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Hindi Course B

Hindi Course A

1. In the given figure if DE || BC, AE = 8 cm, EC = 2 cm and BC = 6 cm, then find DE. [1]Sol. In DADE and DABC,

ÐDAE = ÐBAC [Common]ÐADE = ÐABC [Corresponding angles]By AA axiom of similarityDADE ~ DABC

\AEAC

DEBC

= [CPCT]

Þ8

8 2+DE6

= Þ 10 × DE = 48

Þ DE = 4.8 cm Ans.

2. Evaluate :2

21 cot 45º10 · .1 sin 90º-+

[1]

Sol.2

21 cot 4510·1 sin 90- °+ °

2

21 (1)10·1 (1)-=+

010. 02

= = Ans.

3. If cosec q5= ,4 find the value of cot q. [1]

Sol. We know that,cot2 q = cosec2 q – 1

25 14æ ö= -ç ÷è ø

25 25 16 9116 16 16

-= - = =

Þ cot2 q9

16=

i.e. cot q34

= Ans.

Section – A

A

B

D E

C6 cm

8 cm

2 cm

MATHEMATICS 2014 TERM ISET I

Time allowed : 3 hours Maximum marks : 90

Arundeep’s Solved Papers Mathematics 2014 (Term I)1

https://www.arundeepselfhelp.info/index.php?route=product/product&path=180&product_id=395


4. Following table shows sale of shoes in a store during one month :

Size of shoe 3 4 5 6 7 8Number of pairs sold 4 18 25 12 5 1

Find the model size of the shoes sold. [1]Sol. Maximum number of pairs sold = 25 (size 5)\ Model size of shoes = 5 Ans.

Section – B

5. Find the prime factorisation of the denominator of rational number expressed as 6.12 insimplest form. [2]

Sol. Let x 6.1212...........= ...(i)Þ 100x 612.1212...........= ...(ii)

Subtracting eq. (i) from (ii), we get99x = 606

Þ x606 20299 33

= =

\ Denominator = 33and Prime factorisation of 33 be = 3 × 11 Ans.

6. Find a quadratic polynomial, the sum and product of whose zeroes are 3 and13

respectively. [2]

Sol. Given, sum of zeroes, (S) 3=

and Product of zeroes, (P)13

=

Thus, quadratic polynomial is given as f(x) = x2 – Sx + P

\ f(x) 2 133

x x= - +23 3 1

3x x- += 21 ( 3 3 1)

3x x= - + Ans.

7. Complete the following factor tree and find the composite number x. [2]

Sol. y = 5 × 13 = 65x = 3 × 195 = 585 Ans.

3 3311 11

1

3 5853 1955 65

13 131


Arundeep’s Solved Papers Mathematics 2014 (Term I)38. In a rectangle ABCD, E is middle point of AD. If AD = 40 m and AB = 48 m, then find EB.

[2]Sol. Given, E is the mid-point of AD

\ AE40 m 20 m

2= =

Also ÐA = 90º [Angle of a rectangle]\ In rt. DBAE, we have

EB2 = AB2 + AE2 [Pythagoras’ theorem]= (48)2 + (20)2 = 2304 + 400 = 2704

EB 2704= = 52 52´ = 52 m Ans.9. If x = p sec q + q tan q and y = p tan q + q sec q, then prove that x2 – y2 = p2 – q2. [2]

Sol. L.H.S. = x2 – y2

= (p sec q + q tan q)2 – (p tan q + q sec q)2

= (p2 sec2 q + q2 tan2 q + 2pq sec q tan q) – (p2 tan2 q + q2 sec2 q + 2pq sec q tan )= p2 sec2 q + q2 tan2 q + 2pq sec q tan q – p2 tan2 q – q2 sec2 q – 2pq sec q tan q= p2 (sec2 q – tan2 q) – q2 (sec2 q – tan2 q)

= p2 – q2 [Q sec2 q – tan2 q = 1]

= R.H.S. Hence Proved.10. Given below is the distribution of weekly pocket money received by students of a class.

Calculate the pocket money that is received by most of the students. [2]

Pocket Money (in ` ) 0–20 20–40 40–60 60–80 80–100 100–120 120–140No. of Students 2 2 3 12 18 5 2

Sol. Pocket Money Number of(in ` ) Students0–20 220–40 240–60 360–80 12 f080–100 18 f1 (Maximum)100–120 5 f2120–140 2

Here maximum frequency is 18 which lies in class 80 – 100.\ Modal class = 80 – 100

Thus, mode 1 0

1 0 22f f

l hf f f-

= + ´- -

18 1280 2036 12 5-= + ´- -



680 2019

= + ´ 1208019

= + = 80 + 6.32

= 86.32 (approx.)\ Required pocket money = ` 86.32 (approx.)

Ans.

Section – C

11. Prove that 3 + 2 3 is an irrationalnumber. [3]

Sol. Let us assume to the contrary, that 3 2 3+is rational.So that we can find co-prime positiveintegers a and b (b ¹ 0), such that

3 2 3 ab

+ =

Rearranging the equation, we get

2 333a a b

b b-= - =

Þ 33 3

2 2 2a b a b

b b b-= = -

Þ 33

2 2ab

= -

Since a and b are integer, and b ¹ 0,

So3

2 2ab- is rational number and hence

3 is rational.

But this contradicts the fact that 3 is anirrational number.

So we conclude that 3 2 3+ is anirrational number.

Hence Proved.12. Solve by elimination :

3x = y + 55x – y = 11 [3]

Sol. Given equations are,3x = y + 5 ...(i)5x – y = 11 ...(ii)

On subtracting eq. (i) and (ii), we get

3x – y = 55x – y = 11– + –

– 2x = –6Þ x = 3

Putting the value of x in eq. (i) ; we have3 (3) – y = 5

Þ 9 – 5 = y Þ y = 4\ x = 3, y = 4 Ans.

13. A man earns ` 600 per month more thanhis wife. One-tenth of the man’s salaryand one-sixth of the wife’s salary amountto ` 1,500, which is saved every month.Find their incomes. [3]

Sol. Let wife’s monthly income = ` xThen man’s monthly income = ` (x + 600)According to the question, we have

1 1( 600) ( )10 6

x x+ + = 1,500

Þ3( 600) 5

30x x+ +

= 1,500

Þ 3x + 1,800 + 5x = 45,000Þ 8x = 45,000 –1,800

Þ x43, 200 5, 400

8= =

Thus, wife’s income = ` x = ` 5,400and Man’s income = ` (x + 600) = ` (5400 + 600)

= ` 6,000 Ans.14. Check whether polynomial x – 1 is a

factor of the polynomial x3 – 8x2 + 19x – 12.Verify by division algorithm. [3]

Sol. Let P (x) = x3 – 8x2 + 19x – 12Put x = 1, we haveP (1) = (1)3 – 8 (1)2 + 19 (1) – 12= 1 – 8 + 19 – 12= 20 – 20= 0

\ (x – 1) is a factor of P (x).


Arundeep’s Solved Papers Mathematics 2014 (Term I)5Verification :

Since remainder = 0.Thus, (x – 1) is a factor of P (x).

Hence Verified.15. If the perimeters of two similar triangles

ABC and DEF are 50 cm and 70 cmrespectively and one side of DABC = 20 cm,then find the corresponding side ofDDEF. [3]

Sol.

D

Given, DABC ~ DDEF,Perimeter of DABC = 50 cmPerimeter DDEF = 70 cmOne side of DABC = 20 cmLet AB = 20 cmDABC ~ DDEF [Given]

\Peri ( ABC)Peri ( DEF)

DD

ABDE

=

Þ5070

20DE

=

\ 50 × DE = 1400Þ DE = 28 m

The corresponding side of DDEF = 28 cm. Ans.

16. In the figure if DE || OB and EF || BC,then prove that DF || OC. [3]

Sol. Given, In DABC, DE || OB and EF || BCTo Prove : DF || OCProof : In DAOB, we haveDE || OB

\AEEB

AD=DO ...(i)

[Thales’ Theorem]Similarly, in DABC, we haveEF || BC

\AEEB

AF=FC ...(ii)

[Thales’ Theorem]From (i) and (ii), we have

ADDO

AF=FC

\ DF || OC[By Converse of Thales’ Theorem]

Hence Proved.17. Prove the identity :

(sec A – cos A) · (cot A + tan A) = tan A · sec A.[3]

Sol. L.H.S. = (sec A – cos A) (cot A + tan A)

1 cos A sin Acos Acos A sin A cos A

=æ ö æ ö- +ç ÷ ç ÷è ø è ø

2 2 21 cos A cos A sin Acos A sin A cos A

æ ö æ ö- += ç ÷ ç ÷è ø è ø

2sin A 1cosA sin A cos A

= ´

[Qcos2 A + sin2 A = 1]



sin A 1cos A cos A= ´

= tan A · sec A = R.H.S. Hence Proved.

18. Given 2 cos 3q = 3, find the value of q. [3]

Sol. Given, 2 cos 3q 3=

Þ cos 3q3

2= Þ cos 3q = cos 30º Þ 3q = 30º Þ q = 10º Ans.

19. For helping poor girls of their class, students saved pocket money as shown in the followingtable :

Money saved (in ) 5–7 7–9 9–11 11–13 13–15Number of students 6 3 9 5 7

Find mean and median for this data. [3]Sol. The table of values is given as under :

Money No. of x idi =

fidi c.f.

saved Studentsx – 10

2i

(in ` ) (fi)5–7 6 6 –2 –12 67–9 3 8 –1 –3 99–11 9 a=10 0 0 1811–13 5 12 1 5 2313–15 7 14 2 14 30

Sfi = 30 Sfidi = 4

where h = 2

(i) Then by step deviation method, we have,

Mean i i

i

f da h

få

= + ´å

Þ x410 2

30= + ´ = 10 + 0.27 = ` 10.27Ans.

(ii) Now, N = Sfi = 30

\N2

30= 15,2

= Clearly the cummlative frequency just greater than 15 be 18 which lies in class

9 – 11.


Arundeep’s Solved Papers Mathematics 2014 (Term I)7\ Median class is 9–11.

Thus, median

N F2l h

f

-= +

15 99 29-= + ´

69 29

= + ´

= 9 + 1.33 = ` 10.33 Ans.20. Monthly pocket money of students of a class is given in the following frequency distribution:

[3]

Pocket money (in ) 100–125 125–150 150–175 175–200 200–225Number of students 14 8 12 5 11

Find mean pocket money using step deviation method.Sol. The table of values is given as under :

Pocket No. ofx i di = fidiMoney Students x – 162.5

25i

(in ` ) (fi)

100–125 14 112.5 –2 –28125–150 8 137.5 –1 –8150–175 12 0 0175–200 5 187.5 1 5200–225 11 212.5 2 22

Sfi = 50 Sfidi = –9

Then big step deviation method, we have

Mean i i

i

f da h

få

= + ´å

9162.5 2550-= + ´ = 162.5 – 4.5 = ` 158 Ans.

Section – D

21. If two positive integers x and y are expressible in terms of primes as x = p2q3 and y = p3q,what can you say about their LCM and HCF. Is LCM a multiple of HCF ? Explain. [4]

Sol. Given, x = p2q3 = p × p × q × q × qAnd y = p3q = p × p × p × q

\ HCF = product of the smallest power of each common prime factor in the numbers x and y = p × p × q = p2q


Arundeep’s Solved Papers Mathematics 2014 (Term I)8and LCM product of the greatest power ofeach prime factor in the numbers x and y= p × p ×p × q × q × q = p3q3

Þ LCM = pq2 × p2q = pq2 × HCFYes, LCM is a multiple of HCF.Explanation :Let a = 12 = 22 × 3b = 18 = 2 × 32

\ HCF = 2 × 3 = 6 ...(i)LCM = 22 × 32 = 36LCM = 6 × 6LCM = 6 (HCF) [From (i)]Here LCM is 6 times the HCF. Ans.

22. Sita Devi wants to make a rectangularpond on the road side for the purpose ofproviding drinking water for streetanimals. The area of the pond will bedecreased by 3 square feet if its length isdecreased by 2 ft. and breadth isincreased by 1 ft. Its area will beincreased by 4 square feet if the lengthis increased by 1 ft. and breadth remainssame. Find the dimensions of the pond.What motivated Sita Devi to providewater point for street animals ? [4]

Sol. Let length of rectangular pond = x ft.and breadth of rectangular pond = y ft.

\ Area of rectangular pond = xyAccording to the question, we have(x – 2) (y + 1) = (xy – 3)

Þ xy + x – 2y – 2 = xy – 3Þ x – 2y = – 1 ...(i)

according to second given condition, wehave(x + 1) y = (xy + 4)

Þ xy + y = xy + 4i.e. y = 4 ...(ii)

Putting the value of y in eq. (i), we getx – 2 (4) = – 1

Þ x – 8 = – 1Þ x = – 1 + 8 = 7

Thus, length of rectangular pond = 7 ft.and Breadth of rectangular pond = 4 ft.

Values :1. Water is essential for the survival of all

living things including street animals.2. Water is the base of life and no one can

live without it.23. If a polynomial x4 + 5x3 + 4x2 – 10x – 12

has two zeroes as – 2 and – 3, then findthe other zeroes. [4]

Sol. Given, polynomial isf(x) = x4 + 5x3 + 4x2 – 10x – 12.Since two zeroes of f(x) are – 2 and – 3

\ (x + 2) (x + 3) = x2 + 3x + 2x + 6= x2 + 5x + 6

Dividing the polynomial f(x) withx2 + 5x + 6,

By division algorithm, we have

\ x4 + 5x3 + 4x2 – 10x – 12= (x2 + 5x + 6) (x2 – 2)

= (x + 2) (x + 3) ( 2) ( 2)x x- +

Other zeroes : 2 0x - = or 2 0x + =

2x = or 2x = -The zeroes of the polynomial are – 2, – 3,

2 and 2.- Ans.

24. Find all the zeroes of the polynomial8x4 + 8x3 – 18x2 – 20x – 5, if it is given

that two of its zeroes are52 and

5 .2

-

Sol. Given polynomial is given byf(x) = 8x4 + 8x3 – 18x2 – 20x – 5



Since two zeroes are52 and

52

-

\ 25 5 5( )2 2 2

x x x2æ ö æ ö æ ö

- ´ + = -ç ÷ ç ÷ ç ÷ç ÷ ç ÷ ç ÷è ø è ø è ø2 5

2x= -

Dividing the polynomial f(x) with 2 52

x -

By division algorithm, we have

\ f(x) = 8x4 + 8x3 – 18x2 – 20x – 5

2 25 (8 8 2)2

x x xæ ö= - + +ç ÷è ø

2 25 .2 (4 4 1)2

x x xæ ö= - + +ç ÷è ø

2 2(2 5) (4 2 2 1)x x x x= - + + +

2(2 5) [2 (2 1) 1(2 1)]x x x x= - + + +

f(x) 2(2 5) (2 1) (2 1)x x x= - + +

all zeroes of given polynomial x given byf(x) = 0

Þ 2x2 – 5 = 0 or 2x + 1 = 0 or 2x + 1 = 0

i.e. x =52

± or x =1

2-

or x =1

2-

All the zeroes are5 5 1 1, , and .2 2 2 2

- --

Ans.25. In the figure, there are two points D and

E on side AB of DABC such thatAD = BE. If DP || BC and EQ || AC, thenprove that PQ || AB. [4]

Sol.

Q

In DABC, we haveDP || BC (Given)

ÞADDB

AP=PC ...(i)

[Thales’ Theorem]Also, EQ || AC (Given)

ÞBEEA

BQ=QC [Thales’ Theorem]

ÞADDB

BQ=QC ...(ii)

[Q AD = BE ;\ EA = DB]From eq. (i) and (ii) ; we have

APPC

BQ=PC

\ PQ || AB (Inverse of Thales theorem)Hence proved.


Arundeep’s Solved Papers Mathematics 2014 (Term I)1026. In DABC, altitudes AD and CE intersect each other at the point P. Prove that

(i) DAPE ~ DCPD (ii) AP × PD = CP × PE(iii) DADB ~ DCEB (iv) AB × CE = BC × AD [4]

Sol. Given, In DABC, AD ^ BC and CE ^ AB(i) In DAPE and DCPD, we haveÐ1 = Ð4 [Each 90º]Ð2 = Ð3 [Vertically opposite angles]By AA axiom of similarity, we haveDAPE ~ DCPDHence proved.

(ii) DAPE ~ DCPD [Proved above]

\APCP

PE=PD [CPCT]

Þ AP × PD = CP × PEHence Proved

(iii) In DADB and DCEB, we haveÐ5 = Ð7 (Each 90º)Ð6 = Ð6 (Common)By AA axiom, of similarityDADB ~ DCEB Hence Proved.

(iv) DADB ~ DCEB [Proved Above]

ABCB

AD=CE [cpct]

Þ AB × CE = BC × AD Hence Proved27. Prove that :

(cot A + sec B)2 – (tan B – cosec A)2 = 2 (cot A . sec B + tan B . cosec A). [4]Sol. L.H.S.

= (cot A + sec B)2 – (tan B – cosec A)2

= (cot2 A + sec2 B + 2 cot A sec B) – (tan2 B + cosec2 A – 2 tan B cosec A)= cot2 A + sec2 B + 2 cot A sec B – tan2 B – cosec2 A + 2 tan B cosec A= (sec2 B – tan2 B) – (cosec2 A – cot2 A) + 2 (cot A sec B + tan B cosec A)= 1 – 1 + 2 (cot A sec B + tan B cosec A) [Q sec2 B – tan2 B = 1, cosec2 A – cot2 A = 1]= 2 (cot A sec B + tan B cosec A) = R.H.S. Hence Proved.

28. Prove that :(sin q + cos q + 1) · (sin q – 1 + cos q) · sec q · cosec q = 2. [4]

Sol. L.H.S. = (sin q + cos q + 1) · (sin q – 1 + cos q) · sec q cosec q= [(sin q + cos q) + 1] · [(sin q + cos q) – 1)] · sec q · cosec q

= [(sin q + cos q)2 – (1)2] sec q cosec q [Q (a + b) (a – b) = a2 – b2]= [sin2 q + cos2 q + 2 sin q cos q – 1] · sec q cosec q


Arundeep’s Solved Papers Mathematics 2014 (Term I)11= (1 + 2 sin q · cos q – 1) · sec q cosec q [Q sin2 q + cos2 q = 1]

= 2 (sin q cos q) ·1 1·

cos sinq q= 2 = R.H.S. Hence Proved.

29. If tan (20º – 3a) = cot (5a – 20º), then find the value of a and hence evaluate :sin a · sec a · tan a – cosec a · cos a · cot a. [4]

Sol. Given, tan (20º – 3a) = cot (5a – 20º)Þ tan (20º – 3a) = tan [90 – (5a – 20º)] [Q cot q = tan (90º – q)]Þ 20º – 3a = 90º – 5a + 20º Þ – 3a + 5a = 90º + 20º – 20ºÞ 2a = 90ºÞa = 45º

Now, sin a · sec a · tan a – cosec a · cos a · cot a= sin 45º · sec 45º · tan 45º – cosec 45º · cos 45º · cot 45º

1 12 1 2 12 2

= ´ ´ - ´ ´ = 1 – 1 = 0 Ans.

30. The frequency distribution of weekly pocket money received by a group of students isgiven below :

Pocket More More More More More More More More More Moremoney than than than than than than than than than thanin ( ) or or or or or or or or or or

equal equal equal equal equal equal equal equal equal equalto 20 to 40 to 60 to 80 to 100 to 120 to 140 to 160 to 180 to 200

Number

90 76 60 55 51 49 33 12 8 4of

students

Draw a ‘more than type’ ogive and from it, find median. Verify median by actualcalculations. [4]

Sol.

PQ

R


Arundeep’s Solved Papers Mathematics 2014 (Term I)12Taking pocket money (in Rs.) along x-axis and number of students along y-axis, plot the points(20, 90), (40, 76), (60, 60), (80, 55), (100, 51), (120, 49), (140, 33), (160, 12), (180, 8) and (200,4) on graph paper and join all these point by free hand gives the required more than type ogive.Now mark a point representing 45 on y-axis by P, from P draw a line || to x-axis meeting ogive atQ. From Q, draw QR perpendicular to x-axis.Then abscissa of point R gives the required median

\ Md = 125.The table of values is given as under :

Pocket money No. ofc.i. fi c.f(in ` ) Students

More than or equal to 20 90 20–40 14More than or equal to 40 76 40–60 30More than or equal to 60 60 60–80 60–55 = 5 35More than or equal to 80 55 80–100 55–51 = 4 39

More than or equal to 100 51 100–120 51–49 = 2 41More than or equal to 120 49 120–140 57More than or equal to 140 33 140–160 33–12=21 78

More than or equal to 160 12 160–180 82More than or equal to 180 8 180–200

12 – 8 = 486

More than or equal to 200 4 200–2208–4 = 4

90

N = 90 å i

49–33=16

4

Here, N = 90

\N2

90 452

= =

The cumulative frequency just greater than 45 be 51 and corresponding median class is 120–140

\ Median

N2

cfl h

f

-= + ´

45 41120 2016-= + ´ 4 20120

16´= + = 120 + 5 = ` 125

Hence Verified31. Cost of living Index for some period is given in the following frequency distribution :

Index 1500–1600 1600–1700 1700–1800 1800–1900 1900–2000 2000–2100 2100–2200No. of weeks 3 11 12 7 9 8 2

Find the mode and median for above data. [4]


Arundeep’s Solved Papers Mathematics 2014 (Term I)13The table of values is given as under :

Index Number of weeks (f i) cf1500–1600 3 31600–1700 11 f0 141700–1800 12 f1 261800–1900 7 f2 331900–2000 9 422000–2100 8 502100–2200 2 52

N = Sfi = 52

Here N = 52

\N2

52 262

= = , now commulative frequency which is just greater than or equal to 26 be

1700 – 1800.\ Median class is 1700–1800, Here l = 1700 ; c.f. = 14 ; f = 12 ; h = 100

\ Median

N2

cfl h

f

-= + ´

26 141700 10012-= + ´

121700 100 180012

= + ´ =

Also, Maximum frequency is 12 which lies in 1700 – 1800.\ Modal class is 1700–1800, Here l = 1700 ; f1 = 12 ; f0 = 11 ; f2 = 7

Thus, mode 1 0

1 0 22f f

l hf f f-

= + ´- -

12 111700 10024 11 7-= + ´- -

11700 1006

= + ´ = 1700 + 16.67

= 1716.67 Ans.


General Instructions :(i) All questions are compulsory.

(ii) The question paper consists of 34 questions divided into four sections – A, B, C and D.(iii) Section A contains 8 questions of 1 mark each, which are multiple choice type questions, Section

B contains 6 questions of 2 marks each, Section C contains 10 questions of 3 marks each andSection D contains 10 questions of 4 marks each.

(iv) Use of calculators is not permitted.Section – A

Question numbers 1 to 8 carry 1 mark each. For each of these questions four alternative choices havebeen provided of which only one is correct. Select the correct choice.

1. The first three terms of an APrespectively are 3y – 1, 3y + 5 and 5y + 1.Then y equals :(a) – 3 (b) 4(c) 5 (d) 2

Sol. (c) Since 3y – 1, 3y + 5 and 5y + 1 are in A.P.\ 2 (3y + 5) = 3y – 1 + 5y + 1

[If a, b, c are in A.P., b – a = c – bÞ 2b = a + c]

Þ 6y + 10 = 8yÞ 10 = 2yÞ y = 52. In Fig. 1, QR is a common tangent to

the given circles, touching externally atthe point T.

The tangent at T meets QR at P. IfPT = 3.8 cm, then the length ofQR (in cm) is:(a) 3.8 (b) 7.6(c) 5.7 (d) 1.9

Sol. (b) Given, PT= 3.8 cmWe know that the lengths of the tangentsdrawn to a circle from a point outside thecircle are equal.

\ QP = PT = 3.8 cmand PR = PT = 3.8 cm

\ QR = QP + PR = (3.8 + 3.8) cm = 7.6 cm3. In Fig. given below PQ and PR are two

tangents to a circle with centre O. IfÐQPR = 46º, then ÐQOR equals :

MATHEMATICS 2014 TERM - II (OUTSIDE DELHI)SET I


Arundeep’s Solved Papers Mathematics 2014 (Outside Delhi)14


Arundeep’s Solved Papers Mathematics 2014 (Outside Delhi)15(a) 67º (b) 134º(c) 44º (d) 46º

Sol. (b) We know that tangent to a circle isperpendicular to the radius at the point ofcontact.

\ ÐOQP = Ð ORP = 90ºNow, in quadrilateral ORPQ, we haveÐQOR + 90º + 46º + 90º = 360º

Þ ÐQOR + 226º = 360ºÞ ÐQOR = 360º – 226º = 134º4. A ladder makes an angle of 60º with the

ground when placed against a wall. Ifthe foot of the ladder is 2 m away fromthe wall, then the length of the ladder(in metres) is :

(a)43 (b) 4 3

(c) 2 2 (d) 4

Sol. (d) Suppose AB is the ladder of length x m.\ OA = 2 m, ÐOAB = 60º

In right DAOB, sec 60º 2x=

Þ 2 2x= Þ x = 4 m

Hence the required length of ladder be4 metre.

5. If two different dice are rolled together,the probability of getting an evennumber on both dice, is :

(a)136 (b)

12

(c)16 (d)

14

Sol. (d) Number of possible even numbers onboth dice = 9, i.e. (2, 2), (2, 4), (2, 6),(4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6).Total possibilities when two dice are rolled = 36{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

\ Probability of getting an even number on

both dice9 1

36 4= =

6. A number is selected at random fromthe numbers 1 to 30. The probability thatit is a prime number is :

(a)23 (b)

16

(c)13 (d)

1130

Sol. (c) Total outcomes of selecting a numberfrom 30 numbers = 30Favourable numbers (prime numbers) = 10,i.e. (2, 3, 5, 7, 11, 13, 17, 19, 23, 29)

\ Required probability of selecting a prime

number10 1 .30 3

= =

7. If the points A (x, 2), B (– 3, – 4) andC (7, – 5) are collinear, then the valueof x is :(a) – 63 (b) 63(c) 60 (d) – 60

Sol. (a) If points A (x, 2), B (– 3, – 4) andC (7, – 5) are collinear, then area of triangle


Arundeep’s Solved Papers Mathematics 2014 (Outside Delhi)16formed by these points is 0.[if A(x1, y1) ; B(x2, y2) and C(x3, y3) arevertices of DABCThen area of DABC

=12 |x1(y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)|

Þ12 [x (– 4 + 5) – 3 (– 5 – 2) + 7 (2 + 4)] = 0

Þ12 [x + 21 + 42] = 0 Þ x + 63 = 0

Þ x = – 638. The number of solid spheres, each of

diameter 6 cm that can be made bymelting a solid metal cylinder of height45 cm and diameter 4 cm, is :(a) 3 (b) 5(c) 4 (d) 6

Sol. Given diameter of sphere = 6 cm

\ radius of sphere =62 cm = 3 cm = r

\ volume of each solid sphere = 343

rp

= 34 3 363p ´ = p

Also, diameter of solid cylinder = R =42 cm = 2cm

and height of solid cylinder = h = 45 cm\Volume of solid cylinder =pR2h =p × 22 × 45 =180p

\ Required no. of solid sphere made

=volume of cylinder 180

volume of each sphere 36p=p = 5

\ Ans (b)

Section – B

Question numbers 9 to 14 carry 2 marks each.9. Solve the quadratic equation

2x2 + ax – a2 = 0 for x.

Sol. Consider the equation2x2 + ax – a2 = 0

Þ 2x2 + 2ax – ax – a2 = 0Þ 2x (x + a) – a (x + a) = 0Þ (2x – a) (x + a) = 0Þ 2x – a = 0 or x + a = 0

Þ 2ax = or x = – a

10. The first and the last terms of an APare 5 and 45 respectively. If the sum ofall its terms is 400, find its commondifference.

Sol. Let total no. of terms of given A.P. be n.Given, a = 5, an = 45 and Sn = 400

We know, Sn ( )2 nn a a= +

Þ 400 (5 45)2n= + Þ 400 = 50

2n ´

Þ 400 = 25n Þ n =40025

Þ n = 16\ a16 = 45Þ a + 15d = 45 Þ 5 + 15d = 45

Þ 15d = 40 Þ 40 815 3

d = =

Hence the required common difference of

given A.P. be83 .

11. Prove that the line segment joining thepoints of contact of two parallel tangentsof a circle, passes through its centre.

Sol. Given : PQ and RS are two parallel tangentsto a circle at B and A respectively. O is thecentre of the circle.To prove : AB passes through O.Construction : Join OA and OB.Proof : OB is perpendicular to PQ.[Tangent is perpendicular to radius at thepoint of contact.]



Now, PQ || RS Þ BO (Produced to RS) isperpendicular to RS. ...(i)[A line perpendicular to one of the twoparallel lines is perpendicular to other linealso]Also, OA is perpendicular to RS[Reason as above] ...(ii)From (i) and (ii), OA and OB must coincideas only one line can be drawn perpendicularfrom a point outside the line to the line.

\ A, O, B are collinear.Þ AB Passes through O, the centre of the

circle.12. If from an external point P of a circle

with centre O, two tangents PQ and PRare drawn such that ÐQPR = 120º, provethat 2PQ = PO.

Sol. Given : PQ and PR are tangents from pointP to circle with centre O.Also, ÐQPR = 120ºTo Prove : 2PQ = OPConstruction : Join OQ, OP and ORProof : In triangles OQP and ORP, we haveOQ = OR (radii)OP = OP (common)PQ = PR[tangents drawn from a point outside thecircle to the circle are equal in length]

\ DOQP @ DORP (SSS axiom of congruency)\ ÐOPQ = ÐOPR = 60º (c.p.c.t)

In right-angled triangle OQP, we have

OPPQ = sec 60º = 2 Þ OP = 2PQ

13. Rahim tosses two different coinssimultaneously. Find the probability ofgetting at least one tail.

Sol. When two different coins are tossedsimultaneously, then total possibilities = 4,i.e. (H, H), (H, T), (T, H), (T, T)Number of favourable outcomes for at leastone tail = 3, i.e. (H, T), (T, H), (T, T).

\ Required probability of getting at least one

tail3= .4

14. In given below, a square OABC isinscribed in a quadrant OPBQ of a circle.If OA = 20 cm, find the area of theshaded region. (Use p = 3.14)

Sol. OA = 20 cm.

\ OB 2= × side

(Q OB is a diagonal of the square)

Þ OB 20 2 cm=



\ Shaded area = Area of sector OPBQ– Area of square OABC

2 290 (20 2) (20)360º

°= ´ p -

3.14400 2 1 400 14 2pé ù é ù= ´ - = -ê ú ê úë û ë û

= 400 [1.57 – 1] = 400 × 0.57 = 228 cm2

Section – CQuestion numbers 15 to 24 carry 3 marks each.

15. Solve the equation4 53 ;

2 3x x- =

+

3 ,2

x ¹ 0, - for x.

Sol. Given equation be,4 53 ,

2 3x x- =

+

30,2

x ¹ -

Þ4 3x

x- 5

2 3x=

+Þ (4 – 3x) (2x + 3) = 5xÞ 8x + 12 – 6x2 – 9x = 5xÞ 6x2 + 6x – 12 = 0Þ 6 (x2 + x – 2) = 0Þ x2 + x – 2 = 0Þ x2 + 2x – x – 2 = 0Þ x (x + 2) – 1 (x + 2) = 0Þ (x + 2) (x – 1) = 0Þ Either x + 2 = 0 or x – 1 = 0Þ x = – 2, 1

16. If the seventh term of an AP is19 and

its ninth term is1 ,7 find its 63rd term.

Sol. Let a be the first term and d be the commondifference of the given AP.

Given, a719

= Þ a + 6d19

= ...(i)

[_ an = a + (n – 1) d]

Also, a917

= Þ a + 8d17

= ...(ii)

Subtracting (i) from (ii), we get

a + 8d – a – 6d1 17 9

= -

Þ 2d9 7 2

63 63-= = Þ

163

d =

Substituting the value of d in (i), we get

1663

a + ´ 19

=

Þ a 1 6 7 – 6 19 63 63 63

= - = =

\ a163

= and d163

=

Now, a63 = a + 62d1 16263 63

= + ´

1 62 63 163 63+= = =

17. Draw a right triangle ABC in whichAB = 6 cm, BC = 8 cm and ÐB = 90º.Draw BD perpendicular from B on ACand draw a circle passing through thepoints B, C and D. Construct tangentsfrom A to this circle.

Sol. Steps of construction :1. BC = 8 cm is drawn.2. ÐABC = 90º is drawn.3. AB = 6 cm is cut and AC is joined.4. Triangle ABC is the required triangle, right

angled at B.5. From point B, an arc is drawn cutting AC

at X and Y.6. Arc XY is bisected, by drawing two arcs

of same radius cutting at N.



7. BN is joined meeting AC at D.8. BD is perpendicular to AC.9. As ÐBDC = 90º, therefore, BC is

hypotenuse of DBDC.10. Take point O as mid-point of BC.11. With O as centre, a circle is drawn passing

through B, D and C.12. AO is joined.13. With AO as diameter, a circle is drawn

cutting the circle BDC at R and B

(Q ÐABO = 90°) join AR.

14. AB and AR are the tangents from A to circleBDC.

18. If the point A (0, 2) is equidistant fromthe points B (3, p) and C (p, 5), find p.Also find the length of AB.

Sol. Q A (0, 2) is equidistant from the pointsB (3, p) and C (p, 5).

\ AB = AC Þ AB2 = AC2

Þ (0 –3)2 + (2 – p)2 = (0 – p)2 + (2 – 5)2

Þ 9 + 4 – 4p + p2 = p2 + 9Þ 4p = 4 Þ p = 1\ Coordinates of point B be (3, 1)

\ AB 2 2(0 3) (2 1)= - + -

9 1 10= + = units

19. Two ships are there in the sea on eitherside of a light house in such a way thatthe ships and the light house are in thesame straight line. The angles ofdepression of two ships as observed fromthe top of the light house are 60º and45º. If the height of the light house is200 m, find the distance between the two

ships. [Use 3 1.73]=

Sol. Let AB be the light house of height 200 m.C and D are two ships on either sides oflight house with angles of depression 60ºand 45º respectively.

\ ÐACB = 60º and ÐADB = 45º[Alternate angles in both cases]

In right-angled triangle ABC,

BCAB = cot 60º

Þ BC1200×3

= 200 m3

= ...(i)

In right-angled triangle ABD,

BDAB = cot 45º

Þ BD = 200 × 1 = 200 m ...(ii)

\ Distance between ships = CD = CB + BD

200 200 3= 200 20033

+ = +



200 1.73 2003´= + 346 200

3= + = (115.33 + 200) m

= 315.33 m.20. If the points A (– 2, 1), B (a, b) and C (4, – 1) are collinear and a – b = 1, find the values

of a and b.

Sol. Q Points A (– 2, 1), B (a, b) and C (4, – 1) are collinear..

\ Area of triangle formed by these points is zero.

[if A(x1, y1) ; B(x2, y2) and C(x3, y3) are vertices of DABC

Then area of DABC =12

|x1(y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)|]

Þ12 [– 2 (b + 1) + a (– 1 – 1) + 4 (1 – b)] = 0

Þ – 2b – 2 – 2a + 4 – 4b = 0 Þ 2a + 6b = 2

Þ a + 3b = 1 ...(i)

Also, a – b = 1 Þ a = b + 1 ...(ii)

Putting the value of a from (ii) in (i), we get

b + 1 + 3b = 1 Þ 4b = 0 Þ b = 0

\ from eqn. (ii), a = 1

Thus, a = 1, b = 021. In Fig. given below a circle is inscribed in an equilateral

triangle ABC of side 12 cm. Find the radius of inscribedcircle and the area of the shaded region.

[Use p = 3.14 and 3 1.73]=

Sol. Given : An equilateral triangle of side 12 cm.Let radius of incircle be r. Join OA, OB and OC.Also, join OD, OE and OF.Here, AB, BC and AC are the tangents to the circle.

\ OD ^ BC, OE ^ AC and OF ^ AB.[Radius is perpendicular to the tangent at the point of contact]Now, ar (ABC) = ar (AOB) + ar (BOC) + ar (COA)

Þ 23 · (12)4

1 112 122 2

r r= ´ ´ + ´ ´ 1 122

r+ ´ ´ [_ area of equilateral DABC =3

4 (side)2]

Þ 36 3 = 18r

Þ r 2 3 cm= = 2 × 1.73 = 3.46 cm


Arundeep’s Solved Papers Mathematics 2014 (Outside Delhi)21Thus, required radius of incircle = 3.46 cmArea of shaded portion = Area of equilateraltriangle DABC – area of circle

2 23 (12) (2 3)4

= - p

36 3 12= - p = 12 (3 × 1.73 – 3.14) cm2

= 12 (5.19 – 3.14) cm2 = 12 × 2.05 cm2

= 24.60 cm2

22. In Fig. given below PSR, RTQ and PAQare three semicircles of diameters 10cm, 3 cm and 7 cm respectively. Findthe perimeter of the shaded region.

[Use p = 3.14]

Sol. Diameters of Semicircle PSR, PAQ andQTR are 10 cm, 7 cm and 3 cm and hence

corresponding radii are102 cm,

72 cm and

32 cm i.e. 5 cm, 3. 5 cm and 1.5 cm

respectively.Perimeter of shaded region= length of arc PSR + length of arc PAQ +length of arc QTR= [p (5) + p (3.5) + p (1.5)] cm= p × 10 cm = 3.14 × 10 cm = 31.4 cm

23. A farmer connects a pipe of internaldiameter 20 cm from a canal into acylindrical tank which is 10 m indiameter and 2 m deep. If the waterflows through the pipe at the rate of 4km per hour, in how much time will thetank be filled completely ?

Sol. Internal diameter of pipe = 20 cm

\ Internal radius of pipe = 10 cm1 m

10=


Arundeep’s Solved Papers Mathematics 2014 (Outside Delhi)22In 1 h, 4 km length of water flows into thetank.

\ In 1 hour, the volume of water which flowsout

2110

æ ö= p ç ÷è ø × 4 × 1000 m3

[_ 1 km = 1000 m]Volume of cylindrical tank = p (5)2 × 2 m2

[_ diameter of tank = 10 m, height = 2 m]\ Required time taken to fill the tank

Volume of tankVolume of water flows in 1 hour

=

25 2 100 h4 1000

p ´ ´ ´=p ´ ´

5 h4

= = 1 h 15 min.

24. A solid metallic right circular cone 20cm high and whose vertical angle is 60º,is cut into two parts at the middle of itsheight by a plane parallel to its base. Ifthe frustum so obtained be drawn into

a wire of diameter1 cm,

12 find the

length of the wire.Sol. Cone is cut by plane PB and PQDB is a

frustrum of cone.OA = AC = 10 cm, AB = r1 m ; CD = r2

Vertical angle = 60º, so Semi vertical angle = 30º

In right-angled triangle OAB, we have

ABOA = tan 30º Þ 1 1

10 3r

=

Þ r110= cm.

3In right angled triangle OCD, we have

CDOC = tan 30º Þ 2 1

20 3r

=

Þ r220= cm

3

Frustum is drawn into wire of diameter1 cm

12 and length x cm (say)

\ Volume of wire = volume of frustum

Þ pr2x 2 21 2 1 2

1 [ · ]3

h r r r r= p + +

Þ21

24xæ öp ´ç ÷è ø

103

p ´=2 210 20 10 20·

3 3 3 3

é ùæ ö æ ö+ +ê úç ÷ ç ÷ê úè ø è øë û

Þ1

576x´

10 100 400 2003 3 3 3

é ù= + +ê úë û

Þ 576x 10 700

3 3= ´

Þ x7000 576

9= ´ = 7000 × 64 cm

= 448000 cm

\ length of wire448000 km100000

= = 4.48 km

[_ 1 km = 1000 m = 1000 × 100 cm

Þ 1 cm =1 km

100000 ]



Section – AQuestion numbers 25 to 34 carry 4 marks each.

