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MATHEMATICS–X

TERM-I

Marks : 90

UNITS MARKS

I. NUMBER SYSTEMS 11

II. ALGEBRA 23

III. GEOMETRY 17

IV. TRIGONOMETRY 22

V. STATISTICS AND PROBABILITY 17

TOTAL 90

UNIT I : NUMBER SYSTEMS

1. REAL NUMBERS (15) Periods

Euclid’s division lemma, Fundamental Theorem of Arithmetic - statements afterreviewing work done earlier and after illustrating and motivating through examples,

Proofs of results - irrationality of 2 3 5, , , decimal expansions of rational numbers

in terms of terminating/non-terminating recurring decimals.

UNIT II : ALGEBRA

1. POLYNOMIALS (7) Periods

Zeros of a polynomial. Relationship between zeros and coefficients of quadraticpolynomials. Statement and simple problems on division algorithm for polynomialswith real coefficients.

2. PAIR OF LINEAR EQUATIONS IN TWO VARIABLES (15) Periods

Pair of linear equations in two variables and their graphical solution. Geometricrepresentation of different possibilities of solutions/inconsistency.

Algebraic conditions for number of solutions. Solution of a pair of linear equations intwo variables algebraically—by substitution, by elimination and by cross multiplication.Simple situational problems must be included. Simple problems on equations reducibleto linear equations may be included.

UNIT III : GEOMETRY

1. TRIANGLES (15) Periods

Definitions, examples, counter examples of similar triangles.

1. (Prove) If a line is drawn parallel to one side of a triangle to intersect the other twosides in distinct points, the other two sides are divided in the same ratio.

2. (Motivate) If a line divides two sides of a triangle in the same ratio, the line isparallel to the third side.

Syllabus.p65 5/10/2014, 10:23 AM1

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3. (Motivate) If in two triangles, the corresponding angles are equal, theircorresponding sides are proportional and the triangles are similar.

4. (Motivate) If the corresponding sides of two triangles are proportional, theircorresponding angles are equal and the two triangles are similar.

5. (Motivate) If one angle of a triangle is equal to one angle of another triangle andthe sides including these angles are proportional, the two triangles are similar.

6. (Motivate) If a perpendicular is drawn from the vertex of the right angle of a righttriangle to the hypotenuse, the triangles on each side of the perpendicular are similarto the whole triangle and to each other.

7. (Prove) The ratio of the areas of two similar triangles is equal to the ratio of thesquares on their corresponding sides.

8. (Prove) In a right triangle, the square on the hypotenuse is equal to the sum of thesquares on the other two sides.

9. (Prove) In a triangle, if the square on one side is equal to sum of the squares on theother two sides, the angles opposite to the first side is a right triangle.

UNIT IV : TRIGONOMETRY1. INTRODUCTION TO TRIGONOMETRY (10) Periods

Trigonometric ratios of an acute angle of a right-angled triangle. Proof of their existence(well defined); motivate the ratios, whichever are defined at 0° & 90°. Values (withproofs) of the trigonometric ratios of 30°, 45° & 60°. Relationships between the ratios.

2. TRIGONOMETRIC IDENTITIES (15) PeriodsProof and applications of the identity sin2A + cos2A = 1. Only simple identities to begiven. Trigonometric ratios of complementary angles.

UNIT V : STATISTICS AND PROBABILITY1. STATISTICS (18) Periods

Mean, median and mode of grouped data (bimodal situation to be avoided). Cumulativefrequency graph.

