x = 0 for x = 1

(c) logax > 0 for all x satisfying 0 < x < 1

(d) x > y logax < log

a y i.e. log

a x is a decreasing

function.

graph of y = loga x, 0 < a < 1.

x'

y'

x0(1,0)

y = log x, 0 < a < 1. a

y

Let m & n are arbitrary positive numbers, a > 0,

a 1, b > 0, b 1 and is any real number then ;

(i) loga (mn) = log

a m + log

an

[Where m and n are +ve numbers]

in general loga(x

1 x

2 ......x

n) = log

ax

1 + log

a x

2 + ........+

loga x

n

(ii) loga

nm

= logam � log

an

(iii) loga(m)n = n log

am

(iv)ablog

mblogmalog

(v) logam . log

ma = 1

(vi) If �a� is a positive real number and �n� is a positive

rational number, then

na nloga

(vii) If �a� is a positive real number and �n� is a positive

rational number, then

qalog np nlog

qp

a

(viii) plogqlog aa qp

Ex.1 If log3a = 4, find value of a.

Sol. log3a = 4

a = 34

a = 81.

id22910421 pdfMachine by Broadgun Software - a great PDF writer! - a great PDF creator! - http://www.pdfmachine.com http://www.broadgun.com

59PAGE # 59

Ex.2 Find the value of 43

log3227

log�89

log

Sol. Given 43

log3227

log�89

log 43

log3227

89

log

43

2732

89

log

= log1 = 0. [ loga1 = 0]

Ex.3 If 2log4x = 1 + log

4(x � 1), find the value of x.

Sol. Given 2log4x = 1 + log

4(x � 1)

log4x2 � log

4(x � 1) = 1

log4 1�x

x2

= 1

41 = 1�x

x2

x2 = 4x � 4

x2 � 4x + 4 = 0

(x � 2)2 = 0 x = 2

Ex.4 Evaluate : 5log�2 33 .

Sol. Given, 5log�2 33 = 5log�2 33.3 [am + n = am.an]

= 9.1�

3 5log3

= 9 × 5�1

= 59

.

Ex.5 Find the value of log25

125 � log84.

Sol. Given log25

125 � log84

= 22

35

2log�5log 32

= 23

log55 �

32

log2 2

= 23�

32

[ 1aloga ]

= 65

.

Ex.6 If A = log27

625 + 13log117 and B = log9125 + 7log1113 ,

then find the relation between A and B.

Sol. A = log27

625 + 13log117 = 4

35log 3 + 13log117

or, A = 34

log35 + 13log117 ...(i)

and, B = log9125 + 7log1113

or, B = 33

5log 2 + 13log117

or, B = 23

log35 + 13log117 ...(ii)

By (i) and (ii) we have,

A � 34

log35 = B �

23

log35

34

log35 <

23

log35

A < B.

Ex.7 Find the value of the following :

(i) log48 (ii)

22log 1024

(iii) log1/49

343 (iv))32(

log

)32(

(v) log0.001

0.0001

Sol. (i) Let x = log48 4x = 8

22x = 23 x = 3/2.

(ii) Let x = 22log 1024

x)22( = 1024

23x/2 = 210 x = 20/3.

(iii) Let x = log1/49

343 7�2x = 73 x = � 3/2.

(iv) Let x = )32(

log

)32(

x)32( = 32

x)32( =

32

)32()32(

= 1)32(

x = �1.

(v) Let x = log0.001

0.0001 (0.001)x = 0.0001

10�3x = 10�4 x = 4/3.

Ex.8 Find the value of the following :

(i)

30seclog)60cos60(sin 22

(ii) 45cotlog 45tan (iii) log20

1

Sol. (i) For (i) and (ii) log is not defined because base isequal to 1.(iii) log

201 = 0.

Fundamental Logarithmic Identity :

Nlogaa = N, a > 0, a 1 & N > 0

REMARKS :

(i) loga1 = 0

(ii) loga a = 1 [As a1 = a]

(iii) loga 0 = not defined

[As an = 0 is not possible, where n is any number]

(iv) loga (�ve no.) = not defined.

[As in loga N, N will always be (+ ve)]

(v) log1/a

a = 1 (vi) logba =

blog1

a

(vii) ax = anxe

Ex.9 Find the value of the following

(i) 3log1

581 (ii)

31

121log 32

8

Sol. (i) Let x = 3log1

581 = 5log4 3)3(

= 5log33 = 54 = 625.

(ii) Let x =

31

121log 32

8 = 1121log3 322

= 121log22.2 = 2(121) = 242.

60PAGE # 60

Ex.10 Find the value of

(i) log3 4. log

4 5. log

5 6. log

67 . log

78. log

89.

(ii) log tan1° + log tan2° + log tan3° + .... + log tan89°.

Sol. (i) log3 4. log

4 5. log

5 6. log

67 . log

78. log

89

3log4log

. 4log5log

. 5log6log

. 6log7log

. 7log8log

. 8log9log

= 3log9log

= 2.

(ii) log tan1° + log tan2° +.......+ log tan89°

= log(tan1°. tan2°. tan3° .... tan45° ..... tan89°)

= log (tan1° tan2° tan3° ... tan45° ... cot3° cot2° cot1°)

= log (1.1.1 ....... 1) = log 1 = 0.

Ex.11 Find the value of :

(i) 5log32 � 2log35

(ii) 1675

log2 � 95

log2 2 + 24332

log2

(iii) 1.0log3

Sol. (i) 5log32 � 2log35

Here 5log32 = 3log5log

2

2

2

= 2log5log 322 = 2log35

Hence 5log32 � 2log35 = 0.

(ii) 1675

log2 � 95

log2 2 + 24332

log2

=

2581

24332

.1675

log2 = log22 = 1

(iii) 1.0log3 = log31/9 = log

33�2 = � 2

Ex.12 If log615 = a, log

1218 = b, then find log

2524 in terms

of a and b.

Sol. log615 = 6log

15log

3

3 = 2log1

5log1

3

3

= a ...(i)

log12

18 = 12log

18log

3

3 = 2log21

2log2

3

3

= b ...(ii)

From (i) and (ii)

log32 =

1b2b2

And log35 =

1b21b2aab

log25

24 = 25log

24log

3

3 = 5log2

2log31

3

3

=

1b21b2aab

2

1b2b2

31

= )1abb2a(2b5

.

The equality loga x = log

a y is possible if and only if

x = y i.e. loga x = loga y x = y

Always check that the solutions should satisfy x > 0,

y > 0, a > 0, a 1.

Ex.13 Solve the following equations :

(i) log3(x + 1) + log

3 (x + 3) = 1

(ii) log2 log

4 log

5 x = 0

(iii) log3 [5 + 4 log

3 (x � 1)] = 2

Sol. (i) log3(x + 1) + log

3 (x + 3) = log

3 (x + 1) (x + 3) = 1

(x + 1) (x + 3) = 3

x2 + 4x = 0

x = 0, � 4.

Also x + 1 > 0 or x > � 1

And x + 3 > 0 or x > � 3

Hence solution is x = 0.

(ii) log2 log

4 log

5x = 0

log4 log

5x = 1

log5x = 4

x = 54 = 625.

(iii) log3 [5 + 4log

3 (x � 1)] = 2

5 + 4 log3 (x � 1) = 9

log3 (x � 1) = 1

x � 1 = 3 x = 4.

Ex.14 Find the total number of digits in the number

6100 (Given log10

2 = 0.3010 ; log10

3 = 0.4771).

Sol. Let x = 6100

log10

x = 100 log10

6

log10

x = 100 [log10

2 + log10

3]

= 100 [0.3010 + 0.4771]

= 100 (0.7781)

= 77.81

The number of digits = 78.

Ex.15 Prove that log35 is irrational.

Sol. Let log35 is rational.

log35 =

qp

[where p and q are co-prime numbers]

3p/q = 5 3p = 5q.

which is not possible, hence our assumption is

wrong and log35 is irrational.

1. Given log2 = 0.3010, then log 16 is :

(A) 2.4080 (B) 1.2040

(C) 0.2408 (D) 1.9030

2. If logxy = 100 and log

2x = 10, then the value of y is :

(A) 21000 (B) 2100

(C) 22000 (D) 210000

61PAGE # 61

3. The value of [log10

(5 log10

100)]2 is :

(A) 0 (B) 1

(C) 2 (D) 10

4. log10

p + log10

q = log10

(p�q), then :

(A) p = q = 0 (B) p = q1q

(C) p = q = 1 (D) p = q1q

5. If 33 log3Mlog31

N = 1+ log0.008

5, then :

(A) N9

M9 (B) M9

N9

(C) N3

M3 (D)

M3

N9

6. The value of x, when log3(log

2 x) + 2 log

9(log

7 8) = 2, is :

(A) 243 (B) 27

(C) 343 (D) 64

7. Find all the real values of x, which satisfy the equation

2x 4

5xlogxlog

32

22

2

.

(A) 2

1,2 (B) 2,

2

1 and

22

1

(C) 1,22

1,

2

1(D) 2

8. If ap = bq = cr = ds, then loga (bcd) is equal to :

(A)

s1

r1

q1

p (B) 1

(C) s1

r1

q1

(D) 0

9. Va lue o f t he f o l l owing s um m at i on

xlog1

...xlog

1xlog

1

4332

i s eq ua l t o :

(A) xlog1

e(B)

e1

(C) xlog1

)!43((D) elog

!43

x

10. Solve for �x� if 2logx 7 + log

7x7+3log

49x7 = 0.

(A) 34

x (B) 2/17x

(C) 3/47x (D) (B) or (C) only

11. Find the number of different values of x that satisfy the

equation: 2 xloglog 4327 = (log4x)2 + 5(log

4 x) + 2.

(A) 1

(B) 2

(C) 3

(D) No real value of x exists

12. If log30

3 = x and log30

5 = y, then log8 30 is equal to :

(A) )yx1(31

(B) 3 (1 - x - y)

(C) )yx1(3

(D) 3

yx1

13. If logyx = 10, then 3x

log (y6) is equal to :

(A) 5 (B) 51

(C) 6 (D) 61

14. If log 2 = a and log3 = b then [log(1) + log(1 + 3) + log

(1 + 3 + 5) + .......+ .....+ log (1 + 3 + 5 + 7 + ..... + 19)]

� 2[log 1 + log2 + log3 + ....... log7] = p + qa + rb where

p,q,r are constants. What is the value of p + 2q + 3r if all

logs are in base 10 ?

(A) 12

(B) 26

(C) 18

(D) Cannot be determined

15. The sum to 2n terms of the series,

log3 1 � log

33 + log

3 9 � log

3 27 ...... is :

(A) 1n

1n2

(B)

1n2

1n

(C) 1 � n (D) � n

16. When the curves y = log10

x and y = x � 1 are drawn in the

x-y plane. How many times do they intersect for values

x 1 ?

(A) Never (B) Once

(C) Twice (D) More than twice

17. Solve : log (x + 4) +log (x � 4) = log 3.

(A) x = � 19 (B) x = + 19

(C) x = 19

(D) Cannot be determined

18. Find x from the equation ax c� 2x = b3x + 1.

(A) blog3clog2alogblog2

(B) blog3clog2alogblog

(C) blog3clog2alogblog

(D) Cannot be determined

62PAGE # 62

19. If log 2 = 0.3010 and log 3 = 0.4771, then find thenumber of digits in 617 .(A) 12 (B) 14(C) 18 (D) 17

20. Express 3 log 10

5 � 4 log10

x + log10

y2 as the log of asingle number.

(A) log10

4

2

x

y25(B) log

10

4

3

x

y125

(C) log10

3

2

x

y125(D) log

10

4

2

x

y125

21. Express the log

25

3

bc

a in terms of log a, log b and

log c.

(A) 23

log a � 5 log c � 2 log b

(B) 3 log a � 5 log c � 2 log b

(C) 23

log a � 5 log c � 32

log b

(D) 23

log a � 5 log c � log b

22. Find x if log10

1250 + log10

80 = x.(A) 5 (B) 4(C) 8 (D) 7

23. If logx2 = a, log

x3 = b, log

x5 = c, then find the value of the

following in terms of a, b and c.(i) log

x16 (ii) log

x75 (iii) log

x60

(A) 4a, 2a + b, 2a + b + c(B) 4a, 2c + b, 2a + b + c(C) 4a, 2c + b, 2c + b + a(D) 4a, 2c + b, 2a + 3b + c

24. Find the value of log (1 + 2 + 3) =(A) log1 + log2 � log3 (B) log1 + log2 + log3(C) log1 � log2 + log3 (D) � log1 + log2 + log3

25. log418 is :

(A) an irrational number (B) a rational number(C) natural number (D) whole number

26. If log2

yx =

21

(log x + log y), then :

(A) x = y (B) x = � y

(C) x = 2y (D) 2x = yZ

27. Find the value of (yz)logy � logz × (zx)logz � logx × (xy)logx � logy = ?(A) 3 (B) 2(C) 0 (D) 1

28. The number of solutions for the equation inlog

ex + x � 1 = 0 is :

(A) 1 (B) 2(C) 4 (D) None of the above

29. If 5.2Nlog 10 then, find out total number of digits in

N.(A) 3 (B) 4(C) 5(D) Cannot be determined

30. log10

21

1 + log10

31

1 + log10

41

1 + ... +

log10

19991

1 . When simplified has the value equal to

(A) 1 (B) 3(C) 10 (D) 100

31. [(log x � log y)(logx2 + logy2)] [(logx2 � logy2)(logx + log y)]is equal to :(A) 0 (B) 1(C) log x/y (D) log xy

32. If log10

[log10

(log10

x)] = 0.(A) x = 103 (B) x = 1010

(C) x = 155 (D) None

33. The number of solutions for real x, which satisfy the

equation 2log2 log

2 x + log

1/2 log

2 )x22( = 1 :

(A) 1 (B) 2(C) 4 (D) none of these

34. Find the number of integral pairs for (x, y) from thefollowing equations :

log100

|x + y| = 21

& log10

y � log10

|x| = log100

4

(A) 0 (B) 1(C) 2 (D) none of these

35. How many positive real numbers x are there such that

xxx xxx ? [KVPY 2007]

(A) 1 (B) 2(C) 4 (D) infinite

36. Let logab = 4, log

cd = 2 where a, b, c, d are natural

numbers. Given that b � d = 7, the value of c � a is :

[KVPY 2009](A) 1 (B) � 1

(C) 2 (D) � 2

37. The value of log10

(3/2) + log10

(4/3) + ------ up to 99terms. [IAO 2008](A) 0 (B) 2(C) 2.5 (D) none of the above

38. If a, b 1, ab > 0, a b and logba = log

ab, then ab = ? [IAO 2008]

(A) 1/2 (B) 1(C) 2 (D) 10

39. Find x, if 100.3010 = 2, 100.4771 = 3 and 10x = 45. [IAO 2009]

(A) 0.6532 (B) 1.6532(C) 1.6570 (D) 1.7781

40. If x < 0 and log7 (x2 � 5x � 65) = 0, then x is :

[IJSO-2011] (A) �13 (B) �11

(C) � 6 (D) � 5

PAGE # 63

DIGESTIVE SYSTEM

It includes alimentary canal and digestive glands.Alimentary canal starts from mouth and ends intoanus.

� Mouth or Buccal Cavity : An adult has 16 teeth oneach jaw.In each half of jaws starting from middle to backwardthese are, incisors - 2, canine - 1, premolar - 2, molars- 3 (2 + 1 + 2 + 3).Dental formula is 2123/2123.Last molars are called wisdom teeth.Milk teeth start erupting after 6 months of birth andappear between 6 to 24 months.Dental formula of milk teeth is 2102/2102.

� Stomach : Inner mucosa of stomach is raised intolarger number of longitudinal folds called gastricrugae.

NUTRITION

It helps in mechanical churning and chemicaldigestion of food.The stomach of ruminant (cud chewing) animals havecompound stomach and consists of 4 chambersrumen, reticulum; omasum and abomasum.Rumen and reticulum harbour numerous bacteriaand Protozoa which help in digestion of cellulosecalled symbiotic digestion.Abomasum is true stomach as it secretes gastric juicecontaining HCl and pepsin.

� Small Intestine : Villi and microvilli increase thesurface area of digestion and absorption of food.Pancreatic duct release few enzymes which act oncarbohydrates, fats and proteins.