25. The difference of two natural numbersis 5 and the difference of their

reciprocals is1 .

10 Find the numbers.

Sol. Let numbers be x and y, such that x > y.According to question,x – y = 5 ...(i)

If x > y, then1 1y x

>

Now,1 1y x

- 110

= ...(ii)

From (i) and (ii), we have

1 15y y

-+

110

=

Þ5

( 5)y yy y

+ -+

110

=

Þ 50 = y2 + 5yÞ y2 + 5y – 50 = 0Þ y2 + 10y – 5y – 50 = 0Þ y (y + 10) – 5 (y + 10) = 0Þ (y + 10) (y – 5) =0Þ Either y + 10 = 0 or y – 5 = 0Þ y = – 10 (rejected)

Since y be any natural number.Thus, y = 5.

\ from (i), x = 10\ Required natural numbers are 10 and 5.

26. Prove that the length of the tangentsdrawn from an external point to a circleare equal.

Sol. Given : A circle C (O, r). P is a pointoutside the circle and PA and PB aretangents to a circle.To Prove : PA = PBConstruction : Draw OA, OB and OP.

Proof : Consider triangles OAP and OBP.ÐOAP = ÐOBP = 90º ...(i)[Radius is perpendicular to the tangent atthe point of contact]

OA = OB (radii) ...(ii)OP is common ...(iii)

\ DOAP @ DOBP (RHS)[from (i), (ii), (iii)]

Þ AP = BP (cpct)27. The angles of elevation and depression

of the top and the bottom of a tower fromthe top of a building, 60 m high, are 30ºand 60º respectively. Find the differencebetween the heights of the building andthe tower and the distance betweenthem.

Sol. Let AB be the height of the tower and CDbe the building of height 60 m.ÐACE = 30º, ÐECB = 60º



Let difference between heights of buildingand tower be y m and distance between thetower and building by x m.

\ CD = EB = 60 m, CE = BD = x m.In right-angled triangle CEB.

CEBE = cot 60º

Þ CE1= 60 m3

´ 60 3 m3

=

Þ BD = CE 20 3 m=\ Distance between tower and building is

20 3 m.

In right-angled triangle CEA,

AECE = tan 30º Þ y 120 3 20 m

3= ´ =

\ Difference between the height of tower andbuilding is 20 m.

28. A bag contains cards numbered from 1to 49. A card is drawn from the bag atrandom, after mixing the cardsthoroughly. Find the probability that thenumber on the drawn card is :

(i) an odd number(ii) a multiple of 5

(iii) a perfect square(iv) an even prime numberSol. Total cards in the bag = 49.

\ Total number of outcomes = 49, when onecard is drawn.

(i) No. of favourable outcomes for oddnumber = 25, i.e. 1, 3, 5, .... 49.

\ Probability of getting an odd number25 .49

=

(ii) No. of favourable outcomes for multipleof 5 = 9, i.e., 5, 10, 15, .... 45.

\ Probability of getting a multiple of 59 .49

=

(iii) Favourable outcomes for a perfect square= 7, i.e., 1, 4, 9, 16, 25, 36 and 49.

\ Probability of getting a card with perfect

square number7 1 .49 7

= =

(iv) Favourable outcomes for an even primenumber = 1, i.e., 2.

\ Probability of getting a card numbered even

prime1 .49

=

29. Find the ratio in which the point P (x, 2)divides the line segment joining thepoints A (12, 5) and B (4, – 3). Also findthe value of x.

Sol. Let P (x, 2) divides the join of A (12, 5) andB (4, – 3) in the ratio k : 1.Then by section formula, we have

\ Point of division is P

4 12 3 5,1 1

k kk k

+ - +æ öç ÷+ +è ø = (x, 2)

Þ4 12

1kk

++ = x and

3 51

kk

- ++ = 2 ...(i)

Consider,3 5

1k

k- +

+ = 2

Þ –3k + 5 = 2k + 2

Þ – 5k = – 3 Þ k35

=

\ Required ratio is k : 1, i.e., 3 :15

i.e. 3 : 5

Now, substituting35

k = in (i), we get

x34 12 12 60 725 9

3 3 5 815

´ + += = = =++


Arundeep’s Solved Papers Mathematics 2014 (Outside Delhi)2530. Find the values of k for which the

quadratic equation(k + 4) x2 + (k + 1) x + 1 = 0 has equalroots. Also find these roots.

Sol. Given equation is(k + 4) x2 + (k + 1) x + 1 = 0 ...(i)On comparing with ax2 + bx + c = 0, a ¹ 0Here, a = k + 4 ; b = k + 1 ; c = 1first we note k + 4 ¹ 0 Þ k ¹ – 4Here D = b2 – 4ac = (k + 1)2 – 4 · (k + 4)= k2 + 2k + 1 – 4k – 16= k2 – 2k – 15

Q roots are equal, therefore, D = 0Þ k2 – 2k – 15 = 0Þ k2 – 5k + 3k – 15 = 0Þ k (k – 5) + 3 (k – 5) = 0Þ (k + 3) (k – 5) = 0Þ k + 3 = 0 or k – 5 = 0 Þ k = – 3, 5

when k = – 3, then from (i), we get(– 3 + 4) x2 + (– 3 + 1) x + 1 = 0

Þ x2 – 2x + 1 = 0Þ (x – 1)2 = 0Þ (x – 1) · (x – 1) = 0Þ x = 1, 1

when k = 5, then from (i), we get(5 + 4) x2 + (5 + 1) x + 1 = 0

Þ 9x2 + 6x + 1 = 0Þ (3x + 1)2 = 0Þ (3x + 1) · (3x + 1) = 0

Þ x1 1,3 3

= - -

\ Roots are 1, 1 or1 1, .3 3

- -

31. In an AP of 50 terms, the sum of first10 terms is 210 and the sum of its last15 terms in 2565. Find the A.P.

Sol. Let a be the first term and d be the commondifference of the given AP.

\ According to question, we havea1 + a2 + .... + a10 = 210and a36 + a37 + .... + a50 = 2565

Consider, a1 + a2 + .... + a10 = 210

Þ 1 1010 [ ]2

a a+ = 210

[using S ( )2nn a l= + where l is the last term]

Þ 5 [a + a + 9d] = 210Þ 2a + 9d = 42 ...(i)

Again consider,a36 + a37 + .... + a50= 2575

Þ 36 5015 [ ]2

a a+ = 2565

Þ a + 35d + a + 49d = 171 × 2Þ 2a + 84d = 171 × 2Þ a + 42d = 171 ...(ii)

eqn. (i) – 2 × eqn. (ii), we have2a + 9d – 2a – 84d Þ 42 – 342

Þ – 75d = – 300 Þ d = 4\ From (ii), a + 42 × 4 = 171Þ a = 171 – 168 = 3\ a = 3 and d = 4.

Hence required AP is a + a + d, a + 2d,a + 3d, ......i.e. 3, 3 + 4, 3 + 8, 3 + 12, .....i.e. 3, 7, 11, 15, ....

32. Prove that a parallelogramcircumscribing a circle is a rhombus.

Sol. Given : A parallelogram ABCD, whichcircumscribes a circle.To prove : ABCD is a rhombus.Proof : We know that the lengths of thetangents drawn from a point outside thecircle to the circle are equal in length. Usingthis result, we getAP = AS, BP = BQ, CQ = CR and DR = DS

...(i)Consider, AB + CD = AP + BP + CR + DR= AS + BQ + CQ + DS [using (i)]= (AS + DS) + (BQ + CQ)= AD + BC



Þ AB + AB = BC + BC[Q AB = CD ; AD = BC, opp. sides of ||gmare equal]

Þ 2AB = 2BC Þ AB = BCAs adjacent sides of parallelogram ABCDare equal, hence, parallelogram ABCD is arhombus.

33. Sushant has a vessel, of the form of aninverted cone, open at the top, of height11 cm and radius of top as 2.5 cm and isfull of water. Metallic spherical ballseach of diameter 0.5 cm are put in the

vessel due to which2 th5 of the water in

the vessel flows out. Find how many ballswere put in the vessel. Sushant madethe arrangement so that the water thatflows out irrigates the flower beds. Whatvalue has been shown by Sushant ?

Sol. Radius of base of the cone = R = 2.5 cmand height of cone = h = 11 cm.

\ Volume of water in cone

= 2R3

hp

2 31 (2.5) 11cm3

= p ´

2 th5 of the volume of water in cone

2 32 (2.5) 11cm5 3

p= ´ ´ ...(i)

Given diameter of spherical ball = 0.5 cm.

\ Radius of spherical ball = r =0.52 cm

\ Volume of spherical ball

= 343

rp3

24 0.5 cm3 2

æ ö= p ç ÷è øLet x be the number of spherical balls bedropped.

We know volume of water displaced isequal to volume of body immersed.

\3

24 0.5 2 (2.5) 113 2 5 3

x pæ ö´ p = ´ ´ç ÷è ø

Þ x 23

2 1 8(2.5) 115 4(0.5)

= ´ ´ ´

Þ x4 2.5 2.5 115 0.5 0.5 0.5

´ ´ ´=´ ´ ´

44 25 25 10 44025 25

´ ´ ´= =´

440 balls were put in the vessel. As the waterdisplaced was used for irrigation of flowerbed, it shows Sushant is concerned for theconservation and best use of naturalresources.

34. From a solid cylinder of height 2.8 cmand diameter 4.2 cm, a conical cavity ofthe same height and same diameter ishollowed out. Find the total surface area



of the remaining solid.22Take7

é ùp =ê úë ûSol. The shaded conical cavity is hollowed out.

For cylinder :

Radius of base (r)4.2 cm2

= = 2.1 cm

height (h) = 2.8 cm

For cone :

Radius of base (r)4.2 cm2

= = 2.1 cm,

height (h) = 2.8 cm

Slant height (l) = 2 2r h+ 2 2(2.1) (2.8)= +

4.41 7.84= + 12.25= = 3.5 cm

Total surface Area of remaining solid= curved surface area of cylinder + Areaof top circular base + curved surface areaof cone= 2prh + pr2 + prl = pr [2h + r + l]= p × 2.1 [2 × 2.8 + 2.1 + 3.5] cm2

22 2.1 [5.6 5.6]7

= ´ ´ +

227

= × 2.1 × 11.2 cm2 = 73.92 cm2

14. The first and the last terms of an APare 7 and 49 respectively. If sum of allits terms is 420, find its commondifference.

Sol. Let a be the first term, d be the commondifference and an be the last term of theAP, with n terms and Sn be the sum to nterms of given A.P.Given, a = 7, an = 49, Sn = 420

\ Sn ( )2 nn a a= + Þ 420 (7 49)

2n= +

Þ 840 = 56n Þ n = 15\ a15 = 49Þ a + 14d = 49 Þ 7 + 14d = 49Þ 14d = 42 Þ d = 3\ Required common difference of given A.P.

is 3.

22. Solve the equation3 1 2 ;

1 2 3 1x x- =

+ -11, ,3

x x¹ - ¹ for x.

Sol. Consider the equation3 1 2 ,

1 2 3 1x x- =

+ -

where x ¹ – 1,13

Þ6 ( 1)2( 1)

xx

- ++

23 1x

=-

Þ5

2 ( 1)x

x-+

23 1x

=-

Þ (3x – 1) (5 – x) = 4 (x + 1)Þ 15x – 3x2 – 5 + x = 4x + 4Þ 3x2 – 12x + 9 = 0

Set-II (Uncommon Questions to Set-I)NOTE : Except for the following questions, all the

remaining questions have been asked in previous set.



Þ 3 (x2 – 4x + 3) = 0Þ x2 – 4x + 3 = 0Þ x2 – 3x – x + 3 = 0Þ x (x – 3) – 1 (x – 3) = 0Þ (x – 1) (x – 3) = 0Þ Either x – 1 = 0 or x – 3 = 0Þ x = 1, 3

23. Points A (– 1, y) and B (5, 7) lie on acircle with centre O (2, – 3y). Find thevalues of y. Hence find the radius of thecircle.

Sol. As points A (– 1, y) and B (5, 7) lie on acircle with centre O (2, – 3y).

\ OA = OB (radii) Þ OA2 = OB2

Þ (2 + 1)2 + (– 3y – y)2 = (2 – 5)2

+ (– 3 y – 7)2

Þ 9 + 16y2 = 9 + 9y2 + 49 + 42yÞ 7y2 – 42y – 49 = 0Þ 7 (y2 – 6y – 7) = 0Þ y2 – 7y + y – 7 = 0Þ y (y – 7) + 1 (y – 7) = 0Þ (y + 1) (y – 7) = 0Þ Either y + 1 = 0 or y – 7 = 0Þ y =– 1 or 7

when y = – 1

then centre O is (2, 3) and point A is (– 1, – 1)

\ Radius of circle 2 2(2 1) (3 1)= + + +

9 16 25= + = = 5 units

When y = 7,then centre O is (2, – 21) and point A is(– 1, 7)

\ Radius of circle 2 2(2 1) ( 21 7)= + + - -

9 784 793 units= + =

24. If the points P (– 3, 9), Q (a, b) and R(4, – 5) are collinear and a + b = 1, findthe values of a and b.

Sol. Since the points P (– 3, 9), Q (a, b) and R(4, – 5) are collinear, therefore, area oftriangle formed by these points is zero.

[if A(x1, y1) ; B(x2, y2) and (x3, y3) arevertices of DABCThen area of DABC

=12 |x1(y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)|]

Þ12 [– 3 (b + 5) + a (– 5 – 9) + 4 (9 – b)] = 0

Þ –3b – 15 – 14a + 36 – 4b = 0Þ 14a + 7b – 21 = 0 ...(i)Þ 2a + b = 3 ...(ii)

Also, a + b = 1On subtracting, we get a = 2Substituting the value of a in (i), we get4 + b = 3 Þ b = – 1

\ a = 2 b = – 131. The difference of two natural numbers

is 3 and the difference of their

reciprocals is3 .

28 Find the numbers.

Sol. Let the numbers be x and y, where x > y.Given, x – y = 3 ....(i)

If x > y, then1 1 .y x

>

\1 1 3

28y x- =



Þ1 1

3y y-

+3

28= [using eqn. (i)]

Þ3

( 3)y yy y

+ -+

328

=

Þ 23

3y y+3

28=

Þ y2 + 3y = 28Þ y2 + 3y – 28 = 0Þ y2 + 7y – 4y – 28 = 0Þ y (y + 7) – 4 (y + 7) = 0Þ (y – 4) (y + 7) = 0Þ Either y – 4 = 0 or y + 7 = 0Þ y = 4 or – 7

Since – 7 is rejected as it is not a naturalnumber, therefore, y = 4Thus, from (i) ; x – 4 = 3 Þ x = 7

\ Required numbers are 7 and 4.32. Prove that the tangent at any point of a

circle is perpendicular to the radiusthrough the point of contact.

Sol. Given : A circle with centre O, line l istangent to the circle at A.To prove : Radius OA is perpendicular tothe tangent at A.Construct : Take a point P, other then A,on tangent l. Join OP, meeting the circle at R.Proof : We know that tangent to the circletouches, the circle at one point and all otherpoints on the tangent lie in the exterior of acircle.

\ OP > OR (radius of circle)Þ OP > OA

(Q OR = OA, radius of circle)

Þ OA < OPÞ OA is the smallest segment, from O to a

point on the tangent.We know that smallest line segment from a

point outside the circle to the line isperpendicular segment.

l

Hence, OA is ^ to tangent l.Thus, tangent at any point of a circle isperpendicular to the radius through the pointof contact.

33. All the black face cards are removedfrom a pack of 52 playing cards. Theremaining cards are well shuffled andthen a card is drawn at random. Findthe probability of getting a :(i) face card(ii) red card(iii) black card(iv) king

Sol. In total 52 cards, 6 cards are black facecards which have been removed. [Facecards are ; 2 black Jack, 2 black Queen, 2black King]

\ Remaining cards = (52 – 6) = 46Total possibilities of drawing a card = 46.

(i) Favourable outcomes for a face card (king,queen, jack) are 6 (2 red kings + 2 redqueens + 2 red jacks)

\ Required probability of getting a face card

6 346 23

= =

(ii) Favourable outcomes for a red card are 26.\ Required probability of getting a red card

26 13 .46 23

= =

(iii) Favourable outcomes for a black card are(26 – 6) = 20, as all black face cards havebeen removed.



\ Required probability of getting a black card

20 10 .46 23

= =

(iv) Favourable outcomes for a king are 2(2 redkings)

\ Probability of getting a king2 1 .

46 23= =

34. Find the values of k for which thequadratic equation(3k + 1) x2 + 2 (k + 1) x + 1 = 0 has equalroots. Also find the roots.

Sol. Consider the given equation(3k + 1) x2 + 2 (k + 1) x + 1 = 0 ...(i)

First we notice 3k + 1 ¹ 0 Þ13

k ¹ -

On comparing given eqn. with ax2 + bx + c = 0Here a = 3k + 1 ; b = 2(k + 1) ; c = 1Here discriminant = b2 – 4ac= [2 (k + 1)]2 – 4 (3k + 1)= 4 [k2 + 2k + 1 – 3k – 1] = 4 [k2 – k]

Q roots are equal, therefore, D = 0Þ 4 (k2 – k) = 0 Þ k2 – k = 0Þ k (k – 1) = 0Þ Either k = 0 or k – 1 = 0Þ k = 0 or k = 1

When k = 0, then from (i), we get(0 + 1) x2 + 2 (0 + 1) x + 1 = 0

Þ x2 + 2x + 1 = 0 Þ (x + 1)2 = 0Þ (x + 1) (x + 1) = 0 Þ x = – 1, – 1\ Required roots are – 1, – 1.

When k = 1, then from (i), we get(3 + 1) x2 + 2 (1 + 1) x + 1 = 0

Þ 4x2 + 4x + 1 = 0 Þ (2x + 1)2 = 0Þ (2x + 1) (2x + 1) = 0

Þ x1 1,2 2

= - -

\ Roots are1 1,2 2

- -

Hence, roots are – 1, – 1 or1 1, .2 2

- -

Set-III (Uncommon Questions to Set-I and Set-II)NOTE : Except for the following questions, all the


14. The first and the last terms of an APare 8 and 65 respectively. If the sum ofall its terms is 730, find its commondifference.

Sol. Let a be the first term, d be the commondifference, an be the last term and Sn bethe sum to first n terms of the given AP.Given : a = 8, an = 65, Sn = 730

Now, Sn ( )2 nn a a= +

Þ 730 (8 65)2n= +

Þ 73073

2n= Þ n = 20

\ a20 = 65Þ a + 19d = 65 Þ 8 + 19d = 65Þ 19d = 57 Þ d = 3\ Required common difference of given A.P.

be 3.22. If the points A (– 1, – 4), B (b, c) and

C (5, – 1) are collinear and 2b + c = 4,find the values of b and c.

Sol. Since the points A (– 1, – 4), B (b, c) andC (5, – 1) are collinear, therefore, area oftriangle formed by these points is zero.



[if A(x1, y1) ; B(x2, y2) and (x3, y3) arevertices of DABCThen area of DABC

=12 |x1(y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)|]

Þ12 [– 1 (c + 1) + b (– 1 + 4) + 5 (– 4 – c)] = 0

Þ – c – 1 + 3b – 20 – 5c = 0Þ 3b – 6c = 21Þ b – 2c = 7 ...(i)

Also, 2b + c = 4 ...(ii) [given]Multiplying (ii) by 2 and then adding to (i),we getb – 2c + 4b + 2c = 7 + 8

Þ 5b = 15Þ b = 3

Substituting the value of b in (i), we get3 – 2c = 7 Þ 2c = – 4 Þ c = – 2

\ b = 3, c = – 223. If the point P (2, 2) is equidistant from

the points A (– 2, k) and B (– 2k, – 3),find k. Also find the length of AP.

Sol. Point P (2, 2) is equidistant from the pointsA (– 2, k) and B (– 2k, – 3).

\ AP = BPÞ AP2 = BP2

Þ (2 + 2)2 + (2 – k)2 = (2 + 2k)2 + (2 + 3)2

Þ 16 + 4 + k2 – 4k = 4 + 4k2 + 8k + 25Þ – 3k2 – 12k – 9 = 0Þ – 3 (k2 + 4k + 3) = 0Þ k2 + 4k + 3 = 0Þ k2 + 3k + k + 3 = 0Þ k (k + 3) + 1 (k + 3) =0Þ (k + 3) (k + 1) = 0Þ Either k + 3 = 0 or k + 1 = 0Þ k = – 3 or – 1

When k = – 1, then point A is (– 2, – 1)

\ AP 2 2(2 2) (2 1)= + + +

16 9 25= + = = 5 unitsWhen k = – 3, then point A is (– 2, – 3)

\ AP 2 2(2 2) (2 3)= + + +

16 25 41 units= + =

24. Solve the equation14 51 ;

3 1x x- =

+ +x ¹ – 3, – 1, for x.

Sol. Consider the given equation,14 1

3x-

+5 ,

1x=

+ where x ¹ – 3, – 1

Þ14 3

3x

x- -

+5

1x=

+

Þ11

3x

x-+

51x

=+

Þ (11 – x) (x + 1) = 5 (x + 3)Þ 11x + 11 – x2 – x = 5x + 15Þ – x2 + 5x – 4 = 0 Þ x2 – 5x + 4 = 0Þ x2 – 4x – x + 4 = 0Þ x (x – 4) – 1 (x – 4) = 0Þ (x – 1) (x – 4) = 0Þ Either x – 1 = 0 or x – 4 = 0 Þ x = 1, 4

31. Find the value of p for which thequadratic equation(2p + 1) x2 – (7p + 2) x + (7p – 3) = 0 hasequal roots. Also find these roots.

Sol. Given equation is(2p + 1) x2 – (7p + 2) x + (7p – 3) = 0 ...(i)

First we notice 2p + 1 ¹ 0 Þ12

p ¹ -

On comparing given eqn. with ax2 + bx + c = 0We have a = 2p + 1 ; b = – (7p + 2) and

c = 7p – 3\ Discriminant D = b2 – 4ac\ D = [– (7p + 2)]2 – 4 (2p + 1) (7p – 3)



= (7p + 2)2 – 4 (14p2 – 6p + 7p – 3) = 49p2 + 4 + 28p – 4 (14p2 + p – 3)= 49p2 + 4 + 28p – 56p2 – 4p + 12 = – 7p2 + 24p + 16For equal roots, D = 0 Þ – 7p2 + 24p + 16 = 0

Þ 7p2 – 24p – 16 = 0 Þ 7p2 – 28p + 4p – 16 = 0Þ 7p (p – 4) + 4 (p – 4) = 0 Þ (7p + 4) (p – 4) = 0

Þ Either 7p + 4 = 0 or p – 4 = 0 Þ p 47

= - or 4

When4 ,7

p = - then from eqn. (i), we get

28 28 281 2 37 7 7

x xæ ö æ ö æ ö- + - - + + - -ç ÷ ç ÷ ç ÷è ø è ø è ø = 0

Þ – x2 + 14x – 49 = 0 Þ x2 – 14x + 49 = 0 Þ (x – 7)2 = 0 Þ x = 7, 7When p = 4, then from (i), we gt(8 + 1) x2 – (28 + 2) x + (28 – 3) = 0

Þ 9x2 – 30x + 25 = 0 Þ (3x – 5)2 = 0 Þ x =5 5,3 3

\ Roots are 7, 7 or5 5, .3 3

32. Cards numbered from 11 to 60 are kept in a box. If a card is drawn at random from thebox, find the probability that the number on the drawn card is :

(i) an odd number(ii) a perfect square number

(iii) divisible by 5(iv) a prime number less than 20Sol. Cards numbered 11 to 60 are kept in a box. Total cards are 50.

\ Total no. of outcomes of drawing a card = 50.(i) Favourable outcomes for odd number = 25, (i.e., 11, 13, 15, ...., 59).

\ Probability of drawing an odd numbered card25 1 .50 2

= =

(ii) Favourable outcomes for a perfect square number = 4 (i.e. 16, 25, 36, 49).

\ Probability of drawing a perfect square numbered card4 2 .

50 25= =

(iii) Favourable outcomes for a card number divisible by 5 = 10, (i.e., 15, 20, 25, 30, .... 60).

\ Probability of drawing a card numbered divisible by 510 1 .50 5

= =

(iv) Favourable outcomes for a prime number less than 20 = 4 (i.e. 11, 13, 17, 19).



\ Probability of drawing a prime number less than 204 2 .

50 25= =

34. The difference of two natural numbers is 5 and the difference of their reciprocals is5 .

14Find the numbers.

Sol. Let two natural numbers be x and y, where x > y. Hence x – y = 5 ...(i)

If x > y, then1 1y x

>

Hence,1 1 5

14y x- =

Þ1 1

5y y-

+5

14= [from (i)]

Þ5

( 5)y yy y

+ -+

514

= Þ 25

5y y+5

14= Þ y2 + 5y = 14

Þ y2 + 5y – 14 = 0 Þ y2 + 7y – 2y – 14 = 0 Þ y (y + 7) – 2 (y + 7) = 0Þ (y – 2) (y + 7) = 0Þ y = 2, y = – 7 (rejected) as – 7 is not a natural number.\ y = 2

From (i), x – 2 = 5 Þ x = 7\ Required numbers are 7 and 2.



(ii) The question paper consists of 34 questions divided into four sections – A, B, C and D.(iii) Section A contains 8 questions of 1 mark each, which are multiple choice type questions, Section

B contains 6 questions of 2 marks each, Section C contains 10 questions of 3 marks each andSection D contains 10 questions of 4 marks each.

(iv) Use of calculators is not permitted.Section – A

Question numbers 1 to 8 carry 1 mark each. For each of these questions four alternative choices havebeen provided of which only one is correct. Select the correct choice.

1. If k, 2k – 1 and 2k + 1 are threeconsecutive terms of an A.P., the valueof k is(a) 2 (b) 3(c) – 3 (d) 5

Sol. (b) If k, 2k – 1, 2k + 1 are three consecutiveterms of an A.P., then(2k – 1) – k = (2k + 1) – (2k – 1)

Þ 2k – 1 – k = 2 Þ k – 1 = 2 Þ k = 3[if a, b, c are in A.P. then b – a = c – b]

2. Two circles touch each other externallyat P. AB is a common tangent to thecircles touching them at A and B. Thevalue of ÐAPB is(a) 30º (b) 45º(c) 60º (d) 90º

Sol. (d) We have, AT = TP and TB = TP[Lengths of the tangents from ext. point Tto the circles]

Þ ÐTAP = ÐTPA = x (say)and ÐTBP = ÐTPB = y (say)[equal sides have equal angle opposite to it]Also, in triangle APB, x + x + y + y = 180º

Þ 2x + 2y = 180ºÞ x + y = 90ºÞ ÐAPB = 90º

3. In a right triangle ABC, right-angled atB, BC = 12 cm and AB = 5 cm. Theradius of the circle inscribed in thetriangle (in cm) is(a) 4 (b) 3(c) 2 (d) 1

Sol. (c) In DABC,

AC 2 2AB BC= + 2 25 12= +

25 144 169= + = = 13 cmNow, ar (ABC) = ar (AOB) + ar (BOC)

+ ar (COA)

MATHEMATICS 2014 TERM II ( DELHI)SET I


Arundeep’s Solved Papers Mathematics 2014 (Delhi)34



Þ1 5 122

´ ´ 1 15 122 2

r r= ´ ´ + ´ ´ 1 132

r+ ´ ´

[_ area of triangle =12 × base × altitude]

Þ 60 = r (5 + 12 + 13)Þ 30r = 60 Þ r = 2

Alternate method :BP = OR = r, BR = OP = r

[BPOR is rectangle as all angles are 90º]\ PC = 12 – r = CQ and AR = 5 – r = AQ

Also, AC 2 25 12= +

25 144 169= + = = 13

Þ AQ + QC = 13Þ 5 – r + 12 – r = 13Þ 2r = 4 Þ r = 24. In a family of 3 children, the probability

of having at least one boy is

(a)78 (b)

18

(c)58 (d)

34

Sol. (a) In a family of three children,Total possibilities are : BBB, BGB, BBG,GBB, GGB, GBG, BGG, GGG, therefore,total no. of possible outcomes = 8Favourable outcomes for at least one boyare :

BGG, GBG, GGB, BBG, BGB, GBB, BBB,i.e. 7.

\ Required probability of having at least one

boy7 .8

=

5. The angle of depression of a car parkedon the road from the top of a 150 m hightower is 30º. The distance of the car fromthe tower (in metres) is

(a) 50 3 (b) 150 3

(c) 150 2 (d) 75

Sol. (b) Let AB be the tower of height 150 m.C is car and angle of depression is 30º.Let the required distance of the car fromthe tower be x m.

Therefore, ÐACB = 30º (alternate angle)In right-angled triangle ABC, we have

BCAB = cot 30º

Þ 150x

= 3

Þ x 150 3 m.=

Thus the required distance of the car from

the tower is 150 3 m.

6. The probability that a number selectedat random from the numbers 1, 2, 3, ...,15 is a multiple of 4, is

(a)4

15 (b)2

15



(c)15 (d)

13

Sol. (c) Total no. of possible outcomes = 15

(Q 15 numbers are given)Favourable outcomes for a multiple of 4= 3 (i.e. 4, 8, 12)

\ Probability of selecting a number which is

a multiple of 43 1

15 5= =

7. ABCD is a rectangle whose threevertices are B (4, 0), C (4, 3) andD (0, 3). The length of one of itsdiagonals is(a) 5 (b) 4(c) 3 (d) 25

Sol. (a) Both the diagonals are equal\ Length of diagonal BD

2 2(4 0) (0 3)= - + -

16 9 25 5= + = =

8. A chord of a circle of radius 10 cmsubtends a right angle at its centre. Thelength of the chord (in cm) is

(a) 5 2 (b) 10 2

(c)52 (d) 10 3

Sol. (b) AB is a chord. ÐAOB = 90º,AO = OB = 10 cm

\ Triangle AOB is right angled at O.Using pythagoras theorem, we have

\ AB 2 2OA OB= + 2 2(10) (10)= +

= 100 100+ = 100 2´

10 2 cm.=

Section BQuestion numbers 9 to 14 carry 2 marks each.

9. Find the values of p for which thequadratic equation 4x2 + px + 3 = 0 hasequal roots.

Sol. Given quadratic equation be 4x2 + px + 3 = 0On comparing with ax2 + bx + c = 0,We have a = 4 ; b = p ; c = 3Since given quadratic eqn. has equal roots,therefore, D = 0 Þ b2 – 4ac = 0

Þ (p)2 – 4 × 4 × 3 = 0Þ p2 = 48

Þ p 48 4 3= ± = ±

10. Find the number of natural numbersbetween 101 and 999 which are divisibleby both 2 and 5.

Sol. Numbers between 101 and 999 which aredivisible by both 2 and 5 (i.e. by L.C.M. of2 and 5 i.e. 10) are 110, 120, 130, ... 990.This forms an A.P. with a = 110, d = 10and an = 990Now, an = a + (n – 1) d

Þ 990 = 110 + (n – 1) 10Þ 880 = (n – 1) 10Þ 88 = n – 1Þ n = 89\ There are 89 natural numbers which are

divisible both by 2 and 5.


Arundeep’s Solved Papers Mathematics 2014 (Delhi)3711. In given below common tangents AB and

CD to the two circles with centres O1and O2 intersect at E. Prove thatAB = CD.

Sol. In the given figure, AB and CD are commontangents to the two given circles withcentres O1 and O2.We know that the lengths of the tangentsdrawn from a point outside the circle tothe circle are equal in length.

\ AE = EC and EB = EDÞ AE + EB = CE + EDÞ AB = CD.

12. The incircle of an isosceles triangle ABC,in which AB = AC, touches the sides BC,CA and AB at D, E and F respectively.Prove that BD = DC.

Sol. We know that the lengths of the tangentsdrawn from a point outside the circle tothe circle are equal in length.

\ AF = AE, BF = BDand CD = CE ...(i)Given AB = AC

Þ AF + FB = AE + ECÞ FB = EC [using (i), AF= AE]

Þ BD = CD[using (i), BF = BD and CD = CE]

13. Two different dice are tossed together.Find the probability

(i) that the number on each die is even.(ii) that the sum of numbers appearing on

the two dice is 5.Sol. Two different dice are tossed. Therefore,

total outcomes are 36.Here simple space S = {(1, 1), (1, 2),(1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2),(2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2),(3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2),(4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2),(5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2),(6, 3), (6, 4), (6, 5), (6, 6)}

(i) Favourable outcomes for even number onboth dice = 9, {(2, 2) (2, 4), (2, 6), (4, 2),(4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}

\ Probability of getting even number on both

dice9 1

36 4= =

(ii) Favourable outcomes that the sum of thenumbers appearing in two dice is 5 are(1, 4), (2, 3), (3, 2), (4, 1), i.e. 4.

\ Probability of getting sum of numbers

appearing on two dice is 54 1 .

36 9= =

14. If the total surface area of a solidhemisphere is 462 cm2, find its volume.

22Take7

é ùp =ê úë û


Arundeep’s Solved Papers Mathematics 2014 (Delhi)38Sol. Let r cm be the radius of hemisphere

Then total surface area of a solid hemisphere= 2pr2 + pr2 = 3pr2

Now, 3pr2 = 462 (given)

Þ r2 462 7 493 22

´= =´ = 72 Þ r = 7

\ Required volume of hemisphere 323

r= p

32 22 (7)3 7

= ´ ´ 23

= × 22 × 49 cm3

= 718.67 cm3

Section – C

Question numbers 15 to 24 carry 3 marks each.15. Solve for x :

16 151 ;1x x

- =+ x ¹ 0, – 1

Sol. Given equation be,16 1x

- 15 ,1x

=+ where

x ¹ 0, – 1

Þ16 x

x- 15

1x=

+

Þ (16 – x) (x + 1) = 15xÞ 16x + 16 – x2 – x = 15xÞ x2 = 16 = (± 4)2

Þ x = ± 416. The sum of the 5th and the 9th terms

of an AP is 30. If its 25th term is threetimes its 8th term, find the AP.

Sol. Let a be the first term and d be the commondifference of a given A.P.Given, a5 + a9 = 30

Þ a + 4d + a + 8d = 30Þ 2a + 12d = 30Þ a + 6d = 15 ...(i)

Also, a25 = 3a8Þ a + 24d = 3 (a + 7d)Þ a + 24d = 3a + 21dÞ 3d = 2a ...(ii)

From (i) and (ii), we havea + 4a = 15 Þ 5a = 15 Þ a = 3

\ From (ii), we get, 3d = 2 × 3 Þ d = 2\ a = 3, d = 2

Hence required A.P. is a, a + d, a + 2d,a + 3d, .... i.e. 3, 3 + 2, 3 + 4, 3 + 6, .....i.e. 3, 5, 7, 9, ....

17. Construct a triangle with sides 5 cm,5.5 cm and 6.5 cm. Now construct

another triangle, whose sides are35

times the corresponding sides of thegiven triangle.

Sol.

Steps of construction :1. A triangle with sides BC = 6.5 cm, AB = 5

cm and AC = 5.5 cm is constructed.2. ÐCBX is drawn below BC.


Arundeep’s Solved Papers Mathematics 2014 (Delhi)393. On BX, A1, A2, ..., A5 are marked, such

that BA1 = A1A2 = ... = A4A5.4. A5 and C are joined.5. C¢A3 is drawn parallel to A5C which meets

BC at C¢.6. A¢C¢ is drawn parallel to AC meeting AB at

A¢.7. DA¢BC¢ is the required triangle.

18. The angle of elevation of an aeroplanefrom a point on the ground is 60º. Aftera flight of 30 seconds the angle ofelevation becomes 30º. If the aeroplaneis flying at a constant height of

3000 3 m, find the speed of theaeroplane.

Sol. From the point of observation (O),

plane is at A, AL 3000 3 m= andÐAOL = 60º.After 30 seconds, plane is at B, therefore,

BM 3000 3 m= and ÐBOM = 30º.

Distance AB is covered in 30 seconds.In right-angled triangle OLA, we have

OLAL = cot 60º Þ OL = AL cot 60°

Þ OL13000 3 3000 m3

= ´ = ...(i)

In right-angled triangle OMB, we have

OMBM = cot 30º Þ OM = BM cot 30°

Þ OM 3000 3 3= ´ = 9000 m ...(ii)\ AB = LM = OM – OL

= (9000 – 3000) m = 6000 m[from (i) and (ii)]

Now in 30 s, distance covered = 6000 m\ In 1 hour (3600 s), distance covered by

plane 6000 3600 km30 1000

= ´

= 720 km\ Required speed of the aeroplane = 720 km/h.

19. If the point P (k – 1, 2) is equidistantfrom the points A (3, k) and B (k, 5),find the values of k.

Sol. Point P (k – 1, 2) is equidistant from thepoints A (3, k) and B (k, 5).


Þ (k – 1 – 3)2 + (2 – k)2 = (k – 1 – k)2

+ (2 – 5)2

Þ (k – 4)2 + (2 – k)2 = (– 1)2 + (– 3)2

Þ k2 – 8k + 16 + 4 – 4k + k2 = 1 + 9Þ 2k2 – 12k + 10 = 0Þ k2 – 6k + 5 = 0Þ k2 – 5k – k + 5 = 0Þ k (k – 5) – 1 (k – 5) = 0Þ (k – 1) (k – 5) = 0Þ Either k – 1 = 0 or k – 5 = 0Þ k = 1 or 5

20. Find the ratio in which the line segmentjoining the points A (3, – 3) and B (– 2, 7)is divided by x-axis. Also find thecoordinates of the point of division.

Sol. Let point P (x, 0) on x-axis divides the joinof A (3, – 3) and B (– 2, 7) in the ratio k : 1

Then by section formula, the coordinatesof P are

2 3 7 3,1 1

k kk k

- + -æ öç ÷+ +è ø ...(i)


Arundeep’s Solved Papers Mathematics 2014 (Delhi)40If point P lies on x-axis, then y = 0

Þ7 3

1kk

-+ = 0 Þ 7k – 3 = 0 Þ k

37

=

\ Required ratio is3 :1,7 i.e. 3 : 7.

Substituting37

k = in (i), we get the

coordinates of point of division are ;

6 37P , 0 ,3 1

7

æ ö- +ç ÷ç ÷ç ÷+ç ÷è ø

i.e. 3P , 0 .2

æ öç ÷è ø

21. In Figure given below two concentriccircles with centre O, have radii 21 cmand 42 cm. If ÐAOB = 60º, find the area

of the shaded region.22Use7

é ùp =ê úë û

Sol. Given, two concentric circles with centre Oand ÐAOB = ÐCOD = 60º = q (say)OC (r1) = 21 cm, OA (r2) = 42 cm

\ Area of shaded region

2 22 1

(360º ) ( )360º

r r- q= ´ p -

[_ area of sector of circle = pr2 × 360q

° ]

2 2(360º 60º ) 22 (42 21 )360º 7

-= ´ -

2300 22 (21) (4 1)360 7

°= ´ ´ -°

2 25 22 (21) 3cm6 7

= ´ ´ ´ = 3465 cm2.

22. The largest possible sphere is carved outof a wooden solid cube of side 7 cm. Findthe volume of the wood left.

22Use7

é ùp =ê úë ûSol. As largest possible sphere is carved out of

a solid wooden cube of side equal to sideof cube i.e. 7 cm, so, diameter of the sphereis eqnal to side of cube i.e. 7 cm.