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Questions Paper Designs (Math).pmd 5/27/2014, 2:08 PM1

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● Sample Question Paper–I & II (Solved) S-1–S-36(Based on the Latest CBSE Guidelines)

● Chapterwise Important Questions (Solved) 1–107

1. Real Numbers 1–10

2. Polynomials 11–24

3. Pair of Linear Equations in Two Variables 25–45

4. Triangles 46–69

5 & 6. Introduction to Trigonometry& Trigonometric Identities 70–87

7. Statistics 88–107

● Value Based Questions (Solved) 108–116

● Model Question Papers for Practice

Model Question Paper–I M-1

Model Question Paper–II M-5

Model Question Paper–III M-9

Model Question Paper–IV M-13

Model Question Paper–V M-16

Model Question Paper–VI M-20

Model Question Paper–VII M-24

Model Question Paper–VIII M-28

Model Question Paper–IX M-32

Model Question Paper–X M-35

● Answers M-39–M-42

CONTENTS

Content.pmd 6/3/2014, 3:45 PM1

Time Allowed : 3 Hours Maximum Marks : 90

General Instructions :

1. All questions are compulsory.

2. The question paper consists of 31 questions divided into four sections — A, B,

C and D. Section A comprises of 4 questions of 1 mark each, section B comprises

of 6 questions of 2 marks each, section C comprises of 10 questions of 3 marks

each and section D comprises of 11 questions (including one VBQ) of 4 marks

each.

3. There is no internal or external choice.

4. Use of calculators is not permitted.

SECTION — A

Q.1. A positive integer n when divided by 9, gives 7 as remainder. Find the remainderwhen (3n – 1) is divided by 9.

Sol. Here n can be written as 9k + 7, where k ∈ N

Now 3n – 1 = 3 (9k + 7) – 1

= 27 k + 20

Applying Euclid’s division lemma on (27k + 20) and 9, we have

27k + 20 = 9 × (3k + 2) + 2; where k ∈ N

Thus, 2 is the remainder.

Q.2. If x = 3m – 1 and y = 4 is a solution of the equation x + y = 6, then find the valueof m.

Sol. Given x + y = 6 ...(i)

Putting x = 3m – 1, y = 4 in (i)

(3m – 1) + 4 = 6

⇒ 3m = 3

⇒ m = 1.

Q.3. In ΔΔΔΔΔABC shown alongside, find the measureof angle ∠∠∠∠∠B.

A

B C

17 cm8 cm

15 cm

S-1

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Sample-Q-Paper (Solved) Term-I.p65 5/10/2014, 10:58 AM1

MATHEMATICS

Class–X

TERM—I

SAMPLE QUESTION PAPER-I (SOLVED)(Based on the Latest CBSE Guidelines)

MBD SURE SHOT SAMPLE PAPERS (X C.B.S.E)S-2

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Sol. Here, AB = 8 cm, BC = 15 cm, AC = 17 cmObserve 172 = 152 + 82

i.e. AC2 = BC2 + AB2

∴ By converse of Pythagoras theoremΔABC is right angled at B∴ ∠B = 90°.

Q.4. If the curves for ‘more than ogive’ and ‘less than ogive’ of a given grouped datameet at (28.2, 40) then find the median of the data.

Sol. We know that x-coordinate of point of intersection of two ogives namely ‘lessthan’ ogive and ‘more than’ ogive gives median of data.

∴ Median = 28.2.

SECTION — B

Q.5. Find the zeroes of the polynomial − −2 11 23 3

7y y .

Sol. Let p(y) = − −2 11 27

3 3y y

= 2 14 27

3 3y y y− + −

=⎛ ⎞ ⎛ ⎞− + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 27 1

3 3y y y

= ( )⎛ ⎞+ −⎜ ⎟⎝ ⎠2

7 13

y y

Zeroes of p(y) are given by p(y) = 0

⇒ ( )⎛ ⎞+ −⎜ ⎟⎝ ⎠2

7 13

y y = 0

⇒ 7y + 1 = 0 or −23

y = 0

⇒ y =1 2

,7 3

−

∴ and1 27 3

− are zeroes of given polynomial.

Q.6. Find the value of k if the equations 2x + 3y – 5 = 0 and 4x + ky – 10 = 0 have aninfinite number of solutions.