� Large Intestine : In some herbivores (like horse andass), caecum is large and is a site of microbialdigestion of cellulose.In man caecum very small vestigial organ and is calledappendix.

Table : Vitamins Necessary for Normal Cell Functioning

Deficiency Disease Deficient Nutrient Deficiency Disease Deficient Nutrient

1. Xerophthalmia Vitam in A / Retinol10. Megaloblastic anaemia

Folic acid and Vitam in B12

2. Night-blindness Vitam in A11. Pernicious anaemia

Vitam in B12 (Cyanocobalam ine)

3. Rickets (in children) Vitam in D/ Sun-Shine Vitam in 12. Scurvy Vitam in C/ Ascorbic acid

4. Osteomalacia (adults) Vitam in D 13. Tetany Calcium

5. Sterility Vitam in E (Tocopherol) 14. Anaemia Iron

6. Bleeding disease Vitam in K (Phylloquinone) 15. Goitre Iodine

7. Beri beri Vitam in B1 (Thiam ine) 16. Fluorosis Excess of fluorine

8. Cheilosis Vitam in B2 (Riboflavin) 17. Kwashiorkor Proteins

9. Pellagra Vitam in B3 (Niacin) 18. Marasmus Proteins and food calories

� Knowledge Boosters :

(1) Salivary Glands : It produces saliva. In man onlythree pairs of salivary glands are present.

(a) Parotid glands : largest glands present just belowthe external ear. In this glands, virus causes mumpsdisease.

(b) Submaxillary glands : these lie beneath the jaw-angles.

(c) Sublingual glands : smallest glands which liebeneath the tongue and open at the floor of buccalcavity.

(2) Gastric Glands: Present in the mucosa of thestomach. These are of 3 types :

(a) Cardiac glands : secrete an alkaline mucus.

(b) Pyloric glands : secrete an alkaline mucus.

(c) Fundic glands : each gland has 4 types of cells.

1. Peptic / Zymogen cells - secrete pepsinogen,prorennin

2. Oxyntic cells - secrete HCI

3. Goblet cells - secrete mucus

4. Argentaffin cells - produces serotonin, somatostatinand histamine

(3) Liver : It is the largest gland and consists of alarge right lobe, a small left lobe and two small lobes.On the right lobe lies gall bladder, which, stores bilejuice secreted by the liver.

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PAGE # 64

Bile juice contains no enzyme but possesses bilesalts and bile pigments (bi lirubin-yellow andbiliverdin-green).Formation of glucose from excess organic acids.

� Storage of vitamins : A, D, E, K� Synthesis of vitamin A from carotene.

� Secretions of blood anticoagulant named heparin.� Synthesis of blood or plasma proteins, (fibrinogen

and prothrombin)� Secretion of bile, detoxification of harmful chemicals.� Elimination of pathogens and foreign particles through

phagocytic cells called Kupffer�s cells.

Table : Various Enzymes Involved in Digestion

EXERCISE

1. In human digestion is :(A) Intercellular (B) Intracellular(C) Extracellular (D) Both A & B

2. The enzyme rennin converts(A) Proteins to proteoses (B) Fats to fatty acids(C) Casein to paracasein (D) Proteins to peptones

3. Digestion in Amoeba is :(A) Intercellular (B) Intracellular(C) Both of these (D) None of these

4. Statements true or false is:(i) No absorption of food takes place in mouth andoesophagus(ii) Absorption of H

2O, alcohol, simple salts, glucose

and chloride takes place in the stomach to slightextent.(iii) Whole protein particles can be absorbed bypinocytosis.(A) (i) and (ii) true (iii) rarely true(B) All true(C) (i) and (iii) true B false(D) (i) and (ii) true (iii) false

5. Stomach is protected from HCl and gastric juicebecause :(A) The two are dilute(B) Epithelial lining is resistant to them(C) Wall has neutralizing agents(D) Stomach lining is covered by mucus

6. In a villus, some of the glycerol and fatty acids arecombined to form fats, coated with proteins, and thentransported to the :(A) Lacteal(B) Capillaries(C) Lumen of the small intestine(D) Lumen of the large intestine

7. Among mammals, an herbivore has :(A) More teeth than a carnivore(B) Fewer teeth than a carnivore(C) Flatter teeth than a carnivore(D) Teeth that are more pointed than a carnivore

8. Gastric juice has a pH of about :(A) 1 (B) 2(C) 6 (D) 10

9. Softness and deformities of bones like bowlegs andpigeon-chest are symptoms of the disease :(A) Rickets (B) Osteomalacia(C) Goitre (D) Beri-Beri

PAGE # 65

10. In rabbit when a dilute solution of glucose reachessmall intestine is :(A) Absorbed rapidly by the active transport withsodium ions(B) Absorbed by facilitated diffusion(C) Lost to outside with undigested food(D) Absorbed rapidly by active transport independentof sodium ions

11. Chymotrypsin acts on :(A) Carbohydrates (B) Proteins(C) Fats (D) Starch

12. The vermiform appendix is made up of :(A) Striated muscles (B) Excretory tissue(C) Lymphatic tissue (D) Stem cells

13. Digestion is completed in :(A) Ileum (B) Duodenum(C) Stomach (D) Rectum

14. One of the following is not an enzyme of digestivesystem :(A) Trypsin (B) Amylase(C) Enterogastrone (D) Enterokinase

15. Fat soluble vitamins are :(A) A, D and E (B) B, C and D(C) B,D and E (D) A, B and C

16. Sunshine vitamin is :(A) Vitamin A (B) Vitamin D(C) Vitamin K (D) Vitamin E

17. Which of the vitamins is essential for normal vision :(A) Folic acid (B) Biotin(C) Riboflavin (D) Retinol

18. B.M.R. for a normal human adult is :(A) 1600 Kcal/day (B) 2000 Kcal/day(C) 2500 Kcal/day (D) 3000 Kcal/day

19. Cholesterol rich diet is undesirable because it :(A) Makes person obese(B) Causes heart ailments(C) Is difficult to remove it from body(D) All of the above

20. If all bacteria of intestine die, then what will be theeffect on body:(A) Man will feel tired all the body(B) It will cause blindness(C) There will be no synthesis of vitamin -B Complex(D) Excretion will be effected

21. The gall bladder is involved in[KVPY 2011]

(A) synthesizing bile(B) storing and secreting bile(C) degrading bile(D) producing insulin

22. In the 16th century, sailors who travelled longdistances had diseases related to malnutrition,because they were not able to eat fresh vegetablesand fruits for months at a time scurvy is a result ofdeficiency of [KVPY 2011](A) carbohydrates (B) proteins(C) vitamin C (D) vitamin D

23. Several mineral such as iron, iodine, calcium andphosphorous are important nutrients. Iodine isfound in [KVPY 2011](A) thyroxine (B) adrenaline(C) insulin (D) testosterone

PAGE # 66

SERIES COMPLETION

Series completion problems deals with numbers,alphabets and both together. While attempting to

solve the question, you have to check the pattern ofthe series. Series moves with certain mathematicaloperations. You have to check the pattern.

Type of questions asked in the examination :(i) Find the missing term(s).(ii) Find the wrong term(s).

NUMBER SERIES

(a) Some Important Patterns :

(i) a, a ± d, a ± 2d, a ± 3d.......(Arithmetic Progression)

(ii) a, ak, ak2, ak3, ................(Geometric Progression)

(iii) a, ka

, 2k

a, 3k

a, .............(Geometric Progression)

(iv) Series of prime numbers � i.e. 2, 3, 5, 7, 11, ......

(v) Series of composite numbers �i.e. 4, 6, 8, 9, 10, 12, .................

Directions : (1 to 10) Find the missing numbers :

Ex 1. 16, 19, 22, 25, ?(A) 27 (B) 28

(C) 29 (D) 25Sol. (B) As per series a, a + d, a + 2d,.........

a = 16

d = 3a + 4d = 16 + 4 × 3

Ex 2. 9, 18, 36, ?, 144(A) 70 (B) 56

(C) 54 (D) 72Sol. (D) As per series, a, ak, ak2, ak3, ........

a = 9, k = 2

ak3 = 9 × 23 = 72

Ex 3. 2, 6, 14, 26, ?(A) 92 (B) 54(C) 44 (D) 42

Sol. (D) The pattern is +4, +8, +12, +16, .......

Ex 4. 240, ? , 120, 40, 10, 2(A) 120 (B) 240(C) 40 (D) 10

Sol. (B) The pattern is ×1, ×21

, ×31

, ×41

, ×51

missing term = 240 × 1 = 240

Ex 5. 8, 12, 21, 46, 95, ?(A) 188 (B) 214(C) 148 (D) 216

Sol. (D) The pattern is + 22, + 32, + 52, + 72, .......missing number = 95 + 112 = 216

Ex 6. 3, 9, 36, 180, ?(A) 1080 (B) 900(C) 720 (D) None of these

Sol. (A) Each term is multiplied by 3, 4, 5 and so onrespectively. Therefore, the next term would be180 × 6 = 1080.

(b) Multiple Series :

A multiple series is a mixture of more than oneseries :

Ex 7. 4, 7, 3, 6, 2, 5, ?(A) 0 (B) 1(C) 2 (D) 3

Sol. (B) The sequence is a combination of two seriesI 4, 3, 2, ?II 7, 6, 5The pattern followed in I is � 1, � 1, � 1

missing number = 2 � 1 = 1

Ex 8. 14, 15, 12, 16, 9, 18, 4, 21, ?(A) 2 (B) 3(C) � 3 (D) � 5

Sol. (C) The sequence is a combination of two series. 14, 12, 9, 4, (....) and 15. 16, 18, 21The pattern followed in is � 2, � 3, � 5, .......

missing number = 4 � 7 = � 3

Ex 9. 1, 1, 4, 8, 9, ? ,16, 64(A) 21 (B) 27(C) 25 (D) 28

Sol. (B) (i) 1, 4, 9, 16 [12, 13, 22, 23, 32, 33.............](ii) 1, 8, __, 64 mixed combination

Ex 10. 3, 6, 24, 30, 63, 72, ?, 132(A) 58 (B) 42(C) 90 (D) 120

Sol. (D) The difference between the terms is givenbelow as :

3 6 24 30 63 72 ? 132

3 18 6 33 9 48 ?

3 15 3 15 ?

Difference

Difference

Therefore, alternate difference between thedifference is 3 and 15 respectively.Hence, the next term would be 72 + 48 = 120.

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PAGE # 67

Directions : (11 to 12) Find the wrong term(s) �

Ex 11. 5, 8, 10, 12, 15, 18, 20, 23(A) 8 (B) 12(C) 15 (D) 18

Sol. (B)

Therefore, number 12 is wrong and should bereplaced by 13.

Ex 12. 1, 3, 8, 19, 42, 88, 184(A) 3 (B) 8(C) 19 (D) 88

Sol. (D)

1 1848 19 42 893

2 5 11 23 47 95

3 6 12 24 48

Hence, number 88 is wrong and should bereplaced by 89.or 1 × 2 + 1, 3 × 2 + 2, 8 × 2 +3, 19 × 2 + 4, 42 × 2 + 5,

89 × 2 + 6

Directions : (13 to 14) In each of the following questions, anumber series is given. After the series, below it inthe next line, a number is given followed by (P), (Q),(R), (S) and (T). You have to complete the seriesstarting with the number given following thesequence of the given series. Then answer thequestion given below it.

Ex 13. 12 28 64 14037 (P) (Q) (R) (S) (T)Which number will come in place of (T) ?(A) 1412 (B) 164(C) 696 (D) 78

Sol. (A)

Similarly (P) (Q) (R) (S) (T)

37 78 164 340 696 1412

×2+4 ×2+8 ×2+12 ×2+16 ×2+20

Therefore, the number 1412 will come in place of (T).

Ex 14. 2 9 57 3373 (P) (Q) (R) (S) (T)Which number will come in place of (Q) ?(A) 113 (B) 17(C) 3912 (D) 8065

Sol. (A)

Similarly, (P) (Q) (R) (S)

3 17 113 673

×8�7 ×7�6 ×6�5 ×5�4

3361

Therefore, the number 113 will come in place of (Q).

ALPHABET SERIES

(a) Pattern of Alphabets Show Variation Based on :

(i) Position of the letters (ii) Difference between the alphabets

(i) Position of alphabets :

Alphabets in order :

Alphabets in reverse order :

Directions : (15 to 24) Find the missing term(s) :

Ex 15. B, E, H, ?(A) K (B) L(C) J (D) M

Sol. (A) In the given series, every letter is moved threesteps forward to obtain the corresponding lettersof the next term. So, the missing term is K.

Ex 16. Q, N, K, ?, E(A) H (B) I(C) J (D) G

Sol. (A) In the given series, every letter is moved threesteps backward to obtain the corresponding lettersof the next term. So, the missing term is H.

Ex 17. A, Y, D, W, G, U, J, ?(A) R (B) T(C) S (D) P

Sol. (C) The given sequence consists of two series :. A, D, G, J in which each letter is moved threesteps forward to obtain the next term. Y, W, U, ? in which each letter is moved two stepsbackward to obtain the next term.So, the missing term would be S.

Ex 18. AG, LR, WC, HN, ?(A) SX (B) RY(C) SY (D) TX

PAGE # 68

Sol. (C) The first letter of each group and the secondletter of each group differs by 11 letters betweenthem.

Alphabeticalpositions

Difference inAlphabeticalpositions

A 1

L 12

W 23

H 8

11 11 11

Similarly,

Alphabeticalpositions

Difference inAlphabeticalpositions

G 7

R 18

C 3

N 14

11 11 11

Therefore, the next group of letter would be SY.

11

H SAnd

11

N Y

Ex 19. AD, EI, JO, PV, ?(A) VD (B) WC(C) WD (D) VE

Sol. (C) The first letter of subsequent groups have adifference of 4, 5 and 6 places respectively, whereasthe second letter of the subsequent groups has adifference of 5, 6, and 7 places respectively,Therefore, on following the same pattern, we get�WD� as the next term which would replace the

question mark.

Ex 20. AB, BA, ABD, DBA, PQRS, ?(A) SRQP (B) SRPQ(C) SQRP (D) RSQP

Sol. (A) The first term is reversed to get second term.The third term is reversed to get the fourth term.Similarly, to get the sixth term, we reverse the fifthterm. So, the missing term would be SRQP.

Ex 21. HEJ, JGL, LIN, NKP, ?(A) MOR (B) PNS(C) PMR (D) NPT

Sol. (C) First letter of each group differs by 2 letters.Second letter of each group differs by 2 letters.Third letter of each group differs by 2 letters. All theletters differ in the forward direction. Hence, thenext choice would be PMR.

Ex 22. XYQ, ZAR, BCS, DET, ?(A) GFU (B) FUG(C) FZU (D) FGU

Sol. (D) Here, first two terms of every group of lettersare in continuation, like XY, ZA, BC, DE, and thethird letter of each group is again in forwardcontinuation,i.e. Q, R, S, T. Hence, the term replacing thequestion mark would be FGU.

Ex 23. 17Z5, 15X4, 13V3, ?, 9R1(A) 11S2 (B) 11T2(C) 11U2 (D) 11T3

Sol. (B) The first number & second letter of every termis moved two steps backward & the third numberof every term is moved one step backward. So, themissing term would be 11T2.

Ex 24. (ABC) � 6, (DEF) � 15, (GHI) � 24, ?

(A) (IJK) � 33 (B) (JKM) � 33

(C) (IJK) � 32 (D) (JKL) � 33

Sol. (D) In a given seriesLet A = 1, B = 2, C = 3, D = 4, E = 5, F = 6, and so on

6CBA321

, 15FED654

, 24IHG

987

So, the missing term would be 33LKJ121110

Directions : (25 to 27) Find the wrong term (s) :

Ex 25. ABD, DGK, HMS, NTB, SBL, ZKW

(A) NTB (B) DGK(C) SBL (D) ZKW

Sol. (A) First letter of first, second, third,.........terms ismoved three, four, five, ........steps forwardrespectively. Similarly, second letter is moved five,six, seven,......steps forward respectively and thirdletter is moved seven, eight, nine,........stepsforward respectively. Hence, NTB is the wrong termand should be replaced by MTB.