\ Volume of the wood left = Volume of thecube – Volume of the sphere

33 34 7(7) cm

3 2

é ùæ ö= - pê úç ÷è øë û[_ volume of cube = a3

and volume of sphere = 343

rp]

3 4 22 1(7) 13 7 8

é ù= - ´ ´ê úë û= 343 [1 – 0.52] = 343 × 0.48 cm3

= 164.64 cm3.23. Water in a canal, 6 m wide and 1.5 m

deep, is flowing at a speed of 4 km/h.How much area will it irrigate in 10minutes, if 8 cm of standing water isneeded for irrigation ?

Sol. Given ; width of canal = 6 m,and depth of canal = 1.5 m and in 1 hour,


Arundeep’s Solved Papers Mathematics 2014 (Delhi)41length of water flows out by canal = 4 km

\ Volume of water flows out in 1 hour= 6 × 1.5 × 4000 m3 = 36000 m3

\ Volume of water flows out in 10 minutes

36000 1060

= ´ = 6000 m3 ...(i)

Suppose this water irrigates x m2 of areaand we require 8 cm of standing water.

\ Volume of water required 38 m100

x= ´

From (i) and (ii) we get

8 6000100

x ´ =

Þ x6000 100

8´= = 75000 m2

\ 75000 m2 of area is irrigated.24. In Figure given below ABCD is a

trapezium of area 24.5 sq. cm. In it,AD || BC, ÐDAB = 90º, AD = 10 cm andBC = 4 cm. If ABE in a quadrant of acircle, find the area of the shaded

region.22Take7

é ùp =ê úë û

Sol. Area of trapezium = 24.5 sq. cmAD = 10 cm, BC = 4 cmLet AB = h cm

\ Area of trapezium =12

(AD + BC) × AB

Þ 24.51 (10 4)2

h= + ´

Þ h24.5 3.5cm

7= =

Now, area of quadrant ABE

290º (3.5) sq. cm360º

= ´ p

21 22 (3.5) sq. cm4 7

= ´ ´

= 9.625 sq. cm ...(ii)\ Area of shaded region

= area of trap – area of quadrantABE= (24.5 – 9.625) sq. cm= 14.875 sq. cm [from (i), (ii)]

Section – D

Question numbers 25 to 34 carry 4 marks each.25. Solve for x :

2 4 10 ;3 5 3

x xx x

- -+ =- - x ¹ 3, 5

Sol. Given equation be,

2 43 5

x xx x

- -+- -

10 ,3

= x ¹ 3, 5

Þ( 3) 1 ( 5) 1

3 5x x

x x- + - ++

- -103

=



Þ1 11 1

3 5x x+ + +

- -103

=

Þ1 1

3 5x x+

- -10 23

= -

Þ5 3

( 3) ( 5)x xx x

- + -- -

43

=

Þ 22 8

8 15x

x x-

- +43

=

Þ 4x2 – 32x + 60 = 6x – 24Þ 4x2 – 38x + 84 = 0Þ 2x2 – 19x + 42 = 0Þ 2x2 – 12x – 7x + 42 = 0Þ 2x (x – 6) – 7 (x – 6) = 0Þ (2x – 7) (x – 6) = 0Þ Either 2x – 7 = 0 or x – 6 = 0

Þ x7 , 6.2

=

26. In a school, students decided to planttrees in and around the school to reduceair pollution. It was decided that thenumber of trees, that each section ofeach class will plant, will be double of theclass in which they are studying. If thereare 1 to 12 classes in the school and eachclass has two sections, find how manytrees were planted by the students.Which value is shown in this question ?

Sol. According to question, each section of :Class 1 will plant 2 trees, class II will plant4 trees, class III will plant 6 trees and soon.. class 12 will plant 24 trees and eachclass has 2 sections.

\ Number of trees planted= 4 + 8 + 12 + ... + 48This forms an A.P. with a = 4, d = 8 – 4 = 4and n = 12

\ Required number of trees planted

S1212 (4 48)2

= + S ( )2nn a lé ù= +ê úë û

Q

= 6 × 52 = 312

Value promted :

Students are concerned about safe andpollution free environment.

27. The angle of elevation of the top of atower at a distance of 120 m from a pointA on the ground is 45º. If the angle ofelevation of the top of a flagstaff fixedat the top of the tower, at A is 60º, thenfind the height of the flagstaff.

[Use 3 = 1.73]

Sol. Let BC be the tower and BD is flagstaff ofheight h m.

Let BC = x m.

AC = 120 m, ÐBAC = 45º

and ÐDAC = 60º

In right-angled triangle ACB, we have

ACBC = cot 45º Þ

120 1x

=

Þ x = 120 ...(i)

In right angled triangle ACD, we have

CDAC = tan 60º Þ 3

120h x+ =



Þ h + x 120 3= Þ h 120 3 120= - [using (i), x = 120]

Þ h 120[ 3 1]= - = 120 [1.73 – 1] m = 120 × 0.73 = 87.6 m

\ Required height of the flagstaff is 87.6 m.28. Red queens and black jacks are removed from a pack of 52 playing cards. A card is drawn

at random from the remaining cards, after reshuffling them. Find the probability thatthe drawn card is

(i) a king (ii) of red colour (iii) a face card (iv) a queenSol. Red queens and black jacks, i.e. 2 + 2 = 4 cards are removed from a pack of 52 playing cards.

Remaining cards are 52 – 4 = 48.\ Possible number of outcomes of drawing one card from 48 cards is 48.(i) Favourable outcomes for drawing a king are 4.

\ Probability of drawing a king4 1

48 12= =

(ii) Favourable outcomes for a card of red colour are 24 as 2 red queens have already been removedand in total there are 24 red cards.

\ Probability of drawing a red card24 1 .48 2

= =

(iii) Favourable outcomes for a face card (4 kings, 2 queens, 2 jacks) = 8

\ Probability of drawing a face card8 148 6

= =

(iv) Favourable outcomes for drawing a queen are 2, as 2 red queens have been removed.

\ Probability of drawing a queen2 1 .

48 24= =

29. If A (– 3, 5), B (– 2, – 7), C (1, – 8) and D (6, 3) are the vertices of a quadrilateral ABCD,find its area.

Sol. Area of quadrilateral ABCD= Area of triangle ABC + Area of triangle ACD ...(i)Since we know that if A(x1, y1), B(x2, y2) and C(x3, y3) are vertices of DABC

Then area of DABC =12

|x1(y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)|

Now, ar (ABC)12

= [– 3 (– 7 + 8) – 2 (– 8 – 5) + 1 (5 + 7)]

12

= [– 3 + 26 + 12]

35 sq. units2

= ...(ii)


Arundeep’s Solved Papers Mathematics 2014 (Delhi)44Also, ar (ACD)

12

= [– 3 (– 8 – 3) + 1 (3 – 5) + 6 (5 + 8)

12

= [33 – 2 + 78]

1092

= sq. units ...(iii)

from (i), (ii), (iii), we get

ar (ABCD)35 109 1442 2 2

= + = = 72 sq. units

30. A motorboat whose speed in still water is 18 km/h, takes 1 hour more to go 24 kmupstream than to return downstream to the same spot. Find the speed of the stream.

Sol. Speed of motor boat in still water = 18 km/hLet speed of the stream = x km/h

\ Speed upstream = (18 – x) km/hSpeed downstream = (18 + x) km/h

Thus, time taken by motor boat to covered a distance of 24 km stream =24 hr

18 x-

and Time taken by motor boat to covered a distance of 24 km downstream =24 km

18 x+According to question, we have

Þ24 24

18 18x x-

- + = 1 Þ24 (18 ) 24 (18 )

(18 ) (18 )x xx x

+ - -- + = 1

Þ 2432 24 432 24

324x x

x+ - +

- = 1 Þ 48x = 324 – x2

Þ x2 + 48x – 324 = 0 Þ x2 + 54x – 6x – 324 = 0Þ x (x + 54) – 6 (x + 54) = 0 Þ (x – 6) (x + 54) = 0Þ Either x – 6 = 0 or x + 54 = 0 Þ x = 6 or x = – 54 (rejected)\ Required speed of the stream is 6 km/h.

31. In Figure given below PQ is a chord of length 16 cm, ofa circle of radius 10 cm. The tangents at P and Q intersectat a point T. Find the length of TP.

Sol. Given : PQ is chord of length 16 cm, TP and TQ are thetangents to a circle with centre O, radius 10 cm.To find : TP.Solution : Join OP and OQ.In triangles OTP and OTQ, we have



OT is commonOP = OQ (radii)TP = TQ [length of the tangents drawnfrom a point outside the circle to the circleare equal]

\ DOPT @ DOQT(SSS axiom of congruency)

\ ÐPOT = ÐQOT (by cpct) ...(i)Consider, triangles OPR and OQR, we haveOP = OQ (radii)OR is commonÐPOR = ÐQOR [from (i)]

\ DOPR @ DOQR(SAS axiom of congruency)

\ PR = RQ1 162

= ´ = 8 cm ...(ii)

ÐORP = ÐORQ = 90º ...(iii)In right angled triangle TRP, we haveTR2 = TP2 – (8)2 = TP2 – 64Also OT2 = TP2 + (10)2

Þ (TR + 6)2 = TP2 + 100

[Q OR 100 64= - = 6]

Þ TR2 + 12TR + 36 = TP2 + 100Þ TP2 – 64 + 12TR + 36 = TP2 + 100Þ 12TR = 128

Þ TR32 cm3

=

From (iv) ; we have

2323

æ öç ÷è ø

= TP2 – 64

Þ TP2 1024 649

= +

1024 576 16009 9+= = =

2403

æ öç ÷è ø

\ TP40 cm3

=

32. Prove that the tangent at any point of acircle is perpendicular to the radiusthrough the point of contact.

Sol. Given : A circle with centre O, line l istangent to the circle at A.To Prove : Radius OA is perpendicular tothe tangent at A.Construct : Take a point P, other then A, ontangent l. Join OP, meeting the circle at R.Proof : We know that tangent to the circletouches, the circle at one point and all otherpoints on the tangent lie in the exterior of acircle.

\ OP > OR (radius of circle)

Þ OP > OA (Q OR = OA, radius of circle)Þ OA < OPÞ OA is the smallest segment, from O to a

point on the tangent.We know that smallest line segment from apoint outside the circle to the line isperpendicular segment.



Hence, OA is ^ to tangent l.Þ tangent at any point of a circle is

perpendicular to the radius through the pointof contact.

33. 150 spherical marbles, each of diameter1.4 cm, are dropped in a cylindricalvessel of diameter 7 cm containing somewater, which are completely immersedin water. Find the rise in the level ofwater in the vessel.

Sol. Let rise in water in cylinder be h cm when150 spherical marbles, each of diameter1.4 cm are dropped and fully immersed.Then volume of 150 spherical marbles= volume of water raised in cylinder

Þ2

34 7150 (0.7)3 2

hæ ö´ p = p ç ÷è ø

Þ4 7 7 7 7 71503 1000 4

h´ ´ ´´ ´ = ´

Þ h4 4 7 28 cm

20 5´ ´= = = 5.6 cm

\ Required rise in water level by 5.6 cm.

34. A container open at the top, is in theform of a frustum of a cone of height24 cm with radii of its lower and uppercircular ends as 8 cm and 20 cmrespectively. Find the cost of milk whichcan completely fill the container at the

rate of Rs. 21 per litre.22Use7

é ùp =ê úë û

Sol. Given radius of lower end (r1) = 8 cm

and Radius of upper end (r2) = 20 cm

Height of frustum = 24 cm

Volume of the container

2 21 2 1 2[ ]

3h r r r rp= + +

2 222 24 [(8) (20) 8 20]7 3

= ´ + + ´

22 8 [64 400 160]7

= ´ + +

322 8 624 cm7

= ´ ´ = 15689.14 cm3

= 15.68914 L

[Q 1L = 1000 cm3 Þ 1 cm3 =1

1000 L]

\ Cost of milk which can completely fill thecontainer = Rs. 21 × 15.68914

= Rs. 329.47.



14. Find the values of k for which thequadratic equation 9x2 – 3kx + k = 0 hasequal roots.

Sol. Given equation be, 9x2 – 3kx + k = 0on comparing with ax2 + bx + c = 0where a ¹ 0We have a = 9 ; b = –3k ; c = kFor equal roots, D = 0

\ b2 – 4ac = 0Þ (– 3k)2 – 4 × 9 × k = 0Þ 9k2 – 36k = 0 Þ 9k (k – 4) = 0Þ Either k = 0 or k – 4 = 0 i.e. k = 0 or 4

22. The sum of the 2nd and the 7th termsof an AP is 30. If its 15th term is 1 lessthan twice its 8th term, find the AP.

Sol. Let a be the first term and d be the commondifference of a given A.P.Given, a2 + a7 = 30

Þ a + d + a + 6d = 30 [_ an = 2a + (n – 1)d]Þ 2a + 7d = 30 ...(i)

Also, a15 = 2a8 – 1Þ a + 14d = 2 (a + 7d) – 1Þ a + 14d = 2a + 14d – 1 Þ a = 1

Substituting the value of a in (i), we get2 + 7d = 30 Þ 7d = 28 Þ d = 4

\ a = 1, d = 4Hence, required A.P. is a, a + d, a + 2d,a + 3d, .....i.e. 1, 1 + 4, 1 + 2, 1 + 12, .....i.e. 1, 5, 9, 13, ...

23. Draw a line segment AB of length 8 cm.Taking A as centre, draw a circle ofradius 4 cm and taking B as centre, drawanother circle of radius 3 cm. Constructtangents to each circle from the centreof the other circle.

Sol. Steps of construction :1. AB = 8 cm is taken. With centre A, a circle

of radius 4 cm is drawn and with centre B,a circle of radius 3 cm is drawn.

2. With AB as diameter, a circle is drawnmeeting circle with centre A at S and T andcircle with centre B at P and Q.

3. Then AP and AQ are tangents from A tocircle with centre B and BS and BT aretangents from B to circle with centre A.

24. Prove that the diagonals of a rectangleABCD, with vertices A (2, – 1), B (5, – 1),C (5, 6) and D (2, 6), are equal and bisecteach other.

Sol. Given : A (2, – 1), B (5, – 1), C (5, 6) andD (2, 6) are the vertices of a rectangleABCD.

\ AC 2 2(5 2) (6 1)= - + +

9 49 58= + =

BD 2 2(5 2) ( 1 6)= - + - -

9 49 58= + =

\ AC = BD, i.e. diagonals are equal

Set-II (Uncommon Questions to Set-I)NOTE : Except for the following questions, all the



Arundeep’s Solved Papers Mathematics 2014 (Delhi)48Now, the coordinates of mid-point of AC

are2 5 6 1, ,

2 2+ -æ ö

ç ÷è ø i.e.7 5,2 2

æ öç ÷è ø

The coordinates of mid-point of BD are

5 2 1 6, ,2 2+ - +æ ö

ç ÷è ø i.e.7 5,2 2

æ öç ÷è ø

As the coordinates of the mid-points of ACand BD are same, hence diagonals bisecteach other.

31. Solve for x :

3 5 10 ;4 6 3

x xx x

- -+ =- - x ¹ 4, 6

Sol. Consider the equation3 54 6

x xx x

- -+- -

10 ,3

=

where x ¹ 4, 6

Þ( 4) 1 ( 6) 1

4 6x x

x x- + - ++

- -103

=

Þ1 11 1

4 6x x+ + +

- -103

=

Þ1 1

4 6x x+

- -10 23

= -

Þ6 4

( 4) ( 6)x xx x

- + -- -

43

=

Þ 22 10

10 24x

x x-

- +43

=

Þ 25

10 24x

x x-

- +23

=

Þ 2x2 – 20x + 48 = 3x – 15Þ 2x2 – 23x + 63 = 0Þ 2x2 – 14x – 9x + 63 = 0Þ 2x (x – 7) – 9 (x – 7) = 0Þ (2x – 9) (x – 7) = 0Þ Either 2x – 9 = 0 or x – 7 = 0

i.e. x9 or 7.2

=

32. All the red face cards are removed froma pack of 52 playing cards. A card isdrawn at random from the remainingcards, after reshuffling them. Find theprobability that the drawn card is(i) of red colour(ii) a queen(iii) an ace(iv) a face card

Sol. Face cards are taken as (4 jacks, 4 queensand 4 kings) :As all the red face cards are removed

\ 6 cards (2 jacks, 2 queens and 2 kings) areremoved from 52 cards.

\ remaining no. of cards = 52 – 6 = 46 cards.\ Total no. of outcomes of drawing a card

from 46 cards = 46.(i) Favourable outcomes of drawing a red card

= 20 (as 6 red cards from 26 have beenremoved).

\ Probability of drawing a red card

20 1046 23

= =

(ii) Favourable outcomes of drawing a queen= 2 (as 2 red queens have already beenremoved out of 4).

\ Probability of drawing a queen2 1

46 23= =

(iii) Favourable outcomes of drawing an ace = 4

\ Probability of drawing an ace4 2

46 23= =

(iv) Favourable outcomes of drawing a facecard = 6 (as 2 jacks, 2 queens, 2 kingshave already been removed).

\ Required probability of drawing a face card

6 3 .46 23

= =

33. A (4, – 6), B (3, – 2) and C (5, 2) are thevertices of a DABC and AD is its median.Prove that the median AD divides DABCinto two triangles of equal areas.


Arundeep’s Solved Papers Mathematics 2014 (Delhi)49Sol. A (4, – 6), B (3, – 2) and C (5, 2) are the vertices of DABC.

Q AD is the median, so, D is the mid-point of BC.\ Coordinates of D are by using mid point formula we have

3 5 2 2, (4, 0)2 2+ - +æ ö= =ç ÷è ø

Now, ar (ABD)

1 [3(0 6) 4 ( 6 2) 4( 2 0)]2

= + + - + + - -

1 1[18 15 8] | 6 |2 2

= - - = - = 3 sq. units

[if A(x1 y1) ; B (x2, y2) and C(x3, y3) are the vertices of DABC.

Then area of DABC =12 |x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)|]

Thus, ar (ADC)1 [4 (2 6) 5( 6 0) 4(0 2)]2

= + + - - + -

1 1[32 30 8] | 6 |2 2

= - - = - = 3 sq. units

\ ar (ABD) = ar (ADC)34. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary

angles at the centre of the circle.Sol. Given : ABCD is a quadrilateral, circumscribing a circle with centre O and touches the quadrilateral

at P, Q, R and S.To prove :

(i) ÐAOB + ÐCOD = 180º(ii) ÐBOC + ÐAOD = 180º

Construction : Join OP, OQ, OR and OS.Proof : Consider, triangles APO and ASO,AP = AS [Lengths of the tangents drawn from a point outside the circle to the circle are equal]OS = OP (radii)OA is common

\ DAPO @ DASO [SSS axiom of congrency rule]\ ÐOAP = ÐOAS = x (say) [c.p.c.t.]

Similarly, ÐOBP = ÐOBQ = y (say)ÐOCQ = ÐOCR = z (say)and ÐODR = ÐODS = w (say)We have,



ÐDAB + ÐABC + ÐBCD + ÐCDA = 360ºÞ 2x + 2y + 2z + 2w = 360ºÞ x + y + z + w = 180º ...(i)

Consider, ÐAOB + ÐCOD= [180º – x – y] + [180º – w – z]

[Sum of angles of a triangle is 180º]= 360º – (x + y + z + w) = 360º – 180º

[using (i)]

\ ÐAOB + ÐCOD = 180ºAgain consider, ÐBOC + ÐAOD= [180º – y – z] + [180º – x – w]

[Sum of angles of a triangle is 180º]= 360º – (x + y + z + w)= 360º – 180º = 180º [using (i)]Hence proved.

Set-III (Uncommon Questions to Set-I and Set-II)NOTE : Except for the following questions, all the


14. Find the value of p so that the quadraticequation px (x – 3) + 9 = 0 has equal roots.

Sol. Consider the equation px (x –3) + 9 = 0\ px2 – 3px + 9 = 0 ...(i)

on comparing eqn. (i) with ax2 + bx + c = 0We have a = p ; b = –3p ; c = 9For equal roots, discriminant D = 0

\ b2 – 4ac = 0Þ (– 3p)2 – 4 × p × 9 = 0Þ 9p2 – 36p = 0 Þ 9p (p – 4) = 0Þ Either p = 0 or p – 4 = 0Þ p = 0 or p = 4

But from (i), we notice p ¹ 0 \ p = 422. The sum of the first seven terms of an

AP is 182. If its 4th and the 17th termsare in the ratio 1 : 5, find the AP.

Sol. Let a be the first term and d be the commondifference of a given A.P.According to question, S7 = 182

Þ7 [2 (7 1) ]2

a d+ - = 182

[_ Sn = 2n

{2a + (n – 1) d}]Þ a + 3d = 26 ...(i)

Also,4

17

aa

15

= Þ3

16a da d

++

15

=

[_ an = a + (n – 1) d]

Þ 5a + 15d = a + 16d Þ 4a = d ...(ii)From (i) and (ii), we geta + 12a = 26 Þ 13a = 26 Þ a = 2From (ii), we get d = 8

\ a = 2 and d = 8\ Required A.P. is

a, a + d, a + 2d, .....2, 2 + 8, 2 + 16, .....2, 10, 18, ....

23. From the top of a 60 m high building,the angles of depression of the top andthe bottom of a tower are 45º and 60ºrespectively. Find the height of the

tower. [Take 3 1.73]=

Sol. Let AB be 60 m high building and CD bethe tower of height h. Angles of depressionfrom top of building, to the top and bottomof tower are 45º and 60º.

\ ÐACE = 45º andÐADB = 60º [using alternate angles]Let BD = CE = x m and BE = CD = h m

\ AE = (60 – h) mIn right-angled triangle ABD, we have

BDAB = cot 60º



Þ 60x 1

3= Þ x

60 20 33

= = ...(i)

In right angled triangle AEC, we have

AECE = tan 45º

Þ60 h

x-

= 1 Þ 60 – h = x

Þ h = 60 – x

Þ h 60 20 3 20 [3 3]= - = -= 20 [3 – 1.73] = 20 × 1.27 = 25.4 m

[using (i)]\ required height of the tower is 25.4 m.

24. Find a point P on the y-axis which isequidistant from the points A (4, 8) andB (– 6, 6). Also find the distance AP.

Sol. Let point P (0, y) on y-axis is equidistantfrom the points A (4, 8) and B (– 6, 6)


Þ (4 – 0)2 + (8 – y)2 = (– 6 – 0)2 + (6 – y)2

Þ 16 + 64 – 16y + y2 = 36 + 36 – 12y + y2

Þ – 4y = – 8 Þ y = 2\ Coordinates of point P are (0, 2)

\ Required distance AP 2 2(4 0) (8 2)= - + -

16 36 52 2 13 units= + = =

31. Solve for x :

4 6 10 ;5 7 3

x xx x

- -+ =- -

x ¹ 5, 7

Sol. Consider the equation

4 65 7

x xx x

- -+- -

10 ,3

= where x ¹ 5, 7

Þ( 5) 1 ( 7) 1

5 7x x

x x- + - ++

- -103

=

Þ1 11 1

5 7x x+ + +

- -103

=

Þ1 1

5 7x x+

- -10 23

= -

Þ7 5

( 5) ( 7)x xx x

- + -- -

43

=

Þ 22 12

12 35x

x x-

- +43

=

Þ 26

12 35x

x x-

- +23

=

Þ 2x2 – 24x + 70 = 3x – 18Þ 2x2 – 27x + 88 = 0Þ 2x2 – 11x – 16x + 88 = 0Þ x (2x – 11) – 8 (2x – 11) = 0Þ (x – 8) (2x – 11) = 0Þ Either x – 8 = 0 or 2x – 11 = 0

Þ x118,2

=

32. Five cards – the ten, jack, queen, kingand ace of diamonds, are well shuffledwith their faces downwards. One card isthen picked up at random.

(a) What is the probability that the drawncard is the queen ?

(b) If the queen is drawn and put aside, anda second card is drawn, find theprobability that the second card is (i) anace (ii) a queen.

Sol. Five cards : a ten, jack, queen, king and


Arundeep’s Solved Papers Mathematics 2014 (Delhi)52ace of diamonds are shuffled with their facesdownwards.

\ Total no. of possibilities of selecting onecard = 5

(a) Favourable outcomes for drawing a queen = 1.\ Required probability of drawing a queen

15

=

(b) When queen is kept aside, then remainingcards = 4.So, total outcomes of drawing a card = 4

(i) Favourable outcomes of drawing an ace is 1.

\ Probability of drawing an ace1 .4

=

(ii) Favourable outcomes of drawing a queen = 0(as there is no card of queen).

\ Required probability of drawing a queen

0 ,4

= i.e. 0.

33. If A (4, 2), B (7, 6) and C (1, 4) are thevertices of a DABC and AD is its median,prove that the median AD divides DABCinto two triangles of equal areas.

Sol. Given ; A (4, 2), B (7, 6) and C (1, 4) arethe vertices of a triangle ABC and AD is themedian.

\ D is the mid-point of BC.\ The coordinates of D are given by using

mid point formula

7 1 6 4, ,2 2+ +æ ö

ç ÷è ø i.e. (4, 5)

[if A(x, y) ; B (x2, y2) and C(x3, y3) arevertices of DABC

Then area of DABC =12 |x1 (y2 – y3) + x2

(y3 – y1) + x3 (y1 – y2)]Now, ar (DABD)

1 [4 (6 5) 7 (5 2) 4 (2 6)]2

= - + - + -

1 9[4 21 16] sq. units2 2

= + - =

ar (DADC)

1 [4 (5 4) 4 (4 2) 1(2 5)]2

= - + - + -

1 9[4 8 3] sq. units2 2

= + - =

\ ar (ABD) = ar (ADC)34. In Figure given below, a triangle ABC

is drawn to circumscribe a circle ofradius 4 cm, such that the segments BDand DC are of lengths 8 cm and 6 cmrespectively. Find the sides AB and AC.

4 cm

4 cm

Sol. We know that the lengths of the tangentsdrawn from a point outside the circle tothe circle are equal in length.

\ AF = AE = x, (say)BF = BD = 8 cmCE = CD = 6 cm


Arundeep’s Solved Papers Mathematics 2014 (Delhi)53\ AB = (x + 8) cm

AC = (x + 6) cmand BC = 14 cm.Now, 2S = x + 8 + x + 6 + 14 = 28 + 2x

Þ S = 14 + xThus by Heron’s formula, we have

\ Area of DABC

(14 ) (14 8) (14 6)(14 14)x x x x x x= + + - - + - - + -

(14 ) 6 8x x= + ´ ´ ´ ...(i)

Also, the area of DABC = ar (BOC) + ar (BOA) + ar (AOC)

1 114 4 (8 ) 42 2

x= ´ ´ + ´ + ´ 1 (6 ) 42

x+ + ´

= 28 + 16 + 2x + 12 + 2x = 56 + 4x ...(ii)From (i) and (ii), we get

48 (14 )x x+ = 56 + 4x

On squaring both sides, we get48x (14 + x) = (56 + 4x)2

Þ 48x (14 + x) = 16 (14 + x)2

Þ 3x (14 + x) – (14 + x)2 = 0Þ (14 + x) (3x – 14 – x) = 0Þ (14 + x) (2x – 14) = 0Þ 14 + x = 0 or 2x – 14 = 0Þ x = – 14 (rejected) or x = 7\ AB = (7 + 8) cm =15 cm.

and AC = (7 + 6) cm = 13 cm.


1. In DDEW, AB || EW. If AD = 4 cm, DE = 12 cm and DW = 24 cm, then find the value of DB.[1]

Sol. Let BD = x cm,

DW = 24 cm,

Then, BW = (24 – x) cm,

AE = 12 – 4 = 8 cm

In DDEW, AB || EW

\ADAE

BD=BW [using Thales’ Theorem]

Þ48 24

xx

=- Þ 8x = 96 – 4x Þ 12x = 96

Þ x 96 812

= =

\ DB = 8 cm Ans.

2. In DABC is right angled at B, what is the value of sin (A + C). [1]

Sol. ÐB = 90º [Given]

We know that in DABC,

ÐA + ÐB + ÐC = 180º [_ Angle sum property of a D]

Þ ÐA + ÐC + 90º = 180º

Þ ÐA + ÐC = 180º – 90º = 90º\ sin (A + C) = sin 90º = 1 Ans.

SECTION – A

C B

A

E W

B

D

A

4cm

12cm

24 cm

MATHEMATICS 2015 TERM ISET I





3. If 3 sin cos ,q = q find the value of23cos 2cos .

3 cos 2q + qq +

[1]

Sol. Given, 3 sin q = cos q

Þsincosqq

13

= or tan q13

=

Þ tan q = tan 30ºÞ q = 30º

Now,23cos 2 cos

3cos 2q + qq +

cos (3cos 2)(3cos 2)q q +=

q +

= cos q = cos 30° =3

2[_ q = 30°]

4. From the following frequency distribution, find the median class :

Cost oflivingindex

1400-1550 1550-1700 1700-1850 1850-2000

Numberof

weeks8 15 21 8

[1]

Sol. The table of values is given as under :

Cost of livingindex

No. of weeks( f )

c.f.

1400-1550 8 81550-1700 15 231700-1850 21 441850-2000 8 52

Sf = 52N =

Here, N = 52

ÞN2

52 26,2

= =

Since C.F. which is just greater than 26 be 44 and corresponding class interval 1700-1850.\ Median class is 1700-1850. Ans.


Arundeep’s Solved Papers Mathematics 2015 (Term I)56SECTION – B

5. Show that 3 7 is an irrational number..[2]

Sol. Let us assume, to the contrary, that 3 7be a rational numberThat is, we can find co-prime a and

b (b ¹ 0) such that 3 7 ab

=

Rearranging, we get 73ab

=

Since 3, a and b are integers, 3ab can be

written in the form of ,pq so 3

ab is a

rational number and so 7 is rational.

But this contradicts the fact that 7 isirrational number.Hence our supposition is wrong.

So, we conclude that 3 7 is irrational.

Hence Proved.6. Explain why (17 × 5 × 11 × 3 × 2 + 2 × 11)

is a composite number ? [2]Sol. 17 × 5 × 11 × 3 × 2 + 2 × 11

= 17 × 5 × 3 × 22 + 22= 22 (17 × 5 + 3 × 1)= 22 (255 + 1) = 2 × 11 × 256

\ Given expression is divisible by 2, 11 and256, which means it has more than 2 primefactors.

\ (17 × 5 × 11 × 3 × 2 + 2 × 11) is a compositenumber. Ans.

7. Find whether the following pair of linearequations is consistent or inconsistent :3x + 2y = 86x – 4y = 9 [2]

Sol. Given lines are 3x + 2y – 8 = 0 ...(1)and 6x – 4y – 9 = 0 ...(2)

on comparing (1) and eqn. (2) with a1 x + b1y + c1 = 0 and a2x + b2 y + c2 = 0, we havea1 = 3 ; b1 = 2 ; c1 = –8a2 = 6 ; b2 = –4 ; c2 = –9

Here,1

2

3 1 ,6 2

aa

= = 1

2

2 14 2

bb

-= =-

\ 1 1

2 2,

a ba b¹ which will give a unique

solution.Hence, given pair of linear equations isconsistent. Ans.

8. X and Y are points on the sides AB andAC respectively of a triangle ABC such

that =AX 1 ,AB 4 AY = 2 cm and YC = 6 cm.

Find whether XY || BC or not. [2]

Sol. Given,AXAB

1=4

i.e., AX = 1K, AB = 4K(where K be constant)

\ BX = AB – AX = 4K – 1K = 3K

A

B C

6 cm

YX2 cm

Now,AXXB

1K 1=3K 3

=

And,AYYC

2 1=6 3

= ÞAXXB

AY=YC

\ XY || BC Ans.(By converse of Thales’ theorem)

9. Prove the following identity :3 3sin cos

sin cosq + qq + q

= 1 – sin q · cos q [2]



Sol. L.H.S.3 3sin cos

sin cosq + q=q + q

2 2(sin cos ) (sin cos sin ·cos )(sin cos )

q + q q + q - q q=q + q [_ a3 + b3 = (a + b) (a2 + b2 – ab)]

= 1 – sin q · cos q = R.H.S. [Q sin2 q + cos2 q = 1] Hence Proved.10. Show that the mode of the series obtained by combining the two series S1 and S2 given

below is different from that of S1 and S2 taken separately :S1 : 3, 5, 8, 8, 9, 12, 13, 9, 9S2 : 7, 4, 7, 8, 7, 8, 13 [2]

Sol. Mode of S1 series = 9 [_ observation 9 repeated max. no. of times i.e. 3 times]Mode of S2 series = 7 [_ observation 7 repeated max no. of times i.e. 3 times]After combining S1 and S2, the new series will be ; 3, 5, 8, 8, 9, 12, 13, 9, 9, 7, 4, 7, 8, 7, 8, 13.Mode of combined series = 8 (maximum times i.e. 4 times)Mode of (S1, S2) is different from mode of S1 and mode of S2 separately. Hence Proved.

SECTION – C

11. The length, breadth and height of a room are 8 m 50 cm, 6 m 25 cm and 4 m 75 cmrespectively. Find the length of the longest rod that can measure the dimensions of theroom exactly. [3]

Sol. To find the length of the longest rod that can measure the dimensions of the room exactly, wehave to find HCF of length, breadth and heightGiven length of room L = 8 m 50 cm = 850 cm= 21 × 52 × 17Breadth of room B = 6 m 25 cm = 625 cm = 54

Height of room H = 4 m 75 cm = 475 cm = 52 × 19\ HCF of L, B and H is 52 = 25 cm

[HCF of given numbers be the product of smallest power of each common prime factors in givennumbers]

\ Length of the longest rod = 25 cm Ans.12. Solve by elimination :

3x – y = 72x + 5y + 1 = 0 [3]

Sol. Given equations are : 3x – y = 7 ...(i)2x + 5y = – 1 ...(ii)Multiplying equation (i) by 5 and solving it with equation (ii), we get2x + 5y = – 115x – 5y = 3517x = 34 (on Adding)

Þ x34 217

= =


Arundeep’s Solved Papers Mathematics 2015 (Term I)58Putting the value of x in (i), we have3 (2) – y = 7

Þ 6 – y = 7Þ – y = 7 – 6 Þ y = – 1\ x = 2, y = – 1 Ans.

13. Find a quadratic polynomial, the sumand product of whose zeroes are 0 and

35- respectively. Hence find the zeroes.

[3]Sol. Given sum of zeroes = 0

and product of zeroes =3

5-

Required quadratic polynomial = f(x)= x2 – (Sum of zeroes) x + Product of zeroes

2 23 3(0)5 5

x x x-æ ö= - + = -ç ÷è ø

22 3( )

5x

æ ö= - ç ÷

è ø

3 35 5

x xæ ö æ ö= - +ç ÷ ç ÷è ø è ø

2 2By applying( ) ( ) ( )a b a b a bé ùê ú- = + -ë û

Zeroes are given by putting f(x) = 0

35

xæ ö-ç ÷ç ÷è ø

3 05

xæ ö

+ =ç ÷ç ÷è ø

Þ x35

= or35

x = -

Þ x3 55 5

= ´ or3 55 5

x = - ´

Þ155

x = or15

5x -= Ans.

14. The sum of the digits of a two digitnumber is 8 and the difference betweenthe number and that formed by reversingthe digits is 18. Find the number. [3]

Sol. Let unit digit = xand Tens digit = y

So, original number = unit digit + 10 × tens digit= x + 10yAccording to question, we haveSum of digits of required number = 8so, x + y = 8 ...(i)On reversing the digits, unit digit = yTens digit = xso, New number = 10x + yAccording to question, we have

Þ x + 10y – (10x + y) = 18Þ x + 10y – 10x – y = 18Þ 9y – 9x = 18Þ y – x = 2 ...(ii)

By adding eq. (i) and (ii)

2y = 10 Þ y 102

=

Þ y = 5Put the value of y in eq. (i) ; we have

Þ x + 5 = 8 Þ x = 8 – 5Þ x = 3\ Required original number = 10y + x

= 10 × 5 + 3 = 50 + 3 = 53 Ans.15. In given figure, EB ^ AC, BG ^ AE and

CF ^ AE. Prove that :

AB

C

D

F

G

E


Arundeep’s Solved Papers Mathematics 2015 (Term I)59(i) DABG ~ DDCB

(ii)BC BE=BD BA [3]

Sol.

AB

C

D

F

G

E

1 6 5 4

32

7

Given : EB ^ AC, BG ^ AE and CF ^ AETo prove :(i) DABG ~ DDCB

(ii)BCBD

BEBA

=

Proof : (i) In DABG and DDCB, BG || CFas corresponding angles are equal.Ð2 = Ð5 [Each 90º]Ð6 = Ð4 [Corresponding angles]

\ DABG ~ DDCB Hence Proved.[By AA axiom of similarity]

Ð1 = Ð3 [CPCT](ii) In DABE and DDBCÐ1 = Ð3 [Proved above]ÐABE = Ð5

[Each is 90º, EB ^ AC (Given]DABE ~ DDBC[By AA axiom of similarity]In similar triangles, corresponding sides areproportional.

\BCBD

BE=BA Hence Proved.

16. In triangle ABC, if AP ^ BC andAC2 = BC2 – AB2, then prove thatPA2 = PB × CP. [3]

Sol. AC2 = BC2 – AB2 [Given]AC2 + AB2 = BC2

\ ÐBAC = 90º

C

BA

P

[By converse of Pythagoras theorem]DAPB ~ DCPA[If a perpendicular is drawn from the vertexof the right angle of a triangle to thehypotenuse then triangles on both sides ofthe perpendicular are similar to the wholetriangle and to each other]

ÞAPCP

PB=PA

[In similar triangle, corresponding sides areproportional]

Þ PA2 = PB · CP Hence Proved.

17. If sin q12= ,13 0º < q < 90º, find the value

of :

2 2

2sin cos 12 sin ·cos tanq - q ´q q q [3]

Sol. Given, sin q1213

=

Þ sin q =PH

12=13

Let, P = 12K, H = 13KP2 + B2 = H2 [Pythagoras theorem]

Þ (12K)2 + B2 = (13K)2

144K2 + B2 = 169K2


Arundeep’s Solved Papers Mathematics 2015 (Term I)60B2 = 169K2 – 144K2 = 25K2 Þ B = 5K

\ cos qB 5K 5H 13K 13

= = =

\ tan qP 12K 12B 5K 5

= = =

Now,2 2

2sin cos 1

2sin ·cos tanq - q ´q q q

2 2

2

12 5113 13

12 5 12213 13 5

æ ö æ ö-ç ÷ ç ÷è ø è ø= ´æ ö æ ö æ öç ÷ ç ÷ ç ÷è ø è ø è ø

144 2525169

120 144169

-

= ´

119 25 595120 144 3456= ´ = Ans.