Sol. For infinite solutions = =1 1 1

2 2 2

a b ca b c

Here, a1 = 2, b1 = 3, c1 = – 5a2 = 4, b2 = k, c2 = –10

Sample-Q-Paper (Solved) Term-I.p65 4/22/14, 12:20 PM2

MATHEMATICS S-3

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∴ 2 34 k

= =−

−5

10

⇒3 1

2k= ⇒ k = 6.

Q.7. In the given figure if AD = x, DB = x – 2, AE = x + 2and EC = x – 1 and also DE || BC, find the value of x.

Sol. In Δ ABC, DE || BCBy Basic Proportionality theorem

��

��=

��

��

⇒+−

21

xx

=− 2x

x

⇒ (x + 2) (x – 2) = x (x – 1)⇒ x2 – 4 = x2 – x⇒ x = 4.

Q.8. Evaluate : 2 cosec2 30° + 3 sin260° – 34

tan230°

Sol. We know that

cosec 30° = 2, sin 60° = 23

, tan 30° = 13

∴ 2 cosec2 30° + 3 sin2 60° – 34

tan2 30°

= 2 × (2)2 + 3 × ⎛ ⎞ ⎛ ⎞− × ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

2 23 3 12 4 3

= 2 × 4 + 3 × − ×3 3 14 4 3

= 8 + −9 14 4

= 10.

Q.9. Prove that : (tan θθθθθ + 2) (2 tan θθθθθ + 1) = 5 tan θθθθθ + 2 sec2θθθθθSol. L.H.S. = (tan θ + 2) (2 tan θ + 1)

= 2 tan2θ + 4 tan θ + tan θ + 2= 2 tan2θ + 2 + 5 tan θ= 2 (tan2θ + 1) + 5 tan θ= 2 sec2θ + 5 tan θ = R.H.S. [Hence Proved]

A B

C

D

E



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Q.10. The following are the ages of 300 patients getting medical treatment in ahospital on a particular day :

Age (in years) 10–20 20–30 30–40 40–50 50–60 60–70

No. of patients 60 42 55 70 53 20

Form “less than type” cumulative frequency distribution.

Sol. Age No. of Age Cumulative(in years) Patients (in years) frequency

“Less than”10–20 60 20 6020–30 42 30 60 + 42 = 10230–40 55 40 102 + 55 = 15740–50 70 50 157 + 70 = 22750–60 53 60 227 + 53 = 28060–70 20 70 280 + 20 = 300

SECTION — CQ.11. Prove that square of any positive integer is of the form 4m or 4m + 1 for some

integer ‘m’.Sol. Let ‘a’ be any positive integer.

Applying Eudid’s division lemma to a and 4a = 4q + r ; 0 ≤ r < 4

i.e. a = 4q or 4q + 1 or 4q + 2 or 4q + 3When a = 4q :

a2 = (4q)2 = 16q2

= 4 (4q2)= 4m, where m = 4q2 is some integer ...(i)

When a = 4q + 1 :a2 = (4q + 1)2

= 16q2 + 8q + 1= 4q (4q + 2) + 1= 4m + 1, where m = q (4q + 2) is some integer ...(ii)

When a = 4q + 2 :a2 = (4q + 2)2

= 16q2 + 16q + 4= 4 (4q2 + 4q + 1)= 4m, where m = (4q2 + 4q + 1) is some integer ...(iii)

When a = 4q + 3 :a2 = (4q + 3)2


MATHEMATICS S-5

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= 16q2 + 24q + 9= (16q2 + 24q + 8 + 1)= 4 (4q2 + 6q + 2) + 1= 4m + 1, where m = 4q2 + 6q + 2 is some integer

...(iv) (i), (ii) and (iii) ⇒ square of any positive integer is either of the form 4m or 4m + 1.Q.12. If d is the H.C.F. of 45 and 27 find x, y satisfying d = 27x + 45y.Sol. Let us first find H.C.F. of 45 and 27.

Applying Euclid’s division algorithm to 45 and 2745 = 27 × 1 + 1827 = 18 × 1 + 918 = 9 × 2 + 0

⇒ H.C.F. (27, 45) = 9Now, 9 = 27 – 18

= 27 – (45 – 27)= 27 – 45 × 1 + 27= 54 – 45= 27 × 2 + 45 × (–1)

Comparing R.H.S. with 27x + 45y, we getx = 2, y = –1.