Ex 26. EPV, FQW, GRX, HTY, ITZ(A) FQW (B) GRX(C) HTY (D) ITZ

Sol. (C) In every term, first second and third letter is inalphabetical order to its next term respectively.Fourth term is not following the same rule. Hence,HTY is the wrong term and should be replaced byHSY.

Ex 27. D4V, G10T, J20R, M43P, P90N

(A) P90N (B) G10T

(C) J20R (D) D4V

Sol. (B) First letter of every term is moved three steps

forward in each next term. Second number of every

term of the pattern × 2 + 1, × 2 + 2, × 2 + 3,............and

third letter of every term is moved two steps

backward. Hence, G10T is the wrong term and

should be replaced by G9T.

LETTER REPEATING SERIES

Pattern of such questions is that some letters insequence are missing.

(i) The letters may be in cyclic order (clockwise oranti-clockwise).(ii) To solve a problem, we have to select one of thealternative from the given alternatives. Thealternative which gives a sequence form of lettersis the choice.

Directions : (28 to 32) Find the missing term(s) :

Ex 28. a a _ b a a _ b b b _ a(A) baa (B) abb(C) bab (D) aab

PAGE # 69

Sol. (A) we proceed step by step to solve the aboveseries:Steps :

1. The first blank space should be filled in by 'b' sothat we have two a's followed by two b's.

2. Second blank space should be filled in by 'a' sothat we have three a's followed by three b' s.

3. The last blank space must be filled in by 'a' to keepthe series in sequence.

Ex 29. _ bca _ ca _ c _ b _(A) aabbc (B) abbbc(C) aabcc (D) abbac

Sol. (D)

Series is abc/ abc/ abc/ abc. So, pattern abc isrepeated.

Ex. 30 a _ abb _ aa _ ba _ a _ b(A) ababa (B) aabba(C) aabab (D) aaabb

Sol- (C) Series is aaabb/ aaabb/ aaabb. So, patternaaabb is repeated.

Ex 31. a _ c _ abb _ ca _ a(A) baca (B) bbca(C) bacc (D) bacb

Sol- (A) Series is abc/ aabbcc/ aaa

Ex 32. bc _ b _ c _ b _ ccb(A) cbcb (B) bbcb(C) cbbc (D) bcbc

Sol- (A) Series is bccb / bccb / bccb. So, pattern bccb isrepeated

Directions : (33 to 34) The question given below is basedon the letter series, In series, some letters aremissing. Select the correct alternative. If more thanfive letters are missing, select the last five lettersof the series.

Ex 33. xyzu _ yz _ v _ _ uv _ _ _ _ _ _ _(A) uvxyz (B) vuzyx(C) uvzyx (D) vuxyz

Sol. (A) The series is x y z u v / y z u v x/ z u v x y/u v x y zThus the letters are written in a cyclic order.

EX 34. abcd _ bc _ e _ _ de _ _ _ _ _ _ _(A) deabc (B) edcba(C) decba (D) edabc

Sol. (A) The series is a b c d e / b c d e a / c d e a b / b e a b cThus the letters are written in a cyclic order.

Direction : (35 to 36) There is a letter series in the first rowand a number series in the second row. Eachnumber in the number series stands for a letter inthe letter series. Since in each of that series someterm are missing you have to find out as to whatthose terms are, and answer the questions basedon these as given below in the series.

Ex 35. a _ h _ _ c _ n e _ h _ e a c _ _ _ _ _2 1 _ 4 3 _ 5 _ _ 2 5 4 _ _ _ _ _ _ _ _The last five terms in the series are(A) 32524 (B) 43215(C) 25314 (D) 32541

Sol. (B) By taking a = 2, c = 1, n = 4, h = 5 and e = 3, thenumbers series runs as 21543 15432 5432143215. If first digit of a group of five digits is placedas the last digit, we obtain the second group of fivedigits and so on.

Ex 36. _ m y e _ _ y l x _ y l m _ _ l _ _ _ _4 6 _ 5 8 6 _ _ _ 5 7 _ 6 5 8 _ _ _ _ _The last five terms of the number series are(A) 46758 (B) 74658(C) 76485 (D) 46785

Sol. (D) By taking e = 5, l = 4, m = 6, y = 7 and x = 8 thenumber series runs as 46758 67485 74658 46785.By taking the digits in the groups of five, we findthat first digit of the first group (i.e. 4) is the thirddigit of the second group and the last two digitshave interchanged their positions. The same ruleapplies in others groups also.

Direction : (37) In the following question, three sequencesof letter/numbers are given which correspond toeach other in some way. In the given question, youhave to find out the letter/numerals that come inthe vacant places marked by (?). These are givenas one of the four alternatives under the question.Mark your answer as instructed.

Ex 37. C B _ _ D _ B A B C C B_ _ 2 3 5 4 _ _ ? ? ? ?p _ p q _ r _ q _ _ _ _(A) 4 5 5 4 (B) 4 3 3 4(C) 4 2 2 4 (D) 2 5 5 2

Sol. (C) Comparing the positions of the capital letters,numbers and small letters, we find p correspondsto C and 2 corresponds to p. So, p and 2 correspondto C. q corresponds to A and 3 corresponds to q.So, q and 3 corresponds to A. Also, 5 correspondsto D. So, the remaining number i.e., 4 correspondsto B. So, BCCB corresponds to 4, 2, 2, 4.

MISSING TERMS IN FIGURES

Directions : (38 to 47) Find the missing number(s) :

Ex 38.

6 9 15

8 12 20

4 6 ?

(A) 5 (B) 10(C) 15 (D) 21

Sol. (B) In the first row, 6 + 9 = 15In the second row, 8 + 12 = 20 In the third row, missing number = 4 + 6 = 10.

PAGE # 70

Ex 39.

(A) 11 (B) 6(C) 3 (D) 2

Sol. (C) Clearly, in the column, 83

46

In the column, 272

318

We take x in place of ?

Similarly in the column, 95

15

x , x 315

59

Ex 40.

3C 27D 9E

7I 21K 3M

4D ? 7J

(A) 11E (B) 28G(C) 35I (D) 48F

Sol. (B) The letters in the first row form a series C, D, E(a series consecutive letters). The letters in thesecond row form a series I, K, M (a series ofalternate letters). Similarly, the letters in the thirdrow will form the series D, G, J (a series in whicheach letter is three steps ahead of the previousone). So, the missing letter is G. Also, thenumber in the second column is equal to theproduct of the numbers in the first and thirdcolumns. So, missing number is (4 × 7) i.e. 28.

Thus, the answer is 28G.

Ex 41.

6

4 5

41

? 5

1

2

7

(A) 16 (B) 9(C) 85 (D) 112

Sol. (C) Hint ; 42 + 52 = 16 + 25 = 4112 + 22 = 1 + 4 = 562 + 72 = 36 + 49 = 85

Ex 42.84 81 8814 18 ?12 9 11

(A) 16 (B) 21(C) 61 (D) 81

Sol. (A) In first figure, 2

1412 = 84.

In second figure, 2

189 = 81.

Let the missing number In third figure be x.

Then, 2x

11 = 88 or x = 11

288 = 16.

Ex 43.

3 5 5

10 30 ?2 3 56 9 6

4 5 2

(A) 15 (B) 20(C) 25 (D) 40

Sol. (B) ClearlyIn first figure] 6 × 3 � 4 × 2 = 18 � 8 = 10

In second figure] 9 × 5 � 5 × 3 = 45 � 15 = 30

In third figure] 6 × 5 � 2 × 5 = 30 � 10 = 20

Ex 44. 26 ? 294 3 35 6 8

6 4 5

(A) 32 (B) 22(C) 18 (D) 27

Sol. (B) In first figure] 5 × 4 + 6 = 26

In second figure] 8 × 3 + 5 = 29

missing number in third figure] 6 × 3 + 4 = 22

Ex 45.174 336 ?

8 5 3

3 2 9

6 7 5 3

2 7 9

2 5

6 4 5(A) 140 (B) 150(C) 200 (D) 180

Sol. (B) In first figure] 8 × 5 × 3 + 3 × 2 × 9 = 120 + 54 =

174In second figure] 6 × 7 × 5 + 2 × 7 × 9= 210 + 126

= 336

missing number in third figure] 3 × 2 × 5 + 6 × 4 × 5 = 30 + 120 = 150

Ex 46.

11 9 15 7 25 21

40 176 ?

(A) 184 (B) 210(C) 241 (D) 425

Sol. (A) The number at the bottom is the difference ofsquares of two numbers given at topIn first figure] 112 � 92 = 121 � 81 = 40

In second figure] 152 � 72 = 225 � 49 = 176

In third figure] 252 � 212 = 625 � 441 = 184

Ex 47. 33

3 5

6 3

7

48

5 4

4 3 5

45

?

(A) 47 (B) 45(C) 37 (D) 35

Sol. (D) In first figure, 6 × 3 + 3 × 5 = 33

In second figure, 5 × 4 + 4 × 7 = 48

In third figure, 5 × 4 + 3 × 5 = 35

PAGE # 71

EXERCISE-1

Directions : (1 to 25) Find the missing numbers :

1. 2, 8, 18, 32, ?(A) 62 (B) 60(C) 50 (D) 46

2. 16, 54, 195, ?(A) 780 (B) 802(C) 816 (D) 824

3. 14, 316, 536, 764, ?(A) 981 (B) 1048(C) 8110 (D) 9100

4. 8, 11, 15, 22, 33, 51, ?, 127, 203(A) 80 (B) 53(C) 58 (D) 69

5. 2, 3, 6, 18, ?, 1944(A) 154 (B) 180(C) 108 (D) 452

6. 7,19, 55, 163, ?(A) 387 (B) 329(C) 527 (D) 487

7. 1, 2, 9, 4, 25, 6, ?(A) 51 (B) 49(C) 50 (D) 47

8. 16, 33, 67, 135, ?(A) 371 (B) 175(C) 271 (D) 287

9. 8, 24, 16, ?, 7, 14, 6, 18, 12, 5, 5, 10(A) 14 (B) 10(C) 7 (D) 5

10. 2, 12, 36, 80, 150, ?(A) 194 (B) 210(C) 252 (D) 258

11. 101, 100, ?, 87, 71, 46(A) 92 (B) 88(C) 89 (D) 96

12. 100, 50, 52, 26, 28, ? 16, 8(A) 30 (B) 36(C) 14 (D) 32

13. 6, 24, 60, 120, 210, 336, ?, 720(A) 496 (B) 502(C) 504 (D) 498

14. 3, 1, 4, 5, 9, 14, 23, ?(A) 32 (B) 37(C) 41 (D) 28

15. 3, 6, 18, 72, 360, ?(A) 720 (B) 1080(C) 1600 (D) 2160

16. 78, 79, 81, ?, 92, 103, 119(A) 88 (B) 85(C) 84 (D) 83

17. 0, 6, 20, 42, 72, ?(A) 106 (B) 112(C) 110 (D) 108

18. 2, 9, 28, 65, ?(A) 121 (B) 195(C) 126 (D) 103

19. 1, 11, ?, 11, 11, 11, 16, 11(A) 1 (B) 11(C) 6 (D) 192

20. 137, 248, 359, 470, ?(A) 582 (B) 581(C) 571 (D) 481

21. 3, 15, 35, ?, 99, 143(A) 63 (B) 77(C) 69 (D) 81

22. 9, 16, 30, 58, ?(A) 104 (B) 114(C) 116 (D) 118

23. 3, 12, 27, 48, 75, 108, ?(A) 192 (B) 183(C) 162 (D) 147

24. 1, 4, 12, 30, ?(A) 60 (B) 62(C) 64 (D) 68

25. 94, 166, 258, ?, 4912(A) 3610 (B) 1644(C) 1026 (D) 516

Directions : (26 to 28) In each of the following questions, anumber series is given. After the series, below it inthe next line, a number is given followed by (P), (Q),(R), (S) and (T). You have to complete the seriesstarting with the number given following thesequence of the given series. Then answer thequestion given below it.

26. 2 3 8 275 (P) (Q) (R) (S) (T)Which of the following numbers will come in placeof (T) ?(A) 184 (B) 6(C) 925 (D) 45

27. 5 18 48 1127 (P) (Q) (R) (S) (T)Which number will come in place of (S) ?(A) 172 (B) 276(C) 270 (D) 376

28. 15 159 259 3237 (P) (Q) (R) (S) (T)Which of the following numbers will come in placeof (R) ?(A) 251 (B) 315(C) 176 (D) 151

PAGE # 72

Directions : (29 to 35) Find the wrong term(s) �

29. 9, 11, 15, 23, 39, 70, 135(A) 23 (B) 39(C) 70 (D) 135

30. 3, 9, 36, 72, 216, 864, 1728, 3468(A) 3468 (B) 1728(C) 864 (D) 216

31. 2, 5, 11, 20, 30, 47, 65(A) 5 (B) 20(C) 30 (D) 47

32. 121, 143, 165, 186, 209(A) 143 (B) 165(C) 186 (D) 209

33. 9, 15, 24, 34, 51, 69, 90

(A) 15 (B) 24

(C*) 34 (D) 51(A) 15 (B) 24(C) 34 (D) 51

34. 9, 13, 21, 37, 69, 132, 261

(A) 21 (B) 37

(C) 69 (D) 132

35. 105, 85, 60, 30, 0, � 45, � 90

(A) 85 (B) � 45

(C) 105 (D) 0

EXERCISE-2

Directions : (1 to 24) Find the missing term(s) :

1. X, U, S, P, N, K, I, ?(A) J (B) K(C) M (D) F

2. Z, X, U, Q, L, ?(A) F (B) K(C) G (D) E

3. A, H, N, S, W, ?(A) A (B) Y(C) B (D) Z

4. Q, T, V, Y, A, ?(A) B (B) C(C) D (D) F

5. X, A, D, G, J, ?(A) N (B) O(C) M (D) P

6. Z, L, X, J, V, H, T, F, ?, ?(A) R, D (B) R, E(C) S, E (D) Q, D

7. AZ, YB, CX, WD, ?(A) VE (B) UE(C) EU (D) EV

8. DFK, FEL, HDM, JCN, ?(A) KBN (B) KBO(C) LBO (D) LBN

9. JXG, HTJ, FPN, ?, BHY(A) EKS (B) ELS(C) DLR (D) DLS

10. CYD, FTH, IOL, LJP, ?(A) PET (B) OET(C) OEY (D) PEV

11. ZGL, XHN, VIQ, TJU, ?(A) RKX (B) RKY(C) RLZ (D) RKZ

12. MTH, QRK, UPN, YNQ, ?(A) CKT (B) ELT(C) CLT (D) EKT

13. ZSD, YTC, XUB, WVA, ?(A) VZZ (B) ZVX(C) VWZ (D) VZX

14. RML, VIJ, ZFH, DDF, ?(A) HDC (B) CHI(C) HCD (D) DIC

15. LRX, DJP, VBH, NTZ, ?(A) ELS (B) FMR(C) GKS (D) FLR

16. MAD, OBE, SCH, YDM, ?(A) HET (B) HES(C) GET (D) UAE

17. 2B, 4C, 8E, 14H, ?(A) 22L (B) 24L(C) 22K (D) 2M

18. 1 BR, 2 EO, 6 HL, 15 KI, ?(A) 22 NF (B) 31 NF(C) 31 NE (D) 28 NF

19. P3C, R5F, T8I, V12L, ?(A) Y17O (B) X17M(C) X17O (D) X16O

20. Z 15 A, W 13 C, ?, Q 9 G, N 7 I(A) T 12 E (B) R 11F(C) T 11E (D) R 13 D

21. B3M, E7J, H15G, K31D, ?(A) N65A (B) O63A(C) N63A (D) N63Z

22. 5X9, 8U12, 11R15, 14O18, ?(A) 17L21 (B) 17K21(C) 17M21 (D) 17L23

23. 6C7, 8F10, 11J14, 15O19, ?(A) 19U24 (B) 20U25(C) 19U25 (D) 20U24

24. B2E, D5H, F12K, H27N, ?(A) J58Q (B) J56Q(C) J57Q (D) J56P

PAGE # 73

Directions : (25 to 30) Find the wrong term(s) :

25. ECA, JHF, OMK, TQP, YWU(A) ECA (B) JHF(C) TQP (D) YWU

26. DKY, FJW, HIT, JHS, LGQ(A) FJW (B) LGQ(C) JHJ (D) HIT

27. DVG, FSI, HPK, JNM, LJO(A) DVG (B) JNM(C) HPK (D) LJO

28. CDF, DEG, EFH, FHI(A) CDF (B) DEG(C) FHI (D) EFH

29. ZLA, BMY, CNW, FOU, HPS(A) ZLA (B) BMY(C) FOU (D) CNW

30. G4T, J10R, M20P, P43N, S90L

(A) G4T (B) J10R

(C) M20P (D) P43N

EXERCISE-3

Directions : (1 to 15) Which sequence of letters when placedat the blanks one after the other will complete thegiven letter series ?