Aliter : Given sin q =1213 \ cos q =

22 12 1441 sin 1 1

13 169æ ö- q = - = -ç ÷è ø

25 5169 13

= =

\ tan q =sin 12 / 13 12cos 5 / 13 5q = =q

Thus, given expression =

2 2

2

12 5 144 251 2513 13 169

12 5 120 14412213 13 1695

æ ö æ ö --ç ÷ ç ÷è ø è ø ´ = ´æ ö´ ´ç ÷è ø

=119 25 595120 144 3456´ =

18. If sec q + tan q = p, prove that sin qpp

2

211-=+ [3]

Sol. R.H.S.2

211

pp-=+

2

2(sec tan ) 1(sec tan ) 1q + q -=q + q +

2 2

2 2sec tan 2sec tan 1sec tan 2sec tan 1q + q + q q -=q + + q q +

[By (a + b)2 = a2 + b2 + 2ab]

2 2

2 2(sec 1) tan 2sec tansec (1 tan ) 2sec tan

q - + q + q q=q + + q + q q

H

B

P



2 2

2 2tan tan 2 sec tansec sec 2sec tanq + q + q q=q + q + q q

2 2

2 2sec 1 tan

and sec 1 tané ùq - = qê úq = + që û

Q

2

22 tan 2 sec tan2sec 2sec tan

q + q q=q + q q

2 tan (tan sec ) tan2sec (sec tan ) sec

q q + q q= =q q + q q

sincos

1cos

qq=

q= sin q = L.H.S.Aliter : Given sec q + tan q = p ...(1)

\ sec2q – tan2q = 1Þ (sec q – tan q) (sec q + tan q) = 1

Þ sec q – tan q =1p ....(2)

on adding (1) and (2) ; we have

2 sec q = p +1p =

2 1pp+Þ cos q = 2

21

pp +

\ sin q =2

2221 cos 1

1p

pæ ö

- q = - ç ÷+è ø

=( )

2 2 2 2 2 2

2 2 2 22

( 1) 4 ( 1) 1( 1) 11

p p p pp pp

+ - - -= =+ ++

Hence Proved.19. Find the mean of the following distribution by Assumed Mean Method : [3]

Class interval Frequency10-20 820-30 730-40 1240-50 2350-60 11

1386

60-7070-8080-90

90-100 12


Arundeep’s Solved Papers Mathematics 2015 (Term I)62Sol. The table of values is given as under :

Classinterval

Frequency( )fi

xd =

ii xi –55

f di i

8 15 -40 -32020-30 7 25 -30 -21030-40 12 35 -20 -24040-50 23 45 -10 -23050-60 11 55 0 060-70 13 65 10 13070-80 8 75 20 16080-90 6 85 30 180

90-100 12 95 40 480Sfi Sf di i = –50

where A = 5510-20

Mean 50A 55100

i i

i

f df

S -æ ö= + = + ç ÷S è ø

5055100

= - = 55 – 0.5 = 54.5 Ans.

20. The average score of boys in the examination of a school is 71 and that of the girls is 73.The average score of the school in the examination is 71.8. Find the ratio of number ofboys in the number of girls who appeared in the examination. [3]

Sol. Let the number of boys = n1and number of girls = n2

Average boys’ score = 71 1X= (Let)

Average girls’ score = 73 2X= (Let)

Combined mean 1 1 2 2

1 2

X Xn nn n

+=

+

Þ 71.8 1 2

1 2

(71) (73)n nn n

+=

+Þ 71n1 + 73n2 = 71.8n1 + 71.8n2

Þ 71n1 – 71.8n1 = 71.8n2 – 73n2 Þ – 0.8n1 = – 1.2n2

Þ1

2

nn

1.20.8

= Þ1

2

nn

32

= Þ n1 : n2 = 3 : 2

\ No. of boys : No. of girls = 3 : 2. Ans.


Arundeep’s Solved Papers Mathematics 2015 (Term I)63SECTION – D

21. Find HCF of numbers 134791, 6341 and6339 by Euclid’s division algorithm. [4]

Sol. First we find HCF of 6339 and 6341 byEuclid’s division method

63416339 1

2

1

3169

2

63396339

220

6321312

1918

6341 > 6339 Þ 6341 = 6339 × 1 + 2Also, 6339 = 2 × 3169 + 12 = 1 × 2 + 0

\ HCF of 6341 and 6339 is 1.Now, we find the HCF of 134791 and 1134791 = 1 × 134791 + 0

\ HCF of 134791 and 1 is 1.Hence, HCF of given three numbers is 1.

Ans.22. Draw the graph of the following pair of

linear equations :x + 3y = 6 and 2x – 3y = 2Find the ratio of the areas of the twotriangles formed by first line, x = 0, y = 0and second line, x = 0, y = 0. [4]

Sol. First Line Second Linex + 3y = 6 2x – 3y = 12Þ x = 6 – 3y Þ 2x = 12 + 3y

Þ x12 3

2y+=

6 3 00 1 2

xy

6 3 00 2 4

xy - -

draw the points (6, 0), (3, 1), (0, 2) on graphpaper and join them by straight line to getthe graph of x + 3y = 6 and draw the points(6, 0), (3, –2), (0, –4) on graph paper andjoin them by line to give the graph of2x – 3y = 2.

1 2 3 4 5 6–1–2–3–4–5

1

2

3

4

–1

–2

–3

–4

–5

X' X

Y'

Y

B(0, 2)(3, 1)

A(6, 0)

x y+ 3 = 6

(3, –2)

C (0, –4)

2– 3

= 12

xy

O

Since, area of triangle

12

= × base × corresponding altitude

\Area of AOBArea of AOC

DD

1 OA OB21 OA OC2

´ ´=´ ´

=OBOC

2 1=4 2

=

\ Required ratio = 1 : 2 Ans.


Arundeep’s Solved Papers Mathematics 2015 (Term I)6423. If the polynomial (x4 + 2x3 + 8x2 + 12x + 18) is divided by another polynomial

(x2 + 5), the remainder comes out to be (px + q), find the values of p and q. [4]

Sol. By long division method, we have\ Remainder = 2x + 3

Also given remainder = px + qi.e., px + q = 2x + 3

\ p = 2, q = 3 Ans.

24. What must be subtracted from p (x) = 8x4 + 14x3 – 2x2 + 8x – 12 so that 4x2 + 3x – 2 isfactor of p (x) ? This question was given to group of students for working together. Do youthink teacher should promote group work ? [4]

Sol. By long division method, we haveFor 4x2 + 3x – 2 be a factor of p(x)

\ remainder must be equal to 0.Thus, polynomial to be subtracted is (15x – 14).

Value : Yes, as it increases confidence and team spirit among students. Ans.25. Prove “If a line is drawn parallel to one side of a triangle to intersect the other two sides in

distinct points, the other two sides are divided in the same ratio.” [4]Sol. Given, In DABC, DE || BC.

To prove :ADDB

AE=BC

Construction : Draw EM ^ AB and DN ^ AC. Join B to E and C to D.Proof : In DADE and DBDE, we have

ar ( ADE)ar ( BDE)DD

1 AD EM AD21 DBDB EM2

´ ´= =´ ´

...(i)

x x x2 4 3 2

2

8 1 2 1 82 3

+ + + + ++ +

+ x4 +5x2

– –2x3 + 3x2 + 12 x + 18

+ 2x3 + 10x– –

3x 2 + 2x + 18+ 3x 2 + 15– –

2 x + 3

4 3 2 8 14 2 8 122 2 1

3 2

2

x x x x x xx x

+ – + – + –+ -

8 +x4 6x3 – 4 x 2

– – +8x3 + 2x2 + 8x – 12

+ 8x3 + 6x2 – 4 x– – +

– 4x2 + 12 x –12– 4x2 –3x + 2+ + –

15x – 14

A

M N

ED

B C



[Area of D12

= × base × corresponding

altitude]In DADE and DCDE, we have

ar ( ADE)ar ( CDE)DD

1 AE DN AE21 ECEC DN2

´ ´= =´ ´

...(ii)

Since, DE || BC [Given]\ ar (DBDE) = ar (DCDE) ...(iii)

[Ds on the same base and between the sameparallel sides are equal in area]From eq. (i), (ii) and (iii) ; we have

ADDB

AE=EC Hence Proved.

26. In the given figure, AD = 3 cm, AE = 5 cm,BD = 4 cm, CE = 4 cm, CF = 2 cm, BF =2.5 cm, then find the pair of parallel lineand hence their lengths. [4]

A

3 cm 5 cm

ED

4 cm 4 cm

CB2.5 cm 2 cmF

Sol.ECEA

4=5

andCFFB

2 4=2.5 5

=

ÞECEA

CF=FB

In DABC, EF || AB[Converse of Thales’ Theorem]

Also,CECA

4 4=4 5 9

=+ ...(i)

andCFCB

2 2 4=2 2.5 4.5 9

= =+

ECEA

CF=CB

ÐECF = ÐACB [Common]DCFE ~ DCBA [SAS similarity]

ÞEFAB

CE=CA

[In similar D’s, corresponding sides areproportional]

ÞEF7

4=9 [Q AB = 3 + 4 = 7 cm]

\ EF28 cm9

= and AB = 7 cm Ans.

27. If tan (A + B) = 3 and tan (A – B)1= ,3

where 0 < A + B < 90º, A > B, find Aand B. Also calculatetan A . sin (A + B) + cos A . tan (A – B).

[4]Sol. Given,

tan (A + B) 3,= and tan (A – B)13

=

Þ tan (A + B) = tan 60ºÞ (A + B) = 60º ...(i)

and, tan (A – B) = tan 30º\ (A – B) = 30º ...(ii)

On adding eq. (i) and (ii) ; we have2A = 90º

Þ A90º 45º2

= =

From eq. (i), A + B = 60ºÞ 45º + B = 60ºÞ B = 15º\ A = 45º, B = 15º

Now,tan A . sin (A + B) + cos A . tan (A – B)= tan 45º . sin (60º) + cos 45º . tan (30º)



3 1 112 2 3

= ´ + ´ 3 1 6 3 62 2 66 6

= + ´ = + 3 3 66+= Ans.

28. Prove that :

(1 + cot A + tan A) · (sin A – cos A)3 3

2 2sec A cosec A=sec A ·cosec A

- [4]

Sol. L.H.S. = (1 + cot A + tan A) (sin A – cos A)

cos A sin A1sin A cos A

æ ö= + +ç ÷è ø (sin A – cos A)2 2sin A cos A cos A sin A

sin A.cos Aæ ö+ += ç ÷è ø

(sin A – cos A)A)

3 3sin A cos Asin A .cos A

-= [Using a3 – b3 = (a – b) (a2 + ab + b2)]

3 3

3 3 3 3

3 3

sin A cos Asin A.cos A sin A.cos A

sin A cos Asin A.cos A

-= [Dividing Num. and Deno. by sin3 A . cos3 A]

3 3

2 2sec A cosec Asec A.cosec A

-= = R.H.S. Hence Proved.

29. Prove the identity :

sin A cos A sin A cos Asin A cos A sin A cos A

+ -+- + 2

21 2cos A

=-

[4]

Sol. L.H.S.sin A cos A sin A cos Asin A cos A sin A cos A

+ -= +- +

2 2(sin A cos A) (sin A cos A)(sin A cos A) (sin A cos A)

+ + -=- +

2 2

2 2

2 2

sin A cos A 2sin A cos Asin A cos A 2sin A cos A

sin A cos A

+ ++ + -=

-

2 21 1

1 cos A cos A+=

- -[Q sin2 A + cos2 A = 1, sin2 A = 1 – cos2 A]

22 R.H.S.

1 2 cos A= =-

Hence Proved.

30. The following table gives the daily income of 50 workers of a factory. Draw both types(“less than type” and “greater than type”) ogives.



Daily income (in )` No. of workers100 – 120 12120 – 140 14140 – 160 8160 – 180 6180 – 200 10

[3]Sol.

Less than ogive More than ogive

Daily In-come (in )`

No. ofwork-

ers(c.f.)

Less than 120 12Less than 140 26Less than 160 34Less than 180 40Less than 200 50

Daily Income(in )`

No.of

work-ers

(c.f.)More than 100 50More than 120 38More than 140 24More than 160 16More than 180 10

For less than ogive : Plot the points (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50) ongraph paper and join them by free hand to gives the required less than ogive.For more than ogive : plot the points (100, 50), (120, 38), (140, 24), (160, 16), (180, 10) ongraph paper and join them by free hand to give the required more than ogive.

31. In a class test, marks obtained by 120 students are given in the following frequencydistribution. If it is given that mean is 59, find the missing frequencies x and y. [4]



Marks No. of students0 – 10 110 – 20 320 – 30 730 – 40 1040 – 50 1550 – 60 x60 – 70 970 – 80 2780 – 90 18

90 – 100 y

Sol. The table of values is given as under :

Marks

No. ofStudents

fixi di =

xi - 5510

f idi

0-10 1 5 –5 –5

–73

10-20 3 15 –4 –1220-30 7 25 –3 –2130-40 10 35 –2 –2040-50 15 45 –1 –15

117+

4

50-60 x A= 55 0 060-70 9 65 1 970-80 27 75 2 5480-90 18 85 3 54

90-100 y 95 4 4y

Sf =i 90+ x + y

f di i = 44 + 4y

Since, Sfi = 90 + x + y



But Sfi = 120 [Given]

\ 90 + x + y = 120

x = 120 – 90 – y = 30 – y ...(i)

Also, mean A i i

i

f dh

fS

= + ´S

[using step deviation method]

Þ 59 44 455 10120

y+æ ö= + ´ç ÷è ø[A = 55, h = 10, Sfi = 120]

Þ 59 – 554 (11 )

12y+= Þ 4 × 3 = 11 + y Þ y = 12 – 11 = 1

From eq. (i), x = 30 – 1 = 29

\ x = 29, y = 1 Ans.


1. If the quadratic equation2 2 5 15 0px px- + = has two equal

roots, then find the value of p.Sol. Given quadratic eqn. be

px2 – 2 5 px + 15 = 0

On comparing with ax2 + bx + c = 0

we have a = p : b = – 2 5 p ; c = 15For equal roots, D = 0

Þ b2 – 4ac = 0

Þ 2( 2 5) 4 15p- - ´ ´ = 0

Þ 20 – 60p = 0

Þ13

p =

2. In Figure given below a tower AB is 20m high and BC, its shadow on the

ground, is 20 3 m long. Find the Sun’ssaltitude.

Sol. Let Sun’s altitude = q = ÐACBIn right angled DABC, we have

\ tan qABBC

=

Þ tan q20 1

20 3 3= = = tan 30° Þ q = 30º

3. Two different dice are tossed together. Findthe probability that the product of the twonumbers on the top of the dice is 6.

Sol. Total number of outcomes = 36Here sample space = {(1, 1), (1, 2) .... (1, 6)

(2, 1), (2, 2) ......(2, 6)(3, 1), (3, 2) .... (3, 6)(4, 1), (4, 2) .... (4, 6)(5, 1), (5, 2) ..... (5, 6)(6, 1), (6, 2) ...... (6, 6)

Let A be the event such thatA = Product of the numbers on the top ofthe dice is 6.

\ Favourable outcomes (1, 6), (2, 3), (3, 2),(6, 1)

\ No. of favourable out comes = 4

\ Required probability = P (A)4 1

36 9= =

4. In Figure given below PQ is a chord of acircle with centre O and PT is a tangent.If ÐQPT = 60º, find ÐPRQ.

Sol. ÐXPQ + ÐQPT = 180º (linear pair)Þ ÐXPQ + 60º = 180ºÞ ÐXPQ = 120º

Now ÐXPQ = ÐPRQ [Angle made bya chord with a tangent is equal to anglesubtended by the chord in alternate segment]

\ ÐPRQ = 120º

MATHEMATICS 2015 TERM II (OUTSIDE DELHI)SET I


General Instructions – As usual.




5. In Figure given below two tangents RQand RP are drawn from an external pointR to the circle with centre O. If ÐPRQ =120º, then prove that OR = PR + RQ.

Sol. OR bisects ÐPRQ\ ÐPRO = ÐQRO = 60º

In right DOPR (Q OP ^ PR)

PROR = cos 60º

1=2

Þ OR = 2PR ...(i)

SimilarlyQROR

1=2

Þ OR = 2QR ...(ii)(i) + (ii), we get2OR = 2PR + 2QRThus, OR = PR + RQ

6. In Figure given below a triangle ABC isdrawn to circumscribe a circle of radius3 cm, such that the segments BD and DCare respectively of lengths 6 cm and 9 cm.If the area of DABC is 54 cm2, then findthe lengths of sides AB and AC.

Sol. Let AF = x cmQ AF = AE [tangents from A]

[Q tangents drawn from a point outside thecircle are of equal length.]

\ AE = xAlso BD = DF = 6 cmand CD = CE = 9 cm

\ AB = (6 + x) cm and AC = (9 + x) cm

Area DABC = Area DBOC + Area DCOA+ Area DAOB

Þ 5412

= BC × OD +12 AAC × OE +

12 AB ×OF

Þ 54 × 2 = 15 × 3 + (9 + x) × 3 + (6 + x) 3Þ 108 = 45 + 27 + 3x + 18 + 3xÞ 6x = 18 Þ x = 3\ AB = 6 + x = 6 + 3 = 9 cm

and AC = 9 + x = 9 + 3 = 12 cm


Arundeep’s Solved Papers Mathematics 2015 (Outside Delhi)727. Solve the following quadratic equation

for x :4x2 + 4bx – (a2 – b2) = 0

Sol. Given quadratic eqn. be,4x2 + 4bx – (a2 – b2) = 0On comparing with ax2 + bx + c = 0We have a = 4 ; b = 4b ; c = –(a2 – b2)

\ Discriminant D = b2 – 4acD = (4b)2 – 4 × 4 × [– (a2 – b2)]= 16b2 + 16a2 – 16b2 = 16a2

\ x =D

2b

a- ±

\ x24 16 4 4

2 4 8b a b a- ± - ±= =

´

4 ( )8 2b a b a- ± - ±= =

8. In an AP, if S5 + S7 = 167 and S10 = 235,then find the AP, where Sn denotes thesum of its first n terms.

Sol. Let Ist term of the AP = aand common difference = dand required A.P. be a, a + d, a + 2d, ....Now S5 + S7 = 167

{2 ( 1) }2nnS a n dé ù= + -ê úë û

Q

Þ5 7(2 4 ) (2 6 )2 2

a d a d+ + + = 167

Þ 5a + 10d + 7a + 21d = 167Þ 12a + 31d = 167 ...(i)

Also S10 = 235

Þ102 (2a + 9d) = 235 Þ 2a + 9d = 47

Þ 6 (2a + 9d) = 6 × 47Þ 12a + 54d = 282 ...(ii)

eqn. (i) – eqn. (ii) gives ;– 23d = – 115 Þ d = 5

When d =5, eqn. (i) becomes12a + 155 = 167 Þ a = 1Thus required AP is 1, 6, 11, ....

9. The points A (4, 7), B (p, 3) and C (7, 3)are the vertices of a right triangle, right-angled at B. Find the value of p.

Sol. Using distance formula, we have

Now AC 2 23 ( 4)= + - = 5

AB 2( 4) 16p= - +

C(7, 3) B( , 3)p

A(4, 7)

BC 2( 7) 0p= - +

Now AC2 = AB2 + BC2

(using pythagoras theorem)Þ 25 = (p – 4)2 + 16 + (p – 7)2

Þ 25 = p2 – 8p + 16 + 16 + p2 – 14p + 49Þ 2p2 – 22p + 56 = 0 Þ p2 – 11p + 28 = 0Þ p2 – 4p – 7p + 28 = 0Þ p(p – 4) – 7 (p – 4) = 0Þ (p – 4) (p – 7) = 0 Þ p = 4 or p = 7

If p = 7 then B = (7, 3)It coincide with C \ p ¹ 7Hence, p = 4

10. Find the relation between x and y if thepoints A (x, y), B (– 5, 7) and C (– 4, 5)are collinear.

Sol. Q A, B and C are collinearThus, A, B and C lies on same line

\ area of DABC = 0\ x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) = 0Þ x (7 – 5) + (– 5) (5 – y) + (– 4) (y – 7) = 0Þ 2x – 25 + 5y – 4y + 28 = 0


Arundeep’s Solved Papers Mathematics 2015 (Outside Delhi)73Þ 2x + y + 3 = 0

gives the required relation.11. The 14th term of an AP is twice its 8th

term. If its 6th term is – 8, then find thesum of its first 20 terms.

Sol. Let 1st term of AP = a andcommon difference = dAccording to given condition, we havea14 = 2a8

Þ a + 13d = 2 (a + 7d) Þ a = – d[_ an = a + (n – 1) d]

Also a6 = 8 Þ a + 5d = – 8Þ – d + 5d = – 8 Þ 4d = – 8 Þ d = – 2\ a = 2

We know that Sn =2n

{2a + (n – 1)d}

S2020 (2 2 19 ( 2)2

= ´ + ´ -

= 10 × (– 34) = – 34012. Solve for x :

23 2 2 2 3 0x x- - =

Sol. Given eqn. be 23 2 2 2 3x x- - = 0

Þ 23 3 2 2 2 3x x x- + - = 0

Þ 3 ( 6) 2 ( 6)x x x- + - = 0

Þ ( 6) ( 3 2)x x- + = 0

either x – 6 = 0 or 3x + 2 = 0

26,3

x -=

13. The angle of elevation of an aeroplanefrom a point A on the ground is 60º. Aftera flight of 15 seconds, the angle ofelevation changes to 30º. If the aeroplaneis flying at a constant height of

1500 3 m, find the speed of the plane inkm/hr.

Sol. Let plane is at P. After 15 seconds it reaches

at Q\ Distance covered in 15 seconds = PQ

In right DPBA,PBAB = tan 60º

1500 3AB 3= Þ AB = x = 1500 m

In right DQCA, we have

QCAC = tan 30º Þ

1500 3x y+

13

=

Þ x + y = 4500 m\ BC = AC – AB = 3000 m

Also PQ = BC\ PQ = 3000 m

Thus required speedDistance covered

Time taken=

3000 m/s15

= = 200 m/s

3600200 km/hr1000

= ´ = 720 km/hr

[_ 1 km = 1000 m and 1 hr = 3600s]

PQ

C

14. If the coordinates of points A and B are(– 2, – 2) and (2, – 4) respectively, findthe coordinates of P such that

AP3= AB,7 where P lies on the line

segment AB.

Sol. Given AP3 AB7

=



Þ AP =37 (AP + PB)

Þ 7AP = 3AP + 3PBÞ 4AP = 3PB Þ AP : PB = 3 : 4

A(–2, –2) P B(2, –4)3 : 4

\ P divides AB in the ratio 3 : 4Then by section formula, we have

\ x coordinates of P

3 2 4 ( 2) 23 4 7

´ + ´ - -= =+

and y coordinate of P

3 ( 4) 4 ( 2) 203 4 7

´ - + ´ - -= =+

\ Required coordinates of P are2 20,

7 7- -æ ö

ç ÷è ø15. The probability of selecting a red ball at

random from a jar that contains only

red, blue and orange balls is1 .4 The

probability of selecting a blue ball at

random from the same jar is1 .3 If the

jar contains 10 orange balls, find the totalnumber of balls in the jar.

Sol. Let number of red balls in the jar = xand number of blue balls in the jar = y

\ Total number of balls in the jar = x + y = 10[We have given no. of orange balls in Jar = 10]

Given probability of selecting red ball14

=

Also prob. of selecting red ball

= 10x

x y+ +

Þ 10x

x y+ +14

= Þ 4x = x + y + 10

Þ 3x – y = 10 ...(i)Prob. of drawing a blue ball from jar

= 10y

x y+ +Also given prob. of drawing a blue ball

from jar =13

Also13 10

yx y

=+ +

Þ x + y + 10 = 3y Þ x = 2y – 10 ...(ii)Putting x = 2y – 10 in equation (i) ; we have3 (2y – 10) – y = 10

Þ 6y – 30 – y = 10 Þ 5y Þ y = 8When y = 8, equation (ii) becomesx = 2 × 8 – 10 Þ x = 6

\ Total number of balls in the jar = 6 + 8 + 10 = 2416. Find the area of the minor segment of a

circle of radius 14 cm, when its centralangle is 60º. Also find the area of thecorresponding major segment.

22Use π =7

é ùê úë û

Sol. In DAOB, ÐAOB = 60ºAlso AO = BO = radii of circleSince equal sides have equal angle oppositeto it.

\ ÐOAB = ÐOBAAlso ÐOAB + ÐOBA + 60º = 180°

\ 2ÐOAB = 120° Þ ÐOBA = ÐOAB = 60°\ DAOB is an equilateral D.

\ Area of equilateral DAOB3 14 14

4= ´ ´

249 3 cm=and Area of sector AOB

= pr2× 360q

°60 14 14

360°= ´ p ´ ´°

1 22 14 146 7

= ´ ´ ´ 2308 cm3

=



A B

O

60°14cm

14 cm

Minor segment

Major segment

Area of minor segment

2308 49 3 cm3

æ ö= -ç ÷è ø

Also, Area of circle = pr2 22 14 147

= ´ ´

= 616 cm2

Area of major segment

2308616 49 3 cm3

é ùæ ö= - -ç ÷ê úè øë û

= 21540 49 3 cm3

æ ö+ç ÷è ø17. Due to sudden floods, some welfare

associations jointly requested thegovernment to get 100 tents fixedimmediately and offered to contribute50% of the cost. If the lower part of eachtent is of the form of a cylinder ofdiameter 4.2 m and height 4 m with theconical upper part of same diameter butof height 2.8 m, and the canvas to be usedcosts ` 100 per sq. m, find the amount,the associations will have to pay. Whatvalues are shown by these associations ?

22Use π =7

é ùê úë û

Sol. Given Height of cylinder = H = 4 mRadius of cylinder = R = 2.1 m

\ Curved surface area of cylinder = 2pRH

222 2.1 47

= ´ ´ ´ = 52.8 m2

Radius of cone = R = 2.1 m,height of cone h = 2.8 m.

4.2 m

2.8 m

4 m

Let slant height of conical part = l

\ l 2 2(2.1) (2.8)= + = 3.5 m

\ Curved surface area of cone = prl

22 2.1 3.57

= ´ ´ = 23.1 m2

Area of canvas required for one tent= 52.8 + 23.1 = 75.9 m2

Canvas required for 100 tents = 100 × 75.9 = 7590 m2

\ Total cost of canvas = ` (7590 × 100) = ` 759000

Thus, amount to be paid by association

50 759000100

= ´ = ` 379500

So, values promoted are care for the society.18. A hemispherical bowl of internal

diameter 36 cm contains liquid. Thisliquid is filled into 72 cylindrical bottlesof diameter 6 cm. Find the height of theeach bottle, if 10% liquid is wasted inthis transfer.

Sol. Radius of hemispherical bowl = R = 18 cm

Volume of liquid in bowl 32 R3

= p



23

= × p × 18 × 18 × 18 = 3888 p cm3

\ Quantity of Liquid wasted = 10% of total

volume of liquid 310 3888 cm100

= ´ p

\ Volume of Liquid transferred into bottles

= 3888p388810

- p

334992 cm10

= p

Given Radius of bottle = r = 3 cmLet height of each bottle = x cm

\ Volume of one bottle = pr2h= p × 3 × 3 × x= 9px cm3

Thus, Volume of 72 bottles = 72 × 9px cm3

= 648px cm3

according to given condition, we have

648px 3499210

= p

x34992 5.4 cm

10 648= =

´\ required height of each bottle be 5.4 cm.

19. A cubical block of side 10 cm issurmounted by a hemisphere. What isthe largest diameter that the hemispherecan have ? Find the cost of painting thetotal surface area of the solid so formed,at the rate of ` 5 per 100 sq. cm.

[Use p = 3.14]Sol. Largest diameter of hemisphere = side of

cubical block = 10 cm\ Radius of hemisphere = 5 cm

Total surface area of the solid = Surfacearea of cube + Curved surface area ofhemisphere – Area of base of hemisphere= 6r2 + 2pr2 – pr2

= 6 × 102 + 2 × 3.14 × 5 × 5 – 3.14 × 5 × 5= 600 + 78.5 = 678.5 cm2

\ Total cost of painting the solid @ the rate

of Rs. per 100 eqn. cm = `678.5 5

100´

= ` 33.9220. 504 cones, each of diameter 3.5 cm and

height 3 cm, are melted and recast into ametallic sphere. Find the diameter of thesphere and hence find its surface area.

22Use π =7

é ùê úë û

Sol. Given radius of cone r =3.52 cm

and height of cone = h = 3 cm

\ Volume of one cone 213

r h= ´ p

31 3.5 3.5 3 cm3 2 2

= p ´ ´ ´ 312.25 cm4

= p

\ Volume of 504 cones

312.25504 cm4

= ´ p

= 1543.5 p cm3

Since it is given that 504 cones are meltedand recast into a metallic sphere let R bethe radius of sphere

Then volume of sphere = 34 R3p

\ Volume of sphere = 1543.5 p cm3

34 R3

p = 1543.5 p

Þ R3 1543.5 34

´= = 1157.625

R 3 1157.625 10.5 cm= =\ Surface area of sphere = 4pR2



224 10.5 10.5 cm7

= ´ ´ ´ 2 = 1386 cm2

21. The diagonal of a rectangular field is16 metres more than the shorter side. Ifthe longer side is 14 metres more thanthe shorter side, then find the lengths ofthe sides of the field.

Sol. Let shorter side of rectangular field = x cm\ Diagonal of rectangular field = (x + 16) m

and longer side of rectangular field= (x + 14) m

Then by pythagoras theorem, we haveNow (x + 16)2 = (x + 14)2 + x2

Þ x2 + 32x + 256 = x2 + 28x + 196 + x2

Þ x2 – 4x – 60 = 0Þ x2 – 10x + 6x – 60 = 0Þ x(x – 10) + 6 (x – 10) = 0Þ (x – 10) (x + 6) = 0Þ x = 10 or x = – 6

Rejecting x = – 6,Since length of side of rectangular fieldcan’t be negativeThus x = 10

\ Shorter side of rectangular field = 10 m,length of diagonal = 26 mand longer side of rectanglar field = 24 m

22. Find the 60th term of the AP 8, 10, 12,..., if it has a total of 60 terms and hencefind the sum of its last 10 terms.

Sol. Given, AP is 8, 10, 12, ...Here, a = 8, d = 2a60 = a + 59d [_ an = a + (n – 1) d]= 8 + 59 × 2= 8 + 118 = 126

[we know that Sn = 2n

(a + an)]

\ S60 6060 ( )2

a a= + = 30 (8 + 126)

= 30 × 134 = 4020

and S5050 (2 49 )2

a d= + = 25 (16 + 49 × 2)

= 25 (114) = 2850\ Sum of last 10 terms = S60 – S50

= 4020 – 2850= 1170

23. A train travels at a certain average speedfor a distance of 54 km and then travelsa distance of 63 km at an average speedof 6 km/h more than the first speed. If ittakes 3 hours to complete the totaljourney, what is its first speed ?

Sol. Let average speed of the train = x kmDistance covered by train = 54 km

\ Time taken by train to covered a distance

of 54 km54 hrx

=

Also distance to be covered by train = 63 kmand speed = (x + 6) km/hr

\ Time taken by train to covered a distance

of 63 km63 hr

6x=

+According to given condition, we have

54 636x x

++ = 3

Þ54 ( 6) 63

( 6)x xx x+ ++ = 3

Þ 54x + 324 + 63x = 3 (x2 + 6x)Þ 117x + 324 = 3x2 + 18xÞ 3x2 – 99x – 324 = 0Þ x2 – 33x – 108 = 0Þ x2 – 36x + 3x – 108 = 0Þ (x – 36) (x + 3) = 0Þ x(x – 36) +3 (x – 36) = 0


Arundeep’s Solved Papers Mathematics 2015 (Outside Delhi)78Þ x = 36 or x = – 3\ First speed of train = 36 km/hr

24. Prove that the lengths of the tangentsdrawn from an external point to a circleare equal.

Sol. Given : A circle C (O, r), P is a point outsidethe circle and PA and PB are tangents to acircle.To Prove : PA = PBConstruction : Draw OA, OB and OP.Proof : Consider triangles OAP and OBP.ÐOAP = ÐOBP = 90º ...(i)

O

B

A

P

[Radius is perpendicular to the tangent atthe point of contact]OA = OB (radii) ...(ii)OP is common ...(iii)

\ DOAP@ DOBP (RHS axiom of congruency)[from (i), (ii), (iii)]

Þ AP = BP (cpct)25. Prove that the tangent drawn at the mid-

point of an arc of a circle is parallel tothe chord joining the end points of thearc.

Sol. Given : AB is arc of the circle C (O, r), P ismid-point of arc AB and XY is tangent tothe circle at P.To Prove : AB || XYJoin OA and OBHereÐAOP = ÐBOP (angle subtended byequal arc)OA = OB, OC = OC

\ DACO @ DBCO(SAS axiom of congurency)

Þ AC = BC (CPCT)Þ OC ^ AB(line joining mid-point of chord

with centre of the circle is perpendicular tothe chord)Also ÐOPY = 90º

Þ ÐOCB=ÐOPYthese are corresponding angles\ AB || XY

26. Construct a DABC in which AB = 6 cm,ÐA = 30º and ÐB = 60º. Constructanother DAB¢C¢ similar to DABC withbase AB¢ = 8 cm.

Sol.

Steps of construction :(i) draw in line segment AB = 6 cm

(ii) at A, construct ÐBAQ = 30°(iii) at B, construct ÐABP = 60°(iv) Let ÐABP and ÐBAQ intersects at C

Join CA and CB, to get the DABC.(v) Produce AB to B¢ s.t. AB¢ = 8 cm

(vi) at B¢, construct ÐAB¢Z = 60°making ÐBAQ at C¢ s.t. B¢C¢ || BCThen DAB¢C¢ is similar to DABC.


Arundeep’s Solved Papers Mathematics 2015 (Outside Delhi)7927. At a point A, 20 metres above the level

of water in a lake, the angle of elevationof a cloud is 30º. The angle of depressionof the reflection of the cloud in the lake,at A is 60º. Find the distance of the cloudfrom A.

R

x

ED

x

Sol. Let C is cloud and R is its reflection.ÐDAC = 30º, ÐDAR = 60º, let CD = x m

\ Height of the cloud above the lake= (x + 20) m

\ ER = (20 + x) m.Now In right DADC, we have

CDAD = tan 30º

Þ ADx 1

3=

Þ AD 3x= ...(i)In right DADR, we have

DRAD = tan 60º

Þ DE ERAD+ 3=

Þ 20 203

xx

+ +3= (using (i))

Þ 40 + x = 3xÞ x = 20 m

Now In right DADC, we have

ACCD = cosec 30º

ÞAC20 = 2

Þ AC = 40 m\ Required distance of the cloud from A = 40m

28. A card is drawn at random from a well-shuffled deck of playing cards. Find theprobability that the card drawn is(i) a card of spade or an ace.(ii) a black king.(iii) neither a jack nor a king(iv) either a king or a queen

Sol. Total no. of outcomes = 52A = Card is spade or an aceCards favourable to A = 13 + 3 = 16[Since there are 13 spade cards and 3 acecards, 1 ace of spade is already counted inspade cards.]

\ Required prob. P (A)16 4=52 13

=

(ii) B = Card is black king\ No. of black kings = 2 = total no. of

favourable cases

\ P (B)2 1

52 26= =

(iii) C = Card is neither a jack nor a kingNo. of favourable cards to C = 52 – 4 – 4 = 44

\ Required prob. P (C)44 1152 13

= =

(iv) D = Card is either a king or a queenNo. of cards favourable to event D = 4 + 4 = 8

\ Required prob. P (D)8 2

52 13= =


Arundeep’s Solved Papers Mathematics 2015 (Outside Delhi)8029. Find the values of k so that the area of the triangle with vertices (1, – 1), (–4, 2k)

and (– k, – 5) is 24 sq. units.Sol. Area of D = 24

Þ12 | x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) | = 24 | 1 (2k + 5) – 4 (– 5 + 1) – k (– 1 – 2k) | = 48

Þ | 2k + 5 + 16 + k + 2k2 | = 48 Þ | 2k2 + 3k + 21 | = 48Þ 2k2 + 3k + 21 = ± 48 Þ 2k2 + 3k + 21 = 48 or 2k2 + 3k + 21 = – 48Þ 2k2 + 3k – 27 = 0 or 2k2 + 3k + 69 = 0

Case - I : 2k2 + 3k – 27 = 0Þ 2k2 + 9k – 6k – 27 = 0Þ k (2k + 9) – 3 (2k + 9) = 0 Þ (2k + 9) (k – 3) = 0

Þ k9

2-= or k = 3 Þ k

92

-= or k = 3

Case - II : 2k2 + 3k + 21 + 48 = 0Þ 2k2 + 3k + 69 = 0

on comparing with ax2 + bx + c = 0we have a = 2 ; b = 3 ; c = 69Here, D = b2 – 4ac = 32 – 4 × 2 × 69 = 9 – 552 = – 543 < 0

\ given quadratic eqn. has no. soln. in this case.

Thus, k =9

2-

or 3

30. In Figure given below PQRS is a square lawn with sidePQ = 42 metres. Two circular flower beds are there on thesides PS and QR with centre at O, the intersection of itsdiagonals. Find the total area of the two flower beds(shaded parts).

Sol. Here PR2 = PQ2 + QR2 [using pythagoras theorem]PR2 = (42)2 + (42)2

Þ PR 42 2 m=

Þ PO42 2 21 2 m

2= = [Since diagonals of square bisect each other at right angles]

Area of sector POS 290º (21 2)360º

= ´ p

1 22 21 21 24 7

= ´ ´ ´ ´ = 693 m2

\ Area of DPOS12

= PO × OS (Q PO ^ OS)



1 21 2 21 22

= ´ ´ = 441 m2

\ Area of one flower bed = 693 – 441 = 252 m2

Thus, Area of two flower bed = 2 × 252 =504 m2

31. From each end of a solid metal cylinder,metal was scoped out in hemisphericalform of same diameter. The height of thecylinder is 10 cm and its base is of radius4.2 cm. The rest of the cylinder is meltedand converted into a cylindrical wire of1.4 cm thickness. Find the length of the

wire.22Use π =7

é ùê úë û

Sol. Given Radius of hemisphere = 4.2 cm = r

\ Volume of hemisphere 323

r= p

3 32 (4.2) cm3

= p ´

= 49.392 p cm3

\ Volume of 2 hemispheres= 2 × 49.392 p cm3

= 98.784 p cm3

Given Height of cylinder = 10 cm = hand Radius of cylinder = 4.2 cm = r

\ Volume of cylinder = pr2h= p × (4.2)2 × 10 = 176.4 p

\ Volume of metal left= 176.4 p – 98.784 p= 77.616 p cm2

Given Radius of wire = 0.7 cmlet length of wire = x

\ Volume of wire = p × 0.7 × 0.7 × x= 0.49px cm3

Since the rest of the cylinder in melted andconverted into cylinderical wire.

Þ 0.49px = 77.616 pÞ x = 158.4 cm\ Required length of wire = 158.4 cm

10. If A (4, 3), B (– 1, y) and C (3, 4) are thevertices of right triangle ABC, right-angled at A, then find the value of y.

Sol. \ BC2 = (3 + 1)2 + (4 – y)2 = 16 + (4 – y)2

AB2 = (– 1 – 4)2 + (y – 3)2 = 25 + (y – 3)2

and AC2 = (4 – 3)2 + (3 – 4)2 = 2Also BC2 = AB2 + AC2

[using pythagoras theorem]Þ 16 + (4 – y)2 = 25 + (y – 3)2 + 2Þ 16 + 16 + y2 – 8y = 25 + y2 – 6y + 9 + 2Þ – 2y = 4Þ y = – 2

18. All the vertices of a rhombus lie on acircle. Find the area of the rhombus, Ifthe area of the circle is 1256 cm2.

[Use p = 3.14]Sol. Diagonal of a rhombus are perpendicular

bisector of each other.\ Each diagonal of rhombus is diameter of

the circle.

Now area of circle = 1256 cm2

C(3, 4)

A(4, 3) B(–1, )y

SET-II [UNCOMMON QUESTIONS TO SET-I]


Arundeep’s Solved Papers Mathematics 2015 (Outside Delhi)82Þ pr2 = 1256

Þ r2 1256=p

Þ r2 1256 4003.14

= = = 202

Þ r = 20 cm\ Diameter of the circle = 40 cm\ Each diagonal of the rhombus = 40 cm

Thus Area of rhombus

1 212

d d= ´ 1 40 402

= ´ ´ = 800 cm2

19. Solve for x : 22 6 3 60 0x x+ - =

Sol. Given eqn. be, 22 6 3 60x x+ - = 0

Þ 22 10 3 4 3 60x x x+ - - = 0

Þ 2 ( 5 3) 4 3 ( 5 3)x x x+ - + = 0

Þ ( 5 3) (2 4 3)x x+ - = 0

5 3x + = 0 or 2 4 3x - = 0

x 5 3= - or4 3 2 3.