Q.13. Find k so that the polynomial x2 + 2x + k is a factor of the polynomial 2x4 + x3

– 14x2 + 5x + 6. Also, find all the zeroes of the two polynomials.Sol. Let, p(x) = 2x4 + x3 – 14x2 + 5x + 6

g(x) = x2 + 2x + kLet us divide p(x) by g(x)

x2 + 2x + k 2x4 + x3 – 14x2 + 5x + 6 2x2 – 3x – (8 + 2k)2x4 + 4x3 + 2kx2

– – –

– 3x3 – (14 + 2k)x2 + 5x + 6– 3x2 – 6x2 – 3kx + + +

– (8 + 2k)x2 + (3k + 5) x + 6– (8 + 2k)x2 – (16 + 4k)x – (8k + 2k2)

+ + +

(7k + 21)x + (2k2 + 8k + 6)

Here, remainder, r(x), = (7k + 21)x + (2k2 + 8k + 6) for g(x) to be factor of p(x), r(x) = 0⇒ 7k + 21 = 0 and (2k2 + 8k + 6) = 0 or 2 (k2 + 4k + 3) = 0

⇒ k = – 3 ...(i)

or 2(k2 + 3k + k + 3) = 0



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A

D

B

C

2 (k + 1) (k + 3) = 0⇒ k = – 1 or k = – 3 ...(ii)(i) and (ii) ⇒ k = – 3.

Q.14. Solve for x and y :

+ = − = ≠ ≠2 3 1 1 1

2, , 0, 02 3

x yx y x y

Sol. Taking =1

ux

and =1v

y the given system of equations gets reduced to

2u + 3v = 2 ...(i)

u – 12

v =13

...(ii)

Multiplying (ii) by 2 and subtracting from (i) 2u + 3v = 2

2u – v =23

– + –

4v = −2

23

4v =43

⇒ v = 13

Putting the value of v in (ii), we get

u – ×1 12 3

=13

⇒ u = + =1 1 13 6 2

Now, u =12

⇒ =1 1

2x ⇒ x = 2

and v =13

⇒ =1 13y

⇒ y = 3.

Q.15. In the given figure, if AD ⊥⊥⊥⊥⊥ BC, prove that AB2 + CD2 = BD2 + AC2

Sol. In rt. angled triangle, ΔABDAB2 = BD2 + AD2 ...(i) [By Pythagoras theorem]

Also, in rt. angled triangle, ΔADCAC2 = AD2 + CD2 [By Pythagoras theorem]

⇒ CD2 = AC2 – AD2 ...(ii)Adding (i) and (ii)

AB2 + CD2 = BD2 + AD2 + AC2 – AD2

⇒ AB2 + CD2 = BD2 + AC2 [Hence Proved]


MATHEMATICS S-7

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Q.16. In the given figure if LM || CB and LN || CD. Prove that AM AN=

AB AD.

A

M

N

D

B

CL

Sol. In ΔABC, LM || CB

⇒��

MB=

ALLC

...(i) [Basic Proportionality theorem]

Also, in ΔADC, LN || CD

⇒ ND��

=ALLC ...(ii) [Basic Proportionality theorem]

(i) and (ii) ⇒��

MB=

ANND

orMB��

=NDAN

Adding 1 to both sides

�� + 1AM

= +ND

1AN

⇒��

AM=

��

AN

⇒ABAM

=ADAN

orAMAB

=ANAD

[Hence Proved]

Q.17. Prove that : (1 + tan A tan B)2 + (tan A – tan B)2 = Sec2A Sec2BSol. L.H.S. = (1 + tan A tan B)2 + (tan A tan B)2

= 1 + tan2A tan2B + 2 tan A tan B + tan2A + tan2B – 2 tan A tan B= 1 + tan2A + tan2A tan2B + tan2B= (1 + tan2A) + tan2B (tan2A + 1)= (1 + tan2A) (1 + tan2B)= sec2A sec2B= R.H.S. [Hence Proved]



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Q.18. If 7 sin2θθ + 3 cos2θθ = 4, show that tan θθ = �

�± .