1. a _ b a a _ a a _ _ a b(A) a a a a (B) b a a a(C) b b a a (D) a b b a

2. _ a a b b _ a _ a b _ b(A) b b a a (B) b a b a(C) b a a b (D) a b a b

3. a a b _ a a a _ b b a _(A) b a a (B) a b b(C) b a b (D) a a b

4. a _ _ b _ a _ a b _ a a(A) a b a a b (B) b b a b a(C) b b a b b (D) b a a b a

5. abc _ d _ bc _ d _ b _ cda(A) bacdc (B) cdabc(C) dacab (D) dccbd

6. a _ bbc _ aab _ cca _ bbcc(A) bacb (B) acba(C) abba (D) caba

7. _ b c _ _ b b _ a a b c(A) acac (B) babc(C) abab (D) aacc

8. _ b c c _ ac _ a a b b _ a b _ c c(A) aabca (B) abaca(C) bacab (D) bcaca

9. a _ cab _ a _ c _ b c(A) bbac (B) abab(C) abba (D) bcba

10. ba _ cb _ b _ bab _(A) acbb (B) bcaa(C) cabb (D) bacc

11. a _ bc _ a _ bcda _ ccd _ bcd _(A) abddbd (B) acbdbb(C) adbbad (D) bbbddd

12. cc _ ccdd _ d _ cc _ ccdd _ dd(A) dcdcc (B) dcddc(C) dccdd (D) None of these

13. a _ b a a _ b a a _ b a(A) a a b (B) b a b(C) b b a (D) b b b

14. b a b b b _ b _ b _ b b(A) b b a (B) b a a(C) a b a (D) a a a

15. m _ l _ ml _ m _ llm(A) lmmm (B) lmlm(C) lmml (D) mllm

Directions : (16 to 19) The questions given below are basedon the letter series, In each of these series, someletters are missing. Select the correct alternative. Ifmore than five letters are missing, select the lastfive letters of the series.

16. _ _ r _ ttp _ _ s _ tp _ _ _ s _ _ _(A) rstqp (B) tsrqp(C) rstpq (D) None

17. _ _ x _ zbxazyxabyz _ _ _ _ _(A) abxzy (B) abzxy(C) abxyz (D) bxayz

18. x _ xxy _ x _ xy _ yxx _ _ yy _ y(A) xyyyy (B) xxyyx(C) yxxyx (D) xyxyx

19. _ _ r _ tqrptsrpqst _ _ _ _ _(A) pqrts (B) pqtrs(C) pqrst (D) qrpst

Directions : (20 to 23) There is a letter series in the first rowand a number series in the second row. Eachnumber in the number series stands for a letter inthe letter series. Since in each of that series someterm are missing you have to find out as to whatthose terms are, and answer the questions basedon these as given below in the series.

20. a b _ c d _ a _ a b d _ d b a _1 _ 3 _ 3 2 _ 1 _ _ _ 4 _ _ _ _The last four terms in the series are(A) 1234 (B) 3112(C) 3211 (D) 4312

PAGE # 74

21. _ b n t _ _ n a m _ n a b _ _ a _ _ _ _1 3 _ 2 5 3 _ _ 5 2 4 _ 3 2 5 _ _ _ _ _The last five terms in the series are(A) 13425 (B) 41325(C) 34125 (D) 13452

22. n _ g f _ t _ f h t n _ _ t _ b _ f1 3 _ 2 4 5 0 _ 4 _ _ 3 _ _ _ _ _ _The last five terms of the number series are(A) 50123 (B) 40321(C) 40231 (D) 51302

23. _ m i a x _ i r x a _ _ m a _ _ _ _ _ _4 _ 5 _ 7 3 _ _ _ 6 _ _ _ _ _ _ _ _ _ _The last five term of the letter series are(A) r m x i a (B) x m r a i(C) x r m a i (D) r m i x a

Directions : (24 to 26) In each of the following questions,three sequences of letter/numbers are given whichcorrespond to each other in some way. In eachquestion, you have to find out the letter/numeralsthat come in the vacant places marked by (?). Theseare given as one of the four alternatives under thequestion. Mark your answer as instructed.

24. _ A C _ B D _ C D C D2 _ 4 1 _ 1 4 _ _ _ _r s _ q r _ p ? ? ? ?(A) p q p q (B) p r p r(C) r q r q (D) r s r s

25. A _ B A C _ D _ B C D C_ 4 _ 3 _ 2 _ 5 ? ? ? ?d c _ _ b a c b _ _ _ _(A) 2 4 5 4 (B) 2 5 4 5(C) 3 4 5 4 (D) 4 5 2 5

26. _ A D A C B _ _ B D C C2 4 _ _ 2 3 5 3 _ _ _ _p _ _ q _ _ r s ? ? ? ?(A) p r s s (B) p s r r(C) r p s s (D) s r p p

EXERCISE-4

Directions : (1 to 39) Find the missing term in the givenfigures

1.

(A) 36 (B) 9(C) 25 (D) 64

2.

(A) 14 (B) 18(C) 11 (D) 13

3.

(A) 112 (B) 92(C) 82 (D) 102

4.

(A) 235 (B) 141(C) 144 (D) 188

5.

18 3012

6

3216 40

8

36 18 27

?

(A) 18 (B) 12(C) 9 (D) 6

6. 12 21 ?6 7 85 6 4

4 5 10

(A) 14 (B) 22(C) 32 (D) 320

7.

5 9 8

5 15 ?

3 5 6

(A) 12 (B) 11(C) 16 (D) 26

8.

(A) 72 (B) 18(C) 9 (D) 19

9.

(A) 1 (B) 18(C) 90 (D) 225

10.

(A) 20 (B) 22(C) 24 (D) 12

PAGE # 75

11.

7 11 4912 8 5415 4 ?

(A) 36 (B) 7(C) 25 (D) 0

12.

18 24 3212 14 163 ? 472 112 128

(A) 2 (B) 3(C) 4 (D) 5

13. C26

5

3

2

4 H70

4

5

4

10 J90

6

?

8

6

(A) 1 (B) 3(C) 4 (D) 5

14. 80 70 80

29 29 5927 30 40

33 31 10

43 44 2045 43 39

39 42 ?

(A) 69 (B) 49(C) 50 (D) 60

15.

101 48

35 5615 184

43 3438 ?

(A) 127 (B) 142(C) 158 (D) 198

16.

1 7 63 3 ?5 4 8

35 74 104

(A) 1 (B) 2(C) 3 (D) 4

17.

(A) 33 (B) 145(C) 135 (D) 18

18.

(A) 28 (B) 36(C) 81 (D) 49

19. 72 6 140

5

8

4 10 7

?

2

6 2

4 314 12 3

6

6 8

(A) 16 (B) 14(C) 20 (D) 22

20.5 3 94 8 4

20 24 ?9 11 13

(A) 117 (B) 36(C) 32 (D) 26

21.

(A) 26 (B) 25(C) 27 (D) 30

22.

12 16 3630 40 3418 32 18

30 44 ?

(A) 48 (B) 9(C) 44 (D) 64

23.

5

16 109 2

6

21

22 53 19

15

51

17 ? 48

13

(A) 25 (B) 129(C) 7 (D) 49

24.

4 5 4

33 54 ?3 4 32 2 5

2 3 6

(A) 78 (B) 82(C) 94 (D) 86

PAGE # 76

25.

2

4 5

3

28

5

7 4

3

38

1

2 3

7

?

(A) 14 (B) 18(C) 11 (D) 26

26.

(A) 9 (B) 11(C) 10 (D) 12

27. BIG - 792 HCA - 138 FED - 456 E?H - 87?

(A) G, 6 (B) I, 9(C) G, 5 (D) I, 5

28. 26 21 ?

36 9 25

25 16 36

64 25 14449 81 64

(A) 19 (B) 23(C) 25 (D) 31

29. 2 1 ?8 6 83 4 12

2 3 6

6 8 4(A) 3 (B) 4(C) 5 (D) 6

30. 80 70 80

29 29 5927 30 40

33 31 10

43 44 2045 43 39

39 42 ?

(A) 69 (B) 49(C) 50 (D) 60

31.

(A) 0 (B) 2(C) 3 (D) 1

32.

(A) 12 (B) 9(C) 14 (D) 10

33 Find the missing letters from left to right.

(A) JSN (B) JNS(C) JRS (D) KRS

34.3 8 10 2 ? 1

6 56 90 2 20 0

(A) 0 (B) 3(C) 5 (D) 7

35. 80 65 ?15 9 132 7 16

5 4 116 6 8

(A) 48 (B) 72(C) 35 (D) 120

36.

(A) 38 (B) 64(C) 4 (D) 16

37.

101 48

35 5615 184

43 3438 ?

(A) 127 (B) 142(C) 158 (D) 198

38.

18 32 1830 40 2712 16 36

6 8 ?(A) 18 (B) 12(C) 9 (D) 6

39.4 9

6

9 16

12

16 ?

20

(A) 60 (B) 50(C) 21 (D) 25

40. Find the value of X in the following figure :

15 4

33 2

27 2

36 8

32 X

18 9

22 11

12 3

(A) 3 (B) 4(C) 8 (D) 12

PAGE # 77

Directions : (1 to 5) Read the following information carefullyand answer the questions given below it.(i). Five professors (Dr. Joshi, Dr. Davar, Dr.Natrajan, Dr. Choudhary and Dr. Zia) teach fivedifferent subjects (zoology, physics, botany, geologyand history) in four universities ( Delhi, Gujarat,Mumbai, and Osmania). Do not assume anyspecific order.(ii). Dr. Choudhary teaches zoology in MumbaiUniversity .(iii). Dr. Natrajan is neither in Osmania Universitynor in Delhi University and he teaches neithergeology nor history.(iv). Dr. Zia teaches physics but neither in MumbaiUniversity nor in Osmania University.(v). Dr. Joshi teaches history in Delhi University.(vi). Two professors are from Gujarat University.(vii). One professor teaches only one subject andin one University only.

Ex 1. Who teaches geology ?(A) Dr Natrajan (B) Dr. Zia(C) Dr. Davar (D) Dr. Joshi

Ex 2. Which university is Dr. Zia from ?(A) Gujarat (B) Mumbai(C) Delhi (D) Osmania

Ex 3. Who teaches botany ?(A) Dr. Zia (B) Dr. Davar(C) Dr. Joshi (D) Dr. Natrajan

Ex 4. Who is from Osmania University ?(A) Dr. Natrajan (B) Dr. Davar(C) Dr. Joshi (D) Dr. Zia

Ex 5. Which of the following combinations is correct ?(A) Delhi University - Dr. Zia(B) Dr. Choudhary - geology(C) Dr. Davar - Mumbai University(D) Dr. Natranjan - Gujarat University

Sol. : (1 to 5)From the given information in the question :From II, we get Dr. Choudhary teaches zoology inMumbai University.

From III, We get Dr. Natrajan is neither in Osmanianor in Delhi University. Therefore, he will be eitherat Mumbai or Gujarat University. Similarly, as heteaches neither geology nor history, therefore, hemust be teaching physics or botany. ..........(1)From IV, Dr. Zia Physics but as he is not teachingin either Mumbai or Osmania University, he mustbe teaching either in Delhi or Gujarat University...(2)Form V, we get Dr Joshi teaches history in DelhiUniversity Form (1) and (2), we conclude that DrNatarajan teaches botany. And from (1), (2) and VI,we get both Natarajan and Zia teach in GujaratUniversity. Finally, On summarisation we canprepare the following table.

PUZZLE TEST

Names University SubjectDr. Joshi Delhi HistoryDr. Davar Osmania GeologyDr. Natrajan Gujarat BotanyDr. Choudhary Mumbai Zoology

Dr. Zia Gujarat Physics

On the basis of the above table, rest of the questionscan be solved very easily.

1. (C) Dr. Davar teaches geology.

2. (A) Dr. Zia is from Gujarat university.

3. (D) Dr. Natrajan teaches botany.

4. (B) Dr. Davar is from Osmania University.

5. (D) Dr. Natranjan - Gujarat University is the correctcombination.

Ex 6. Ramesh is taller than Vinay who is not as tall asKaran. Sanjay is taller than Anupam but shorterthan Vinay. Who among them is the tallest ?(A) Ramesh (B) Karan(C) Vinay (D) Cannot be determined

Sol. (D) In this question ranking of Karan is not defined.Consequently, either Ram or Karan occupies thetop position with regard to height. Hence,option (d) is the correct choice.

Directions : (7 to 11) Read the following information carefullyand answer the questions given below it :There are five men A, B, C, D and E and six womenP, Q, R, S, T and U. A, B and R are advocates; C, D,P, Q and S are doctors and the rest are teachers.Some teams are to be selected from amongstthese eleven persons subject to the followingconditions :A, P and U have to be together.B cannot go with D or R.E and Q have to be together.C and T have to be together.D and P cannot go together.C cannot go with Q.

Ex 7. If the team is to consist of two male advocates, twolady doctors and one teacher, the members of theteam are(A) A B P Q U (B) A B P U S(C) A P R S U (D) B E Q R S

Sol. (B) The male advocates are A and B, lady doctorsare P, Q and S ; teachers are E, T and U.Now, A and B will be selected.A, P and U have to be together. Now, we have toselect one lady doctor more. It can be Q or S. But Qand E have to be together. Since E is not selected,so S will be selected. Thus, the team is A B P U S.

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PAGE # 78

Ex 8. If the team is to consist of one advocate, two

doctors, three teachers and C may not go with T,

the members of the team are :

(A) A E P Q S U (B) A E P Q T U

(C) B E Q S T U (D) E Q R S T U

Sol. (B) The advocates are A, B and R ; doctors are

C, D, P, Q, S ; teachers are E, T and U. The team

consists of 3 teachers i.e. E, T, U. Now, A, P and U

have to be together. E and Q have to be together.

Thus, the team is A E P Q T U.

Ex 9. If the team is to consist of one male advocate, one

male doctor, one lady doctor and two teachers, the

members of the team are :

(A) A C P T U (B) A D E P T

(C) A D E P U (D) B C E Q U

Sol. (A) The male advocates are A and B ; male doctors

are C and D ; lady doctors are P, Q and S ; teachers

are E, T and U. If A is selected, P and U will be

selected. D and P cannot go together. So, a male

doctor C will be selected. C and T have to be

together. Thus, the team is A C P T U. If B is

selected, D will not be selected. So, male doctor C

will be chosen. C and T have to be together. Now,

the second teacher to be selected is E or U. But, U

cannot go without A. So, E will be selected. E and Q

have to be together. Thus, the team can also be

B C E Q T.

Ex 10. If the team is to consist of one advocate, three

doctors and one male teacher, the members of

the team are:

(A) A D P S U (B) C D R S T

(C) D E Q R S (D) D E Q R T

Sol. (C) The advocates are A, B and R ; the doctors are

C, D, P, Q and S ; male teacher is E. Clearly, E will

be selected. E and Q have to be together. C and Q

cannot be together. So, C will not be selected. P

also cannot be selected because U is not selected.

So, two other doctors D and S will be selected. P is

not selected, so A will not be selected. D is

selected, so B cannot be selected. Thus, the team

is D E Q R S.

Ex 11. If the team is to consist of two advocates, two

doctors, two teachers and not more than three

ladies, the members of the team are :

(A) A B C P T U (B) A C P R T U

(C) A E P Q R T (D) B C E Q R T

Sol. (A) A C P R T U and A E P Q R T are wrong because

each of these combinations consists of four ladies.

B C E Q R T is incorrect because B and R cannot

go together.