2x = =

20. The 16th term of an AP is five times itsthird term. If its 10th term is 41, thenfind the sum of its first fifteen terms.

Sol. Let the 1st term of the AP = aCommon difference = dA.T.Q. ; we havea16 = 5 × a3

Þ a + 15d = 5 (a + 2d) [_ an = a + (n – 1) d]Þ a + 15d = 5a + 10d

Þ 5d = 4a Þ a54

d= ...(i)

Also a10 = 41 Þ a + 9d = 41

Þ5 94

d d+ = 41 (Using eqn. (i))

Þ 41d = 164 Þ d = 4When d = 4, eq. (i) becomes

a5 44

= ´ Þ a = 5

[we know that Sn =2n

{2a + (n – 1) d}]

\ S1515 (2 14 )2

a d= +

15 (2 5 14 4)2

= ´ + ´

15 662

= ´ = 15 × 33 = 495

28. A bus travels at a certain average speedfor a distance of 75 km and then travelsa distance of 90 km at an average speedof 10 km/h more than the first speed. Ifit takes 3 hours to complete the totaljourney, find its first speed.

Sol. Let average speed of the bus = x km/hDistance to be covered by bus = 75 km

_ Time taken by bus to covered a distance of

75 km by bus75 hrx

=

Speed of bus for a distance of 90 km= (x + 10) km/h

\ Time taken by bus to covered a distance of

90 km by bus90 hr

10x=

+

A.T.Q.,75 90

10x x+

+ = 3

Þ75 ( 10) 90

( 10)x xx x

+ ++ = 3

Þ 75x + 750 + 90x = 3x2 + 30xÞ 3x2 – 135x – 750 = 0Þ x2 – 45x – 250 = 0


Arundeep’s Solved Papers Mathematics 2015 (Outside Delhi)83Þ x2 – 50x + 5x – 250 = 0Þ x(x – 50) + 5(x – 50) = 0

Þ (x – 50) (x + 5) = 0

Þ x = 50, x = – 5

\ Required speed of bus = 50 km/h


Sol. Proof : We are given a circle with centre Oand a tangent XY to the circle at a point P.We need to prove that OP is perpendicularto XY.

Take a point Q on XY other than P and joinOQ.

The point Q must lie outside the circle.(Note that if Q lies inside the circle, XYwill become a secant and not a tangent tothe circle). Therefore, OQ is longer thanthe radius OP of the circle. That is,OQ > OP.

Since this happens for every point on theline XY except the point P. OP is theshortest of all the distances of the point Oto the points of XY. So OP is perpendicularto XY.

30. Construct a right triangle ABC withAB = 6 cm, BC = 8 cm and ÐB = 90º.Draw BD, the perpendicular from B onAC. Draw the circle through B, C and Dand construct the tangents from A to thiscircle.

Sol. Steps of Construction :Draw a line segment BC = 8 cmFrom B draw an angle of 90o

Draw an arc BA = 6cm cutting the angleat A.Join AC.DABC is the required D.Draw ^ bisector of BC cutting BC at M.Take M as centre and BM as radius, drawa circle.Take A as centre and AB as radius drawan arccutting the circle at E. Join AE.AB and AE are the required tangents.

C90º

90º90º

A

X

B OM8 cm

90º6cm D E

Justification :

ABCÐ = 90° (Given)Since, OB is a radius of the circle.

\ AB is a tangent to the circle.Also AE is a tangent to the circle.

31. Find the values of k so that the area ofthe triangle with vertices (k + 1, 1),(4, – 3) and (7, – k) is 6 sq. units.



Sol. \ Area of D12

= | x1 (y2 – y3) + x2 (y3 – y1) + x3(y1 – y2) |

Here x1 = k + 1 ; y1 = 1 ; x2 = 4 ; y2 = –3 ; x3 = 7 and y3 = –k

Þ12 | (k + 1) (– 3 + k) + 4 (– k – 1) + 7 (1 + 3) | = 6

Þ | – 3k – 3 + k2 + k – 4k – 4 + 28 | = 12 Þ | k2 – 6k + 21 | = 12Þ k2 – 6k + 21 = ± 12Þ k2 – 6k + 21 = 12 or k2 – 6k + 21 = – 12

k2 – 6k + 9 = 0 or k2 – 6k + 33 = 0Þ (k – 3)2 = 0 Here, D = (– 6)2 – 4 × 1 × 33

k =3 = 36 – 132 = – 96 < 0\ k = 3 \ Given quadratic eqn. has no solution.

10. Solve the following quadratic equationfor x :x2 – 2ax – (4b2 – a2) = 0

Sol. Given quadratic eqn. bex2 – 2ax – (4b2 – a2) = 0on comparing with ax2 + bc + c = 0, wehavea = 1, b = –2a ; c = –(4b2 – a2)

\ Distriminant D = b2 – 4aci.e. D = (– 2a)2 – 4 × [– (4b2 – a2)]

= 4a2 + 16b2 – 4a2

= 16b2

By quadratic formula, we have

\ x =D

2b

a- ±

\ x2( 2 ) 16

2 1a b- - ±=

´

2 42

a b±= = a ± 2b

18. The 13th term of an AP is four times its3rd term. If its fifth term is 16, then find

the sum of its first ten terms.Sol. Let 1st term of the AP = a

Common difference = dA.T.Q. we havea13 = 4 × a3

Þ a + 12d = 4 (a + 2d)[_ an = a + (n – 1)d]

Þ a + 12d = 4a + 8d Þ 3a = 4d

Þ a 43

d= ...(i)

Also a5 = 16 Þ a + 4d = 16

Þ4 43

d d+ = 16 [Using (i)]

Þ 16d = 48 Þ d = 3When d = 3, (i) becomes ;

a4 3 43

= ´ = Þ a = 4

Now, S1010 (2 9 )2

a d= +

{ }S 2 ( 1)2nn a n dé ù= + -ê úë û

Q

SET-III [UNCOMMON QUESTIONS TO SET-I and Set-II]


Arundeep’s Solved Papers Mathematics 2015 (Outside Delhi)85= 5 (2 × 4 + 9 × 3)= 5 × 35 = 175.

19. Find the coordinates of a point P on theline segment joining A (1, 2) and B (6, 7)

such that AP2= AB.5

Sol.

(1, 2) (x, y) (6,7)

A 2 P 3 B

2 : 3

Given, AP2 AB5

=

Þ AP = 2 (AP + PB)5

=

Þ 5AP = 2AP + 2PBÞ 3AP = 2PB

ÞAP 2PB 3

=

\ AP : PB = 2 : 3Thus, P divides AB in the ratio 2 : 3

\ x coordinate of P2 6 3 1 3

2 3´ + ´= =

+

and y coordinate of P2 7 3 2 4

2 3´ + ´= =

+\ Coordinates of P are (3, 4).

20. A bag contains, white, black and red ballsonly. A ball is drawn at random from thebag. If the probability of getting a white

ball is3

10 and that of a black ball is2 ,5

then find the probability of getting a redball. If the bag contains 20 black balls,then find the total number of balls in thebag.

Sol. Let R = getting a red ballB = getting a black ballW = getting a white ball

Now, P (R) + P (B) + P (W) = 1

Þ P (R)2 35 10

+ + = 1

Þ P (R)2 315 10

= - -

10 4 3 310 10

- -= =

Let total number of balls = x

\ P (B)20x

=

A.T.Q. ; we have

20x

25

= Þ x = 50

\ Total no. of balls = 5028. A truck covers a distance of 150 km at a

certain average speed and then coversanother 200 km at an average speedwhich is 20 km per hour more than thefirst speed. If the truck covers the totaldistance in 5 hours, find the first speedof the truck.

Sol. Let average speed of truck = x km/hrDistance covered by truck = 150 km

\ Time taken by truck to covered a distance

of 150 km 150 hrsx

=

A.T.Q., Speed to cover 200 km = (x + 20) km/hr\ Time taken by truck to covered a distance

of 200 km200 hr

20x=

+

Given Total time taken = 5 hrs

Þ150 200

20x x+

+ = 5

Þ150 ( 20) 200

( 20)x xx x

+ ++ = 5



Þ 150x + 3000 + 200x = 5 (x2 + 20x) Þ 350x + 3000 = 5x2 + 100xÞ 5x2 – 250x – 3000 = 0 Þ x2 – 50x – 600 = 0Þ x2 – 60x + 10x – 600 = 0 Þ x (x – 60) + 10 (x – 60) = 0Þ (x – 60) (x + 10) = 0 Þ x = 60 or x = – 10\ Average speed of truck = 60 km/hr.

29. An arithmetic progression 5, 12, 19, ... has 50 terms. Find its last term. Hence find the sumof its last 15 terms.

Sol. Let a be the first term and d be the common difference of given A.P.\ a = 5, d = 7, n = 50\ a50 = a + 49d = 5 + 49 × 7 = 348 [_ an = a + (n – 1)d]

and S50 5050 ( )2

a a= + = 25 (5 + 348) = 8825 ( )S2n nn a aé ù= +ê úë û

Q

\ Sum of last 15 terms = S50 – S35

Also, S3535 (2 5 34 7)2

= ´ + ´ = 4340

\ Sum of last fifteen terms = 8825 – 4340 = 4485.30. Construct a triangle ABC in which AB = 5 cm, BC = 6 cm and ÐABC = 60º. Now construct

another triangle whose sides are57 times the corresponding sides of DABC.

Sol. Steps of construction :(i) Draw a line segment BC = 6 cm.

(ii) Draw a ray BX making an angle of 60º and cut off BA = 5 cm.(iii) Join AC. Then ABC is the required triangle.(iv) Draw a ray BY making an acute angle with BC and cut off

7 equal parts making BB1 = B1B2 = B2B3 = B3B4 = B4B5 =B5B6 = B6B7

(v) Join B7 and C(vi) Draw B5C¢ parallel to B7C and C¢A¢ parallel to CA.

Then DA¢BC¢ is the required triangle.B C60º

B1

B2

B3

B4

B5

B6

B7

C¢

A¢

A

X

6 cm

5 cm

Y



31. Find the values of k for which the points A (k + 1, 2k), B (3k, 2k + 3) and C (5k – 1, 5k) arecollinear.

Sol. If points are collinear then area of triangle having given points are vertices be 0.\ x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) = 0Þ (k + 1) (2k + 3 – 5k) + 3k (5k – 2k) + (5k – 1) (2k – 2k – 3) = 0Þ (k + 1) (– 3k + 3) + 3k × 3k + (5k – 1) (– 3) = 0Þ – 3k2 + 3k – 3k + 3 + 9k2 – 15k + 3 = 0Þ 6k2 – 15k + 6 = 0 Þ 3 (2k2 – 5k + 2) = 0Þ 2k2 – 5k + 2 = 0Þ 2k2 – 4k – k + 2 = 0Þ 2k (k – 2) – 1 (k – 2) = 0 Þ (k – 2) (2k – 1) = 0

Þ k = 2 or 1 .2

k =

Þ k = 2,12

.



(ii) The question paper consists of 31 questions divided into four sections – A, B, C and D.(iii) Section A contains 4 questions of 1 mark each, Section-B contains 6 questions of 2 marks each,

Section-C contains 10 questions of 3 marks each and Section-D contains 11 questions of 4marks each.

(iv) Use of calculators is not permitted.

Þ – k12 3

4-=

Þ – k94

=

Þ k94

= -

2. The tops of two towers of height x and y,standing on level ground, subtend anglesof 30º and 60º respectively at the centreof the line joining their feet, then findx : y.

Sol. Let AB and DC are towers of height x andy respectivelyIn right angled DABE, we have

AB tan 30BE

= °

Þxa = tan 30º

Þxa

13

=

Þ x3

a=

1. If1 ,2

x = - is a solution of the quadratic

equation 3x2 + 2kx – 3 = 0, find the valueof k.

Sol. p (x) = 3x2 + 2kx – 3

Q x12

= - is a solution of p (x)

\12

p æ ö-ç ÷è ø = 0

12

p æ ö-ç ÷è ø

21 13 2 32 2

kæ ö æ ö= - + - -ç ÷ ç ÷è ø è ø

1 13 2 34 2

k -æ ö æ ö= + -ç ÷ ç ÷è ø è ø

3 34

k= - -

Q

12

p æ ö-ç ÷è ø = 0

\3 34

k- - = 0

– k3 31 4= -

MATHEMATICS 2015 TERM II (DELHI)SET I





In right angled DCDE, we have

ÞDC tan 60EC

= ° Þya = tan 60º

Þ ya

3= Þ y 3a=

D

y

CEB

A

60°30°

Now,xaya

33

a

a=

Þx aa y´ 1

3 3a

a= ´

Þxy

13

= Thus x : y = 1 : 3

3. A letter of English alphabet is chosen atrandom. Determine the probability thatthe chosen letter is a consonant.

Sol. Total numbers (letters) of English alphabet= 26No. of consonant letters = 21

\ P (getting consonant letter)2126

=

4. In Fig. given below PA and PB aretangents to the circle with centre O suchthat ÐAPB = 50º. Write the measure ofÐOAB.

Sol. Since we know that tangents drawn from apoint exterior to circle are of equal length\ PA = PBSince equal sides have equal angle oppositeto it

\ ÐPAB = ÐABPSince the sum of all three angles of DPABis 180°.

\ 2ÐPAB = 180° – 50° = 130°ÞÐPAB = 65°Since radius of circle is ^ to the tangentto circle at their point of contact.

\ OA ^ PA Þ ÐOAP = 90°Thus ÐOAB = ÐOAP – ÐPAB

= 90° – 65°= 25°5. In Fig. given below AB is the diameter

of a circle with centre O and AT is atangent. If ÐAOQ = 58º, find ÐATQ.

Sol. ÐAOQ = 2ÐABQ[_ angle made by arc at centre of circle isdouble the angle made by same arc at any

point on circumference of circle]

Þ ÐABQ1 58º 29º2= ´ =

In DABT, we haveÐBAT + ÐABT + ÐATB = 180º

Þ 90º + 29º + ÐATB = 180ºÞ ÐATB = 61º

as ÐATB = ÐATQÞ ÐATQ = 61º6. Solve the following quadratic equation

for x :4x2 – 4a2x + (a4 – b4) = 0

Sol. Given, quadratic eqn. be


Arundeep’s Solved Papers Mathematics 2015 (Delhi)904x2 – 4a2x + (a4 – b4) = 0on comparing with ax2 + bc + c = 0, wehavea = 4, b = – 4a2, c = (a4 – b4)

\ D = b2 – 4ac= (– 4a2)2 – 4(4) (a4 – b4)= 16a4 – 16 (a4 – b4)= 16a4 – 16a4 + 16b4 = 16b4

by quadratic formula, we have

\ x2 4D ( 4 ) 16

2 (2) (4)b a b

a- ± - - ±= =

2 2 2 24 4 4 ( )8 8

a b a b± ±= =

2 2

2a b±=

7. From a point T outside a circle of centreO, tangents TP and TQ are drawn to thecircle. Prove that OT is the right bisectorof line segment PQ.

Sol. In DTPO and DTQO, we haveTP = TQOP = OQ (radii)OT = OT

\ DTPO @ DTQO[by SSS axiom of congruency]

\ Ð1 = Ð2

O R T

P

Q

12

In DOPR and DOQR,OP = OQ (radii)OR = ORÐ1 = Ð2 [Proved]

\ DOPR @ DOQRÞ PR = QR

Þ OR ^ PR [line joining mid point of a chordto the centre of a circle is perpendicular tothe chord]

\ OT is the perpendicular bisector of PQ.8. Find the middle term of the A.P.

6, 13, 20, ..., 216Sol. Given A.P. be, 6, 13, 20, ... 216

Here a = 6, d = 7Let the given A.P. contains n termsthen Tn = 216We know thatTn = a + (n – 1) d

Þ 216 = 6 + (n – 1) 7Þ 216 = 6 + 7n – 7Þ 216 = 7n – 1Þ 216 + 1 = 7nÞ 217 = 7n

Þ217

7 = n

i.e. 31 = ni.e. n = 31

Thus, the given A.P. contains 31 terms.

\ Its middle term1 (31 1)2

= +

1 (32)2

= = 16th term

\ T16 = 6 + (16 – 1)7= 6 + (15) (7) = 6 + 105 = 111Hence, middle term = 111

9. If A (5, 2), B (2, – 2) and C (– 2, t) are thevertices of a right angled triangle withÐB = 90º, then find the value of t.

Sol. AB 2 2(5 2) (2 ( 2))= - + - -

2 2(3) (4) 9 16 25= + = + =

BC 2 2(2 ( 2)) ( 2 )t= - - + - -



2 2(2 2) ( 2)t= + + +

2 2(4) 4 4t t= + + +

216 4 4t t= + + +

2 4 20t t= + +

and AC 2 2(5 2) (2 )t= + + -

2 2(7) 4 4t t= + + -

249 4 4t t= + + -

2 4 53t t= - +by using Pythagoras Theorem, we have(AC)2 = (AB)2 + (BC)2

t2 – 4t + 53 = 25 + t2 + 4t + 20Þ – 4t – 4t = 45 – 53Þ – 8t = – 8

i.e. t 8 18

= =

i.e. t = 110. Find the ratio in which the point

Pæ öç ÷è ø

3 5,4 12 divides the line segment

joining the points Aæ öç ÷è ø

1 3,2 2 and

B (2, – 5).

Sol. Let the given point P3 5,4 12æ öç ÷è ø divides the

line segment joining the point A1 3,2 2æ öç ÷è ø

and B(2, –5) in the ratio k : 1 internally.Then by section formula,

34

1(2) 12

1

k

k

æ ö+ ç ÷è ø=+

3 5P ,4 12æ öç ÷è ø

1 3A ,2 2æ öç ÷è ø

B(2, –5)1k

34

12 4 121 2 ( 1)

k kk k

+ += =+ +

Þ 3 (2) (k + 1) = 4 (4k + 1)Þ 6 (k + 1) = 4 (4k + 1)Þ 6k + 6 = 16k + 4Þ 6k – 16k = 4 – 6 Þ – 10k = – 2

i.e. k2 1

10 5= =

Thus, required ratio be k : 1

i.e.1 :15 i.e. 1 : 5.

11. Find the area of the triangle ABC withA (1, – 4) and mid-points of sides throughA being (2, – 1) and (0, – 1).

Sol. Let P(2, –1) be the mid point of AB andQ(0, –1) be the mid point of AC. Let(x1, y1), (x2, y2) be the coordinates ofvertices B and C of DABC.Since P(2, –1) be the mid point of AB.

\ 1 12

x + = 2

Þ x1 + 1 = 4 Þ x1 = 4 – 1 = 3

and 1 ( 4)2

y + - = – 1


Arundeep’s Solved Papers Mathematics 2015 (Delhi)92Þ y1 – 4 = – 2Þ y1 = – 2 + 4

i.e. y1 = 2Also Q(0, –1) be the mid point of AC.

\ 2 12

x + = 0

x2 + 1 = 0Þ x2 = – 1

and 2 ( 4)2

y + - = – 1

Þ 2 42

y - = – 1

Þ y2 – 4 = – 2i.e. y2 = – 2 + 4 = 2i.e. y2 = 2

\ Vertices of triangle ABC areA (1, – 4), B (3, 2) and C (– 1, 2)

\ Area of DABC12

= | x1 (y2 – y3) +

x2 (y3 – y1) + x3 (y1 – y2) |

12

= | 1 (2 – 2) + 3 (2 + 4) + (– 1) (– 4 – 2) |

12

= | 1 (0) + 3 (6) + (– 1) (– 6) |

12

= | 18 + 6 |12

= | 24 | = 12 sq. unit

12. Find that non-zero value of k, for whichthe quadratic equationkx2 + 1 – 2 (k – 1) x + x2 = 0 has equalroots. Hence find the roots of the equation.

Sol. Given quadratic eqn. be,Þ kx2 + 1 – 2 (k – 1) x + x2 = 0Þ kx2 + x2 – 2 (k – 1) x + 1 = 0

(k + 1) x2 – 2 (k – 1) x + 1 = 0on comparing with ax2 + bx + c = 0, wehavea = k + 1, b = – 2 (k – 1), c = 1Here, D = b2 – 4ac= [– 2 (k – 1)]2 – 4 (k + 1) (1)= 4 (k – 1)2 – 4 (k + 1)

Q Roots are equal\ D = 0\ 4 (k – 1)2 – 4 (k + 1) = 0

4 [(k – 1)2 – (k + 1)] = 0Þ (k – 1)2 – k – 1 = 0Þ k2 + 1 – 2k – k – 1 = 0Þ k2 – 3k = 0Þ k (k – 3) = 0

i.e. k = 0 or k – 3 = 0i.e. k = 0 or k = 3\ Non zero value of k = 3

After putting the value of k = 3 in givenequation, we get3x2 + 1 – 2 (3 – 1) x + x2 = 0

Þ 3x2 + 1 – 2 (2) x + x2 = 0Þ 3x2 + 1 – 4x + x2 = 0Þ 4x2 – 4x + 1 = 0Þ 4x2 – 2x – 2x + 1 = 0Þ 2x (2x – 1) – 1 (2x – 1) = 0Þ (2x – 1) (2x – 1) = 0Þ either 2x – 1 = 0 2x – 1 = 0

i.e. 2x = 1 or 2x = 1

i.e. 12

x = or12

x =

\ Required roots are1 1,2 2 .

13. The angle of elevation of the top of abuilding from the foot of the tower is 30º


Arundeep’s Solved Papers Mathematics 2015 (Delhi)93and the angle of elevation of the top ofthe tower from the foot of the building is45º. If the tower is 30 m high, find theheight of the building.

Sol. Let DC be the building is of height be h m

In right DABC,ABBC

= tan 45º

Þ30BC = 1

Þ BC = 30

In right DACD, we have

CDBC = tan 30º

Þ 30h 1

3=

Þ h30

3=

Þ h30 3 30 3 10 3

33 3= ´ = =

\ h 10 3 m=Thus required height of building DC

10 3 m=14. Two different dice are rolled together.

Find the probability of getting :(i) the sum of numbers on two dice to be 5.

(ii) even numbers on both dice.Sol. When two dice are thrown simultaneously,

all possible outcomes = 36Since, sample space = {(1, 1), (1, 2) .... (1, 6)

(2, 1), (2, 2) ......(2, 6)(3, 1), (3, 2) .... (3, 6)(4, 1), (4, 2) .... (4, 6)(5, 1), (5, 2) ..... (5, 6)(6, 1), (6, 2) ...... (6, 6)}

(i) Let E1 be the event of getting two numberswhose sum is 5. Then favourable outcomesare (1, 4), (2, 3), (3, 2), (4, 1)

\ P (E1)4 1

36 9= =

(ii) Let E2 be the event of getting even numberon both dice, then favourable outcomes are(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6),(6, 2), (6, 4), (6, 6)Number of fevourable outcomes = 9

\ P (E2)9 1

36 4= =

15. If Sn, denotes the sum of first n terms ofan A.P., prove that S12 = 3 (S8 – S4).

Sol. Let first term of an A.P. = aLet common difference = dThen sum of nth term is

Sn [2 ( 1) ]2n a n d= + -

S1212 [2 (12 1) ]2

a d= + -

= 6 [2a + 11d] = 12a + 66d

S882

= [2a + (8 – 1) d] = 4 [2a + 7d]

= 8a + 28d

S442

= [2a + (4 – 1) d] = 2 (2a + 3d)

= 4a + 6d\ RHS = 3 (S8 – S4) = 3 [(8a + 28d) – (4a + 6d)]

= 3 [8a + 28d – 4a – 6d] = 3 [4a + 22d]= 12a + 66d = LHSThus, LHS = RHS

16. In Fig. given below APB and AQO aresemicircle, and AO = OB. If the


Arundeep’s Solved Papers Mathematics 2015 (Delhi)94perimeter of the figure is 40 cm, find the

area of the shaded region.é ùê úë û

22Use π =7

P

Sol. Let AO = OB = rThen perimeter of semicircle APB

22

r rp= = p

Perimeter of semicircle AQO2

22 2

rrp p= =

radius of semicircle AQO2r=æ ö

ç ÷è øQ

\ Perimeter of shaded region

2rr rp= p + +

2 22

r r rp + p +=

But Perimeter is 40 cm. (given)

\2 2

2r r rp + p +

= 40

Þ r (2p + p + 2) = 80Þ r (3p + 2) = 80

Þ223 27

r æ ö´ +ç ÷è ø = 80

Þ66 27

r æ ö+ç ÷è ø = 80

Þ807

r æ öç ÷è ø = 80

i.e. r80 7

80´= = 7 cm

Now, Area of APB2

2rp=

222 7 7 77 cm7 2´ ´= =´

Area of AQO2

2rp=

222 7 7 1 77 cm7 2 2 2 4

= ´ ´ ´ =

\ Area of shaded region77774

= +

308 77 3854 4+= =

= 96.25 cm2

17. In Fig. given below From the top of asolid cone of a height 12 cm and baseradius 6 cm, a cone of height 4 cm isremoved by a plane parallel to the base.Find the total surface area of theremaining solid.

æ öç ÷è ø

22Use π = and 5 = 2.2367

Sol. Let radius of the upper face = BE = xDABE ~ DACD

ÞABAC

BE=CD

Þ4

12 6x= Þ x = 2 cm


Arundeep’s Solved Papers Mathematics 2015 (Delhi)95Remaining solid is a frustum of cone withheight = 12 – 4 = 8 cm = hr1 = 2 cm, r2 = 6 cmLet Slant height = l

\ l 2 22 1( )h r r= + -

l 64 16= +

Þ l 80=

Þ l 4 5 cm=

B

C

E

D

Now, total surface area of frustum of cone(remaining solid)

2 21 2 1 2( )l r r r r= p + + p + p

2 21 2 1 2[ ( ) ]l r r r r= p + + +

22 [4 5 (2 6) 4 36]7

= + + +

22 [32 5 40]7

= ´ +

22 [32 2.236 40]7

= ´ + cm2

= 350.592 cm2

18. A solid wooden toy is in the form of ahemisphere surmounted by a cone ofsame radius. The radius of hemisphereis 3.5 cm and the total wood used in the

making of toy is 35166 cm .6 Find the

height of the toy. Also, find the cost of

painting the hemispherical part of thetoy at the rate of `10 per cm2.

é ùê úë û

22Use π =7

Sol.

Let height of the cone = h cmRadius of the cone = 3.5 cmRadius of the hemisphere = 3.5 cm

Given volume of toy =51666 cm3

\ Volume of the toy 2 31 23 3

r h r= p + p

Þ51666

2 31 22 2 22(3.5) (3.5)3 7 3 7

h= ´ ´ + ´ ´

Þ1001

6269.5 1886.5

21 21h= +

Þ1001 1886.5

6 21- 269.5

21h=

Þ7007 3773

42- 269.5

21h=

Þ3234 2142 269.5

´´ = hÞ h = 6 cm

\ Height of toy = 6 + 3.5 = 9.5 cmThus, curved surface area of hemisphericalpart = 2pr2

222 3.5 3.57

= ´ ´ ´

= 77 cm2



Thus, cost of painting at the rate of ` 10per cm2 = 77 × 10 = ` 770

19. In Fig. given below from a cuboidal solidmetallic block, of dimensions 15 cm × 10cm × 5 cm, a cylindrical hole of diameter7 cm is drilled out. Find the surface area

of the remaining block.é ùê úë û

22Use π =7

Sol. Given radius of cylindrical hole = r =72 cm

and height of cylindrical hole = height ofcuboid = h = 5 cm

\ Surface area of remaining block = totalsurface area of cuboid

+ Curved surface area of cylinder

– Surface area of two circular end ofcylindrical hole

= 2 [lb + bh + lh] + 2prh – 2pr2

= 2 [15 × 10 + 10 × 5 + 15 × 5]

+ 2 ×22 7 22 7 75 27 2 7 2 2´ ´ - ´ ´ ´

= 2 [(150 + 50 + 75) + 110 – 77] cm2

= 616 cm2

20. In Fig. given below find the area of theshaded region. [Use p = 3.14]

Sol. Here PQ = (14 – 6) cm = 8 cm

\ OP = OQ =8 cm2 = 4 cm

and XY = (14 – 6) cm = 8 cmeach side of square EFGH = 4 cm

\ Area of unshaded region = Area of 4semicircles of radius 2cm + Area of squareof side 4 cm

24 side side

2r ×p= ´ +

D C

A B

XH G

Y

QE F

14 cm

O

= 24 3.14 2 2 42

´ ´ ´ +

= 25.12 + 16 = 41.12 cm2

Area of square of side 14 cm = 14 × 14 = 196 cm2

\ Area of shaded region = (196 – 41.12) cm2

= 154.88 cm2

21. The numerator of a fraction is 3 less thanits denominator. If 2 is added to both the


Arundeep’s Solved Papers Mathematics 2015 (Delhi)97numerator and the denominator, thenthe sum of the new fraction and original

fraction is29 .20 Find the original

fraction.Sol. Let denominator of fraction = x

Then, numerator = x – 3

Then, fraction be3x

x-

ATQ,1 32

x xx x- -++

2920

=

Þ( 1) ( 2) ( 3)

( 2) ( )x x x x

x x- + + -

+2920

=

Þ2 2

23 2 62

x x x x xx x

- + - + -+

2920

=

Þ2

22 2 6

2x xx x- -+

2920

=

Þ 20 (2x2 – 2x – 6) = 29 (x2 + 2x)Þ 40x2 – 40x – 120 = 29x2 + 58xÞ 40x2 – 40x – 120 – 29x2 – 58x = 0Þ 11x2 – 98x – 120 = 0Þ 11x2 – 110x + 12x – 120 = 0Þ 11x (x – 10) + 12 (x – 10) = 0

either (x – 10) (11x + 12) = 0x – 10 = 0 or 11x + 12 = 0

x = 10 or12

11x -=

\ x = 10[_ x be a natural number]

Hence, required fraction is given by

3xx- 10 3 7

10 10-= =

22. Ramkali required ` 2500 after 12 weeksto send her daughter to school. She saved` 100 in the first week and increased herweekly saving by ` 20 every week. Findwhether she will be able to send her

daughter to school after 12 weeks.What value is generated in the abovesituation ?

Sol. Ramkali saving per week forms a followingsequence 100, 120, 140, 160...which clearly forms A.P.Whose first term (a) = 100 andcommon difference (d) = 20Then, we know that,

Sn [2 ( 1) ]2n a n d= + -

\ S1212 [2(100) (12 1) 20]2

= + -

= 6 [200 + 220] = 6 [420] = 2520Yes, she will be concern for education ofgirl child.

23. Solve for x :

2 3 23,

1 2 ( 2) 5x x x+ =

+ - x ¹ 0, – 1, 2

Sol. Given eqn. be,2 3

1 2( 2)x x+

+ -235x

=

Þ4 ( 2) 3( 1)

2 ( 1) ( 2)x x

x x- + ++ -

235x

=

Þ 24 8 3 3

2 ( 2 2)x x

x x x- + +- + -

235x

=

Þ 27 5

2 ( 2)x

x x-- -

235x

=

Þ 5x (7x – 5) = 46 (x2 – x – 2)Þ 35x2 – 25x = 46x2 – 46x – 92Þ 35x2 – 25x – 46x2 + 46x + 92 = 0Þ – 11x2 + 21x + 92 = 0Þ 11x2 – 21x – 92 = 0Þ 11x2 – 44x + 23x – 92 = 0Þ 11x (x – 4) + 23 (x – 4) = 0Þ (x – 4) (11x + 23) = 0

either x – 4 = 0 or 11x + 23 = 0


Arundeep’s Solved Papers Mathematics 2015 (Delhi)98\ x = 4 or 11x = – 23

Thus, 2311

x -= , 4


Sol. Proof : We are given a circle with centre Oand a tangent XY to the circle at a point P.We need to prove that OP is perpendicularto XY.Take a point Q on XY other than P and joinOQ.The point Q must lie outside the circle.(Note that if Q lies inside the circle, XYwill become a secant and not a tangent tothe circle). Therefore, OQ is longer thanthe radius OP of the circle. That is,OQ > OP.Since this happens for every point on theline XY except the point P. OP is theshortest of all the distances of the point Oto the points of XY. So OP is perpendicularto XY.

A

O

P Q B

R

25. In Fig. given below tangents PQ and PRare drawn from an external point P to acircle with centre O, such that ÐRPQ =30º. A chord RS is drawn parallel to thetangent PQ. Find ÐRQS.

Sol. In DRQP, QP = RP

\ Ð3 = Ð4Now, Ð3 + Ð4 + 30º = 180º

Þ 2Ð3 = 150ºÞÐ3 = 75ºNow, ÐQOR + 90° + 90° + 30° = 360°

Þ ÐQOR = 150°

\ ÐQSR = Ð1 =12 ÐQOR = 75°

Also SR || QP\ Ð1 = Ð2 [Alternate interior angles]Þ Ð2 = 75º

Now, Ð2 + ÐRQS + Ð3 = 180ºÞÐRQS = 180º – 150º = 30º

26. Construct a triangle ABC with BC = 7 cm,ÐB = 60º and AB = 6 cm. Construct

another triangle whose sides are34 times

the corresponding sides of DABC.Sol. Steps of Construction

(i) Draw a line BC = 7 cm(ii) Draw angle 60º at B. Cut AB = 6 cm

(iii) Join AC.Thus, DABC is obtained.

(iv) Below BC, make an acute ÐCBX.(v) Along BX, mark off four points B1, B2, B3,

B4 such that BB1 = B1B2 = B2B3 = B3B4.(vi) Join B4C.

(vii) From B3, draw B3D || B4C, meeting BC atD.

(viii) From D, draw DE || CA, meeting AB at E.Then, DEBD is the required triangle, each



of whose sides is34 of the corresponding side of

DABC.

27. From a point P on the ground the angleof elevation of the top of a tower is 30ºand that of the top of a flag staff fixedon the top of the tower, is 60º. If thelength of the flag staff is 5 m, find theheight of the tower.

Sol. Let height of the tower BC be hand AC be the flag staff s.t AC = 5 mIn DPBC, we have

BCPB = tan 30º

Þ PBh 1

3=

h

Þ 3h = PB

\ PB 3h= ...(i)In right angled DABP, we have

ABPB = tan 60º

Þ 5PB

h +3=

Þ5

3h +

= PB

or PB5

3h += ...(ii)

From (i) and (ii) ; we have

3h5

3h +=

Þ 3h = h + 5Þ 3h – h = 5Þ 2h = 5

i.e. h5 2.5 m2

= =

Thus required height of tower be 2.5 m.28. A box contains 20 cards numbered from

1 to 20. A card is drawn at random fromthe box. Find the probability that thenumber on the drawn card is(i) divisible by 2 or 3(ii) a prime number.

Sol. Total number of all possible outcomes = 20(i) Let E1 be the event of getting card divisible

by 2 or 3. Then number of favourableoutcomes = 13 {2, 3, 4, 6, 8, 9, 10, 12, 14,15, 16, 18, 20}

\ P (E1)1320

=

(ii) Let E2 be the event of getting card a primenumber.Then number of favourable outcomes = 8

{2, 3, 5, 7, 11, 13, 17, 19}



\ P (E2)8 220 5

= =

29. If A (– 4, 8), B (– 3, – 4), C (0, – 5) and D(5, 6) are the vertices of a quadrilateralABCD, find its area.

Sol. Area of quadrilateral ABCD = Area ofDABC + Area of DACD

We know, Area of D12

= | x1 (y2 – y3) + x2

(y3 – y1) + x3 (y1 – y2) |

Now, Area of DABC12

= | – 4 (– 4 + 5) +

(– 3) (– 5 – 8) + 0 (8 + 4) |

12

= | – 4 (1) + (– 3) (– 13) |12

= | – 4 + 39 |

35= sq. unit2

Now, Area of DACD1=2 | – 4 (– 5 – 6) + 0

(6 – 8) + 5 (8 + 5) |

1=2 | – 4 (– 11) + 0 + 5 (13) |

12

= | 44 + 65 |12

= (109)

109 sq. unit2

=

\ Area of quadrilateral ABCD35 1092 2

= +

35 109 1442 2+= = = 72 sq. units

30. A well of diameter 4 m is dug 14 m deep.The earth taken out is spread evenly allaround the well to form a 40 cm highembankment. Find the width of theembankment.

Sol. Given radius of the well = R4 2 m2

= =

and Depth of the well = h = 14 m\ Volume of the earth dug out = pR2h

= p (2)2 × 14 = p × 4 × 14 = 56 p m2

Let the width of the embankment be r m.Clearly, embankment forms a cylindricalshell whose inner and outer radii are 2 mand (2 + r) m respectively and height40 m

100\ Volume of embankment

= p [(2 + r)2 – (2)2] ×40

100

= p [(4 + r2 + 4r – 4] ×40

100

2 ( 4)5

r rp= +

2r m

2r m

quarding to given condition, we haveVolume of embankment = Volume of the


Arundeep’s Solved Papers Mathematics 2015 (Delhi)101earth dug out

Þ2 ( 4)

5r rp +

= 56p Þ r (r + 4)56 5 140

2p´= =p

Þ r2 + 4r = 140 Þ r2 + 4r – 140 = 0Þ r2 + 14r – 10r – 140 = 0 Þ r (r + 14) – 10 (r + 14) = 0Þ (r + 14) (r – 10) = 0

either r + 14 = 0 or r – 10 = 0i.e. r = – 14 or r = 10But r ¹ – 14, Since width can’t be negative\ r = 10 m

Thus required width of embankment be 10 metre.31. Water is flowing at the rate of 2.52 km/h through a cylindrical pipe into a cylindrical tank,

the radius of whose base is 40 cm, If the increase in the level of water in the tank, in half anhour is 3.15 m, find the internal diameter of the pipe.

Sol. Let internal radius of the pipe = x cmSpeed of water = 2.52 km/h = 2520 m/h

\ Volume of water that flows in half an hour 212

r h= p

1 25202 100 100

x x= p´ ´ ´

23126 m

1000xp=

Given radius of cylindrical tank = R = 40 cm

\ Volume of water in cylindrical tank 340 40 3.15 m100 100

= p´ ´ ´

according to given condition, we have volume of water that flows in half an hour = volume ofwater in cylindrical tank

Þ2126

1000xp 40 40 3.15

100 100= p´ ´ ´

Þ x2 40 40 10003.15100 100 126= ´ ´ ´

Þ x2 = 4Þ x = 2 cm\ Internal diameter of the pipe = 2x = 4 cm



10. Find the middle term of the A.P. 213, 205,197, ..., 37.

Sol. Given A.P be, 213, 205, 197, ...., 37.Let a be the first term and d be the commondifference of given A.P213, 205, 197, ..., 37a = 213 ; d = – 8

and an = 37We know that an = a + (n – 1) d

Þ 37 = 213 + (n – 1) (– 8)Þ 37 = 213 – 8n + 8Þ 37 = 221 – 8nÞ 37 – 221 = – 8n

i.e. – 184 = – 8n

i.e.184

8 = n

Thus, 23 = n\ n = 23

Now, Middle term1 (23 1)2

= +

1 (24) 12th2

= =

Now, a12 = 213 + (12 – 1) (– 8)= 213 + (11) (– 8)= 213 – 88 = 125

18. If the sum of the first n terms of an A.P.

is 21 (3 7 ),2

n n+ then find its nth term.

Hence write its 20th term.

Sol. Given, Sn21 (3 7 )

2n n= +

Where n = 1 ; S112

= [3 (1)2 + 7 (1)]

1 1(3 7) (10) 52 2

= + = =

\ First term (a) = 5

Where n = 2 ; S212

= [3 (2)2 + 7 (2)]

12

= [3 (4) + 14]

1 1(12 14) (26) 132 2

= + = =

\ 2nd term = S2 – S1 = 13 – 5 = 8

Where n = 3 ; S312

= [3 (3)2 + 7 (3)]

1 [3(9) 21]2

= +

1 1(27 21) (48) 242 2

= + = =

\ 3rd term S3 – S2 = 24 – 13 = 11Hence required A.P. is 5, 8, 11...