Sol. 7 sin2θ + 3 cos2θ = 4Dividing throughout by cos2θ

θ θ+θ θ

2 2

2 27 sin 3 cos

cos cos=

θ24

cos

⇒ 7 tan2θ + 3 = 4 sec2θ7 tan2θ + 3 = 4 (1 + tan2θ)7 tan2θ + 3 = 4 + 4 tan2θ

⇒ 3 tan2θ = 1

⇒ tan2θ =13

⇒ tan θ = ± 13

[Hence Proved]

Q.19. Find the median for the following distribution :

Classes 0–10 10–20 20–30 30–40 40–50

Frequencies 6 10 12 8 8

Sol.Classes Frequencies Cumulative

frequency

0–10 6 610–20 10 1620–30 12 2830–40 8 3640–50 8 44

N = 44

Here, N = 44 ∴ =N

222

Now, cumulative frequency just greater than 22 is 28 which corresponds to classinterval 20 – 30.

∴ Median class = 20–30l = 20, c = 16, f = 12, h = 10

We know that median = l +

�⎛ ⎞−⎜ ⎟⎝ ⎠×2

ch

f


MATHEMATICS S-9

K

K

=( )−

+ ×22 16

20 1012

= +60

2012

= 20 + 5 = 25.

Q.20. Calculate the mode of the following frequency distribution :

Marks 0–10 10–20 20–30 30–40 40–50

No. of students 6 10 12 32 20

Sol. Observe that the maximum frequency is 32 and it corresponds to class interval30–40.

∴ Modal class = 30–40l = 30, h = 10, f1 = 32, f0 = 12, f2 = 20We know that

Mode = l + ⎛ ⎞−

×⎜ ⎟− −⎝ ⎠1 0

1 0 22f f

hf f f

=−⎛ ⎞+ ×⎜ ⎟⎝ ⎠× − −

32 1230 10

2 32 12 20

= +−

20030

64 32

= + 20030

32 = 30 + 6.25 = 36.25

SECTION — D

Q.21. Prove that 2 3

5 is irrational.

Sol. Let, if possible 2 3

5 is rational.

So, we can find co-prime numbers p and q (q ≠ 0) such that

2 35

=pq

⇒ 3 =52

pq

Since p and q are integers, therefore 52

pq

is rational.

But = ⇒5

3 32

pq

is rational which is contradiction to the fact that 3 is

irrational number.



K

K

∴ Our supposition is wrong.

Hence, 2 3

5 is irrational number.

Q.22. Divya has pens and pencils which are 60 in number. If she has 25 more pensand 5 less pencils, then the number of pens become three times the number of pencils.Find the original number of each.

Sol. Let x, y be number of pens and pencils respectively.∴ According to problem : x + y = 60 ...(i)

(25 + x) = 3 (y – 5)⇒ x – 3y + 40 = 0 ...(ii)(i) ⇒ y = 60 – xPutting the value of y in (ii), we get

x – 3 (60 – x) + 40 = 0⇒ x – 180 + 3x + 40 = 0⇒ 4x – 140 = 0⇒ 4x – 140 = 0

⇒ x = =140

354

Putting for x in (i)35 + y = 60 ⇒ y = 25

Thus, Divya originally has 35 pens and 25 pencils.

Q.23. If ααααα and βββββ are zeroes of the polynomial f(x) = x2 – 3x – 2 find the quadratic

polynomial whose zeroes are

Sol. Given f(x) = x2 – 3x – 2

∴ α + β =( )

23Coeff.of

31Coeff.of

x

x

− −−= =

αβ = 2Constant term 2

21Coeff.of x

−= = −

Sum of zeroes of required polynomial p(x);

S = +α + β β + α

1 12 2

=( ) ( )( ) ( )

β + α + α + βα + β β + α

2 22 2

=( )

( )α + β

αβ + α + β + αβ2 2

3

4 2


α + β1

2 and

β + α1

2.