Directions : (12 to 15) Read the following paragraphcarefully :Four women A, B, C and D and three men E, F andG play bridge, a game for four players.(i) The group consists of three married couplesand a widow.(ii) Spouses are never partners in a game.(iii) No more than one married couple ever plays inthe same game.(iv) One day they played four games as follows.

A and E versus B and F.A and G versus D and F.B and C versus F and G.C and E versus D and G.

Ex 12. Whom is E married to ?(A) A (B) B(C) C (D) D

Ex 13. Whom is F married to ?(A) A (B) B(C) C (D) D

Ex 14. Whom is G married to ?(A) A (B) B(C) C (D) D

Ex 15. Which of the following is a widow ?(A) A (B) B(C) C (D) D

Sol. : (12 to 15)From (iv), is married either to A or to C. If F is marriedto A, then G is married to B or to C. If G is married toB, then E is married to D ; if G is married to C, thenE is married to B or to D. If F is married to C, then Gis married to B ; then E is married to D. Hence, themarried couples are : FA, GB, ED or FA, GC, EB orFA, GC, ED or FC, GB, ED. Of these, only FA, GB,ED does not contradict any of the statements.

12. (D) E is married to D.

13. (A) F is married to A.

14. (B) G is married to B.

15. (C) C is a widow.

Ex 16. A vagabond runs out of cigarettes. He searches forthe stubs, having learnt that 7 stubs can make anew cigarette, good enough to be smoked, hegathers 49 stubs, If he smokes 1 cigarette everythree - quarters of an hour, how long will his supplylast ?(A) 5.25 hr (B) 6 hr(C) 4.5 hr (D) 3 hr

Sol. (B) He has got = 7749

cigarettes.

The duration of time he will take to smoke these

7 cigarettes = 43

7 hr = 5.25 hr (i.e. 5 hr and 15

min). Now note that after he has smoked these 7cigarettes, he will collect 7 more stubs (one formeach), form which he will be able to make another

cigarette. This will take him another 43

hr (45 min)

to smoke. Therefore, total time taken = 6hr.

PAGE # 79

Directions : (17 to 18) Read the following information andanswer the questions that follow.There are 70 clerks working with M/s. Jha LalKhanna & Co. chartered accountants, of which 30are female.(i) 30 clerks are married.(ii) 24 clerks are above 25 years of age(iii) 19 Married clerks are above 25 years of age;among them 7 are males.(iv) 12 males are above 25 years of age(v) 15 males are married.

Ex 17. How many unmarried girls are there ?(A) 12 (B) 15(C) 18 (D) 10

Ex 18. How many of these unmarried girls are above 25 ?(A) 12 (B) 15(C) 4 (D) 0

Sol. (17 to 18) : From the given data, we can make thefollowing table with the help of which rest of thequestions can be solved very easily.

Male (40) Female (30)

Above 25

Married 7 12

Unmarried 5 0

Below 25

married 8 3

unmarried 20 15

Total 40 30

17. There are 15 unmarried girls.

18. In these 15 unmarried girls no one is above 25.

EXERCISE

Directions : (1 to 5) Study the following information carefullyand answer the questions given below it :There are five friends A, B, C, D and E. Two of themare businessmen while the other three belong todifferent occupations viz. medical, engineer andlegal. One businessman and the lawyer stay inthe same locality S, while the other three stay inthree different localities P, Q and R. Two of thesefive persons are Hindus while the remaining threecome from three different communities viz. Muslim,Christian and Shikh. The lawyer is the oldest inage while one of the businessmen who runs afactory is the youngest. The other businessman isa cloth merchant and agewise lies between thedoctor and the lawyer. D is a cloth merchant andstays in locality S while E is a Muslim and stays inlocality R. The doctor is a Christian and stays inlocality P, B is a Shikh while A is a Hindu and runsa factory.

1. Who stays in locality Q ?(A) A (B) B(C) C (D) E

2. What is E�s occupation ?(A) Business (B) Engineer(C) Lawyer (D) Doctor

3. Agewise who among the following lies between Aand C ?(A) Lawyer (B) Doctor(C) Cloth merchant (D) Engineer

4. What is B�s occupation ?(A) Business (B) Engineer(C) Lawyer (D) Doctor

5. What is C�s occupation ?(A) Doctor (B) Lawyer(C) Engineer (D) Business

Directions : (6 to 10) Study the given information carefullyand answer the questions that follow.There are four people sitting in a row : one eachfrom India, Japan, USA and Germany, but not inthat order,. They are wearing caps of different colours - green,yellow, red and white, not necessarily in that order.II. One is wearing a kurta and one a T-shirt.III. The Indian is wearing a green cap and a jacket.IV. The American is not seated at either end.V. The persons with kurta and T-shirt are sittingnext to each other.VI. The persons with kurta wears a red cap andsits next to the Japanese.VII. The Japanese wears a shirt and is not seatedat either end.VIII. The man with white cap wears T-shirt and isseated at one end.

6. Who wears the T-shirt ?(A) Indian (B) Japanese(C) American (D) German

7. Who is wearing a kurta ?(A) Indian (B) Japanese(C) American (D) German

8. What is the colour of the cap worn by the Japanese?(A) Red (B) Green(C) Yellow (D) White

9. Who precedes the man wearing T-shirt ?(A) Indian (B) Japanese(C) American (D) German

10. Who precedes the man wearing jacket ?(A) Indian (B) German(C) Japanese (D) Cannot say

PAGE # 80

Directions : (11 to 15) Read the following informationcarefully and answer the questions that follow.I. There are six students ( A, B, C, D, E and F) in agroup. Each student can opt for only three choicesout of the six which are music, reading, painting,badminton, cricket and tennis.II. A, C and F like reading.III. D does not like badminton, but likes music.IV. Both B and E like painting and music.V. A and D do not like painting, but they like cricket.VI. All student except one like badminton.VII. Two students like tennis.VIII. F does not like cricket, music and tennis.

11. Which pair of students has the same combinationof choices ?(A) A and C (B) C and D(C) B and E (D) D and F

12. Who among the following students likes bothtennis and cricket ?(A) A and B (B) C(C) B and D (D) D

13. How many students like painting and badminton ?(A) 1 (B) 2(C) 3 (D) 4

14. Who among the following do not like music ?(A) A , C and D (B) A, B and C(C) A, C and F (D) B, D and F

15. Which of the following is the most popular choice?(A) Tennis (B) Badminton(C) Reading (D) Painting

16. R earns more than H but not as much as T, Mearns more than R. Who earns least amongthem?(A) R (B) T(C) H (D) M

17. Harish is taller than Manish but shorter thanSuresh. Manish is shorter than Anil but taller thanRaghu. Who among them is the shortest havingregard to height ?(A) Anil (B) Manish(C) Raghu (D) Cannot be determined

Direction : (18) Examine the following statements :I. Either A and B are of the same age or A is olderthan B.II. Either C and D are of the same age or D is olderthan C.III. B is older than C.

18. Which one of the following conclusions can bedrawn from the above statements ?(A) A is older than B(B) B and D are of the same age(C) D is older than C(D) A is older than C

Directions : (19 to 23) Read the information given belowand answer the questions.The age and height of six children in a class are asfollows :(i) A is taller and older than B but shorter andyounger than C.(ii) D is taller than E who is not as tall as B.(iii) The oldest is the shortest.(iv) The youngest would be fourth if the childrenstood in a line according to their height and onestarted counting from the tallest.(v) D is younger than F but older than E who isolder than C.

19. Who among them is the tallest ?(A) B (B) E(C) C (D) Data inadequate

20. Who is older than B but younger than C ?(A) F (B) D(C) A (D) Data inadequate

21. Which of the following statements is definitely true?(A) D is the most old person(B) B has the max. height(C) A is older than D(D) F is the shortest

22. Which of the following is the correct order of heightin descending order?(A) A, C, D, B, E, F (B) F, D, E, C, A, B(C) D, C, A, B, E, F (D) C, D, A, B, E, F

23. Whose Rank in height cannot be positioneddefinitely ?(A) B (B) D(C) C (D) E

Directions : (24 to 28) Study the information given belowand answer the questions that follow.(i) Six Plays P, Q, R, S, T and U are to be organisedfrom Monday to Saturday i.e. 10 to 15 one play eachday.(ii) There are two plays between R and S and oneplay between P and R.(iii) There is one play between U and T and T is tobe organised before U.(iv) Q is to be organised before P, not necessarilyimmediately.(v) The organisation does not start with Q.

24. The organisation would start from which play ?(A) P (B) S(C) T (D) None

25. On which date is play T to be organised ?(A) 10th (B) 11th

(C) 12th (D) None

26. The organisation would end with which play ?(A) P (B) Q(C) S (D) None

PAGE # 81

27. Which day is play Q organised ?(A) Tuesday (B) Wednesday

(C) Thursday (D) None

28. Which of the following is the correct sequence oforganising plays ?(A) PTRUQS (B) QSTURP

(C) SUTRQP (D) None

Directions : (29 to 30) Read the following informationcarefully and answer the questions given below it.I. Seven books are placed one above the other in a

particular way .II. The history book is placed directly above thecivics book.

III. The geography book is fourth from the bottomand the English book is fifth from the top.IV. There are two books in between the civics and

economics books.

29. To find the number of books between the civicsand the science books, which other extra piece ofinformation is required, from the following ?

(A) There are two books between the geographyand the science books.(B) There are two books between the mathematics

and the geography books .(C) There is one book between the English andthe science books.

(D) The civics book is placed before two booksabove the economics book.

30. To know which three books are kept above theEnglish book, which of the following additional

pieces of information, if any, is required?(A) The economics book is between the Englishand the science books.

(B) There are two books between the English andthe history books.(C) The geography book is above the English book.

(D) No other information is required.

Directions : (31 to 32) A five-member team that includesRama, Shamma, Henna, Reena, and Tina, isplanning to go to a science fair but each of them

put up certain conditions for going .They are asfollows.I. If Rama goes, then at least one amongst

Shamma and Henna must go.II. If Shamma goes, then Reena will not go.III. If Henna will go, then Tina must go.

IV. If Reena goes, then - Henna must go.V. If Tina goes, then Rama must go but Shammacannot go.

VI. If Reena plans not to go the fair, then Rama willalso not go.

31. If it is sure that Henna will go to the fair, then who

among the following will definitely go ?

(A) Rama (B) Shamma

(C) Reena (D) Rama and Reena

32. If Tina does not go to the fair, which of the following

statements must be true ?

(i) Henna cannot go

(ii) Shamma cannot go

(iii) Reena cannot go

(iv) Rama cannot go

(A) (i) and (ii) (B) (iii) and (iv)

(C) (i), (iii) and (iv) (D) (i) and (iv)

Directions : (33 to 37) Read the following paragraph

carefully and choose the correct alternative.

The office staff of XYZ corporation presently

consists of three females A, B, C and five males D,

E, F, G and H. The management is planning to

open a new office in another city using three males

and two females of the present staff. To do so they

plan to separate certain individuals who do not

function well together. The following guidelines

were established

I. Females A and C are not to be together

II. C and E should be separated

III. D and G should be separated

IV. D and F should not be part of a team.

33. If A is chosen to be moved, which of the following

cannot be a team ?

(A) ABDEH (B) ABDGH

(C) ABEFH (D) ABEGH

34. If C and F are to be moved to the new office, how

many combinations are possible ?

(A) 1 (B) 2

(C) 3 (D) 4

35. If C is chosen to the new office, which number of

the staff cannot be chosen to go with C ?

(A) B (B) D

(C) F (D) G

36. Under the guidelines, which of the following must

be chosen to go to the new office ?

(A) B (B) D

(C) E (D) G

37. If D goes to the new office, which of the following

is/are true ?

I. C cannot be chosen

II. A cannot be chosen

III. H must be chosen.

(A) I only (B) II only

(C) I and II only (D) I and III only

PAGE # 82

Directions : (38 to 42) Study the following information

carefully and answer the questions that follow :

A team of five is to be selected from amongst five

boys A, B, C, D and E and four girls P, Q, R and S.

Some criteria for selection are :

A and S have to be together

P cannot be put with R.

D and Q cannot go together.

C and E have to be together.

R cannot be put with B.

Unless otherwise stated, these criteria are

applicable to all the questions below :

38. If two of the members have to be boys, the team

will consist of :

(A) A B S P Q (B) A D S Q R

(C) B D S R Q (D) C E S P Q

39. If R be one of the members, the other members of

the team are :

(A) P S A D (B) Q S A D

(C) Q S C E (D) S A C E

40. If two of the members are girls and D is one of the

members, the members of the team other than D

are :

(A) P Q B C (B) P Q C E

(C) P S A B (D) P S C E

41. If A and C are members, the other members of the

team cannot be :

(A) B E S (B) D E S

(C) E S P (D) P Q E

42. If including P at least three members are girls, the

members of the team other than P are :

(A) Q S A B (B) Q S B D

(C) Q S C E (D) R S A D

Directions : (43 to 44) Read the given information carefully

and answer the questions that follow :

Ratan, Anil, Pinku and Gaurav are brothers of Rakhi,

Sangeeta, Pooja and Saroj, not necessarily in that

order. Each boy has one sister and the names of

bothers and sisters do not begin with the same

letter. Pinku and Gaurav are not Saroj �s or

Sangeeta�s brothers. Saroj is not Ratan�s sister.

43. Pooja�s brother is

(A) Ratan (B) Anil

(C) Pinku (D) Gaurav

44. Which of the following are brother and sister ?

(A) Ratan and Pooja (B) Anil and Saroj

(C) Pinku and Sangeeta (D) Gaurav and Rakhi

Directions : (45 to 49) Read the following informationcarefully and answer the questions given below.(i) There is a family of six persons- L, M, N, O, Pand Q. They are professor, businessman,chartered account, bank manager, engineer andmedical representative, not necessarily in thatorder.(ii) There are two married couples in the family.(iii) O, the bank manager is married to the ladyprofessor.(iv) Q, the medical representative, is the son of Mand brother of P.(v) N, the chartered accountant, is the daughter - inlaw of L.(vi) The businessman is married to the charteredacconuntant.(vii) P is an unmarried engineer.(viii) L is the grandmother of Q

45. How is P related to Q.(A) Brother (B) Sister(C) Cousin (D) Either brother or sister

46. Which of the following is the profession of M ?(A) Professor(B) Chartered accountant(C) Businessman(D) Medical representative

47. Which of the following is the profession of L ?(A) Professor (B) Charted accountant(C) Businessman (D) Engineer

48. Which of the following is one of the couples ?(A) QO (B) OM(C) PL (D) None of these

49. How is O related to Q?(A) Father (B) Grandfather(C) Uncle (D) Brother

Directions : (50 to 54)I. There is a group of six persons P,Q, R, S, T and Ufrom a family. They are Psychologist, Manager,Lawyer, Jeweller, Doctor and Engineer.II. The Doctor is grandfather of U, who is aPsychologist.III. The Manager S is married to P.IV. R, the Jeweller is married to the Lawyer.V. Q is the mother of U and T.VI. There are two married couples in the family.

50. What is the profession of T ?(A) Doctor (B) Jeweller(C) Manager (D) None of these

51. How is P related to T ?(A) Brother (B) Uncle(C) Father (D) Grandfather

52. How many male members are their in the family ?(A) One (B) Three(C) Four (D) Data inadequate

PAGE # 83

53. What is the profession of P ?(A) Doctor (B) Lawyer(C) Jeweller (D) Manager

54. Which of the following is one of the pairs of couplesin the family ?(A) PQ (B) PR(C) PS (D) Cannot be determined

Direction : (55) The ages of Mandar, Shivku, Pawan andChandra are 32, 21, 35 and 29 years, not in order,Whenever asked they lie of their own age but tellthe truth abut others.(i) Pawan says, �My age is 32 and Mandar�s age isnot 35�(ii) Shivku says, �My age is not 2 9 and Pawan�sage in not 21�(iii) Mandar says, �My age is 32.�

55. What is Chandra�s age ?(A) 32 years (B) 35 years(C) 29 years (D) 21 years

Directions : (56 to 57) Answer the questions on the basisof the information given below. 5 friends Nitin,Reema, Jai, Deepti and Ashutosh are playing agame of crossing the roads. In the beginning, Nitin,Reema and Ashutosh are on the one side of theroad and Deepti and Jai are on the other side. Atthe end of the game, it was found that Reema andDeepti are on the one side and Nitin, Jai andAshutosh are on the other side of the road. Rulesof the game are as follows :I. One �movement� means only one person crossesthe road from any side to the other side.II. No two persons can cross the roadsimultaneously from any side to the other side.