\ a = 5, d = 3and nth term= an = a + (n – 1) d= 5 + (n – 1) 3 = 5 + 3n – 3= 3n + 2Thus, a20 = 3 (20) + 2 = 60 + 2 = 62

19. Three distinct coins are tossed together.Find the probability of getting.

(i) at least 2 heads(ii) at most 2 heads

Sol. When 3 coins are tossed simultaneously, allpossible outcomes are{HHH, HHT, HTH, THH, HTT, THT,TTH, TTT}Total number of possible outcomes = 8

(i) Let E1 be the event of getting at least 2 head.Then, favourable outcomes are HHT, HTH,THH, HHH



Arundeep’s Solved Papers Mathematics 2015 (Delhi)103Number of favourable outcomes = 4

\ P (Getting at least 2 heads) = P (E1)48

=

12

=

(ii) Let E2 be the event of getting at most 2heads.Then, E2 = event of getting 0 or 1 or 2 headsSo, the favourable outcomes areTTT, HTT, THT, TTH, HHT, HTH, THHNumber of favourable outcomes = 7

\ P (Getting at most 2 heads) = P (E2)

78

=

20. Find that value of p for which thequadratic equation(p + 1) x2 – 6 (p + 1) x + 3 (p + 9) = 0,p ¹ – 1 has equal roots. Hence find theroots of the equation.

Sol. Given quadratic eqn. be(p + 1) x2 – 6 (p + 1) x + 3 (p + 9) = 0 ...(i)On comparing with ax2 + bx + c = 0, a ¹ 0we have a = p + 1 ; b = –6 (p + 1) andC = 3 (p + 9)

\ Roots are equal \ D = 0Now, D = b2 – 4acD = [– 6 (p + 1)]2 – 4 (p + 1) 3 (p + 9)= [36 (p2 + 1 + 2p) – 12 (p + 1) (p + 9)]= 36p2 + 72p + 36 – 12 (p2 + 10p + 9)= 36p2 + 72p + 36 – 12p2 – 120p – 108= 24p2 – 48p – 72

Q D = 0\ 24p2 – 48p – 72 = 0Þ 24 (p2 – 2p – 3) = 0Þ p2 – 2p – 3 = 0Þ p2 – 3p + p – 3 = 0Þ p (p – 3) + 1 (p – 3) = 0Þ (p – 3) (p + 1) = 0

either p – 3 = 0 or p + 1 = 0

p = 3 or p = – 1\ p ¹ – 1 [Given]

[_ a ¹ 0Þ p + 1 ¹ 0Þ p ¹ –1]\ p = 3\ equation (i) becomes ;

4x2 – 24x + 36 = 0Þ 4(x2 – 6x + 9) = 0Þ x2 – 6x + 9 = 0Þ (x – 3)2 = 0 Þ x = 3.

28. To fill a swimming pool two pipes are tobe used. If the pipe of larger diameter isused for 4 hours and the pipe of smallerdiameter for 9 hours, only half the poolcan be filled. Find, how long it would takefor each pipe to fill the pool separately,if the pipe of smaller diameter takes 10hours more than the pipe of the largerdiameter to fill the pool.

Sol. Let time taken by the pipe of larger diameter= x hr

\ Time taken by the pipe of smaller diameter= (x + 10) hr

ATQ4 1 9

10x x´ +

+12

=

Þ4 40 9

( 10)x xx x

+ ++

12

= Þ 213 40 1

210x

x x+ =

+

Þ 26x + 80 = x2 + 10xÞ x2 – 16x – 80 = 0Þ x2 – 20x + 4x – 80 = 0Þ x (x – 20) + 4(x – 20) = 0Þ (x – 20) (x + 4) = 0Þ x = 20 or x = – 4 (rejected)\ Time taken by larger pipe = 20 hr

and Time taken by smaller pipe = 20 + 10 = 30 hr30. Construct an isosceles triangle whose

base is 6 cm and altitude 4 cm. Thenconstruct another triangle whose sides

are34 times the corresponding sides of


Arundeep’s Solved Papers Mathematics 2015 (Delhi)104the isosceles triangle.

Sol. Steps of construction :

(i) Draw a line segment AB = 6 cm and draw its ^ bisectorDX and cut off DC = 4 cm

(ii) Join AC and BC to get the required DABC(iii) draw a ray AB making on acute angle with AB and cut

off 4 equal parts making AA1 = A1A2 = A2A3 = A3A4(iv) Join A4 B(v) From A3, draw A3 B¢ || to A4B meeting AB at B¢

(vi) Draw B¢C¢ || to BC meeting AC at C¢Then AB¢C¢ is the required triangle.

31. If P (– 5, – 3), Q (– 4, – 6), R (2, – 3) and S (1, 2) are the vertices of a quadrilateral PQRS,find its area.

Sol. \ Area of DPQR12

= | x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) |

12

= | – 5 (– 6 + 3) + (– 4) (– 3 + 3) + 2 (– 3 + 6) |

12

= | 15 + 6 |21 sq. unit2

=

Area of DPRS12

= | – 5 (– 3 – 2) + 2 (2 + 3) + 1 (– 3 + 3) |

12

= | 25 + 10 |352

= sq. units

Area of quadrilateral PQRS = area of DPQR + area of DPRS

21 35 562 2 2

= + = = 28 sq. units.

10. Solve the following quadratic equation for x :9x2 – 6b2x – (a4 – b4) = 0

Sol. Given quadratic eqn. be,9x2 – 6b2x – (a4 – b4) = 0On comparing with ax2 + bx + c = 0, where a ¹ 0

X

Y

S(1, 2)

P(–5, –3) Q(–4, –6)

R(2, –3)

SET-III [UNCOMMON QUESTIONS TO SET-I and Set-II]



a = 9, b = – 6b2, c = – (a4 – b4)\ Discriminant D = b2 – 4ac\ D = (– 6b2)2 – 4 × 9 [– (a4 – b4)]

= 36b4 + 36a4 – 36b4

= 36a4

By quadratic formula, we have

Thus x =D

2b

a- ±

\ x2 4( 6 ) 362 9

b a- - ±=´

2 26 618

b a±=

2 2

3b a±=

18. All red face cards are removed from apack of playing cards. The remainingcards were well shuffled and then a cardis drawn at random from them. Find theprobability that the drawn card is

(i) a red card(ii) a face card

(iii) a card of clubsSol. Total number of cards = 52

Number of red face cards = 6[2 red kings, 2 red queens and 2 red jacks]

\ Number of remaining cards = 52 – 6 = 46

(i) Probability (a red card)20 1046 23

= =

(Since 6 red cards have already beenremoved)

(ii) Probability (face card)6 3

46 23= =

[as it is given that 6 red face cards areremoved]

(iii) Probability (club card)1346

=

19. Find the area of the triangle PQR withQ (3, 2) and the mid-points of the sidesthrough Q being (2, – 1) and (1, 2).

Sol. Q M is the mid point of PQ.

\ 1 32

x + = 2

Þ x1 + 3 = 4Þ x1 = 4 – 3 = 1

Also, 1 22

y + = – 1

y1 + 2 = – 2y1 = – 2 – 2 = – 4

Q N is the mid point of QR.

\ 3 32

x + = 1

Þ x3 + 3 = 2Þ x3 = 2 – 3Þ x3 = – 1

Also, 3 22

y + = 2

Þ y3 + 2 = 4Þ y3 = 4 – 2 Þ y3 = 2

Hence, vertices of DPQR are ;P (1, – 4), Q (3, 2) and R (–1, 2)


Arundeep’s Solved Papers Mathematics 2015 (Delhi)106We know that

Area of DPQR12

= | x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) |

12

= | 1 (2 – 2) + 3 (2 + 4) – 1 (– 4 – 2) |

12

= | 1 (0) + 3 (6) + 6 |12

= | 18 + 6 |12

= | 24 | = 12 sq. units

20. If Sn denotes the sum of first n terms of an A.P., prove that S30 = 3 [S20 – S10]Sol. Let first term of A.P. = a and Common difference of A.P. = d

_ Sn [2 ( 1) ]2n a n d= + -

Now, S30302

= [2a + (30 – 1) d]

= 15 [2a + 29d] = 30a + 435d

S2020 [2 (20 1) ]2

a d= + - = 10 [2a + 19d] = 20a + 190d

S10102

= [2a + (10 – 1) d] = 5 [2a + 9d] = 10a + 45d

R.H.S. = 3 (S20 – S10) = 3 [20a + 190d – (10a + 45d)] = 3 [20a + 190d – 10a – 45d] = 3 [10a + 145d] = 30a + 435d = S30 = L.H.S.Thus, LHS = RHS

28. A 21 m deep well with diameter 6 m is dug and the earth from digging is evenly spread to

form a platform 27 m × 11 m. Find the height of the platform.é ùê úë û

22Use π =7

Sol. Given, radius of well (r) = 3 mand Depth of well (h) = 21 m\ Volume of earth dug out = pr2h

22 9 217

= ´ ´ = 594 m2

Let the required height of platform = hLength of platform = l = 27 mBreadth of platform = b = 11 m

\ Volume of platform = 27 × 11 × hNow, Volume of platform = Volume of earth dug out = 27 × 11 × h = 594



Þ h594 2 m

27 11= =

´Here the required height of platform be 2metre

29. A bag contains 25 cards numbered from1 to 25. A card is drawn at random fromthe bag. Find the probability that thenumber on the drawn card is :

(i) divisible by 3 or 5(ii) a perfect square number

Sol. Total number of all possible outcomes = 25(i) Out of the given numbers, divisible by 3 or

5 are3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25Then the number of favourable outcomes = 12

\ Required probability (that drawn card is

divisible by 3 or 5)1225

=

(ii) Out of the given numbers, perfect squareare 1, 4, 9, 16 and 25.

\ Probability (a card drawn having no is a

perfect square)5 125 5

= =

30. Draw a line segment AB of length 7 cm.Taking A as centre, draw a circle ofradius 3 cm and taking B as centre, drawanother circle of radius 2 cm. Constructtangents to each circle from the centreof the other circle.

Sol. Steps of construction :(i) Draw a line segment AB = 7 cm.

(ii) With centre A and radius 3 cm and with centreB and radius 2 cm, circles are drawn.

(iii) Bisect AB at M.(iv) With centre M and diameter AB, draw a circle

which intersects the two circles at S¢, T¢ andS, T respectively.

(v) Join AS, AT, BS¢ and BT¢.

Then AS, AT, BS¢ and BT¢ are the requiredtangent.

T¢T

A B

SS¢

M 7 cm3 cm 2 cm

31. Solve for x :3 4 29 ;

1 1 4 1x x x+ =

+ - -11, 1,4

x ¹ -


1 1x x+

+ -29

4 1x=

-

Þ3( 1) 4(x 1)

( 1) ( 1)xx x- + ++ -

294 1x

=-

Þ 23 3 4 4

1x x

x- + +-

294 1x

=-

Þ 27 1

1x

x+-

294 1x

=-

Þ (7x + 1) (4x – 1) = 29 (x2 – 1)Þ 28x2 – 7x + 4x – 1 = 29x2 – 29Þ 28x2 – 3x – 1 = 29x2 – 29Þ 28x2 – 3x – 1 – 29x2 + 29 = 0Þ – x2 – 3x + 28 = 0Þ – (x2 + 3x – 28) = 0Þ x2 + 3x – 28 = 0Þ x2 + 7x – 4x – 28 = 0Þ x (x + 4) – 4 (x + 7) = 0Þ (x + 7) (x – 4) = 0Þ x + 7 = 0 or x – 4 = 0Þ x = – 7 or x = 4


1. In DABC, D and E are points AC andBC respectively such that DE || AB. IfAD = 2x, BE = 2x – 1, CD = x + 1 andCE = x – 1, then find the value of x. [1]

Sol.

DE || AB

So,ADCD

BE=EC [By B.P.T.]

21

xx +

2 11

xx-=-

Þ 2x (x – 1) = (x + 1) (2x – 1)Þ 2x2 – 2x = 2x2 + 2x – x – 1Þ – 2x = x – 1Þ 1 = 3x

or x13

= Ans.

2. In A, B and C are interior angles ofDABC, then prove that :

(A + C)sin2

B= cos .2 [1]

Sol. In DABCÐA + ÐB + ÐC = 180ºÐA + ÐC = 180º – ÐBDivide by 2 on both sides

A C2

Ð + Ð 180º B2-Ð=

A C2

Ð + Ð B90º2Ð= -

\A Csin

2Ð + Ðæ öç ÷è ø

Bsin 90º2Ðæ ö= -ç ÷è ø

ÞA Csin

2Ð + Ðæ öç ÷è ø

Bcos2Ð=

Thus,(A C)sin

2+ Bcos

2= Hence proved.

3. If x = 3 sin q and y = 4 cos q, find the

value of 2 216 9 .x y+ [1]

Sol. x = 3 sin qÞ x2 = 9 sin2 q

sin2 q2

9x= ...(i)

and y = 4 cos qÞ y2 = 16 cos2 q

\ cos2 q2

16y= ...(ii)

On adding eq. (i) and eq. (ii) ; we have

sin2 q + cos2 q2 2

9 16x y= +

Þ 12 2

9 16x y= +

Þ 12 216 9144

x y+=

i.e. 16x2 + 9y2 = 144

MATHEMATICS -2016 TERM I





Þ 2 216 9x y+ = 144

\ 2 216 9x y+ = 12 Ans.

4. If empirical relationship between mean,median and mode is expressed as mean= k (3 median – mode), then find thevalue of k. [1]

Sol. Given, mean = k (3 median – mode)As we know, mode = 3 median – 2 mean

\ mean = k [3 median – (3 median – 2 mean)]mean = k [3 median – 3 median + 2 mean]mean = 2k mean

Þ 2k mean – mean = 0i.e. mean [2k – 1] = 02k – 1 = 0 [since mean ¹ 0]

Þ 2k = 0 + 1i.e. k = 1/2 Ans.

SECTION – B

5. Express 23150 as product of its primefactors. Is it unique ? [2]

Sol. Prime factor of 23150 = 2 × 5 × 5 × 463As per the fundamental theorem ofArithmetic every number has a uniquefactorisation.

2 231505 115755 2315

463 4631

Ans.

6. State whether the real number 52.0521is rational or not. If it is rational express

it in the form ,pq where p, q are co-

prime, integers and q ¹ 0. What can yousay about prime factorisation of q ? [2]

Sol.

Þ 52.052152052110000

=

2 100002 50002 25002 12505 6255 1255 25

5Yes, it is rational number.where q = 10000 = 24 × 54

The given decimal expression is aterminating decimal as the factors of qconsist only 2 and 5. Ans.

7. Given the linear equation x – 2y – 6 = 0,write another linear equation in thesetwo variables, such that the geometricalrepresentation of the pair so formed is :(i) coincident lines(ii) intersection lines [2]

Sol. (i) Given, x – 2y – 6 = 0For line to be coincident

1 1 1

2 2 2

a b ca b c

= =

Thus one possible option will be2x – 4y – 12 = 0Here, a1 = 1, b1 = – 2, c1 = – 6a2 = 2, b2 = – 4, c2 = – 12

1

2

1 ;2

aa

= 1

2

2 1 ;4 2

bb

-= =-

1

2

6 112 2

cc

-= =-

Þ1 1 1

2 2 2

a b ca b c

= =

and hence both lines are coincident lines.Ans.


Arundeep’s Solved Papers Mathematics 2016 (Term I)110(ii) Given, x – 2y – 6 = 0

For intersecting lines

1 1

2 2

a ba b¹

Thus, one possible option will be,

2x – 7y – 13 = 0

Here, a1 = 1, b1 = – 2, c1 = – 6

a2 = 2, b2 = – 7, c2 = – 13

Here,1

2

1 ;2

aa

= 1

2

2 27 7

bb

-= =-

Þ1 1

2 2

a ba b¹

So, both lines representing intersectinglines. Ans.

8. In an isosceles DABC right angled at B,prove that AC2 = 2AB2. [2]

Sol. In an isosceles DABC, AB = BC ...(i)

[Q triangle is isosceles]

A

B C

In DABC by pythagoras theorem,

AC2 = AB2 + BC2

Þ AC2 = AB2 + AB2 [From (i)]

i.e. AC2 = 2AB2 Hence Proved.

9. Prove the following identity :

21 tan A1 cot A-é ù

ê ú-ë û = tan2 A : ÐA is acute. [2]

Sol. L.H.S.21 tan A

1 cot A-é ù= ê ú-ë û

2sin A1cos Acos A1sin A

é ù-ê ú= ê úê ú-ê úë û

2cos A sin Acos A

sin A cos Asin A

-é ùê ú

= ê ú-ê úê úë û

2(cos A sin A) sin A(cos A sin A) cos A

-é ù= ê ú- -ë û

2sin Acos A

é ù= -ê úë û

= [– tan A]2

= tan2 A = R.H.S. Hence Proved.

10. Given below is a cumulative frequencydistribution table. Corresponding to it,make an ordinary frequency distributiontable. [2]

x cfMore than or equal to 0 45

More than or equal to 10 38






Arundeep’s Solved Papers Mathematics 2016 (Term I)111Sol. Table of values is given as under :

C.I. Frequency0 – 10 07 (45 – 38)10 – 20 09 (38 – 29)20 – 30 12 (29 – 17)30 – 40 6(17 – 11)40 – 50 5 (11 – 6)50 – 60

SECTION – C11. Find LCM and HCF of 3930 and 1800

by prime factorisation method. [3]Sol. By prime factorization method,

Factors of 3930 and 1800 are,

2 39303 19655 655

131 1311

2 18002 9002 4503 2253 755 255 5

1

So, 3930 = 2 × 3 × 5 × 1311800 = 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5then, HCF = product of the smallest powerof each common prime factor in givennumbers = 2 × 3 × 5 = 30and, LCM = product of the greatest powerof each prime factor involved in givennumbers = 2 × 3 × 5 × 131 × 2 × 2 × 3 × 5= 235800 Ans.

12. Using division algorithm, find thequotient and remainder on dividing f (x)by g (x) where f (x) = 6x3 + 13x2 + x – 2

and g (x) = 2x + 1. [3]Sol. Given, f (x) = 6x3 + 13x2 + x – 2

and g (x) = 2x + 1,By division algorithm, we have

2 + 1 6 + 13 + – 2 3 + 5 – 2x x x x x x3 2 2

6x x3 2 + 3– –

10x x2 + – 210x x2 + 5

– – – 4x – 2

+ +0

) (

– 4x – 2

quotient = 3x2 + 5x – 2, remainder = 0 Ans.13. If three zeroes of a polynomial

x4 – x3 – 3x2 + 3x are 0, 3 and 3,-then find the fourth zero. [3]

Sol. Let P (x) = x4 – x3 – 3x2 + 3x

Given, 0, 3, 3- are three zeroes, so

x, x – 3 and x + 3 are factors of f(x)

[If a be the root of f(x) = 0Then x – a be a factor of f(x).]

Here, ( 3) ( 3)x x x+ - will also be thefactor of P (x).Or, x (x2 – 3) will be the factor of P (x).then by long division, we have

x x3 – 3 x x x x x4 3 2 – – 3 + 3 – 1x x4 2 – 3

– +– x x3 + 3– x x3 + 3+ – 0

) (

\ quotient = (x – 1)



Thus P(x) = q(x) (x) ( )( )3 3 0x x- + +

Thus zeroes of P(x) are given by puttingP(x) = 0So fourth zero be given by x – 1 = 0

Þ x = 1

Hence four zeroes will be 1, 0, 3, 3.-

Ans.14. Solve the following pair of equations by

reducing them to a pair of linearequations : [3]

1 42

x y- =

1 39

x y+ =

Sol. Given, equations are ; 1 4x y- = 2

1 3x y

+ = 9

Let1 1,ux y

= = u

Given equations reduces to,So, u – 4u = 2 ...(i)u + 3u = 9 ...(ii)On solving eq. (i) and eq. (ii) ; we haveu – 4u = 2u + 3u = 9– – –– 7u = – 7

Þ u = 1Putting the value of u in eq. (i) ; we get

Þ u – 4u = 2Þ u – 4 × 1 = 2i.e. u – 4 = 2i.e. u = 2 + 4Þ u = 6

So u = 1Þ 1 1y

= Þ y = 1

and u = 6Þ1 6,x

= Þ16

x =

Hence,16

x = and y = 1 Ans.

15. DABC is a right angled triangle in whichÐB = 90º. D and E are any point on ABand BC respectively. Prove thatAE2 + CD2 = AC2 + DE2. [3]

Sol. In DABC, ÐB = 60º and D, E are point ofAB, BC respectively.

To prove :AC2 + DE2 = AE2 + CD2

In DABC by using Pythagoras theorem,AC2 = AB2 + BC2 ...(i)In DABE by using Pythagoras theoremAE2 = AB2 + BE2 ...(ii)In DBCD by Pythagoras theoremCD2 = BD2 + BC2 ...(iii)In DDBE by Pythagoras theoremDE2 = DB2 + BE2 ...(iv)Adding eq. (i) and eq. (iv)AC2 + DE2 = AB2 + BC2 + BD2 + BE2

= AB2 + BE2 + BC2 + BD2

AC2 + DE2 = AE2 + CD2

[From eq. (ii) and eq. (iii)]Hence Proved.


Arundeep’s Solved Papers Mathematics 2016 (Term I)11316. In the given figure, RQ and TP are

perpendicular to PQ, also TS^ PR provethat ST.RQ = PS.PQ. [3]

Sol.

In DRPQÐ1 + Ð2 + Ð4 = 180ºÐ1 + Ð2 + 90º = 180ºÐ1 + Ð2 = 180º – 90ºÐ1 = 90º – Ð2 ...(i)

Q TP ^ PQ\ ÐTPQ = 90ºÞ Ð2 + Ð3 = 90ºÐ3 = 90º – Ð2 ...(ii)From eq. (i) and eq. (ii)Ð1 = Ð3Now in DRQP and DPST, we haveÐ1 = Ð3 [Proved above]Ð4 = Ð5 [Each 90º]So by AA axiom of similarityDRQP ~ DPST

Thus their corresponding sides areproportional.

STQP

PS=RQ [By c.p.c.t]

Þ ST.RQ = PS.PQ Hence Proved.

17. If sec A2= ,3 find the value of

tan A 1 sin Acos A tan A

++ [3]

Sol. Given, sec A23

=

In DABC, we haveAC2 = AB2 + BC2

Þ 22 2 2( 3) BC= +

Þ 4 = 3 + BC2

Þ BC2 = 4 – 3Þ BC2 = 1Þ BC = 1

So, tan A1 ;3

= cos A3 ;

2= sin A

12

=

\tan A 1 sin Acos A tan A

++

1 113 2

1332

+= +

32 2

133

= + 2 3 33 2

= + 4 9 36

+= Ans.


Arundeep’s Solved Papers Mathematics 2016 (Term I)11418. Prove that : [3]

sec2 q – cot2 (90º – q) = cos2 (90º – q) + cos2 q.Sol. To prove :

sec2 q – cot2 (90º – q) = cos2 (90º – q) + cos2 qL.H.S. = sec2 q – cot2 (90º – q)= sec2 q – [cot (90 – q)]2

= sec2 q – (tan q)2 [_ cot (90° – q) = tan q]= sec2 q – tan2 q [_ sec2 q = 1 + tan2 q]= 1R.H.S. = cos2 (90° – q) + cos2 q= [cos (90° – q)]2 + cos2 q= (sin q)2 + cos2 q [_ cos (90° – q) = sin q]= sin2 q + cos2 q= 1Hence, L.H.S. = R.H.S. Hence Proved

19. For the month of February, a class teacher of Class IX has the following absentee recordfor 45 students. Find the mean number of days, a student was absent.

Number of days of absent 0 – 4 4 – 8 8 – 12 12 – 16 16 – 20 20 – 24Number of students 18 3 6 2 0 1

[3]Sol. The table of values is given as under :

C.I. f xi i (mid-value) d xi i = – A f × di i0 – 4 18 2 – 12 – 2164 – 8 3 6 – 8 – 248 – 12 6 10 – 4 – 2412 – 16 2 A = 14 0 0016 – 20 0 18 4 0020 – 24 1 22 8 08

Sfi = 30 Sf di i = – 256

Mean A i i

i

f df

S= +

S2561430-æ ö= + ç ÷è ø

= 14 – 8.53 = 5.47 Ans.

20. Find the missing frequency (x) of the following distribution, if mode is 34.5 :

Marks obtained 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50Number of students 4 8 10 x 8

[3]



Sol. C.I. Frequency0 – 10 4 Here max frequency = 10 which lies in 20 – 3010 – 20 8 = f0 \ Modal class be 20 – 3020 – 30 10 = f1 \ l = 20, f1 = 10 ; f0 = 8 ; f2 = x ; h = 1030 – 40 x = f240 – 50 8

\ Mode 1 0

1 0 22f f

l hf f f-æ ö

= + ç ÷- -è ø

Þ 34.510 820 10

20 8 x-æ ö= + ç ÷- -è ø Þ 34.5

220 1012 xæ ö= + ç ÷-è ø

Þ14.5

120

12 x=- Þ 20 = 14.5 (12 – x) Þ

2014.5 = 12 – x

Þ4029 = 12 – x Þ x

401229

= - Þ x348 40

29-=

Þ x30829

=

Þ x = 10.62 Ans.

SECTION D

21. Prove that 5 is an irrational number..

Hence show that 3 + 2 5 is also anirrational number. [4]

Sol. Let 5 be a rational number..

So, 5pq

=

On squaring both sides ; we have

52

2pq

=

Þ 5q2 = p2

Þ 5 is a factor of p2

Þ 5 is a factor of p.Now, again let p = 5c.

Þ 5q2 = 25c2

Þ q2 = 5c2

Þ 5 is factor of q2

Þ 5 is a factor of q.Here 5 is a common factor of p, q whichcontradicts the fact that p, q are co-prime.

Hence our assumption is wrong, 5 is anirrational number.

Now we have to show that 3 2 5+ is anirrational number. So let us assume

3 2 5+ is a rational number..

Þ 3 2 5+pq

= ; where p, q Î I and q ¹ 0

and (p, q) = 1

Þ 2 5 3pq= -

Þ 2 53p q

q-=



Þ 53

2p q

q-=

32

p qq-

is in the rational form ofpq so 5 is a rational number but we have already proved that

5 is an irrational number so contradiction arises. Here our supposition is wrong that 3 2 5+is a rational number. So we can say that 3 2 5+ is an irrational number.. Hence Proved.

22. Obtain all other zeroes or the polynomial x4 + 6x3 + x2 – 24x – 20, if two of its zeroes are+ 2 and – 5. [4]

Sol. Given, 2 – 5 are the zeroes of polynomialp (x) = x4 + 6x3 + x2 – 24x – 20 By long division method, we haveSo (x – 2) and (x + 5) are factors of p (x)

Þ (x – 2) (x + 5) is also a factor of p (x)So (x – 2) (x + 5) = x2 + 3x – 10So, by Euclid algorithm, we haveDividend = Divisor × Quotient + Remainderx4 + 6x3 + x2 – 24x – 20 = (x2 + 3x – 10) × (x2 + 3x + 2) + 0= (x2 + 3x – 10) (x2 + 2x + x + 2)= (x2 + 3x – 10) [x (x + 2) + 1 (x + 2)]= (x2 + 3x – 10) (x + 2) (x + 1)So other zeroes are – 2 and – 1. Ans.[We know that if x = a be a root of f(x) = 0Then x – a be a factor of f(x)]

23. Draw graph of following pair of linear equations :y = 2 (x – 1)4x + y = 4Also write the coordinate of the points where these lines meets x-axis and y-axis. [4]

Sol. Given line be, y = 2 (x – 1) ...(1)So,

x 1 2 3 4y 0 2 4 6

Ploting all points (1, 0), (2, 2), (3, 4) on graph paper and join them by straight line to get thegraph for line (1).and for, 4x + y = 4

or y = 4 – 4x ...(2)

x x x x x x x x2 4 3 2 2 + 3 – 10) + 6 + – 24 – 20( + 3 + 2x x x4 3 2 +3 – 10– – +

3x x x2 2 + 11 – 24 – 203x x x3 2 + 9 – 30– – +

2x x2 + 6 – 202x x2 + 6 – 20

– – + 0



x 1 2 1/2y 0 – 4 2

Ploting all the points (1, 0), (2, –4) and (½, 2) on graph paper and join them by line to get a graphfor eqn. (2).

7

6

5

4

3

2

1

0–1

–2

– 4

1 2 3 4 5 6– 4 –3 –2 –1

–3

Y

X

yx

=2(

–1)

y x= 4 – 4

Co-ordinates of point where lines meets axes :Line y = 2 (x – 1) : x-axis (i.e. y = 0) = (1, 0)and y-axis (i.e. x = 0) = (0, – 2)Line 4x + y = 4 : x-axis (i.e. y = 0) = (1, 0)

and y-axis (i.e. x = 0) = (0, 4) Ans.24. A boat goes 30 km upstream and 44 km downstream in 10 hours. The same boat goes 40

km upstream and 55 km downstream in 13 hours. On this information some student guessedthe speed of the boat in still water as 8.5 km/h and speed of the stream as 3.8 km/h. Do youagree with their guess ? Explain what do we learn from the incident ? [4]

Sol. Let the speed of boat = x km/hr.Let the speed of stream = y km/hr.Speed of boat in upstream = (x – y) km/hr.Speed of boat in downstream = (x + y) km/hr.

Time taken to cover 30 km upstream30= hrs.

x y-


Arundeep’s Solved Papers Mathematics 2016 (Term I)118Time taken to cover 44 km downstream

44 hrs.x y

=+

According to question, we haveTotal time taken = 10 hrs.

30 44x y x y

+- + = 10 ...(i)

Now, Time taken to cover 55 km

downstream55 hrs.

x y=

+Time taken to cover 40 km upstream

40 hrs.x y

=-

Given Total time taken = 3 hrs.

40 55x y x y

+- +

= 13 ...(ii)

Solving eq. (i) and eq. (ii)

Let1 ,u

x y=

-1 .

x y=

+u

Then eqn. (i) and (ii) reduces to,30u + 44u = 1040u + 55u = 13or 15u + 22u = 5 ...(iii)

8u + 11u135

= ...(iv)

Multiplying eq. (iii) by 8 and eq. (iv) by15, we get

120u v + 176 = 40120u v + 165 = 39

– – –11v = 1

Þ u1

11=

Putting the value of u in eq. (iii) ; we have15u + 22u = 5

Þ115 22

11u + ´ = 5

Þ 15u + 2 = 5 Þ 15u = 3

Þ u3

15= or u

15

=

Now, u1

11=

Þ1

x y+1

11=

Þ x + y = 11 ...(v)

And u15

=

Þ1

x y-15

=

Þ x – y = 5 ...(vi)On adding eq. (v) and (vi) ; we havex + y + x – y = 11 + 5

Þ 2x = 16 or x = 8Put the value of x in eq. (v) ; we have

Þ 8 + y = 11 Þ y = 11 – 8i.e. y = 3

The speed of boat in still water = 8 km/hr.The speed of stream = 3 km/hr.We learns that the speed of boat is slow inupstream and fast in downstream. Ans.

25. In an equilateral DABC, E is any point

on BC such that BE1= BC.4 Prove that

16AE2 = 13AB2. [4]

Sol. Given BE1 BC4

=

Draw AD ^ BC

In DAED by pythagoras theorem,AE2 = AD2 + DE2 ...(i)

A

B CDE



In DADB ; we haveAB2 = AD2 + BD2

Þ AB2 = AE2 – DE2 + BD2 [From (i)]= AE2 – DE2 + (BE + DE)2

Þ AB2 = AE2 – DE2 + BE2 + DE2 + 2BE.DEÞ AB2 = AE2 + BE2 + 2BE.DE

Þ AB22

2 BC BCAE 2 (BD BE)4 4

æ ö= + + -ç ÷è ø

Þ AB22

2 BC BC BC BCAE16 2 2 4

æ ö= + + -ç ÷è ø

Þ AB22

2 AB AB 2AB ABAE16 2 4

-é ù= + + ê úë û

ABC be an equilateral triangleAB BC AC

Dé ùê ú\ = =ë ûQ

Þ AB22 2

2 AB AB ABAE16 2 4

= + + ´

Þ2 2

2 AB ABAB16 8

- - = AE2

Þ2 2 216AB AB 2AB

16- -

= AE2

Þ 16AB2 – 3AB2 = 16AE2

i.e. 13AB2 = 16AE2 Hence Proved.26. In the figure, ifÐABD =ÐXYD =ÐCDB

= 90º. AB = a, XY = c and CD = b, thenprove that c (a + b) = ab. [4]

Sol. To prove : c (a + b) = abIn DABD and DDXY ; we haveÐB = ÐXYD [Each 90º]ÐXDY = ÐADB [Common]So by AA axiom of similarityDDAB ~ DDXY

\DYDB

XYAB

=

Þ DY (BD)ca

= ...(i)

In DBCD & DBYX ; we haveÐXYB = ÐD [Each 90º]ÐCBD = ÐXBY [Common]So by AA axiom of similarity,DBYX ~ DBDCThus their corresponding sides areproportional.

ÞBYBD

XYCD

=

Þ BY (BD)cb

= ...(ii)

Adding eq. (i) and eq. (ii) ; we have

Þ DY + BY (BD) (BD)c ca b

= +

Þ BD BD c ca bé ù= +ê úë û

ÞBDBD

cb caab+é ù= ê úë û

1( )c a b

ab+=

c (a + b) = ab Hence Proved.


Arundeep’s Solved Papers Mathematics 2016 (Term I)12027. In the DABC (see figure), ÐA = right

angle, AB = x and BC 5.x= +Evaluatesin C · cos C · tan C + cos2 C · sin A [4]

Sol. In DABC, by pythagoras theorem ; we haveBC2 = AC2 + AB2

2( 5)x + 2 2( ) ACx= +

Þ x + 5 = x + AC2

5 = AC2 or AC 5=

\ sin C ;5

xx

=+

cos C5 ;

5x=

+

tan C5x=

and sin A = sin 90º = 1Then, sin C cos C tan C + cos2 C sin A

25 5 ·1

5 5 5 5x x

x x x

æ ö= + ç ÷

+ + +è ø

55 5

xx x

= ++ +

55

xx

+=+ = 1 Ans.

28. Ifcos Bsin A

n= andcos B ,cos A

m= then show

that (m2 + n2) cos2 A = n2. [4]

Sol. Given, ncos B ;sin A

= cos Bcos A

m =

So, n22

2cos B ;sin A

=2

22

cos Bcos A

m =

L.H.S. = (m2 + n2) cos2 A

2 22

2 2cos B cos B cos Acos A sin Aæ ö

= +ç ÷è ø

2 2 2 22

2 2(sin A cos B cos A cos B) cos A

cos Asin A+= ´

2 2 2

2cos B (sin A cos A)

sin A+=

2

2cos Bsin A

=

= n2 = R.H.S. Hence Proved.

29. Prove that :

2sec A 1 sin Asec A 1 1 cos A

- æ ö= ç ÷+ +è ø

= (cot A – cosec A)2 [4]

Sol. L.H.S.sec A 1sec A 1

-=+

1 1 cos A1cos A cos A

1 1 cos A1cos A cos A

--= =

++

1 cos A1 cos A-=+



(1 cos A) (1 cos A)(1 cos A) (1 cos A)- +=+ + [on rationalization]

2

21 cos A

(1 cos A)-=+

2

2sin A

(1 cos A)=

+

2sin A1 cos Aæ ö= ç ÷+è ø

Hence Proved

And,2sin A

1 cos Aæ öç ÷+è ø

sin A (1 cos A)1 cos A (1 cos A)é - ùæ ö= ´ç ÷ê ú+ -è øë û

2

2sin A (1 cos A)

1 cos A-é ù= ê ú-ë û

2

2sin A (1 cos A)

sin A-é ù= ê úë û

21 cos Asin A-é ù= ê úë û

=2

1 cos Asin A sin Aé ù

-ê úë û

= (cosec A – cot A)2 = (– 1)2 [cot A – cosec A]2 = [cot A – cosec A]2 = R.H.S.

Hence Proved.

30. Following table shows marks (out of 100) of students in a class test :

Marks No. of studentsMore than or equal to 0 80More than or equal to 10 77More than or equal to 20 72More than or equal to 30 65More than or equal to 40 55More than or equal to 50 43More than or equal to 60 28More than or equal to 70 16More than or equal to 80 10More than or equal to 90 8More than or equal to

1000

Draw a ‘more than type’ ogive. From the curve, find the median. Also, check the value ofthe median by actual calculation. [4]


Arundeep’s Solved Papers Mathematics 2016 (Term I)122Sol.

More than type C.I. No. of Students Frequency c.f.More than or equal to 0 0 – 10 80 3 3More than or equal to 10 10 – 20 77 5 8More than or equal to 20 20 – 30 72 7 15More than or equal to 30 30 – 40 65 10 25More than or equal to 40 40 – 50 55 12 37More than or equal to 50 50 – 60 43 15 52More than or equal to 60 60 – 70 28 12 64More than or equal to 70 70 – 80 16 06 70More than or equal to 80 80 – 90 10 02 72More than or equal to 90 90 – 100 8 08 80More than or equal to 100 100 – 110 0 00

Taking lower limits x-axis and no. of students along y–axis. Draw the ordered pairs (0, 80),(10, 77), (20, 72), (30, 65), (40, 55), (50, 43), (60, 28), (70, 16), (80, 10), (90, 8), (100, 0) ongraph paper and join them by free hand to give the required more than type ogive.

HereN2 =

802 = 40, So take a point P representing 40 on y-axis, draw a straight line || to x-axis

meeting ogive at Q. From Q draw a line ^ to x-axis meeting x-axis at R, Then abscissa of Rgiven the required median

\ Median = 52

P Q

R

Median by actual calculation :



Here N = 80 (even)\N2

802

= = 40

So c.f. which is just greater than 40 be 52So median class will be 50 – 60and corresponding class interval be50 – 60.Thus median class be 50–60

Here, l = 50, h = 10, f = 15, c.f = 37,

\ Median

N .2

c fl h

f

é ùæ ö-ç ÷ê úè ø= + ´ê úê úë û

(40 37)50 1015-é ù= + ´ê úë û

350 1015

= + ´

= 50 + 2 = 52Hence Verified.

31. From the following data find the medianage of 100 residents of a colony who tookpart in swachch bharat abhiyan : [4]

Age (in yrs.) Morethan or equal to No. of residents

0 5010 4620 4030 2040 1050 3

Sol. First convert the given table into C.I. Table.

C.I. Frequency f c.f.0 – 10 4 (50-46) 410 – 20 6 (46-40) 1020 – 30 20 (40-20) 3030 – 40 10 (20-10) 4040 – 50 7 (10-3) 4750 – 60 50

i

f = N = 503

å i

Here,N2

50= 252

=

and c.f. which just greater than 25 be 30which lies in class interval 20 – 30.

\ Median class be 20 – 30.

Thus l = 20 ; f = 20 ; c.f. = 10, h = 10

\ Median

N .2

c fl h

f

é ùæ ö-ç ÷ê úè ø= + ´ê úê úë û

(25 10)20 1020-é ù= + ê úë û

15202

= +

= 27.5 Ans.



(ii) The question paper consists of 31 questions divided into four sections – A, B, C and D.(iii) Section A contains 4 questions of 1 mark each, Section-B contains 6 questions of 2 marks each.