MATHEMATICS S-11

K

K

=( )

( )2 2

3

2 5

α + β

α + β + αβ

=( )

( ){ }2

3

2 2 5

α + β

α + β − αβ + αβ

=( )

( )23

2

α + β

α + β + αβ

=( )

( ) ( )=

+ −23 3 9

162 3 2

Product of zeroes of required polynomial p(x);

P = ( ) ( )×α + β β + α

1 12 2

= ( )αβ + α + β2 21

5 2

=( ){ }2

1

5 2 2αβ + α + β − αβ

=( ) ( ) ( )

= =+ −α + β + αβ2 2

1 1 1162 3 22

∴ The required polynomial isp(x) = k (x2 – Sx + P)

= ⎛ ⎞− +⎜ ⎟⎝ ⎠2 9 1

16 16k x x

Choosing, k = 16p(x) = 16x2 – 9x + 1.

Q.24. Solve graphically the system of equations x – y + 1 = 0, 3x + 2y – 12 = 0.Determine the coordinates of the vertices of the triangle formed by these lines. Also,find the area of this triangle.

Sol. x – y + 1 = 0 3x + 2y – 12 = 0

x 0 –1 2 x 0 4 2

y 1 0 3 y 6 0 3

ΔABC is bounded by line and x-axisBase AB = 4 – (–1) = 5, height CM = 3



K

K

∴ ar ΔABC =12

× AB × CM

=12

× 5 × 3

=152

sq. units.

Q.25. In the given figure, in ΔΔΔΔΔABC, XY||AC and XYdivides the ΔΔΔΔΔABC into two regions such that

ar (ΔΔΔΔΔBXY) = 2 ar (ACYX)

Determine AXAB

.

Sol. Given : ΔPQR, X, Y are points on PQ, PR respectively. XY||ACTo prove : ar (ΔBXY) = 2 ar (ACYX)Proof : ar (BXY) = 2 ar (ACYX)

ar (BXY) = 2 [ar (BAC) – ar (BXY)]ar (BXY) = 2 ar (BAC) – 2 ar (BXY)

3 ar (BXY) = 2 ar (BAC)

( )( )

ar BXYar BAC

=23

...(i)

In ΔBXY and ΔBAC,∠B = ∠B

A

B

C

X Y


MATHEMATICS S-13

K

K

∠BXY = ∠BAC∴ ΔBXY ~ ΔBAC [By AA similarity]

∴ ( )( )

ar BXYar BAC

=2

2BX

BA

⇒23

=2

2BX

BA

⇒BXBA

=23

⇒ −BX

1BA

= −2

13

⇒ −BA BXBA

=−3 23

⇒AXAB

= 3 2

3

− .

Q.26. Prove that the ratio of the areas of two similar triangles is equal to the ratio ofthe squares of their corresponding sides.

Sol. Given : Two triangles, ΔABC and ΔPQR such that ΔABC ∼ ΔPQR.

To prove : ( )( )ΔΔ

ar ABCar PQR

= ⎛ ⎞⎜ ⎟⎝ ⎠

2ABPQ

= ⎛ ⎞⎜ ⎟⎝ ⎠

2BCQR =

⎛ ⎞⎜ ⎟⎝ ⎠

2CARP

Construction : Draw AM ⊥ BC and PN ⊥ QR.

A

B CM Q N R

P

Proof : ar (ΔABC) =1

BC AM2

× ×

ar (ΔPQR) =1

QR PN2

× ×

( )( )

ar ABCar PQR

ΔΔ

=× ×

× ×

1BC AM

21

QR PN2

=BC AMQR PN

××

...(i)

Comparing ΔABM and ΔPQN :

A

B

C

X Y


MBD Sure Shot CBSE Sample PapersSolved Mathematics Class X

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