III. Two persons from the same side of the roads

cannot move in consecutive �movements�.

IV. If one person crosses the road in a particular

movement, he or she cannot immediately move

back to the other side.

V. Jai and Reema did not take part in first 3

movements.

56. What is the minimum possible number of

movements that took place in the entire game ?

(A) 3 (B) 4

(C) 5 (D) 6

57. If number of movements are minimised in the

game, then which of the following combination of

friends can never be together on one particular

side of the road during the course of the game ?

(A) Nitin, Reema amd Deepti

(B) Nitin, Jai and Deepti

(C) Deepti, Jai and Ashutosh

(D) Ashutosh, Nitin and Deepti

58. You have 12 similar looking coins. 11 of them weigh

the same. One of them has a different weight, but

you don�t know whether it is heavier or lighter. You

also have a scale. You can put coins on both sides

of the scale and it�ll tell you which side is heavier or

will stay in the middle if both sides weigh the same.

What is the minimum number of weighing required

to find out the odd coin.

(A) 3 (B) 4

(C) 5 (D) 6

PAGE # 84

CALENDAR AND CLOCK TEST

We are to find the day of the week on a mentioneddate. Certain concepts are defined as under.

An ordinary year has 365 days.

In an ordinary year, first and last day of the year aresame.

A leap year has 366 days. Every year which isdivisible by 4 is called a leap year. For example1200, 1600, 1992, 2004, etc. are all leap years.

For a leap year, if first day is Monday than last daywill be Tuesday for the same year.

In a leap year, February is of 29 days but in anordinary year, it has only 28 days.

Year ending in 00's but not divisiable by 400 is notconsidered a leap year. e.g., 900, 1000, 1100, 1300,1400, 1500, 1700, 1800, 1900, 2100 are not leapyears.

The day on which calendar started (or the very firstday ) i.e., 1 Jan, 0001 was Monday.

Calendar year is from 1 Jan to 31 Dec. Financialyear is from 1 April to 31 March.

ODD DAYS

The no. of days exceeding the complete no. ofweeks in a duration is the no. of odd days duringthat duration.

COUNTING OF ODD DAYS

Every ordinary year has 365 days = 52 weeks +1 day. Ordinary year has 1 odd day.

Every leap year 366 days = 52 weeks + 2 days. Leap year has 2 odd days.

Odd days of 100 years = 5,Odd days of 200 years = 3,Odd days of 300 years = 1,Odd days of 400 years = 0.

Explanation : 100 years = 76 ordinary years + 24 leap years

( The year 100 is not a leap year)= 76 odd days + 2 × 24 odd days = 124 odd days.

Odd days = 7

124 = 5 odd days.

Similarly, 200 years = 10 odd days = 03 odd days

300 years = 7

15= 1 odd day..

400 years = 7

120 = 0 odd day (1 is added as 400

is a leap year)Similarly, 800, 1200, 1600, 2000, 2400 yearscontain 0 odd days.

After counting the odd days, we find the dayaccording to the number of odd days.

Sunday for 0 odd day, Monday for 1 odd day and soon as shown in the following table.

Table : 1 (Odd days for week days)

0 1 2 3 4 5 6

SaturdayMondaySunday Tuesday WednesdayDays

Odd Days

Thursday Friday

Table : 2 (Odd days for months in a year)

Ordinary Year

DaysOdd Days

Leap year DaysOdd

Days

January 31 3 January 31 3

February 28 0 February 29 1

March 31 3 March 31 3

April 30 2 April 30 2

May 31 3 May 31 3

June 30 2 June 30 2

Total 181 days 6 Total 182 days 0

July 31 3 July 31 3

August 31 3 August 31 3

September 30 2 September 30 2

October 31 3 October 31 3

November 30 2 November 30 2

December 31 3 December 31 3

Total 184 days 1 Total 184 days 2

Table : 3 (Odd days for every quarter)

Ivth three months1 Oct. to31 Dec.

Total year1 Jan to 31 Dec.

Ist three months1 Jan to

31 March

Monthsof

years

IInd three months1 Apr to30 June

IIIrd three months1 July to 30 Sep.

Total days90 / 91

Ord. / Leap91 92 92

365 / 366Ord. / Leap

1Odd day

1 / 2 Ord. / Leap

Odd days6 / 0

Ord. / Leap0

Odd day1

Odd day

Ex 1. If it was Saturday on 17th December 1982 whatwill be the day on 22nd December 1984 ?

Sol. Total number of odd days between 17 Dec.1982to 17 Dec.1984 the number of odd days = 1+2 = 3.From 17 to 22 Dec. number of odd days = 5 3 + 5 = 8 odd days = 1 odd day. Saturday + 1 odd day = Sunday.

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Ex 2. Find the day of the week on 16 January, 1969.Sol. 1600 years have �0� odd day. .....................(A)

300 years have �1� odd day. ......................(B)68 years have 17 leap years and 51 ordinary years.Thus = (17 × 2 + 51 × 1 ) = 85 odd days

' 01' odd day ...(C)

16 January has = ' 02' odd days..(D)Adding (A) + (B) +(C) +(D),We get, 0 + 01 +01 +02 = 04 odd days

Ans. Thursday

Ex 3. Find the day of the week on 18 July, 1776 (leapyear).

Sol. Here 1600 years have �0� odd day.....................(A)100 years have �5� odd days..............................(B)75 years = (18 leap years + 57 ordinary years)= (18 × 2 + 57 × 1)

= 93 odd days= (7 × 13 + 2) = �2� odd days.............................(C)Now, the no. of days from 1st January to 18 July,1776= 182 + 18 = 200 days= (28 × 7 + 4) days = �4� odd days.....................(D)Adding (A) + (B) +(C) +(D),We get, 0 + 5 + 2 + 4 = 04 odd days

Ans. Thursday

Ex 4. On what dates of October, 1975 did Tuesday fall ?Sol. For determining the dates, we find the day on 1st

Oct, 1975.1600 years have �0� odd days.....................(A).300 years have �01� odd days.....................(B).74 years have (18 leap years + 56 ordinary years)2 × 18 + 1 × 56 = 92 odd days

= �01� odd days.............(C)Days from 1st January to 1st Oct., 19751st Jan � 30 June + 1st July to 1st Oct.181 + 31 + 31 + 30 + 1 = 274 days= �01� odd days......(D) (274/7= 01 days)Adding (A) + (B) +(C) +(D) = 0 + 01 +01 +01= '03' odd days

Ans. Wednesday( 1st Oct), hence 7,14,21,28 Oct. willTuesday fall.

Ex 5. Calendar for 1995 will serve for 2006, prove ?Sol. The Calendar for 1995 and 2006 will be the same

,if day on 1st January of both the years is the same.This is possible only if the total odd days between31st Dec. 1994 and 31st Dec.2005 is 0. [one daybefore both the years as we want to know the dayon 1st January of both the years i.e. same]During this period, we have3 leap years(1996, 2000, 2004) and08 ordinary years(1995,1997,1998,1999, 2001, 2002, 2003,2005)Total odd days = (2 × 3 + 1 × 8) = 14 = 0 odd

days (Thus Proved)

Ex 6. The year next to 1996 having the same Calendar

will be -

Sol. 1996 1997 1998 1999 2000 2001 2002 2003

2 1 1 1 2

Total = 2 + 1 + 1 + 1 + 2 = 7= 0 odd days

Hence, year 2001 will have the same calendar as

year 1996.

Ex 7. Prove that last day of a century cannot be Tuesday,

Thursday or Saturday.

Sol. 100 years have = 5 odd days

Last day of st century is Friday

200 years have = 10 odd days

Last day of IInd century is Wednesday

= 3 odd days

300 years have = 15 odd days

Last day of rd century is Monday

= 01 odd day

400 years have = (5 × 4 + 1)

Last day of 4th century is Sunday

= 21 odd days

= 0 odd days

Since the order keeps on cycling, we see that the

last day of the century cannot be Tuesday, Thursday

or Saturday.

Important Notes :

Minute hand and hour hand coincides once in every

hour. They coincide 11 times in 12 hours and 22

times in 24 hours.

They coincide only one time between 11 to 1 O�

clock. at 12 O� clock.

Minute hand and hour hand are opposite once in

every hour. They do it 11 times in 12 hours and 22

times in 24 hours.

They opposite only one time between 5 to 7 O�

clock. at 6 O� clock.

Both hands (minute and hour) are perpendicular

twice in every hour. 22 times in 12 hours and 44

times in 24 hours.

In one minute, hour hand moves 1/2º and minute

hand moves 6º. In one hour, hour hand moves 30º

and minute hand moves 360º.

In an hour, minute hand moves 55 minutes ahead

of hour hand.

PAGE # 86

HANDS COINCIDE

Ex.8 At what time between 3 O�Clock and 4 O�Clock will

the two hands coincide ?Sol. At 3 O�clock the distance between the two hands is

15 minutes when they coincide with each other thedistance between the two hands will be 0 min.So, the time taken (15 + 0 ) = 15 minutes.

Minute hand is 55 min. ahead of hour hand in 60 min. Minute hand is 1 min. ahead of hour hand in

5560 min.

Minute hand is 15 min. ahead of hour hand in

55

1560 =

11180

= 114

16 min.

Hence the right time is 114

16 minute past 3.

HANDS ARE OPPOSITE

Ex.9 At what time between 2 O�clock and 3 O�clock will

the two hands be opposite ?Sol. At 2 O�clock the distance between the two hands is

10 minutes. When they are at 30 minutes distance,they are opposite to each other. The time taken(30 + 10 ) = 40 min.

Minute hand is 55 min. ahead of hour hand in 60 min. Minute hand is 1 min. ahead of hour hand in

5560

min.

Minute hand is 40 minutes ahead of hour hand

in 55

4060 =

11480

= 117

43 min.

Hence, the right time is 117

43 min. past 2.

HANDS ARE PERPENDICULAR

Ex.10 At what time between 4 O�clock and 5 O�clock will

the hands are perpendicular ?Sol. At 4 O�clock the distance between the two hands is

20 min. When they are at 15 minutes distance,they are perpendicular to each other.

Case-I When the time taken (20 � 15) = 5 min.

Minute hand is 55 min. ahead of hour hand in 60 min.Minute hand is 5 min. ahead of hour hand in

55

560 =

1160

= 115

5 min.

Hence, the right time is 115

5 min. past 4.

Case-II When the time taken (20 + 15) = 35 min.

Minute hand is 55 min. ahead of hour hand in

60 min.

Minute hand is 35 min. ahead of hour hand in

55

3560 =

11420

= 112

38 min.

Hence, the right time is 112

38 min. past 4.

MIRROR IMAGE OF CLOCK

If the time is between 1 O�clock to 11 O�clock, then

to find the mirror image, time is subtracted from

11 : 60.

If the time is between 11 O�clock to 1 O�clock, then

to find the mirror image, time is subtracted from

23 : 60.

Ex.11 The time in the clock is 4 : 46, what is the mirror image ?

Sol. (11 : 60) � (4 : 46) = 7 : 14.

Ex.12 The time in the clock is 12 : 35, then find its mirror

image.

Sol. (23 : 60) � (12 : 35) = 11 : 25.

TO FIND THE ANGLE BETWEEN TWO HANDS

Angle are of two types :

Positive angle : It is obtained by moving from hour

hand to minute hand moving in clockwise direction.

Negative angle : It is obtained by moving from

minute hand to hour hand.

Both types of angles are 360º in total. If one angle

is known, other can be obtained by subtracting from

360º.

Ex.13 At 4 : 30, what is the angle formed between hour

hand and minute hand ?

Sol. At 4 O� clock angle between hour and min. hand is

of 120º.

In 30 min. minute hand make an angle of 180º.

So, the resultant angle is 180º � 120º = 60º.

But in 30 min. hour hand will also cover an angle of 15º.

Hence, the final angle between both hands is

60º � 15º = 45º.

Short trick

PAGE # 87

Ex.14 A bus for Delhi leaves every thirty minutes from a busstand. An enquiry clerk told a passenger that the bushad already left ten minutes ago and the next bus willleave at 9.35 A.M. At what time did the enquiry clerkgive this information to the passenger ?

Sol. Bus leaves after every 30 minutes.The next bus will leave at 9 : 35 A.M.The last bus left at 9 : 35 � 0 : 30 = 9 : 05 A.M.

but clerk said that bus had left 10 minutes earlier.9 : 05 + 0 : 10 = 9 : 15 A.M.

EXERCISE

1. Find the day of the week on 26 January, 1950.(A) Tuesday (B) Friday(C) Wednesday (D) Thursday

2. Which two months in a year have the same

calendar ?(A) June, October (B) April, November(C) April, July (D) October, December

3. Are the years 900 and 1000 leap years ?

(A) Yes (B) No(C) Can't say (D) None of these

4. If it was Saturday on 17th November, 1962 whatwill be the day on 22nd November, 1964 ?

(A) Monday (B) Tuesday(C) Wednesday (D) Sunday

5. Sangeeta remembers that her father's birthdaywas certainly after eighth but before thirteenth of

December. Her sister Natasha remembers thattheir father's birthday was definitely after ninth butbefore fourteenth of December. On which date of

December was their father's birthday ?(A) 10th (B) 11th(C) 12th (D) Data inadequate

6. Find the day of the week on 15 August, 1947.

(A) Tuesday (B) Friday(C) Wednesday (D) Thursday

7. Karan was born on Saturday 22nd March 1982. Onwhat day of the week was he 14 years 7 months

and 8 days of age ?(A) Sunday (B) Tuesday(C) Wednesday (D) Monday

8. If on 14th day after 5th March be Wednesday, what

day of the week will fall on 10th Dec. of the sameyear ?(A) Friday (B) Wednesday

(C) Thursday (D) Tuesday

9. If the day before yesterday was Saturday, what day

will fall on the day after tomorrow ?

(A) Friday (B) Thursday

(C) Wednesday (D) Tuesday

10. If February 1, 1996 is Wednesday, what day is March

10, 1996 ?

(A) Monday (B) Sunday

(C) Saturday (D) Friday

11. If the seventh day of a month is three days earlier

than Friday, what day will it be on the nineteenth

day of the month ?

(A) Sunday (B) Monday

(C) Wednesday (D) Friday

12. Mohini went to the movies nine days ago. She goes

to the movies only on Thursday. What day of the

week is today ?

(A) Thursday (B) Saturday

(C) Sunday (D) Tuesday

13. At what time are the hands of a clock together

between 5 and 6 ?

(A) 33113

min. past 5 (B) 28113

min. past 5

(C) 27113

min. past 5 (D) 26113

min. past 5

14. At what time between 9 and 10 will the hands of a

clock be in the straight line, but not together ?

(A) 16 minutes past 9

(B) 114

16 minutes past 9

(C) 116

16 minutes past 9

(D) 119

16 minutes past 9

15. At what time between 5 & 5 : 30 will the hands of a

clock be at right angle ?

(A) 1110

10 minutes past 5

(B) 115

11 minutes past 5

(C) 1110

9 minutes past 5

(D) 119

10 minutes past 5

16. Ajay left home for the bus stop 15 minutes earlier

than usual. It takes 10 minutes to reach the stop.

He reached the stop at 8.40 a.m. What time does

he usually leave home for the bus stop ?

(A) 8.30 a.m. (B) 8.45 a.m.

(C) 8.55 a.m. (D) Data inadequate

PAGE # 88

17. The priest told the devotee, "The temple bell isrung at regular intervals of 45 minutes. The lastbell was rung five minutes ago. The next bell isdue to be rung at 7.45 a.m." At what time did thepriest give this information to the devotee ?(A) 7.40 a.m. (B) 7.05 a.m.(C) 6.55 a.m. (D) None of these

18. There are twenty people working in an office. Thefirst group of five works between 8.00 A.M. and 2.00P.M. The second group of ten works between 10.00A.M. and 4.00 P.M. And the third group of five worksbetween 12 noon and 6.00 P.M. There are threecomputers in the office which all the employeesfrequently use. During which of the following hoursthe computers are likely to be used most ?(A) 10.00 A.M. �� 12 noon

(B) 12 noon �� 2.00 P.M.