MATHEMATICS 2016 TERM II (OUTSIDE DELHI)SET I


1. In given figure, PQ is a tangent at a pointC to a circle with centre O. If AB is adiameter and ÐCAB = 30º, find ÐPCA.

30°

Sol. Construction : Join AO.Given : PQ is tangent. AB is diameterÐCAB = 30º.To Find : ÐPCASolution : In DAOC, AO = CO

(Q Equal radii)

ÐCAO = ÐOCA (Q Angles opposite toequal sides are equal)or ÐCAB = ÐOCABut, ÐCAB = 30º So, ÐOCA = 30º ...(i)Since, OC ^ PQ

(Q Tangent is perpendicular to radius atpoint of contact)

Þ ÐPCO = 90ºÞ ÐOCA + ÐPCA = 90º

Þ 30º + ÐPCA = 90º\ ÐPCA = 60º2. For what value of k will k + 9, 2k – 1 and

2k + 7 are the consecutive terms of anA.P. ?

Sol. Given that k + 9, 2k – 1 and 2k + 7 are inA.P. [if a, b, c are in A.P then b – a = c – b]Then (2k – 1) – (k + 9) = (2k + 7) – (2k – 1)

Þ k – 10 = 8 Þ k = 183. A ladder, leaning against a wall, makes

an angle of 60º with the horizontal. Ifthe foot of the ladder is 2.5 m away fromthe wall, find the length of the ladder.

Sol. Let AB be the ladder of length x and ACbe the wall.

In right DABC,BCx

= cos 60º

Þ2.5x

12

= Þ x = 2 × 2.5 = 5 cm

60°

A

CB 2.5 m

Thus, length of the ladder is 5 m.



Arundeep’s Solved Papers Mathematics 2016 (Outside Delhi)1254. A card is drawn at random from a well

shuffled pack of 52 playing cards. Findthe probability of getting neither a redcard nor a queen.

Sol. Number of total possible outcomes whenone card is drawn = 52Number of favourable outcomes when cardis neither red nor queen = 52 – 28 = 24[we have 26 red cards and 2 black queens]

\ Required probability

Favourable outcomesTotal possible outcomes

=

24 652 13

= =

5. If – 5 is a root of the quadratic equation2x2 + px – 15 = 0 and the quadraticequation p (x2 + x) + k = 0 has equal roots,find the value of k.

Sol. Q – 5 is the root of the quadratic equation2x2 + px – 15 = 0

Þ 2 (– 5)2 + p (– 5) – 15 = 0[_ if a be a root of f(x) = 0 then f(a) = 0]

Þ 50 – 5p – 15 = 0 Þ 35 – 5p = 0Þ 5p = 35 Þ p = 7

Now, given that equation p (x2 + x) + k = 0has equal rootsi.e. 7 (x2 + x) + k = 0 has equal rootsi.e. 7x2 + 7x + k = 0 has equal roots

Þ 72 – 4 × 7 × k = 0

[Q For equal roots, D = 0, i.e. b2 – 4ac = 0Here a = 7, b = 7 and c = k]

Þ 7 (7 – 4k) = 0 Þ 74

k =

6. Let P and Q be the points of trisection ofthe line segment joining the pointsA (2, – 2) and B (– 7, 4) such that P isnearer to A. Find the coordinates ofP and Q.

Sol. Let A (2, – 2), B (– 7, 4) be given points.

Let P (x, y), Q (x¢, y¢) are point of trisection.

A (2, – 2) B (– 7, 4)

P Q1 : 2 2 : 1

Then using section formulaP divides AB in the ratio 1 : 2 internallyCoordinates of P are

2 2 1( 7) ( 2) (2) 1(4),1 2 1 2

´ + - - +æ öç ÷è + + ø

or (– 1, 0)Q is mid point of PB. So using mid pointformula coordinates of Q are

1 7 0 4,2 2

- - +æ öç ÷è ø

or (– 4, 2)

7. In figure, a quadrilateral ABCD is drawnto circumscribe a circle, with centre O,in such a way that the sides AB, BC, CDand DA touch the circle at the points P,Q, R and S respectively. Prove thatAB + CD = BC + DA.

Sol. We know that tangents drawn to a circlefrom an outer points are equal.So, AP = AS, BP = BQ, CR = CQ and DR= DS.Now, considerAP + BP + CR + DR = AS+ BQ + CQ + DS

Þ AP + PB + CP + PD = AS + SD + BQ + QCÞ AB + CD = AD + BC

Hence proved.8. Prove that the points (3, 0), (6, 4) and

(– 1, 3) are the vertices of a right angledisosceles triangle.


Arundeep’s Solved Papers Mathematics 2016 (Outside Delhi)126Sol. Let the triangle be DABC as shown in

figure.

Using distance formula, we have

AB 2 2(3 6) (0 4)= - + -

= 2 2( 3) 4 9 16 25- + = + = = 5

BC 2 2(6 1) (4 3)= + + -

= 2 27 1 49 1 50 25 2+ = + = = ´

5 2=

CA 2 2( 1 3) (3 0)= - - + -

= 2 2( 4) 3- + = 16 9 25+ = = 5

Here, AB = AC Þ ABC is an isoscelestriangleConsider, AB2 + AC2 = (5)2 + (5)2

= 25 + 25 = 50

and, BC2 2(5 2) 50= =

\ Here, AB2 + AC2 = BC2

Þ DABC is a right angled triangle.

[Q In right D, using Pythagoras theorem(H)2 = (P)2 + (B)2]

where H = hypotenuse, B = base,P = perpendiculars]

9. The 4th term of an A.P. is zero. Provethat the 25th term of the A.P. is threetimes its 11th term.

Sol. Let a be first term and d be the commondifference of the given A.P. Thenwe know that an = a + (n – 1) d

Since an = 0\ 0 = a + (4 – 1) dÞ 0 = a + 3d Þ a = – 3d

Now a25 = a + (25 – 1) d = a + 24d= – 3d + 24d = 21d = 3 × 7dHence, a25 = 3 × a11

[Q Since a111 = a + (11 – 1) d

= – 3d + 10d = 7d]10. In given figure, from an external point

P, two tangents PT and PS are drawn toa circle with centre O and radius r. IfPO = 2r, show that ÐOTS = ÐOST = 30º.

Sol. Let ÐTOP = qIn right triangle OTP we have

\ cos qOT 1OP 2 2

rr

= = = = cos 60º

Þ q = 60ºHence ÐTOS = 2 × 60 = 120º

[Q ÐTOP = ÐPOS as angles opposite toequal tangents are equal]In DOTS, we have OT = OS

[Q Equal radii]

Þ ÐOTS = ÐOST

[Q Angle opposite to equal sides are equal]

In DOTS, we haveÐOTS + ÐOST + ÐTOS = 180º

Þ 2ÐOST = 60º\ ÐOST = ÐOTS = 30º Hence proved.

11. In figure, O is the centre of a circle suchthat diameter AB = 13 cm and AC = 12 cm.


Arundeep’s Solved Papers Mathematics 2016 (Outside Delhi)127BC is joined. Find the area of the shaded region. (Take p = 3.14)

Sol. Here, BC2 = AB2 – AC2 = 169 – 144 = 25\ BC = 5\ Area of shaded region = Area of semicircle – Area of right triangle ABC

21 1 AC BC2 2

r= ´ p - ´

[Since diameter of semi-circle = AB = 13 cm]

21 13 13.14 12 52 2 2

æ ö= ´ - ´ ´ç ÷è ø= 66.33 – 30 = 36.33 cm2

12. In figure, a tent is in the shape of a cylinder surmounted by aconical top of same diameter. If the height and diameter ofcylindrical part are 2.1 m and 3 m respectively and the slantheight of conical part is 2.8 m, find the cost of canvas needed tomake the tent if the canvas is available at the rate of ` 500/sq.

metre.22Use π =7

é ùê úë û

Sol. Given radius of cylindrical part = radius of conical part = r =32

m = 1.5 m

and height of cylindrical part = h = 2.1 m\ Slant height of cone = l = 2.8 m

Area for canvas needed = curved surface area of cylinder + curved surface area of cone= 2prh + prl

22 222 1.5 2.1 1.5 2.87 7

= ´ ´ ´ + ´ ´ 22 22[6.3 4.2] 10.57 7

= + = ´ = 33 m2

Thus required cost of canvas needed to make the tent at the rate of `500 per m2

= 33 × 500 = ` 1650013. If the point P (x, y) is equidistant from the points A (a + b, b – a) and B (a – b, a + b). Prove

that bx = ay.Sol. Given, PA = PB Þ PA2 = PB2

Applying distance formula, we haveÞ (a + b – x)2 + (b – a – y)2 = (a – b – x)2 + (a + b – y)2

Þ (a + b)2 + x2 – 2ax – 2bx + (b – a)2 + y2 – 2by + 2ay

2.1 m

3 m

2.8 m

b – a


Arundeep’s Solved Papers Mathematics 2016 (Outside Delhi)128= (a – b)2 + x2 – 2ax + 2bx + (a +b)2

+ y2 – 2ay – 2byÞ 4ay = 4bx Þ ay = bx or bx = ay

Hence proved.14. In figure, find the area of the shaded

region, enclosed between two concentriccircles of radii 7 cm and 14 cm where

ÐAOC = 40º.22Use π =7

é ùê úë û

Sol. Given outer radius of circle = R = 14 cmRadius of inner circle = r = 7 cmHere q¢= 360° – q = 360° – 40° = 320°

\ Area of shaded region

2 2360º (R )360º

r- q= ´ p -

2 2320º [(14) (7) ]360º

= ´ p -

8 22 (196 49)9 7

= ´ - 8 22 1479 7

= ´ ´

12323

= = 410.67 cm2

15. If the ratio of the sum of first n terms oftwo A.P.’s is (7n + 1) : (4n + 27), find theratio of their mth terms.

Sol. Let Sn and Sn¢ be the sum of n terms of twoA.P.’s Let a, a¢ and d, d¢ be first terms andcommon difference of two A.P.’s. Then

SS

n

n¢

[2 ( 1) ]2

[2 ( 1) ]2

n a n d

n a n d

+ -=

¢ + - ¢

17 12

1 4 272

na d nn na d

-æ ö+ ç ÷ +è ø= =- +æ ö¢ + ¢ç ÷è ø

...(i)

Sincem

m

tt ¢

( 1)( 1)

a m da m a

+ -=¢ + - ¢

[Q Let tm, tm be mth terms of two A.P.’s]

So replacing1

2n -

by m – 1,

i.e. n = 2m – 1 in eqn. (i) ; we get

m

m

tt ¢

( 1)( 1)

a m da m d

+ -=¢ + - ¢

7 (2 1) 1 14 64 (2 1) 27 8 23

m mm m

- + -= =- + +

Thus, the ratio of their mth terms is14m – 6 : 8m + 23.

16. Solve for x :

1 1( 1) ( 2) ( 2) ( 3)x x x x

+- - - -

2 ,3

=

x ¹ 1, 2, 3Sol. Given eqn. be

1 1( 1) ( 2) ( 2) ( 3)x x x x

+- - - -

2 ,3

=

x ¹ 1, 2, 3

Þ3 1

( 1) ( 2) ( 3)x x

x x x- + -

- - -23

=

Þ(2 4)

( 1) ( 2) ( 3)x

x x x-

- - -23

=



Þ2 ( 2)

( 1) ( 2) ( 3)x

x x x-

- - -23

=

Þ 3 = (x – 1) (x – 3)Þ x2 – 4x + 3 = 3 [_ x – 2 ¹ 0]Þ x2 – 4x = 0Þ x (x – 4) = 0Þ x = 0 or x = 4

17. A conical vessel, with base radius 5 cmand height 24 cm, is full of water. Thiswater is emptied into a cylindrical vesselof base radius 10 cm. Find the height towhich the water will rise in the

cylindrical vessel.22Use π =7

é ùê úë û

Sol. Given base radius of conical vessel =r = 5 cmand height of vessel = h = 24 cmand Volume of water in conical vessel = h

213

r h= p

31 22 5 5 24 cm3 7

= ´ ´ ´ ´

Let height of water in cylindrical vessel beh then volume of water in cylinder = pR2hWhere R = radius of cylindrical vessel = 10 cm

227

= × 10 × 10 × h cm3

A.T.Q., we haveVolume of water in conical vessel = Volumeof water in cylindrical vessel

Þ1 22 5 5 243 7

´ ´ ´ ´ 22 10 107

h= ´ ´ ´

Þ h5 5 24 2 cm3 10 10

´ ´= =´ ´

\ Required height to which water rises incylindrical vessel = 2 cm

18. A sphere of diameter 12 cm, is droppedin a right circular cylindrical vessel,partly filled with water. If the sphere iscompletely submerged in water, thewater level in the cylindrical vessel rises

by53 cm.9 Find the diameter of the

cylindrical vessel.

Sol. Given radius of sphere = R =122 cm = 6 cm

We know that,

\ Volume of sphere 34 R3

= p

34 (6)3

= p ´ 34 216 cm3

= p ´

Let radius of cylindrical vessel be r cm then,volume of water in cylinder

= pr2 h 2 332 cm9

r= p ´

[Since rise in water level in cylindrical

vessel = h =539 cm =

329 cm

According to question, we havevolume of sphere = volume of cylindericalvessel

4 2163

p ´ 2 329

r= p ´ ´

Þ r2 4 216 93 32

´ ´=´ = 27 × 3 = 81

Þ r2 = 81 = 92

i.e. r = 9 cm\ Required diameter of cylindrical vessel

= 2r = 18 cm19. A man standing on the deck of a ship,

which is 10 m above water level, observesthe angle of elevation of the top of a hillas 60º and the angle of depression of thebase of hill as 30º. Find the distance of


Arundeep’s Solved Papers Mathematics 2016 (Outside Delhi)130the hill from the ship and the height ofthe hill.

Sol. Let ED be the water level and BE = 10 mLet CD be the hill of height h from waterlevel.Let AB = xIn right DDEB, we have

BEDE = tan 30º Þ

10x

13

=

Þ x 10 3 m=

C

B

Ex D

A

( – 10) mh

h m

Now, in DCAB,CA

x = tan 60º

Þ 1010 3h -

3= [From (i)]

Þ h – 10 = 30 Þ h = 40 m

So, distance of hill from ship 10 3 m= andthe height of the hill = 40 m.

20. Three different coins are tossed together.Find the probability of getting(i) exactly two heads(ii) at least two heads(iii) at least two tails.

Sol. Possible outcomes when three coins aretossedHHH, HHT, HTT, TTT, THH, TTH, HTH,THT

(i) Number of exactly two heads are HHT,HTH and THH.

\ P (exactly two heads)38

=

(ii) In case of at least two heads, outcomes areHHT, HTH, THH and HHH.

\ P (at least two heads)4 18 2

= =

(iii) In case of at least two tails, outcomes areTTH, THT, HTT and TTT.

\ P (at least two tails)4 18 2

= =

21. Due to heavy floods in a state, thousandswere rendered homeless. 50 schoolscollectively offered to the stategovernment to provide place and thecanvas for 1500 tents to be fixed by thegovernment and decided to share thewhole expenditure equally. The lowerpart of each tent is cylindrical of baseradius 2.8 m and height 3.5 m, withconical upper part of same base radiusbut of height 2.1 m. If the canvas used tomake the tents costs ` 120 per sq. m, findthe amount shared by each school to setup the tents. What value is generated by

the above problem ?22Use π =7

é ùê úë û

Sol. Given radius of cylindrical part = radius ofconical part = 2.8 m

and h = height of cylindrical part = 3.5 mH = height of conical part = 2.1 m

Slant height of conical part 2 2Hr= +

2 2(2.8) (2.1) 7.84 4.41= + = +

12.25 3.5 m= =\ Area of tents = Curved surface area of

cylindrical part + Curved surface area ofconical part= 2prh + prl



22 222 2.8 3.5 2.8 3.57 7

= ´ ´ ´ + ´ ´

2223 2.8 3.5 92.4 m7

= ´ ´ ´ =

2.1 m

2.8 m

3.5 m

2.8 m

\ Canvas required for 1500 tents= 1500 × 92.4 = 138600 m2

Cost of 1500 tents = (1500 ×92.4) × 120= ` 16632000

[Q Making tents costs ` 120 per sq. m]

Thus, share of each school16632000

50=

= ` 332640School authorities are concerned aboutsafety of children and their families.

22. Prove that the lengths of tangents drawnfrom an external point to a circle areequal.

Sol. Given : A circle C (O, r), P is a point outsidethe circle and PA and PB are tangents to acircle.To Prove : PA = PBConstruction : Draw OA, OB and OP.Proof : Consider triangles OAP and OBP.ÐOAP = ÐOBP = 90º ...(i)

[Radius is perpendicular to the tangent atthe point of contact]

OA = OB (radii) ...(ii)OP is common ...(iii)

\ DOAP @DOBP(RHS axiom of congruency)[from (i), (ii) and (iii)]

Hence, AP = BP (CPCT)23. Draw a circle of radius 4 cm. Draw two

tangents to the circle inclined at an angleof 60º to each other.

Sol. Steps of construction :1. Draw a circle of radius 4 cm with centre O.2. Take point A on circle. Join OA.3. Draw line AP perpendicular to radius OA.4. Draw ÐAOB = 120º at O.5. Join A and B at P, to get 2 tangents. Here

ÐAPB = 60º.

24. In given figure, two equal circles, withcentres O and O¢, touch each other at X.OO¢ produced meets the circle withcentre O¢ at A. AC is tangent to the circlewith centre O at the point C. O¢D isperpendicular to AC. Find the value of

DO .CO

¢



Sol. AC is tangent to the circle with centre O.In DADO¢and DACO,ÐADO¢ = ÐACO (each 90º)ÐDAO = ÐCAO (common)

r r r

\ By AA axiom of similarityDADO¢ ~ DACOThus their corresponding sides areproportional.

\AOAO

¢ DOCO

¢=

[Q corresponding parts of similar triangle]

AO = AO¢ + O¢X + XO = r + r + r = 3r

ÞDOCO

¢3rr

=

\DOCO

¢ 13

=

25. Solve for x :

1 2 4 ,1 2 4x x x

+ =+ + + x ¹ – 1, – 2, – 4


1 2x x+

+ +4

4x=

+

Þ2 2 2

( 1) ( 2)x xx x+ + ++ +

44x

=+

Þ3 4 4

( 1)( 2) ( 4)x

x x x+ =

+ + +

Þ (3x + 4) (x + 4) = 4 (x2 + 3x + 2)Þ 3x2 + 16x + 16 = 4x2 + 12x + 8Þ x2 – 4x – 8 = 0

on comparing with ax2 + bx + c = 0We have a = 1 ; b = –4 and c = –8Then by quadratic formula, we have

x =2 4

2b b ac

a- ± -

Þ x4 16 32

2± +=

Þ x4 4 3 2 2 3

2±= = ±

26. The angle of elevation of the top Q of avertical tower PQ from a point X on theground is 60º. From a point Y, 40 mvertically above X the angle of elevationof the top Q of tower is 45º. Find theheight of the tower PQ and the distancePX. (Use 3 = 1.73)

Sol. Let height of tower PQ be h.Let z be the distance between X and P.

Q XPRY is a rectangle.\ RP = XY = 40 m and PX = YR = z

In right DQPX,PQPX = tan 60º

Þ 3hz

= Þ 3h

= z

( – 40) mh

40 m

z

R h



In right DQRY,QRYR = tan 45º

Þ40h

z-

= 1 Þ h – 40 = z ...(ii)

From (i) and (ii), we get

3h

= h – 40

Þ h 3 40 3h= -

Þ 3h h- 40 3=

Þ ( 3 1)h - 40 3=

Þ h40 3 40 3 ( 3 1)

( 3 1) ( 3 1) ( 3 1)+= =

- - +

Þ h40 (3 3)

2+=

Þ h = 20 (3 + 1.73) = 20 × 4.73 = 94.6 m...(iii)

So, height of the tower PQ = 94.6 mand the distance PX = 94.6 – 49 = 54.6 m[From (ii) and (iii)]

27. The houses in a row are numberedconsecutively from 1 to 49. Show thatthere exists a value of X such that sumof numbers of houses preceding thehouse numbered X is equal to sum of thenumbers of houses following X.

Sol. The A.P. of numbers of houses precedinghouse numbered x is :1 + 2 + 3 + ... + (x – 1)

\ Sum, Sn [2 ( 1) ],2n a n d= + -

where a ® first termd ® common difference

( 1)2

x -= [2 × 1 + (x – 1 – 1) × 1]

\ Sx–1( 1)

2x -= × [2 + x – 2]

( 1)2

x x -=

Now, A.P. of total number of housesfollowing x is :(x + 1) + (x + 2) + ... + 49

\ n = 49 – (x + 1) + 1 = 49 – x

\ Sum of these numbers, Sn [ ],2n a l= +

where l be the last term

S49 – x(49 ) [ 1 49]

2x x-= + +

(49 ) ( 50)2

x x-= +

According to question, we have

( 1)2

x x - (49 ) ( 50)2

x x- +=

Þ x2 – x = 49x + 2450 – x2 – 50xÞ 2x2 = 2450Þ x2 = 1225 = (35)2 Þ x = 35

Justification :Now, A.P. of numbers before housenumbered x = 1 + 2 + .... + 34

\ S3434 [ ]2

a l= + 34 [1 34]2

= ´ + = 17 × 35

= 595Now, A.P. of numbers following housenumbered x = 36 + 37 ... + 49

\ S¢14 [36 49]2

= + = 7 × 85 = 595

Hence, for value of x = 35, the sum ofnumbers of houses preceding housenumbered x is equal to sum of numbers ofhouses following x.

28. In fig. given below the vertices of DABCare A (4, 6), B (1, 5) and C (7, 2). A line-


Arundeep’s Solved Papers Mathematics 2016 (Outside Delhi)134segment DE is drawn to intersect the sides AB and AC at D and E respectively such thatAD AE 1= .AB AC 3

= Calculate the area of DADE and compare it with area of DABC.

Sol. Given :ADAB

13

=

3AD = AB\ 3AD = AD + DB Þ 2AD = DB

ADDB

1=2

Similarly,AEEC

1=2

Thus D divides AB in the ratio 1 : 2 and E divides AC in the ratio 1 : 2.Then by using section formula, we have

Coordinates of D are1(1) 2 (4) 1(5) 2 (6),

1 2 1 2+ +æ ö

ç ÷è + + ø

i.e.173,3

æ öç ÷è ø

Coordinates of E are1(7) 2(4) 1(2) 2(6),

1 2 1 2+ +æ ö

ç ÷è + + ø

i.e.145,3

æ öç ÷è ø

Area of DADE12

= |x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)|

1 17 14 14 174 3 6 6 62 3 3 3 3

é ùæ ö æ ö æ ö= - + - + -ç ÷ ç ÷ ç ÷ê úè ø è ø è øë û

1 4 14 3 52 3 3

é - ùæ ö æ ö= + +ç ÷ ç ÷ê úè ø è øë û1 54 42 3

é ù= - +ê úë û5 sq. units6

=

Area of DABC12

= |[4 (5 – 2) + 1 (2 – 6) + 7 (6 – 5)]|

12

= |[4 × 3 + (– 4) + 7 × 1]|12

= |[12 – 4 + 7]|1 1515 sq. units2 2

= ´ =

Hence, Area ( ADE)Area ( ABC)

DD

55 156

15 6 1212

= = ¸ 5 12 26 15 3

= ´ =

2 2


Arundeep’s Solved Papers Mathematics 2016 (Outside Delhi)13529. A number x is selected at random from

the numbers 1, 2, 3 and 4. Anothernumber y is selected at random from thenumbers 1, 4, 9 and 16. Find theprobability that product of x and y is lessthan 16.

Sol. x can be any one of 1, 2, 3 or 4 and y can beany one of 1, 4, 9 or 16.Total number of cases of xy = 4 × 4 = 16Number of cases when product is less than16 are 1 × 1, 1 × 4, 1 × 9, 2 × 1, 2 × 4, 3 ×1, 3 ×4, 4 × 1, i.e. 8 cases.

\ Total no. of favourable cases = 8\ Required probability

Number of favourable casesTotal number of cases

=8 1

16 2= =

30. In figure, is shown a sector OAP of acircle with centre O, containing Ðq. ABis perpendicular to the radius OA andmeets OP produced at B. Prove that theperimeter of shaded region is

tan sec 1 .180

r pqé ùq + q + -ê úë û

Sol. Length are »AP = 2360 180

rrq p q´ p =° °

Now ABr

= tan q Þ AB = r tan q

OBr

= sec q Þ OB = r sec q

\ PB = OB – r = r sec q – r

\ Perimeter of shaded region = AB + PB + »AP

= r tan q + r sec q – r + 180rp q

tan sec 1180

r pqé ù= q + q - +ê úë û31. A motor boat whose speed is 24 km/h in

still water takes 1 hour more to go 32 kmupstream than to return downstream tothe same spot. Find the speed of thestream.

Sol. Let the speed of the stream be x km/hUpstream case :Speed of boat = (24 – x) km/hTime taken for going 32 km upstream

3224 x

=- hours

Downstream case :Speed of boat = (24 + x) km/hTime taken for going 32 km downstream

3224 x

=+ hours

According to question, we have

32 3224 24x x

-- + = 1

Þ24 2432

(24 ) (24 )x xx x

+ - +é ùê ú- +ë û

= 1

DistanceTime =Speed

é ùê úë ûQ

Þ 2232

576x

xæ öç ÷-è ø

= 1

Þ 64x = 576 – x2 Þ x2 + 64x – 576 = 0Þ x2 + 72x – 8x – 576 = 0Þ x(x + 72) – 8(x + 72) = 0Þ (x + 72) (x – 8) = 0Þ x = 8, x = – 72Þ x = 8 (as speed cannot be negative)

Hence, speed of the stream is 8 km/hr.



10. Solve for x :

2 9x x+ + = 13.

Sol. Given eqn. be 2 9x x+ + = 13

Þ 2 9x + = (13 – x)On squaring both sides, we get

Þ 2x + 9 = (13 – x)2

Þ 2x + 9 = 169 – 26x + x2

Þ x2 – 28x + 160 = 0Þ x2 – 20x – 8x + 160 = 0Þ x(x – 20) – 8 (x – 20) = 0Þ (x – 20) (x – 8) = 0Þ x = 20 or x = 8Þ x = 8[as x = 20 does not satisfy the given equation

Then L.H.S. is positive while R.H.S. isnegative.]

18. The digits of a positive number of threedigits are in A.P. and their sum is 15. Thenumber obtained by reversing the digitsis 594 less than the original number. Findthe number.

Sol. Let the required numbers in A.P. are a – d,a, a + d respectively.Now, a – d + a + a + d = 15

[Q Sum of digits = 15]Þ 3a = 15 Þ a = 5

According to question, number is100 (a – d) + 10a + a + d, i.e. 111a – 99dNumber on reversing the digits is100 (a + d) + 10a + a – d, i.e. 111a + 99dNow, as per given condition in question,(111a – 99d) – (111a + 99d) = 594

Þ – 198d = 594Þ d = – 3\ Digits of number are

[5 – (– 3), 5, (5 + (– 3)] = 8, 5, 2\ Required number is

111 × (5) – 99 (– 3) = 555 + 297 = 85219. If the roots of the quadratic equation

(a – b) x2 + (b – c) x + (c – a) = 0 areequal, prove that 2a = b + c.

Sol. Given quadratic eqn. be(a – b) x2 + (b – c) x + (c – a) = 0On comparing with ax2 + bx + c = 0we have ‘a’ = a – b ;‘b’ = b – c and ‘c’ = c – aFor equal roots, discriminant, D = 0

Þ b2 – 4ac = 0Þ (b – c)2 – 4 (a – b) (c – a) = 0Þ b2 – 2bc + c2– 4 (ac – a2 – bc + ab) = 0Þ b2 – 2bc + c2 – 4ac + 4a2 + 4bc – 4ab = 0Þ 4a2 + b2 + c2 – 4ab + 2bc – 4ac = 0Þ (2a – b – c)2 = 0Þ 2a – b – c = 0Þ 2a = b + c

20. From a pack of 52 playing cards, Jacks,Queens and Kings of red colour areremoved. From the remaining, a card isdrawn at random. Find the probabilitythat drawn card is :(i) a black King.(ii) a card of red colour.(iii) a card of black colour.

Sol. Removed red colour cards = 3 × 2 = 6[Since there are 2 red jacks, 2 red queensand 2 red kings]

\ No. of remaining cards = 52 – 6 = 46(i) Number of black kings = 2 = No. of

favourable cases

\ P (a black king)2 1

46 23= =

(ii) Number of red colour cards = 26



Arundeep’s Solved Papers Mathematics 2016 (Outside Delhi)137\ Remaining no. of red colour cards = 26 – 6 = 20

= No. of favourable cases

\ P (a card of red colour) 20 1046 23

= =

(iii) Number of black cards = 26= No. of favourable cases

\ P (a black colour card)26 1346 23

= =

28. Draw an isosceles DABC in whichBC = 5.5 cm and altitude AL = 3 cm.Then construct another triangle whose

sides are34 of the corresponding sides

of DABC.Sol.

P

A¢

A

A

A

A

C¢

Steps of construction :1. Draw a line segment BC = 5.5 cm.2. Construct AP the perpendicular bisector of

BC meeting BC at L.

3. Along LP cut off LA = 3 cm.4. Join BA and CA. Then DABC so obtained

is the required DABC.5. Draw an acute angle CBY and cut 4 equal

lengths asBA1 = A1A2 = A2A3 = A3A4 and join CA4.

6. Now draw a line through A3 parallel to CA4intersecting BC at C¢.

7. Draw a line through C¢ and parallel to ACintersecting AB at A¢. Thus, A¢BC¢ is therequired triangle.

29. Prove that tangent drawn at any pointof a circle is perpendicular to the radiusthrough the point of contact.

Sol. Given : A circle C (O, r) and a tangent ABat a point P.To prove : OP ^ ABConstruction : Take any point Q other thanP on the tangent AB.Join OQ, intersecting circle at R.Proof : We have, OP = OR [Radii]OQ = OR + RQ

\ OQ > OR Þ OQ > OP

[Q OR = OP = radius]

Thus, OP < OQ i.e. OP is shorter than anyother segment joining O to any point of AB.But among all line segments, joining pointO to point on AB, shortest one isperpendicular from O on AB.Hence, OP ^ AB

30. As observed from the top of a light house,100 m high above sea level, the angles ofdepression of a ship, sailing directly


Arundeep’s Solved Papers Mathematics 2016 (Outside Delhi)138towards it, changes from 30º to 60º. Findthe distance travelled by the ship duringthe period of observation.

(Use 3 = 1.73)

Sol. Let AB be the tower of height 100 m.and C and D are the two positions of shipss.t BC = y and CD = x.

In right DABC,ABBC = tan 60º

A

y x D

Þ100

y 3= Þ y100

3= ...(i)

In right DABD,ABBD = tan 30º

Þ100y x+

13

=

Þ x + y 100 3=

Þ x 100 3 y= -

Þ x 100 300 100 200100 33 3 3

-= - = =

...[From (i)]

Þ x200 3 200 1.73

3 3´= = = 115.33 m

The required distance travelled by the ship is115.33 m.

31. A rectangular park is to be designedwhose breadth is 3 m less than its length.Its area is to be 4 square metres morethan the area of a park that has alreadybeen made in the shape of an isoscelestriangle with its base as the breadthof the rectangular park and of altitude12 m. Find the length and breadth of therectangular park.

Sol. Let length of rectangular park be x m andBreadth be (x – 3) mGiven and base of isosceles triangle

= (x – 3) mand Altitude of triangle = 12 m

x

x – 3

According to question, we haveArea of rectangular park = Area of isoscelestriangle + 4

Þ x (x – 3)1 ( 3) 12 42

x= - ´ +

Þ x (x – 3) = 6 (x – 3) + 4Þ x2 – 3x = 6x – 18 + 4Þ x2 – 9x + 14 = 0Þ (x – 7) (x – 2) = 0Þ x = 7 or x = 2

( – 3)x

12

But x = 2 is rejected otherwise breadth willbe –ve which is not possible.

\ Length of rectangular park is 7 m andbreadth be 4 m.



10. Solve for x :

6 7 (2 7)x x+ - - = 0.

Sol. Given eqn. be, 6 7 (2 7)x x+ - - = 0

Þ 6 7x + = 2x – 7

On squaring both sides, we getÞ 6x + 7 = (2x – 7)2

Þ 6x + 7 = 4x2 – 28x + 49Þ 4x2 – 34x + 42 = 0Þ 2x2 – 17x + 21 = 0Þ 2x2 – 14x – 3x + 21 = 0Þ 2x (x – 7) – 3 (x – 7) = 0Þ (2x – 3) (x – 7) = 0

Þ x = 7 or 32

x =

Þ x = 7

3as does not satisfy the equation, in this2

case L.H.S. is positive while R.H.S. is negative

xé ù=ê úê úë û

18. There are 100 cards in a bag on whichnumbers from 1 to 100 are written. Acard is taken out from the bag at random.Find the probability that the number onthe selected card :(i) is divisible by 9 and is a perfect square.(ii) is a prime number greater than 80.

Sol. Total possible cases = 100(i) Favourable cases when number is a perfect

square and is divisible by 9 are 9, 36 and81.So, number of favourable cases = 3

\ Required probability

Number of favourable casesTotal possible cases

= 3100

=

(ii) Favourable cases the prime numbers greaterthan 80 are 83, 89 and 97

So, number of favourable cases = 3\ Required probability

Number of favourable casesTotal possible cases

= 3100

=

19. Three consecutive natural numbers aresuch that the square of the middlenumber exceeds the difference of thesquares of the other two by 60. Find thenumbers.

Sol. Let the three consecutive natural numbersare x – 1, x and x + 1.According to given condition, we havex2 – [(x + 1)2 – (x – 1)2] = 60

Þ x2 – [(x + 1 – x + 1) (x + 1 + x – 1)] = 60Þ x2 – 4x – 60 = 0Þ x2 – 10x + 6x – 60 = 0Þ x (x – 10) + 6 (x – 10) = 0Þ (x + 6) (x – 10) = 0Þ x = 10 (x = – 6, rejected)

Hence, the required numbers are 9, 10 and11.

20. The sums of first n terms of threearithmetic progressions are S1, S2 and S3respectively. The first term of each A.P. is1 and their common differences are 1, 2and 3 respectively. Prove that S1 + S3 = 2S2.

Sol. Here, sum of n terms of AP is

Sn [2 ( 1) ]2n a n d= + -

\ S1 [2 ( 1)1]2n n= + -

( 1)2

n n += [Q where a = 1, d = 1]

S2 [2 ( 1) 2]2n n= + -

SET-III [UNCOMMON QUESTIONS TO SET-I and Set II]



2(2 )2n n n= = [Q where a = 1, d = 2]

and S3 [2 ( 1) 3]2n n= + -

[2 3 3]2n n= + - [Here a = 1, d = 3]

[3 1]2n n= -

Now, consider S1 + S3

[ ]( 1) 3 12n n n= + + - 4

2n n= ´

22

24 2 2S

2n n= = =

28. Two pipes running together can fill a

tank in1119 minutes. If one pipe takes 5

minutes more than the other to fill thetank separately, find the time in whicheach pipe would fill the tank separately.

Sol. Let one pipe takes x minutes to fill the tank.Then, another pipe takes x + 5 to fill thetank.According to question, we have

1 15x x

++

9100

=1 100119 9

æ ö=ç ÷è øQ

Þ 100 [x + 5 + x] = 9 [x2 + 5x]Þ 200x + 500 = 9x2 + 45xÞ 9x2 – 155x – 500 = 0Þ 9x2 – 180x + 25x – 500 = 0Þ 9x (x – 20) + 25 (x – 20) = 0Þ (9x + 25) (x – 20) = 0

x = 20 or25 (Rejected)9

x = -

Hence, one pipe takes 20 minutes andanother pipe takes 25 minutes to fill thetank.

29. From a point on the ground, the angle ofelevation of the top of a tower is observedto be 60º. From a point 40 m verticallyabove the first point of observation, theangle of elevation of the top of the toweris 30º. Find the height of the tower andits horizontal distance from the point ofobservation.

Sol. Let h be the height of the tower AD and xbe the horizontal distance from the pointof observation.

Q BDEC is a rectangle.\ CB = ED = x and CE = BD = 40 m

In right DABC, tan 30ºABBC

=

Þ13

ABx

=

Þ x AB 3= ...(i)

Now, in right DAED, tan 60ºADDE

=

Þ 3 DEh=

Þ3

hx = ...(ii)

hBC

E

A

Cx

From equation (i) and (ii), we get

AB 33

h=

[Q AB + 40 = h Þ AB = h – 40]



Þ 3 ( 40)h -3

h=

Þ 3 (h – 40) = hÞ 3h – 120 = hÞ 2h = 120Þ h = 60 m

Hence the required height of tower be60 metre.

From (ii), x 3h= Þ 60

3x =

Þ60 3

3x =

Þ x 20 3=Þ x = 34.641 m

Thus, required horizontal distance frompoint of observation be 34.641 metre.

30. Draw a triangle with sides 5 cm, 6 cmand 7 cm. Then draw another triangle

whose sides are45 of the corresponding

sides of first triangle.Sol.

6 cmC¢

5cm

A

A

A

A

A

1

2

3

4

A5

C

7 cm

Z

B¢B

Steps of construction :1. Draw a line segment AB of length 7 cm.

Then using A as centre and distance 5 cmdraw an arc. Also draw an arc using B ascentre and with distance 6 cm, whichintersect earlier drawn arc at C. Join ACand BC.

2. Draw an acute angle BAZ and cut AZ asAA1 = A1A2 = A2A3 = A3A4 = A4A5 andjoin BA5.

3. Through A4 draw a line parallel BA5intersecting AB at B¢.

4. Through B¢ draw a line parallel to BCintersecting AC at C¢.

DAB¢C¢ is the required triangle.

31. A number x is selected at random fromthe numbers 1, 4, 9, 16 and anothernumber y is selected at random from thenumbers 1, 2, 3, 4. Find the probabilitythat the value of xy is more than 16.

Sol. x can be 1, 4, 9 or 16 and y can be 1, 2, 3 or4.

Total number of cases of xy = 4 × 4 = 16.

Number of cases when xy is more than 16are (9 × 2), (9 × 3), (9 × 4), (16 × 2), (16 × 3), (16 × 4), i.e. 6 cases.

P (value of xy more than 16)6 3

16 8= =



(ii) The question paper consists of 31 questions divided into four sections – A, B, C and D.(iii) Section A contains 4 questions of 1 mark each, Section-B contains 6 questions of 2 marks each.



MATHEMATICS 2016 TERM II (DELHI)SET I


1. From an external point P, tangents PAand PB are drawn to a circle with centreO. If ÐPAB = 50º, then find ÐAOB.

Sol. Given, ÐPAB = 50ºÐPAB + ÐOAB = 90º [Q anglebetween radius OA and tangent PA is 90º]

Þ 50º + ÐOAB = 90ºÞ ÐOAB = 90º – 50º = 40º

Now, PA = PB [Q tangents from anexternal point are same]

Þ ÐPBA = ÐPABÞ ÐPBA = 50º

ÐPBA + ÐOBA = 90º[\ angle between radius OB and tangentPB is 180º]

Þ 50º + ÐOBA = 90ºÞ ÐOBA = 90º – 50º = 40º

Now in DAOB we haveÐAOB + ÐABO + ÐBAO = 180º

[Q sum of angles in triangle is 180º]

Þ ÐAOB + 40º + 40º = 180ºÞ ÐAOB = 180º – 80º = 100º2. In Figure, AB is a 6 m high pole and CD

is a ladder inclined at an angle of 60º tothe horizontal and reaches up to a pointD of pole. If AD = 2.54 m, find the lengthof the ladder. (use 3 = 1.73)

Sol. BD = AB – AD= 6 m – 2.54 m = 3.46 m

In right DDBC,BDCD = sin 60º

Þ3.46CD

32

=

Þ CD2 3.46 2 3.46

1.733´ ´= = = 2 × 2 = 4 m

Hence, length of the ladder is 4 m.3. Find the 9th term from the end (towards

the first term) of the A.P. 5, 9, 13, ..., 185.