(C) 1.00 P.M. �� 3.00 P.M.

(D) 2.00 P.M. �� 4.00 P.M.

19. A tired worker slept at 7.45 p.m.. If he rose at 12noon, for how many hours did he sleep ?(A) 5 hours 15 min. (B) 16 hours 15 min.(C) 12 hours (D) 6 hours 45 min.

20. How many times are the hands of a clocksperpendicular in a day ?(A) 42 (B) 48(C) 44 (D) 46

21. If a clock shows 04: 28 then its mirror image willbe ?(A) 07: 42 (B) 07: 32(C) 08: 32 (D) 08: 42

22. A watch, which gains uniformly, is 3 minutes slowat noon on Monday and is 3 minutes 48 secondsfast at 2 p.m. on the following Monday. What time itwas correct ?(A) 2 p.m. On Tuesday(B) 2 p.m. On Wednesday(C) 3 p.m. On Thursday(D) 1 p.m. On Friday.

23. How many times are the hands of a clocks coincidein a day ?(A) 10 (B) 11(C) 12 (D) 22

24. At what time between 2 and 3 O� clock the hands of

a clock will make an angle of 160º ?

(A) 20 minutes past 2 (B) 30 minutes past 2 (C) 40 minutes past 2 (D) 50 minutes past 2

25. Ashish leaves his house at 20 minutes to seven inthe morning, reaches Kunal�s house in 25 minutes,

they finish their breakfast in another 15 minutesand leave for their office which takes another 35minutes. At what time do they leave Kunal�s house

to reach their office ?(A) 7.40 am (B) 7.20 am(C) 7.45 am (D) 8.15 am

26. The train for Lucknow leaves every two and a halfhours from New Delhi Railway Station. Anannouncement was made at the station that thetrain for Lucknow had left 40 minutes ago and thenext train will leave at 18. 00 hrs. At what time wasthe announcement made ?(A) 15.30 hrs (B) 17.10 hrs(C) 16.00 hrs (D) None of these

27. A monkey climbs 30 feet at the beginning of eachhour and rests for a while when he slips back 20feet before he again starts climbing in thebeginning of the next hour. If he begins his ascentat 8.00 a.m., at what time will he first touch a flag at120 feet from the ground ?(A) 4 p.m. (B) 5 p.m.(C) 6 p.m. (D) None of these

28. If the two incorrect watches are set at 12 : 00 noonat correct time, when will both the watches showthe correct time for the first time given that the firstwatch gains 1 min in 1 hour and second watchloses 4 min in 2 hours :(A) 6 pm, 25 days later(B) 12 : 00 noon, 30 days later(C) 12 noon, 15 days later(D) 6 am 45 days later

29. Rajeev and Sanjeev are too close friends Rajeev'swatch gains 1 minute in an hour and Sanjeev'swatch loses 2 minutes in an hour. Once they setboth the watches at 12 : 00 noon, with my correctwatch. When will the two incorrect watches ofRajeev and Sanjeev show the same time together?(A) 8 days later (B) 10 days later(C) 6 days later (D) can't be determined

30. At a railway station a 24 hour watch loses 3 minutesin 4 hours. If it is set correctly on Sunday noonwhen will the watch show the correct time ?(A) 6 pm after 40 days(B) 12 noon after 75 days(C) 12 pm after 100 days(D) 12 noon after 80 days

31. A swiss watch is being shown in a museum whichhas a very peculiar property. It gains as much inthe day as it loses during night between 8 pm to 8am. In a week how many times will the clock showthe correct time ?(A) 6 times (B) 14 times(C) 7 times (D) 8 times

32. A wrist watch which is running 12 minutes late ona Sunday noon is 16 minutes ahead of the correcttime at 12 noon on the next Sunday. When is theclock 8 minutes ahead of time ?(A) Thursday 10 am (B) Friday noon(C) Friday 8 pm (D) Tuesday noon

PAGE # 89

33. A clock loses 2 minutes in a hour and another clockgains 2 minutes in every 2 hours. Both these clocksare set correctly at a certain time on Sunday andboth the clocks stop simultaneously on the nextday with the time shown being 9 am and 10 : 06AM. What is the correct time at which they stopped?(A) 9 : 54 am (B) 9 : 44 pm(C) 9 : 46 am (D) 9 : 44 am

34. David sets his watch at 6 : 10 am on Sunday, whichgains 12 minutes in a day. On Wednesday if thiswatch is showing 2 : 50 pm. What is the correcttime ?(A) 1 : 50 pm (B) 2 : 10 pm(C) 2 : 30 pm (D) 3 : 30 pm

35. Ramu purchased a second hand Swiss watchwhich is very costly. In this watch the minute-hand

and hour hand coincide after every 113

65 minutes.

How much time does the watch lose or gain perday ?(A) 4 min (B) 5 min(C) 4 min, 20 sec (D) none of these

36 My watch was 8 minutes behind at 8 pm on Sundaybut within a week at 8 pm on Wednesday it was 7minutes ahead of time. During this period at whichtime this watch has shown the correct time :(A) Tuesday 10 : 24 am(B) Wednesday 9 : 16 pm(C) It cannot show the correct time during this period(D) None of the above

37. Out of the following four choices which does notshow the coinciding of the hour hand and minute-hand :(A) 3 : 16 : 2 (B) 6 : 32 : 43(C) 9 : 59 : 05 (D) 5 : 27 : 16

38. Kumbhakarna starts sleeping between 1 pm and

2 pm and he wakes up when his watch shows

such a time that the two hands (i.e., hour-hand

and minute-hand) interchange the respective

places. He wakes up between 2 pm and 3 PM on

the same night. How long does he sleep ?

(A) 135

55 min (B) 1310

110 min

(C) 136

54 min (D) None of these

39. A clock loses 3% time during the first week and

then gains 2% time during the next one week. If the

clock was set right at 12 noon on a Sunday, what

will be the time that the clock will show exactly 14

days from the time it was set right ?

(A) 1 : 36 : 48 (B) 1 : 40 : 48

(C) 1 : 41 : 24 (D) 10 : 19 : 12

Direction : (40 to 41) A 12 dial clock has its minute hand

defective. Whenever it touches dial 12, it

immediately falls down to 6 instead of running

smoothly (the hour hand remains unaffected during

that fall). It was set right at 12 �O� clock in the noon.

40. What was the actual time when the minute hand of

the clock touched dial 9 for the 5th time?

(A) 2 : 15 (B) 3 : 00

(C) 5 : 15 (D) 6 : 45

41. If the actual time is 10 : 10, what is the position of

the hour hand in that defective clock ?

(A) Between 2 and 3 (B) Between 4 and 5

(C) Between 10 and 11 (D) Between 3 and 4

PAGE # 90

CUBE AND DICE-TEST

CUBES

A cube is three dimensional figure, having 8corners, 6 surfaces and 12 edges. If a cube ispainted on all of its surfaces with any colour andfurther divided into various smaller cubes, we getfollowing results. Smaller cubes with threesurfaces painted will be present on the corners ofthe big cube.

3 33

3

33

3

3 3

3

3

3

1 1

1 1

11

11

11 1

1

2 2

2 2

2

22

2

22

2

2

2

2

22

22

222

22 2

Smaller cubes with two surface painted will bepresent on the edges of the big cube. Smallercubes with one surface painted will be present onthe surfaces of the big cube. Smaller cubes withno surface painted will be present inside the bigcube.

If a cube is painted on all of its surfaces with acolour and then divided into smaller cubes of equalsize then after separation, number of smaller cubesso obtained will be calculated as under :Number of smaller cubes with three surfacespainted = 8Number of smaller cubes with two surfacespainted = (n � 2) × 12

Number of smaller cubes with one surfacespainted = (n � 2)2 × 6

Number of smaller cubes with no surfaces painted= (n � 2)3

Where n = No of divisions on the surfaces of thebigger cube

= cubesmalleroneofedgeoflengthcubebigofedgeoflength

TYPE I

If a cube is painted on all of its surfaces with singlecolour and then divided into various smaller cubesof equal size.

Directions : ( 1 to 4) A cube of side 4 cm is painted black onall of its surfaces and then divided into varioussmaller cubes of side 1 cm each. The smallercubes so obtained are separated.

Total cubes of obtained = 64111

444

Here n = 41

4

cubesmallofside

cubebigofside

Ex 1. How many smaller cubes have three surfaces

painted ?

(A) 4 (B) 8

(C) 16 (D) 24

Sol. (B) Number of smaller cubes with three surfaces

painted = 8

Ex 2. How many smaller cubes have two surfaces

painted ?

(A) 4 (B) 8

(C) 16 (D) 24

Sol. (D) Number of smaller cubes with two surfaces

painted = (n � 2) × 12 = (4 � 2) × 12 = 24

Ex 3. How many smaller cubes have only one surface

painted ?

(A) 8 (B) 16

(C) 24 (D) 32

Sol. (C) Number of smaller cubes with one surface

painted = (n � 2)2 × 6 = (4 � 2)2 × 6 = 4 × 6 = 24

Ex 4. How many smaller cubes will have no side painted ?

(A) 18 (B) 16

(C) 22 (D) 8

Sol. (D) Number of smaller cubes with no surface

painted = (n � 2)3 = (4 � 2)3 = (2)3 = 8

TYPE II

If a cube is painted on all of its surfaces with

different colours and then divided into various

smaller cubes of equal size.

Directions : ( 5 to 7 ) A cube of side 4 cm is painted black on

the pair of one opposite surfaces, blue on the pair

of another opposite surfaces and red on remaining

pair of opposite surfaces. The cube is now divided

into smaller cubes of equal side of 1 cm each.

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PAGE # 91

Ex 5. How many smaller cubes have three surfacespainted ?(A) 4 (B) 8(C) 16 (D) 24

Sol. (B) Number of smaller cubes with three surfacespainted = 8(These smaller cubes will have all three surfacespainted with different colour blue, black and red.)

Ex 6. How many smaller cubes have two surfacespainted ?(A) 4 (B) 8(C) 16 (D) 24

Sol. (D) Number of smaller cubes with two surfacespainted = 24. And out of this -(a) Number of cubes with two surfaces paintedwith black and blue colour = 8.(b) Number of cubes with two surfaces paintedwith blue and red colour = 8.(c) Number of cubes with two surfaces paintedwith black and red color = 8.

Ex 7. How many smaller cubes have only one surfacepainted ?(A) 8 (B) 16(C) 24 (D) 32

Sol. (C) Number of smaller cubes with one surfacepainted = 24. And out of this -(a) Number of cubes with one surface paintedwith black colour =8.(b) Number of cubes with one surface paintedwith blue colour = 8.(c) Number of cubes with one surface paintedwith red colour = 8.

TYPE III

If a cube is painted on its surfaces in such a waythat one pair of opposite surfaces is left unpainted.

Directions : ( 8 to 11 ) A cube of side 4 cm is painted red onthe pair of one opposite surfaces, green on thepair of another opposite surfaces and one pair ofopposite surfaces is left unpainted. Now the cubeis divided into 64 smaller cubes of side 1 cm each.

Ex 8. How many smaller cubes have three surfacespainted ?(A) 0 (B) 8(C) 16 (D) 20

Sol. (A) Number of smaller cubes with three surfacespainted = 0 (Because each smaller cube at thecorner is attached to a surface which is unpainted.)

Ex 9. How many smaller cubes have two surfacespainted ?(A) 4 (B) 8(C) 16 (D) 24

Sol. (C) Number of smaller cubes with two surfacespainted = Number of cubes present at the corners+ Numbers of cubes present at 4 edges= 8 + (n � 2) × 4 = 8 + 8 = 16

Ex 10. How many smaller cubes have only one surfacepainted ?(A) 8 (B) 16(C) 24 (D) 32

Sol. (D) Number of smaller cubes with one surfacepainted = Number of cubes present at the 8 edges+ number of cubes present at the four surfaces=(n � 2) × 8 + (n � 2)2 × 4

= 2 × 8 + 4 × 4 = 16 + 16 = 32

Ex 11. How many smaller cubes will have no side painted?(A) 18 (B) 16(C) 22 (D) 8

Sol. (B) Number of smaller cubes with no side painted= Number of cubes on the two unpainted surfaces +number of cubes present inside the cube.= (n � 2)2 × 2 + (n � 2)3 = 4 × 2 + (2)3 = 8 + 8 = 16.

TYPE IV

If a cube is painted on its surfaces in such a waythat one pair of adjacent surfaces is left unpainted.

Directions : (12 to 15 )A cube of side 4 cm is painted red onthe pair of one adjacent surfaces, green on thepair of other adjacent surfaces and two adjacentsurfaces are left unpainted. Now the cube is dividedinto 64 smaller cubes of side 1 cm each.

Ex 12. How many smaller cubes have three surfacespainted ?(A) 2 (B) 4(C) 8 (D) 6

PAGE # 92

Sol. (A) Number of smaller cubes with three surfacespainted = Number of smaller cubes at two corners= 2

Ex 13. How many smaller cubes have two surfacespainted ?(A) 4 (B) 8(C) 16 (D) 14

Sol. (D) Number of smaller cubes with two surfacespainted = Number of smaller cubes at four corners+ Number of smaller cubes at 5 edges.= 4 + (n � 2) × 5 = 4 + 2 × 5

= 4 + 10 = 14

Ex 14. How many smaller cubes have only one surfacepainted ?(A) 8 (B) 16(C) 24 (D) 30

Sol. (D) Number of smaller cubes with one surfacepainted = Number of smaller cubes at foursurfaces + Number of smaller cubes at 6 edges +Number of smaller cubes at two corners.= (n � 2)2 × 4 + (n � 2) × 6 + 2

= 4 × 4 + 2 × 6 + 2 = 16 + 12 = 28 + 2 = 30

Ex 15. How many smaller cubes will have no side painted?(A) 18 (B) 16(C) 22 (D) 8

Sol. (A) Number of smaller cubes with no surfacespainted = Number of smaller cubes from insidethe big cube + Number of cubes at two surfaces +Number of cubes at one edge.= (n � 2)3 + (n � 2)2 × 2 + (n � 2)

= (2)3 + (2)2 × + 2

= 8 + 8 + 2 = 18

DICES

Type-I

General Dice : In a general dice the sum of numberson the any two adjacent faces is �7�.Standard Dice : In a standard dice the sum ofnumbers on the opposite faces is '7'.

Ex 16. Which number is opposite 4 in a standard dicegiven below ?

41

5

(A) 1 (B) 3(C) 5 (D) Can�t be determined

Sol. Clearly , from the standard dice the sum ofnumbers on the opposite faces is '7', so numberopposite to 4 is 3.

Type-II

Ex 17. The figures given below show the two differentpositions of a dice. Which number will appearopposite to number 2 ?.

(A) 3 (B) 4(C) 5 (D) 6

Sol. (C) The above question,where only two positions ofa dice are given, can easilybe solved with thefollowing method.

Step I. The dice, when unfolded, will appear as shown inthe figure given on the right side.

Step II. Write the common number to both the dice in themiddle block. Since common number is 4, hencenumber 4 will appear in the central block.

Step III. Consider the figure (i) and write the first number inthe anti-clockwise direction of number 4,(common number) in block I and second numberin block II. Therefore, numbers 3 and 2 being thefirst and second number to 4 in anticlockwisedirections respectively, will appear in block I & IIrespectively.

Step IV. Consider figure (ii) and wire first and secondnumber in the anticlock-wise direction to number4, (common number) in block (III) & (IV). Hencenumbers 6 and 5 will appear in the blocks III and IVrespectively.

Step V. Write remaining number in the remaining block.Therefore, number 1 will come in the remainingblock. Now, from the unfolded figures we find thatnumber opposite to 6 is 3, number opposite to 2 is5 and number opposite to 4 is 1. Therefore, option(C) is our answer.( Short Trick : From the given dice, we will take thecommon number as the base and then in itsrespect move clockwise direction and write asfollows : 4 � 2 � 3

4 � 5 � 6.Here,we find that number opposite to 6 is 3, numberopposite to 2 is 5 and number opposite to 4 isremaining number 1.Therefore, option (C) is our answer. )

Ex 18. On the basis of two figures of dice, you have to tellwhat number will be on the opposite face of number5 ?

(A) 1 (B) 2(C) 4 (D) 6

PAGE # 93

Sol. (D) The above question where only two positionsof a dice are given, can easily be solved with thefollowing method :If in the given dice, there are two numbers common,then uncommon numbers will always be oppositeof each other.Therefore, option (D) is our answer.

Type-III

Ex 19. From the following figures of dice, find whichnumber will come in place of �?�

(A) 4 (B) 5(C) 2 (D) 3

Sol. (D) If the above dice is unfolded, it will look like asthe figure (i) given below.

Figure (i)

Now the number in place of �?� can be obtained by

making a slight change in the figure as given here.Now comparing figure (ii) with third dice as above,we get that number in place of ? is 3.

Figure (ii)

Type-IV

Ex 20. A dice has been thrown four times and producesfollowing results.

Which number will appear opposite to the number3 ?(A) 4 (B) 5(C) 6 (D) 1

Sol. (A) From the figures (i), (ii) and (iv) we find thatnumbers 6, 1, 5 and 2 appear on the adjacentsurfaces to the number 3. Therefore, number 4will be opposite to number 3.

Type-V

Ex 21. Which of the following dices is identical to theunfolded figure as shown here ?

(X)

(A) (B)

(C) (D)

Sol. (A) From the unfolded figure of dice, we find thatnumber opposite to 2 is 4, for 5 it is 3 and for 1 it is6. From this result we can definitely say that figure(B), (C) and (D) can not be the answer figure asnumbers lying on the opposite pair of surfaces arepresent on the adjacent surfaces.

EXERCISE

Directions : (1 to 5) A cube is coloured orange on one face,pink on the opposite face, brown on one face andsilver on a face adjacent to the brown face. Theother two faces are left uncoloured. It is then cutinto 125 smaller cubes of equal size. Now answerthe following questions based on the abovestatements.

1. How many cubes have at least one face colouredpink ?(A) 1 (B) 9(C) 16 (D) 25

2. How many cubes have all the faces uncoloured ?(A) 24 (B) 36(C) 48 (D) 64

3. How many cubes have at least two faces coloured ?(A) 19 (B) 20(C) 21 (D) 23

4. How many cubes are coloured orange on one faceand have the remaining faces uncoloured ?(A) 8 (B) 12(C) 14 (D) 16

5. How many cubes one coloured silver on one face,orange or pink on another face and have fouruncoloured faces ?(A) 8 (B) 10(C) 12 (D) 16

PAGE # 94

Directions : (6 to 11) A cube is painted red on two adjacentsurfaces and black on the surfaces opposite tored surfaces and green on the remaining faces.Now the cube is cut into sixty four smaller cubes ofequal size.

6. How many smaller cubes have only one surfacepainted ?(A) 8 (B) 16(C) 24 (D) 32

7. How many smaller cubes will have no surfacepainted ?(A) 0 (B) 4(C) 8 (D) 16

8. How many smaller cubes have less than threesurfaces painted ?(A) 8 (B) 24(C) 28 (D) 48

9. How many smaller cubes have three surfacespainted ?(A) 4 (B) 8(C) 16 (D) 24

10. How many smaller cubes with two surfacespainted have one face green and one of theadjacent faces black or red ?(A) 8 (B) 16(C) 24 (D) 28

11. How many smaller cubes have at least one surfacepainted with green colour ?(A) 8 (B) 24(C) 32 (D) 56

Directions : (12 to 16) A cube of 4 cm has been painted onits surfaces in such a way that two oppositesurfaces have been painted blue and two adjacentsurfaces have been painted red. Two remainingsurfaces have been left unpainted. Now the cubeis cut into smaller cubes of side 1 cm each.

12. How many cubes will have no side painted ?(A) 18 (B) 16(C) 22 (D) 8

13. How many cubes will have at least red colour onits surfaces ?(A) 20 (B) 22(C) 28 (D) 32

14. How many cubes will have at least blue colour onits surfaces ?(A) 20 (B) 8(C) 24 (D) 32

15. How many cubes will have only two surfacespainted with red and blue colour respectively ?(A) 8 (B) 12(C) 24 (D) 30

16. How many cubes will have three surfaces coloured ?(A) 3 (B) 4(C) 2 (D) 16

Directions : (17 to 21) The outer border of width 1 cm of a

cube with side 5 cm is painted yellow on each side

and the remaining space enclosed by this 1 cm

path is painted pink. This cube is now cut into 125

smaller cubes of each side 1 cm. The smaller

cubes so obtained are now seperated.

17. How many smaller cubes have all the surfaces

uncoloured ?

(A) 0 (B) 9

(C) 18 (D) 27

18. How many smaller cubes have three surfaces

coloured ?

(A) 2 (B) 4

(C) 8 (D) 10

19. How many cubes have at least two surfaces

coloured yellow ?

(A) 24 (B) 44

(C) 48 (D) 96

20. How many cubes have one face coloured pink and

an adjacent face yellow ?

(A) 0 (B) 1

(C) 2 (D) 4

21. How many cubes have at least one face coloured ?

(A) 27 (B) 98

(C) 48 (D) 121

Directions : (22 to 31) A solid cube has been painted yellow,blue and black on pairs of opposite faces. Thecube is then cut into 36 smaller cubes such that32 cubes are of the same size while 4 others areof bigger sizes. Also no faces of any of the biggercubes is painted blue.

22. How many cubes have at least one face paintedblue ?(A) 0 (B) 8(C) 16 (D) 32

23. How many cubes have only one faces painted ?(A) 24 (B) 20(C) 8 (D) 12

24. How many cubes have only two faces painted ?(A) 24 (B) 20(C) 16 (D) 8

25. How many cubes have atleast two faces painted ?(A) 36 (B) 34(C) 28 (D) 24

26. How many cubes have only three faces painted ?(A) 8 (B) 4(C) 2 (D) 0

PAGE # 95

27. How many cubes do not have any of their facespainted yellow ?(A) 0 (B) 4(C) 8 (D) 16

28. How many cubes have at least one of their facespainted black ?(A) 0 (B) 8(C) 16 (D) 20

29. How many cubes have at least one of their facespainted yellow or blue ?(A) 36 (B) 32(C) 16 (D) 0

30. How many cubes have no face painted ?(A) 8 (B) 4(C) 1 (D) 0

31. How many cubes have two faces painted yellowand black respectively ?(A) 0 (B) 8(C) 12 (D) 16

Directions : (32 to 35) Some equalcubes are arranged in theform of a solid block asshown in the adjacentfigure. All the visiblesufaces of the block (exceptthe bottom) are thenpainted.

32. How many cubes do not have any of the facespainted ?(A) 27 (B) 8(C) 10 (D) 12

33. How many cubes have one face painted ?(A) 9 (B) 24(C) 22 (D) 20

34. How many cubes have only two faces painted ?(A) 0 (B) 16(C) 20 (D) 24

35. How many cubes have only three faces painted ?(A) 4 (B) 12(C) 6 (D) 20

Directions : (36 to 40) A cuboid of dimensions(6 cm 4 cm 1 cm) is painted black on both thesurfaces of dimensions (4 cm 1 cm), green on thesurfaces of dimensions (6 cm 4 cm). and red onthe surfaces of dimensions (6 cm 1 cm). Now theblock is divided into various smaller cubes of side1 cm. each. The smaller cubes so obtained areseparated.

36. How many cubes will have all three colours black,green and red each at least on one side?(A) 16 (B) 12(C) 10 (D) 8

37. How many cubes will be formed?(A) 6 (B) 12(C) 16 (D) 24

38. If cubes having only black as well as green colourare removed then how many cubes will be left?(A) 4 (B) 8(C) 16 (D) 30

39. How many cubes will have 4 coloured sides and2 sides without colour?(A) 8 (B) 4(C) 16 (D) 10

40. How many cubes will have two sides with greencolour and remaining sides without any colour?(A) 12 (B) 10(C) 8 (D) 4

41. Which alphabet is opposite D ?

(A) E (B) C(C) F (D) A

42. What should be the number opposite 4 ?

(i) (ii) (iii)

(A) 5 (B) 1(C) 3 (D) 2

43.

(i) (ii)

(iii) (iv)Which letter will be opposite to letter D ?(A) A (B) B(C) E (D) F

Directions : (44 to 45) The figure (X) given below is theunfolded position of a cubical dice. In each of thefollowing questions this unfolded figure is followedby four different figures of dice. You have to selectthe figure which is identical to the figure (X).

PAGE # 96

44. (X)

(A) (B)

(B) (D)

45. (X)

(A) (B)

(C) (D)

Directions : (46 to 48) In each of the following questions,select the correct option for the question asked.

(i) (ii)

46. Which number will come opposite to number 2?(A) 5 (B) 1(C) 6 (D) 3

47. Which number will come opposite to number 6?(A) 1 (B) 5(C) 4 (D) 3

48. Which number will come opposite to number 4?(A) 3 (B) 5(C) 1 (D) 2

49. On the basis of two figures of dice, you have to tell whatnumber will be on the opposite face of number 5?

(i) (ii)

(A) 1 (B) 2(C) 4 (D) 6

50. Which symbol will appear on the opposite surfaceto the symbol x?

(A) (B) =

(C) (D) O

51. Three positions of the same dice are given below.Observe the figures carefully and tell which numberwill come in place of �?�

(i)

16 3

(ii)

35 4

(iii)

42 ?

(A) 1 (B) 6(C) 3 (D) 5

52. On the basis of the following figures you have totell which number will come in place of �?�

(i)

36 1

(ii)

42 6

(iii)

?1 5

(A) 2 (B) 3(C) 6 (D) 4

Directions : (53 to 55) Choose from the alternatives, theboxes that will be formed when figure (X) is folded:

53. (X)

(A) (B)

(C) (D)

54. (X)

+

(A) (B) +

(C) + (D)

PAGE # 97

55. (X)

(A) (B)

(C) (D)

Direction : (56) The six faces of a cube have been markedwith numbers 1, 2, 3, 4, 5 and 6 respectively. Thiscube is rolled down three times. The threepositions are given. Choose the figure that will beformed when the cube is unfolded.

56.

(A) (B)

(C) (D)

57. Which number is opposite 3 in a standard dicegiven below ?

(A) 1 (B) 4(C) 5 (D) Can�t be determined

58. Which number is opposite 4 ?

(A) 5 (B) 3(C) 2 (D) 1

Directions : (59) In the following question four positions ofthe same dice have been shown. You have to seethese figures and select the number opposite tothe number as asked in each question.

59.

(i) (ii)

(iii) (iv)Which number is opposite to number 5?(A) 6 (B) 5(C) 1 (D) 3

Directions : (60 to 64) Choose the cube from the optionsthat will unfold to give the figure on the left

60.X

M

M

M MX

X

(A) (B) (C) (D) (E)

61.

4 1 8

3

7

9

(A) (B) (C) (D) (E)

981 14

7

7 87 48

7

62.

D

8

(A) (B) (C) (D) (E)

8 8 D

63.

B

(A) (B) (C) (D) (E)

B

PAGE # 98

64.

J

(A) (B) (C) (D) (E)

J J

Directions : (65 to 68) In each of the following questions, adiagram has been given which can be folded intoa cube. The entries given in the squares indicatethe entries on the face of the cube. In each questiona number or a letter has been given . Of the fouralternatives given below it, you have to find the onethat would appear on the face opposite to it in thecube.

65. Which letter is opposite Q ?

Q

O P LNM

(A) L (B) M(C) N (D) P

66. Which number/letter is opposite 2 ?

3 I CAB2

(A) A (B) C(C) 1 (D) 3

67. Which number/letter is opposite O?

N M 2

L

I O

(A) L (B) M(C) N (D) 2

68. Which letter is opposite R?

Q R

S P

U T

(A) P (B) S(C) T (D) U

99PAGE # 99

ANSWER KEY

ELECTRICITY(PHYSICS)

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans. A A C C C B A D A B D C A B BQue. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Ans. B A A A D D B C B C B C B B AQue. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

Ans. A D B C D B C D A B D C ABCD CD BDQue. 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Ans. ABD C D C B B C B C C B B D C BQue. 61 62 63 64 65 66 67 68 69 70 71 72 73Ans. B C B B D A A B A A C D D

MOLE CONCEPT(CHEMISTRY)

Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Ans. B C D B C B A D A B B D B C B

Ques. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30Ans. A A A A A B A A D A A A D A D

Ques. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45Ans. D C C B C B C D A B C A C D A

Ques. 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60Ans. B B C C C C B C B A A D C A D

Ques. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75Ans. C D A B A B C A D B C B C B D

Ques. 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90Ans. A A C A A D B D B C B C C D D

NUMBER SYSTEM(MATHEMATICS)

Q. 1 2 3 4 5 6 7 8 9 10Ans. B B A D C B A A A D

Q. 11 12 13 14 15 16 17 18 19 20Ans. A B A A B C C B C A

Q. 21 22 23 24 25 26 27 28 29 30Ans. C B B A D B D B C C

Q. 31 32 33 34 35 36 37 38 39 40Ans. B A A B D D C C D C

Q. 41 42 43 44 45 46 47 48 49 50Ans. C B A A D A D D A A

Q. 51 52 53 54 55 56 57 58 59 60Ans. C C B D C C C C C A

Q. 61 62 63 64 65 66 67 68 69 70Ans. B B D D D A A B A D

Q. 71 72 73 74 75 76 77 78 79 80Ans. C A C C B B B A D C

Q. 81 82 83 84 85 86 87 88 89 90Ans. B B D C D D A B C D

Q. 91 92 93 94 95 96Ans. A&D C&D A D B C

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100PAGE # 100

LOGARITHM(MATHEMATICS)

Q. 1 2 3 4 5 6 7 8 9 10

Ans. B A B D B C B A C D

Q. 11 12 13 14 15 16 17 18 19 20

Ans. A A B B D D B C B D

Q. 21 22 23 24 25 26 27 28 29 30

Ans. A A B B A A D A A B

Q. 31 32 33 34 35 36 37 38 39 40

Ans. B B A B B A D B B C

NUTRITION(BIOLOGY)

Ques. 1 2 3 4 5 6 7 8 9 10

Ans. C C B D D A C B A A

Ques. 11 12 13 14 15 16 17 18 19 20

Ans. B C A C A B D A D C

Ques. 21 22 23Ans. B C A

SERIES COMPLETION(MENTAL ABILITY)

EXERCISE-1 (Number Series)

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans. C D D A C D B C C C D C C B DQue. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Ans. B C C C B A B D D A C B B C AQue. 31 32 33 34 35

Ans. C C C D D

EXERCISE- 2 (Alphabet Series)

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans. D A D C C A D C D B D C C C DQue. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Ans. C A B C C C A B A C D B C D B

EXERCISE- 3 (Letter Repeating Series)

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans. D D A A C B A C D D C B D C BQue. 16 17 18 19 20 21 22 23 24 25 26

Ans. C A A A C D D D A B D

101PAGE # 101

EXERCISE- 4 (Missing Term In Figure)

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans. B D B D C C C D A D D B C A BQue. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Ans. B C A B B A C A D D A C D B A

Que. 31 32 33 34 35 36 37 38 39 40

Ans. B B A C A C B C D B

PUZZLE-TEST(MENTAL ABILITY)

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Ans. A B D C A D C C C C C D C C B

Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30Ans. C C D D C D D B B C A A D C DQue. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45Ans. D C B A B A D A D C D A C B DQue. 46 47 48 49 50 51 52 53 54 55 56 57 58Ans. C A D B D D D A C A A D B

CALENDAR AND CLOCK-TEST(MENTAL ABILITY)

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans. D C B D D B C B C C A B C B A

Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30Ans. B B B B C B C D C B D C B B DQue. 31 32 33 34 35 36 37 38 39 40 41Ans. D B D B A A C A D A C

CUBE AND DICE TEST(MENTAL ABILITY)

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans. D C C D A C C D B B C A C D BQue. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Ans. C D C B A B D D A D C A D C BQue. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45Ans. C D C D C A D C B C B B A D BQue. 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60Ans. D A B C D A B D B D C B A C CQue. 61 62 63 64 65 66 67 68Ans. A D E D C A B B