Arundeep’s Solved Papers Mathematics 2016 (Delhi)143Sol. Reversing the given A.P., we get

185, 181, 174, ..., 9, 5Now, first term (a) = 185Common difference, (d) = 181 – 185 = – 4We know that nth term of an A.P. is givenby a + (n – 1) dNinth term a9 = a + (9 – 1) d= 185 + 8 × (– 4) = 185 – 32 = 153

4. Cards marked with number 3, 4, 5, ...,50 are placed in a box and mixedthoroughly. A card is drawn at randomfrom the box. Find the probability thatthe selected card bears a perfect squarenumber.

Sol. Total no. of possible outcomes when onecard is drawn = 48When the number on drawn card is a perfectsquare, total favourable cases are 4, 9, 16,25, 36, 49, i.e. = 6P (perfect square number)

Number of total possible outcomesNumber of favourable outcomes

=

6 148 8

= =

5. If23

x = and x = – 3 are roots of the

quadratic equations ax2 + 7x + b = 0, findthe values of a and b.

Sol. Given quadratic equation isax2 + 7x + b = 0 ...(i)

2 is the root of equation ( )3

x ié ù=ê úë ûQ

22 273 3

a bæ ö æ ö+ +ç ÷ ç ÷è ø è ø = 0

[_ if a be a root of f(x) = 0 then f(a) = 0]

Þ4 149 3

a b+ + = 0

Þ4 42 9

9a b+ +

= 0

Þ 4a + 9b + 42 = 0 ...(ii)a (– 3)2 + 7 (– 3) + b = 0

[Q x = – 3 is the root of equation (i)]Þ 9a + b – 21 = 0 ...(iii)

Putting the value of b from (iii) in (ii), weget4a + 9 [21 – 9a] + 42 = 0

Þ 4a + 189 – 81a + 42 = 0Þ 231– 77a = 0Þ 77a = 231Þ a = 3

Putting a = 3 in (iii), we get27 + b = 21

Þ b = – 6Hence, a = 3, b = – 6

6. Find the ratio in which y-axis divides theline segment joining the points A (5, – 6)and B (– 1, – 4). Also find the coordinatesof the point of division.

Sol. Let the point on y-axis be P (0, y) and\ Co-ordinates of P is given by using section

formula, we have

5 4 6,1 1

k kk k

- + - -æ öç ÷+ +è øSince point P lies on y-axis

\ x-coordinate of P = 0

Þ5 0

1kk

- + =+

1

Þ5

1k

k-+ = 0 Þ k = 5

Hence the required ratio is 5 : 1.

Thus, y-coordinate of P( 4) (5) (1) ( 6)

5 1- + -=

+

133

-=



Hence point on y-axis is130,3

-æ öç ÷è ø

7. In given figure, a circle is inscribed in aDABC, such that it touches the sides AB,BC and CA at points D, E and Frespectively. If the lengths of sides AB,BC and CA are 12 cm, 8 cm and 10 cmrespectively, find the lengths of AD, BEand CF.

C

EF

A D B

Sol. Given, AB = 12 cm, CA = 10 cm, BC = 8 cm

Let AD = AF = x [Q Tangent drawn fromexternal point to circle are equal]

\ DB = BE = 12 – x and CF = CE = 10 – xBC = BE + EC Þ 8 = 12 – x + 10 – x

Þ 8 = 22 – 2x Þ 2x = 14

C

EF

A D B

10–

x

10–

x12

–x

x 12 – x

x

Þ x = 7\ AD = 7 cm, BE = 5 cm and CF = 3 cm8. The x-coordinate of a point P is twice its

y-coordinate. If P is equidistant fromQ (2, – 5) and R (– 3, 6), find thecoordinates of P.

Sol. Let the required point be (2y, y).Let Q (2, – 5) and R (– 3, 6) are givenpoints.Now, PQ = PR

Þ 2 2(2 2) ( 5)y y- + + 2 2(2 3) ( 6)y y= + + -

[Q using Distance formula]on Squaring both sides we get4y2 + 4 – 8y + y2 + 10y + 25 = 4y2 + 9

+ 12y + y2 – 12y + 36Þ 2y + 29 = 45Þ 2y = 45 – 29 = 16Þ y = 8\ 2y = 16

Hence coordinates of P are (16, 8)9. How many terms of the A.P. 18, 16, 14,

... be taken so that their sum is zero ?Sol. Let the number of terms of given A.P. taken

for sum to be zero be n.Then, sum of n terms (Sn) = 0 (Given)

\ First term (a) = 18and Common difference (d) = – 2

Therefore, Sn [2 ( 1) ]2n a n d= + -

Þ 0 2n= [2 × 18 + (n – 1) (– 2)]

Þ 0 = 38 – 2nÞ n = 19\ Hence, required sum of 19 terms is 0.

10. If given figure, AP and BP are tangentsto a circle with centre O, such thatAP = 5 cm and ÐAPB = 60º. Find thelength of chord AB.


Arundeep’s Solved Papers Mathematics 2016 (Delhi)145Sol. In DAPB we have

AP = BPÞ ÐPAB = ÐPBA

[Q Tangents from an external point areequally inclined to segment joining centreto point].Let ÐPAB = x,then in DAPB, x + x + 60º = 180º

[Q sum of angles of D is 180º]

Þ 2x = 180º – 60º = 120ºi.e. x = 60º

As all three angles of DAPB are 60º. SoDAPB is an equilateral triangle.Hence AP = BP = AB = 5 cm

11. In figure, ABCD is a square of side 14cm. Semi-circles are drawn with eachside of square as diameter. Find the area

of the shaded region.22Use π =7

é ùê úë û

A B

CD

X

Z W

YO

Sol. Area of the square ABCD = 14 × 14= 196 cm2

Since radius of semi-circle =r =14 cm2

= 7cm

Area of semicircle AOB 212

r= ´ p

1 22 7 72 7

= ´ ´ ´ = 77 cm2 142

ræ ö=ç ÷è øQ

Similarly, area of semicircle DOC = 77 cm2

Hence, the area of shaded region = Area ofsquare – Area of two semicircle AOB andCOD= 196 – 154 = 42 cm2

Therefore, area of four shaded parts (i.e.X, Y, W, Z) = (2 × 42) cm2 = 84 cm2

12. In Figure, a decorative block, made upof two solids – a cube and a hemisphere.The base of the block is a cube of side6 cm and the hemisphere fixed on the tophas a diameter of 3.5 cm. Find the total

surface area of the block.22Use π =7

é ùê úë û

Sol. Given Side of the cube = a = 6 cmTotal surface area of cube = 6 × (side)2 = 6a2

= 6 × (6)2 = 216 cm2

Given radius of hemisphere = r =3.5 cm2

Area covered on the face of cube by circularpart of hemisphere = pr2

222 3.5 3.5 cm7 2 2

= ´ ´

Curved surface area of hemisphere= 2 × p × r2

222 3.5 3.52 cm7 2 2

= ´ ´ ´

So, Total surface area of the block = Surfacearea of cube – Area of circular face ofhemisphere+ Curved surface area of hemisphere



22 3.5 3.5 22 3.5 3.5216 27 2 2 7 2 2

= - ´ ´ + ´ ´ ´

22 3.5 3.52167 2 2

= + ´ ´

= 216 + 9.625 = 225.625 cm2

13. In figure ABC is a triangle coordinatesof whose vertex A are (0, – 1). D and Erespectively are the mid-points of thesides AB and AC and their coordinatesare (1, 0) and (0, 1) respectively. If F isthe mid-point of BC, find the areas ofDABC and DDEF.

A (0, –1)

E (0, 1)

C

(1, 0) D

BF

Sol. Let coordinates of B are (x, y). Then usingmid point formula we have

02

x + = 1 Þ x = 2

12

y - = 0 Þ y = 1

Coordinates of B are (2, 1)Let coordinates of C are (p, q)Since E(0, 1) be the mid point of AC.

\0

2p +

= 0 Þ p = 0

and1

2q -

= 1 Þ q = 3

Coordinates of C are (0, 3)\ Area of DABC

=12 [x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]

12

= [0 (1 – 3) + 2 (3 + 1) + 0 (– 1 – 1)]

1 82

= ´ = 4 sq. units

Coordinates of F are2 0 1 3,

2 2+ +æ ö

ç ÷è øi.e. (1, 2)

[Q Using mid-point formula]

and Area of DDEF12

= |1 (1 – 2) + 0 (2 – 0)

+ 1 (0 – 1)|

12

= |– 1 + 0 – 1|

12

= × |– 2| = |– 1| = 1 sq. units

14. In figure, two arcs PAQ and PBQ areshown. Arc PAQ is a part of circle withcentre O and radius OP while arc PBQis a semicircle drawn on PQ as diameterwith centre M. If OP = PQ = 10 cm, showthat area of shaded region is

225 3 cm .6pæ ö-ç ÷è ø

Sol. OP = OQ = 10 cm (radii of circle)PQ = 10 cm (given)So, DOPQ is an equilateral triangle

\ ÐPOQ = 60º


Arundeep’s Solved Papers Mathematics 2016 (Delhi)147Area of segment PAQM = Area of sectorOPAQ – Area of equilateral DOPQ

60 310 10 10 10360 4

°= ´ p ´ ´ - ´ ´°

[_ area of sector = pr2 × 360q

° ]

2100 100 3 cm6 4

æ öp= -ç ÷è ø

and Area of semicircle =2

2rp 1 5 5

2= ´ p ´ ´

225 cm2

= p

\ Area of the shaded region

25 100 100 32 6 4

æ öp= p - -ç ÷è ø

25 50 25 32 3p p= - +

75 100 25 36

p - p= +

2525 36p= -

225 3 cm6pæ ö= -ç ÷è ø

15. If the sum of first 7 terms of an A.P. is 49and that of its first 17 terms is 289, findthe sum of first n terms of the A.P.

Sol. Let a be the first term and d be the commondifference of an A.P.Given : S7 = 49,

where Sn [2 ( 1) ]2n a n d= + -

Þ7 [2 (7 1) ]2

a d+ - = 49

Þ 2a + 6d = 14 Þ a + 3d = 7 ...(i)

Similarly, S17 = 289

Þ17 [2 (17 1) ]2

a d+ - = 289

Þ 2a + 16d = 34 Þ a + 8d = 17 ...(ii)Solving (i) and (ii), we geta = 1 and d = 2

Now Sn [2 ( 1) ]2n a n d= + -

[2 1 ( 1) 2]2n n= ´ + -

[2 2 2]2n n= + - = n × n = n2

16. Solve for x :2 1 3 9

0,3 2 3 ( 3) (2 3)

x xx x x x

++ + =

- + - +

32

x ¹ 3, -

Sol. Given eqn. be

2 1 3 93 2 3 ( 3) (2 3)

x xx x x x

++ +- + - + = 0

Þ2 (2 3) ( 3) (3 9)

( 3) (2 3)x x x x

x x+ + - + +

- + = 0

Þ 4x2 + 6x + x – 3 + 3x + 9 = 0Þ 4x2 + 10x + 6 = 0Þ 2x2 + 5x + 3 = 0Þ 2x2 + 2x + 3x + 3 = 0Þ 2x (x + 1) + 3 (x + 1) = 0Þ (x + 1) (2x + 3) = 0Þ x + 1 = 0 or 2x + 3 = 0

Þ x = – 1 or3

2x -=

3given,2

x -æ ö¹ç ÷è øQ

Q It is given that3 .

2x -¹

Hence, solution of the given equationbe x = – 1.

17. A well of diameter 4 m is dug 21 m deep.The earth taken out of it has been spread


Arundeep’s Solved Papers Mathematics 2016 (Delhi)148evenly all around it in the shape of a circularring of width 3 m to form an embankment.Find the height of the embankment.

Sol. Radius of the well = r = 2m,Height of the well = h = 21 mVolume of the earth dug out = pr2h

= p × 2 × 2 × 21 m3 =22 2 2 217

´ ´ ´

= 264 m3

\ External radius of embankment = Radiusof well + width of embankment= 2 + 3 = 5 m

[_ given width of embankment = 3m]

2

3

Volume of embankment2 2 3

1 2( ) mr h r h= p - p

= [p × (5)2 × h – p (2)2 h] m3

= (p × 25 × h – p × 4 × h) m3

322 21 m7

hæ ö= ´ ´ç ÷è ø = 66 h m3

As per given condition, we haveVolume of earth dug out = Volume ofembankment264m3 = 66h m3

\ Height of embankment, h264 4 m66

= =

18. The sum of the radius of base and heightof a solid right circular cylinder is 37 cm.If the total surface area of the solidcylinder is 1628 sq. cm, find the volume

of the cylinder.22Use π =7

é ùê úë û

Sol. Given base radius of solid cylinder = r cm

and height of solid cylinder = h cmHere r + h = 37 ...(i)Total surface area of cylinder = 2pr (h + r)

= 2prh + 2pr2

Þ 2pr (h + r) = 1628Þ 2pr × 37 = 1628 [using (i)]

Þ222 377

r´ ´ ´ = 1628

Þ r1628 7

2 22 37´=

´ ´ = 7 cm

Given, r + h = 37Þ 7 + h = 37 Þ h = 37 – 7 = 30 cm

Hence, volume of cylinder = pr2h22 7 7 307

= ´ ´ ´ = 4620 cm3

19. The angles of depression of the top andbottom of a 50 m high building from thetop of a tower are 45º and 60ºrespectively. Find the height of the towerand the horizontal distance between thetower and the building. (use 3 = 1.73)

Sol. Let CD is the building of height 50 m andAB be the tower.Let horizontal distance between the towerand building is BC is x metre.

Q BCDE is a rectangleSo, ED = BC and BE = CDAlso, ED = x and BE = 50 mLet AE = y

45°

60°BC

D

50 m

E

A

60° 45°

y m

x m



Now, in DAED,yx = tan 45º Þ 1y

x=

Þ y = x ...(i)

Now, in DABC,ABBC = tan 60º

ÞAE EB

BC+

3= Þ50y

x+

3=

Þ x + 50 3x= [Q y = x, using (i)]

Þ 3x x- = 50 Þ ( 3 1) x- = 50

Þ x 503 1

=-

Þ x 50 ( 3 1)( 3 1) ( 3 1)

+=- +

50 ( 3 1) 50 ( 3 1)3 1 2

+ += =-

Þ x 25( 3 1)= + = 25 (1.73 + 1)

= 25 × 2.73 = 68.25 m\ Height of the tower = 50 + y

= 50 + 68.25 (Q x = y)= 118.25 mHorizontal distance between the tower andthe building = x = 68.25 m

20. In a single throw of a pair of differentdice, what is the probability of getting

(i) a prime number on each dice ?(ii) a total of 9 or 11 ?

Sol. (i) Total possible cases when two dice arethrown together = 6 × 6 = 36

\ Simple space = {(1, 1), (1, 2) .... (1, 6)(2, 1), (2, 2) ......(2, 6)(3, 1), (3, 2) .... (3, 6)(4, 1), (4, 2) .... (4, 6)(5, 1), (5, 2) ..... (5, 6)(6, 1), (6, 2) ...... (6, 6)

Favourable cases when both numbers areprime are (2, 2), (2, 3), (2, 5), (3, 2), (3, 3),(3, 5), (5, 2), (5, 3), (5, 5), i.e. 9 outcomesP (a prime number on each dice)

Favourable casesTotal cases

=

9 136 4

= =

(ii) Favourable cases when sum of numbers are9 or 11 are (3, 6), (4, 5), (5, 4), (5, 6),(6, 3), (6, 5), i.e. 6 outcomesP (a total of 9 or 11)

Favourable outcomesTotal possible outcomes

=

6 136 6

= =

21. A passenger, while boarding the plane,slipped from the stairs and got hurt. Thepilot took the passenger in the emergencyclinic at the airport for treatment. Dueto this, the plane got delayed by half anhour. To reach the destination 1500 kmaway in time, so that the passengerscould catch the connecting flight, thespeed of the plane was increased by 250km/hour than the usual speed. Find theusual speed of the plane.What value is depicted in this question ?

Sol. Let the usual speed of plane be x km/hr.Time taken to cover 1500 km with usual

speed1500 hours.

x=

When the speed of plane is increased, thennew speed = (x + 250) km/hr.Time taken to cover 1500 km with the new

speed (x + 250) km/hr1500

250x=

+According to question, we have



1500x

1500 1250 2x

= ++

Þ1500 1500

250x x-

+12

=

Þ1500 1500 250 1500

( 250)x x

x x+ ´ -

+12

=

Þ 21500 250

250x x´

+12

=

Þ x2 + 250x = 750000Þ x2 + 250x – 750000 = 0Þ x2+ 1000x – 750x – 750000 = 0Þ x (x + 1000) – 750 (x + 1000) = 0Þ (x + 1000) (x – 750) = 0Þ x + 1000 = 0 or x – 750 = 0Þ x = – 1000 or x = 750

Þ x = 750 (Q Speed cannot be negative)

Hence, the usual speed of plane is 750 km/hr.In this question, pilot’s caring behaviourtowards passengers is shown as well asother side pilot is cautious and alert for hisduty to reach destination point at scheduledtime.

22. Prove that the lengths of tangents drawnfrom an external point to a circle areequal.

Sol. Given : A circle C (O, r), P is a point outsidethe circle and PA and PB are tangents to acircle.

O

N

M

A

To Prove : PA = PBConstruction : Draw OA, OB and OP.Proof : Consider triangles OAP and OBP.

ÐOAP = ÐOBP = 90º ...(i)[Radius is perpendicular to the tangent atthe point of contact]OA = OB (radii) ...(ii)OP is common ...(ii)

\ DOAP @ DOBP(RHS axiom of congruency)

[from (i), (ii) and (iii)]Hence, AP = BP (CPCT)

23. Draw two concentric circles of radii 3 cmand 5 cm. Construct a tangent to smallercircle from a point on the larger circle.Also measure its length.

Sol. Now after measuring, PA and PB comes outto be 4 cm.

B

A

P OM

Steps of construction of tangents :1. Take point O. Draw 2 concentric circles of

radii 3 cm and 5 cm respectively.2. Locate point P on the circumference of

larger circle.3. Join OP and bisect it. Let M be mid-point

of OP.4. Taking M as centre and MP as radius, draw

an arc intersecting smaller circle at A andB.Join PA and PB. Thus, PA, PB are requiredtangents.


Arundeep’s Solved Papers Mathematics 2016 (Delhi)15124. In given figure, O is the centre of a circle of radius 5 cm. T is a point such that OT = 13 cm

and OT intersects circle at E. If AB is a tangent to the circle at E, find the length of AB,where TP and TQ are two tangents to the circle.

Sol. In DOPT ; OP2 + PT2 = OT2 [Q Pythagoras theorem]

PT 2 2OT OP= - 169 25= - = 12 cm

and TE = OT – OE = 13 – 5 = 8 cmLet PA = AE = x [tangent from outer point A]

In DTEA, TE2 + EA2 = TA2 [Q Pythagoras theorem]

Þ (8)2 + (x)2 = (12 – x)2 Þ 64 + x2 = (12 – x)2

Þ 64 + x2 = 144 + x2 – 24x Þ 80 = 24x Þ x = 3.3 cm

Thus AB = 2 × 3.3 cm = 6.6 cm [Q AE = EB, as AB is tangent to circle at E]

25. Find x in terms of a, b and c :

2 ,a b cx a x b x c

+ =- - - x ¹ a, b, c

Sol. Given eqn. bea b

x a x b+

- -2c

x c=

-

Þ( ) ( )( ) ( )

a x b b x ax a x b- + -=- -

2cx c

=- Þ 2

( 2 )( )

ax bx abx ax bx ab

+ -- - +

2cx c

=-

Þ (ax + bx – 2ab) (x – c) = 2c (x2 – ax – bx + ab)

Þ ax2 + bx2 – 2abx – acx – bcx + 2abc = 2cx2 – 2acx – 2bcx + 2abc

Þ (a + b – 2c) x2 + (ac + bc – 2ab) x = 0

Þ x {(a + b – 2c) x + (ac + bc – 2ab)} = 0

Þ x = 0 or2

2ab ac bcxa b c

- -=+ -

26. A bird is sitting on the top of a 80 m high tree. From a point on the ground, the angle ofelevation of the bird is 45º. The bird flies away horizontally in such a way that it remained

O 5 cm T

B

AP

E


Arundeep’s Solved Papers Mathematics 2016 (Delhi)152at a constant height from the ground. After 2 seconds, the angle of elevation of the birdfrom the same point is 30º. Find the speed of flying of the bird. (Take 3 = 1.732)

Sol. Let BC be the tree of height 80 metreAfter 2 seconds, position of bird is at point E.Let CE = x

In DCBA,BCAB = tan 45º

Þ80AB = 1 Þ

80y = 1 Þ y = 80 m

In DEDA,EDAD = tan 30º

Þ80

AB BD+13

=

Þ80

80 x+13

=

Þ 80 3 = 80 + x [Q AB = 80 m]

Þ x 80 3 80= - Þ 80 ( 3 1)x = -Þ x = 80 (1.732 – 1) Þ x = 80 × 0.732 Þ x = 58.56 m Þ BD = x = 58.56 m

So, the speed of flying of the bird distance (BD)Time

= 58.562

= = 29.28 m/s

27. A thief runs with a uniform speed of 100 m/minute. After one minute a policeman runsafter the thief to catch him. He goes with a speed of 100 m/minute in the first minute andincreases his speed by 10 m/minute every succeeding minute. After how many minutes thepoliceman will catch the thief.

Sol. Let total time be (n – 1) minutes in which the police catch the thief.Since thief ran 1 minute before police start running.

Q Time taken by thief before he was caught = (n – 1 + 1) = n minuteThen total distance covered by thief = (100 × n) metresTotal distance covered by policeman in (n – 1) minute= 100 + 110 + 120 + .... + (n – 1) terms

( 1)2

n -= [200 + (n – 2) 10] S [2 ( 1) ]2nn a n dé ù= + -ê úë û

Q

According to question,Total distance covered by thief in ‘n’ minutes= total distance covered by policeman in (n – 1) minute

80 m

CE

D Bx

80 m

30°45°y A



100n( 1)

2n -= [200 + (n – 2) 10]

Þ 200n = (n – 1) [200 + 10n – 20] Þ 200n = (n – 1) (10n + 180)Þ 200n = 10n2 + 180n – 10n – 180 Þ 10n2 – 30n – 180 = 0Þ n2 – 3n – 18 = 0 Þ n2 – 6n + 3n – 18 = 0Þ n (n – 6) + 3 (n – 6) = 0 Þ (n – 6) (n + 3) = 0Þ n = 6 or n = – 3 (rejected)

Hence, time taken by policeman to catch the thief is (6 – 1), i.e., 5 minutes.28. Prove that the area of a triangle with vertices (t, t – 2), (t + 2, t + 2) and (t + 3, t) is independent

of t.Sol. Given vertices of triangle are

{t, t – 2), {t + 2, t + 2), {t + 3, t}Let (x1, y1), (x2, y2), (x3, y3) are vertices of the triangle.

Then Area of the triangle12

= |x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)|

12

= |t (t + 2 – t) + (t + 2) {t – t + 2} + (t + 3) {t – 2 – t – 2}|12

= |2t + 2t + 4 – 4t – 12|

12

= × |– 8| = 4 sq. units,

Hence, area of triangle is independent of t.29. A game of chance consists of spinning an arrow on a circular board,

divided into 8 equal parts, which comes to rest pointing at one ofthe numbers 1, 2, 3, ..., 8 which are equally likely outcomes. Whatis the probability that the arrow will point at (i) an odd number(ii) a number greater than 3 (iii) a number less than 9 ?

Sol. (i) Total possible outcomes when the arrow points at one of the numbersare 8.Favourable outcomes when the required number is odd are 1, 3, 5, 7, i.e. 4 outcomes.

\ P (an odd number)

No. of favourable outcomesTotal no. of possible outcomes

= 4 18 2

= =

(ii) Favourable outcomes when the required number is more than 3 are 4, 5, 6, 7, 8, i.e. 5 outcomes.P (a number is more than 3)

No. of favourable outcomesTotal no. of possible outcomes

=

(iii) Favourable outcomes when the required number is less than 9 are 1, 2, 3, 4, 5, 6, 7, 8 i.e. 8outcomes.

3 2

1

8

76

5

4



P(number is less than 9)No. of favourable outcomes

Total no. of possible outcomes=

8 18

= =

30. An elastic belt is placed around the rim of a pulley of radius 5 cm. From one point C on thebelt, the elastic belt is pulled directly away from the centre O of the pulley until it is at P, 10cm from the point O. Find the length of the belt that is still in contact with the pulley. Alsofind the shaded area. (use p = 3.14 and 3 = 1.73)

5 cm

A

CO

B

Pq

Sol. Given : AO = 5 cm and OP = 10 cmIn right DAOP, we have

cos qBase

Hypotenuse=

AO 5 1OP 10 2

= = =

Þ q = 60ºÐAOB = q¢ = 2 × 60 = 120º

Þ Length of AOB360º 2

360ºr- q¢= ´ p (Q q¢ = 120º)

240 2 3.14 5360

°= ´ ´ ´°

23

= × 10 × 3.14 = 20.93 cm

Hence, length of belt in contact = 20.93 cmNow, in right DOAP, we have

\ tan qAPAO

=

Þ tan 60ºAP5

= Þ AP 5 3 cm=

Area of (DOAP + DOBP)12

= × AO × AP +12 × OB × PB

1 15 5 3 5 5 32 2

= ´ ´ + ´ ´ (Q AP = BP and OA = OB)

225 3 cm= = 25 × 1.73 cm2 = 43.25 cm2

Thus, Area of sector OACB 2360

rq¢= ´ p



120360

°=°

× 3.14 × 5 × 513

= × 3.14 × 25 = 26.16 cm2

\ Area of shaded region = Area of (DOAP + DOBP) – Area of sector OACB= 43.25 – 26.16 = 17.09 cm2

31. A bucket open at the top is in the form of a frustum of a cone with a capacity of 12308.8cm3. The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Findthe height of the bucket and the area of metal sheet used in making the bucket. (use p =3.14)

Sol. Let R, r and V be the upper radius, lower radius and volume of the frustum respectively thenR = 20 cm, r = 12 cmand V = 12308.8 cm3

\ Volume of frustum of cone 13

= × p [(R2 + r2 + Rr)] h

Þ 12308.813

= × 3.14 [400 + 144 + 240] h

h12308.8 33.14 784

´=´ Þ h = 15 cm

Now, l (slant height) = 2 2(R – )r h+ 2 2(20 12) 15= - +

64 225 289= + = = 17 cm2

Total area of metal sheet used = Curved Surface Area of cone + Area of base= p (R + r) l + pr2

= p (20 + 12) × 17 + p × 12 × 12 [Q lower radius, r = 12 cm]= p × 32 × 17 + 144p = 688p = 688 × 3.14 = 2160.32 cm2

10. How many terms of the A.P. 27, 24, 21, .... should be taken so that their sum is zero ?Sol. In the given A.P.,

Here, first term (a) = 27and Common difference (d) = 24 – 27 = – 3

Given Sum of n terms (Sn) = 0

We know that, Sn [2 ( 1) ]2n a n d= + -

02n= [2 × 27 + (n – 1) (– 3)]

Þ 54 – 3n + 3 = 0 Þ 3n = 57 Þ n = 19Thus, the sum of 19 terms of the given A.P. is zero.



Arundeep’s Solved Papers Mathematics 2016 (Delhi)15618. Solve for x :

1 21 2

x xx x

+ -+- +

2 34 ;2

xx

+= -- x ¹ 1, – 2, 2

Sol. Given eqn. be,

1 21 2

x xx x

+ -+- +

2 342

xx

+= -- Þ

1 2 2 31 2 2

x x xx x x

+ - ++ +- + - = 4

Þ2( 1) ( 2) ( 2) ( 2) ( 1) (2 3) ( 1) ( 2)

( 1) ( 2) ( 2)x x x x x x x x

x x x+ + - + - - + + - +

- + - = 4

Þ (x + 1) (x2 – 4) + (x – 1) (x2 + 4 – 4x) + (2x + 3) (x2 + x – 2) = 4 (x – 1) (x2 – 4)Þ x3 – 4x + x2 – 4 + x3 + 4x – 4x2 – x2 – 4 + 4x + 2x3 + 2x2 – 4x + 3x2 + 3x – 6

= 4 (x3 – 4x – x2 + 4)Þ x3 + x2 – 4x – 4 + x3 – 5x2 + 8x – 4 + 2x3 + 5x2 – x – 6 = 4 (x3 – x2 – 4x + 4)Þ 4x3 + x2 + 3x – 14 = 4x3 – 4x2 – 16x + 16Þ 5x2 + 19x – 30 = 0 Þ 5x2 + 25x – 6x – 30 = 0Þ 5x (x + 5) – 6 (x + 5) = 0 Þ (x + 5) (5x – 6) = 0

Þ x + 5 = 0 or 5x – 6 = 0 Þ x = – 5 or65

x =

Thus, solutions of given equation are

x = – 5 and65

x =

19. Two different dice are thrown together. Find the probability of :(i) getting a number greater than 3 on each die.

(ii) getting a total of 6 or 7 of the numbers on two dice.Sol. (i) When two dice are thrown together total possible outcomes = 6 × 6 = 36

Sample space = {(1, 1), (1, 2) .... (1, 6)(2, 1), (2, 2) ......(2, 6)(3, 1), (3, 2) .... (3, 6)(4, 1), (4, 2) .... (4, 6)(5, 1), (5, 2) ..... (5, 6)(6, 1), (6, 2) ...... (6, 6)}

Favourable outcomes when both dice have number more than 3 are (4, 4), (4, 5), (4, 6), (5, 4),(5, 5), (5, 6), (6, 4), (6, 5), (6, 6), i.e. 9 outcomes.

\ Required P (a number greater than 3 on each due)Number of favourable outcomes

Number of total possible outcomes=

9 136 4

= =

(ii) Favourable outcomes when sum of the numbers appearing on the dice is 6 or 7 are, i.e. (1, 5),


Arundeep’s Solved Papers Mathematics 2016 (Delhi)157(1, 6), (2, 4), (2, 5), (3, 3), (3, 4), (4, 2),(4, 3), (5, 1), (5, 2), (6, 1).Thus total no. of favourable cases = 11.

\ Required P (a total of 6 or 7)1136

=

20. A right circular cone of radius 3 cm, hasa curved sur face area of 47.1 cm2. Findthe volume of the cone. (use p = 3.14)

Sol. Given Radius of cone (r) = 3 cm\ Curved surface area of cone = prl = 47.1 cm2

\ l47.1 47.1

3.14 3r= =

p ´ = 5 cm,

lh

r

a

where l = slant height of cone

We know that, h 2 2l r= -

2 25 3 25 9 16 4= - = - = =

\ Required volume of cone 213

r h= ´ p

13

= × 3.14 × 3 × 3 × 4

= 3.14 × 3 × 4 = 37.68 cm3

28. The angles of elevation of the top of atower from two points at a distance of 4m and 9 m from the base of the towerand in the same straight line with it are60º and 30º respectively. Find the heightof the tower.

Sol. Let AB be the tower of height ‘h’.Let D and C are the two points on groundwhich is at a distance of 4 m and 9 metre

from bottom of tower respectively.\ BD = 4 m and BC = 9 m

In right DABD,ABBC = tan 60º

Þ 4h

3=

Þ h 4 3=

A

CB60° 30°

D4 m9m

h m

Hence the required height of tower be 4 3metre.

29. Construct a triangle ABC in whichBC = 6 cm, AB = 5 cm and ÐABC = 60º.Then construct another triangle whose

sides are34 times the corresponding

sides of DABC.

Sol. Steps of construction :1. Draw DABC with side BC = 6 cm,

AB = 5 cm, ÐABC = 60º.2. Draw ray BX making an acute angle with

BC on opposite side of vertex A.3. Locate 4 points P1, P2, P3, P4 on line

segment BY.4. Join P4C and draw a line through P3,

parallel to P4C intersecting BC at C¢.5. Draw a line through C¢ parallel to AC

intersecting AB at A¢.

DA¢BC¢ is the required triangle.



30. The perimeter of a right triangle is 60cm. Its hypotenuse is 25 cm. Find thearea of the triangle.

Sol. Here, Perimeter of triangle, a + b + c = 60[Given]

Let c be the hypotenuse of triangle DABCa + b + 25 = 60a + b = 35 cm ...(1)Using Pythagoras theorem, we havea2 + b2 = (25)2 = 625

Now, (a + b)2 = a2 + b2 + 2ab(35)2 = 625 + 2ab [using (1)]

Þ 1225 – 625 = 2abÞ 2ab = 600 Þ ab = 300

A

C B

c = 25 cm

90°a

b

\ Required area DABC12

= × base × height

=12 ab

12

= × 300 = 150 cm2

10. How many terms of the A.P. 65, 60, 55,... be taken so that their sum is zero ?

Sol. In the given A.P.,First term (a) = 65

and Common difference (d) = 60 – 65 = – 5Sum of n terms (Sn) = 0

Since, Sn [2 ( 1) ]2n a n d= + -

0 [2 65 ( 1) ( 5)]2n n= ´ + - -

0 = 130 – 5n + 5Þ – 5n = – 135 Þ n = 27\ Hence, required sum of 27 terms is zero.

18. A box consists of 100 shirts of which 88are good, 8 have minor defects and 4have major defects. Ramesh, ashopkeeper will buy only those shirtswhich are good but ‘Kewal’ anothershopkeeper will not buy shirts with

major defects. A shirt is taken out of thebox at random. What is the probabilitythat(i) Ramesh will buy the selected shirt ?(ii) ‘Kewal’ will buy the selected shirt ?

Sol. (i) When one shirt is taken out, thennumber of total possible outcomes = 100Ramesh will purchase when shirt is good,

\ Favourable outcomes = number of goodshirts = 88P (Ramesh buys shirt)

Number of favourable outcomesNumber of total possible outcomes

=

88 22100 25

= =

(ii) Kewal will buy shirt if a shirt is not havingmajor defect.Number of favourable outcomes = Numberof shirts without major defect = 96

SET-III [UNCOMMON QUESTIONS TO SET-I and Set II]



\ P (Kewal buys a shirt)96 24

100 25= =

19. Solve the following quadratic equation for x : 2 1 0a a bx xa b a

+æ ö+ + + =ç ÷+è ø

Sol. Given eqn. be, 2 1a a bx xa b a

+æ ö+ + +ç ÷+è ø = 0

Þ 2 a a b ax xa b a a b

+æ ö æ ö+ + +ç ÷ ç ÷+ +è ø è øa b

a+æ ö

ç ÷è ø = 0

Þ 2 a a b ax x xa b a a b

+æ ö æ ö æ ö+ + +ç ÷ ç ÷ ç ÷+ +è ø è ø è ø· a b

a+æ ö

ç ÷è ø = 0

Þ a a b ax x xa b a a b

+é ù æ ö é ù+ + +ç ÷ê ú ê ú+ +ë û è ø ë û= 0 Þ

a a bx xa b a

+æ ö æ ö+ +ç ÷ ç ÷+è ø è ø = 0

Þax

a b+

+ = 0 ora bx

a++ = 0 Þ x

aa b

æ ö= - ç ÷+è ø ora bx

a+æ ö= - ç ÷è ø

20. A toy is in the form of a cone of base radius 3.5 cm mounted on a hemisphere of basediameter 7 cm. If the total height of the toy is 15.5 cm, find the total surface area of the toy.

22Use π =7

é ùê úë û

Sol. Here, given thatbase radius of cone = base radius of hemisphere = 3.5 m = rh = height of cone = 15.5 – 3.5 = 12 cm

Also, slant height of cone, l 2 2h r= +

2 2(12) (3.5) cm= + 144 12.25 cm= + 156.25 cm= = 12.5 cm

\ Curved Surface Area of cone = prl22 3.5 12.57

= ´ ´ = 137.5 cm2

\ Curved surface area of hemisphere = 2pr2

2222 (3.5)7

= ´ ´ = 77 cm2

Hence, Total Surface Area of toy = Surface area of hemisphere + Curved Surface Area of cone= 77 + 137.5 = 214.5 cm2

28. The sum of three numbers in A.P. is 12 and sum of their cubes is 288. Find the numbers.Sol. Let the three numbers in A.P. are a – d, a, a + d

Then a – d + a + a + d = 12 [Q Given that, S3 = 12]

h

3.5 cm

l

15.5 cm

3.5 cm


Arundeep’s Solved Papers Mathematics 2016 (Delhi)160Þ 3a = 12 Þ a = 4

Also, (a – d)3 + a3 + (a + d)3 = 288[Q Sum of their cubes = 288]

Þ (4 – d)3 + (4)3 + (4 + d)3 = 288Þ 64 – 48d + 12d2 – d3 + 64 + 64 + 48d +

12d2 + d3 = 288Þ 24d2 + 192 = 288Þ 24d2 = 96Þ d2 = 4 Þ d = ± 2

For d = 2, the numbers will be 2, 4, 6. Ford = – 2, numbers will be 6, 4, 2.Hence, required numbers are 2, 4, 6.


Sol. Given : A circle C (O, r) and a tangent ABat a point P.To prove : OP ^ ABConstruction : Take any point Q other thanP on the tangent AB.Join OQ, intersecting circle at R.Proof : We have, OP = OR [Radii]

O

P

N BA M

OQ = OR + RQ\ OQ > OR Þ OQ > OP

[Q OR = OP = radius]

Thus, OP < OQ, i.e., OP is shorter than anyother segment joining O to any point of AB.But among all line segments, joining pointO to point on AB, shortest one isperpendicular from O on AB.Hence, OP ^ AB.

30. The time taken by a person to cover

150 km was122 hours more than the

time taken in the return journey. If hereturned at a speed of 10 km/hour morethan the speed while going, find the speedper hour in each direction.

Sol. Let t1 and t2 be the time taken in going andreturning respectively.Also, u and u + 10 be the speed in goingand returning respectively.

As, VelocityDistance

Time=

Þ Time DistanceVelocity

=

\ t1150=

u and t2

15010

=+u

According to question, t1 – t2122

=

150 15010

-+u u

52

=

Þ 1 115010

é ù-ê ú+ë ûu u

52

=

Þ1 1 5υ υ 10 2 150

- =+ ´

Þ1 1

10-

+u u

160

=

Þυ 10 υ 1

υ 60+ - =

Þ 60 [u + 10 – u] = u (u + 10)Þ u2 + 10u – 600 = 0Þ u2 + 30u – 20u – 600 = 0Þ u (u + 30) – 20 (u + 30) = 0Þ (u + 30) (u – 20) = 0


Arundeep’s Solved Papers Mathematics 2016 (Delhi)161Þ u = 20 or u = – 30 (rejected)

Hence, velocity in going 20 km/hour and in returning be (u + 10) i.e. 30 km/hour.31. Draw a triangle ABC with BC = 7 cm, ÐB = 45º and ÐA = 105º. Then construct a triangle

whose sides are45 times the corresponding sides of DABC.

Sol. Given, ÐB = 45º, ÐA = 105º\ Sum of all interior angles in D = 180º\ ÐA + ÐB + ÐC = 180ºÞ ÐC = 30º

A¢

X

P1

P2

P3

P4

P5

Y

C¢

Z

Steps of construction :1. Draw a line segment BC = 7 cm2. at B, construct ÐCBX = 45° and at C, construct ÐBCY = 30°3. Let BX and CY intersects at A.4. Join AB and AC we get the DABC.5. Draw a ray BZ making an acute angle with BC on opposite side of vertex A.6. Locate 5 points P1, P2, P3, P4, P5 on BZ, s.t BP1 = P1 P2 = P2P3 = P3P2 = P4P5

7. Join P5C. Draw line through P4 parallel to P5C intersecting BC at C¢.

8. Through C¢, draw line parallel to AC intersecting AB at A¢.DA¢BC¢ is the required triangle.


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