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10th CBSE{SA – I}

Mathematics

Solution Visits: www.pioneermathematics.com/latest_updates

Mock Paper With

Blue Print of Original Paper on Latest Pattern

http://www.pioneermathematics.com/latest_updates.com



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10th CBSE First Term {SA- I}

Blue Print

Topic/Unit MCQs SA(1) SA(II) LA Total

Number System 2(2) 1(2) 2(6) – 5(10)

Algebra 2(2) 2(4) 2(6) 2(8) 8(20)

Geometry 1(1) 2(4) 2(6) 1(4) 6(15)

Trigonometry 4(4) 1(2) 2(6) 2(8) 9(20)

Statistics 1(1) 2(4) 2(6) 1(4) 6(15)

Total 10(10) 8(16) 10(30) 6(24) 34(80)



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General instructions:

Time: 3hrs. M: M: 90

(i) All questions are compulsory.

(ii) The question paper consists of 34 questions divided into sections A, B, C and D.

Section-A comprises of 8 questions of 1 mark each, section-B comprises of 6

questions of two marks each, section-C comprises of 10 questions of three marks

each and section-D comprises of 10 questions of four marks each.

(iii) Question numbers 1 to 8 in section-A are multiple choice questions where you

are required to select one option out of the given four.

(iv) There is no overall choice. However, internal choice has been provided in 1

question of two marks, 3 questions of three marks each and two questions of four

marks each. You have to attempt only one of the alternatives in all such questions.

(v) Use of calculator is not permitted.



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Section-A

Questions number 1 to 8 carry one marks each

1. The pair of equation x = 2 and x = –4 has :

(a) infinitely many solutions (b) no solution

(c) two solutions (d) one solution

Sol : (b)

2. If sides of two similar triangles are in ratio 4 : 9, then area of these triangles are in the

ratio

(a) 2 : 3 (b) 4 : 9 (c) 81 : 16 (d) 16 : 81

Sol : (d)

Ratio of Areas = 2

4

9 =

16

81

3. The x-coordinate of the point of intersection of more than and less then ogive is :

(a) mode (b) mean (c) median (d) Variance

Sol : (c)

4. If LCM (54, 336) = 3024, then HCF (54, 336) is:

(a) 54 (b) 6 (c) 336 (d) 36

Sol : (b)

L.C.M × H.C.F

=Product of two numbers

= 54 × 336 = 3024 × H.C.F

H.C.F = 6



5

5. Value of 5 tan2A–5sec2A is :

(a) 1 (b) 0 (c) –5 (d) 5

Sol : (c)

5 tan2A – 5 × (1 + tan2A) = 5 × (tan2A – 1 – tan2A)

= –5

6. If 2 sin2A = 3 , then A equal to :

(a) 90o (b) 60o (c) 45o (d) 30o

Sol : (d)

sin 2A = 3

2Sin2A = Sin600 A = 300

7. If sec A = q

p, then value of

1

p cos A is:

(a) 1

p (b)

1

q (c)

1

pq (d)

p

q

Sol : (b)

1 1 p 1cos A

p p q q

8. If a positive integer n is divided by 2, then the remainder can be :

(a) 1 or 2 (b) 1, 2, or 3 (c) 0 or 1 (d) 2

Sol : (c)



6

Section-B

Questions number 9 to 14 carry two marks each

9. In the given figures, find measure of X .

Sol :

In PQR and zyx

PQ 4.2 1

ZY 8.4 2

PR 3 3 1

ZX 26 3

QR 7 1

YX 14 2

PQ PR QR

ZY ZX YX

PQR ZYX (By SSS similarity criteria)

PRQ ZXY ..(1) (By CPST)

In PQR

PQR + PRQ + RPQ = 1800 (Angle sum property of )

600 + 700 + PRQ = 1800

0PRQ 50 ..(2)

From (1) & (2)

0ZXY X 50

10. Is 7 × 11 ×13+13 a composite number? Justify your answer.

Sol :

7 × 11 × 13 + 13



7

= 13(7 × 11 × 1 + 1)

= 13(77 + 1)

= 13(78)

= 13 × 2 × 13 × 3

As, the no.7 × 11 × 13 + 13 is having more than 2 prime factors,

The given no. is a composite no.

11. Solve: 3 2 2 5

0 and 19x y x y

Sol :

3 20

x y ..(1)

2 519

x y ..(2)

Put 1 1

a & bx y

in (1) & (2)

3a – 2b = 0 ..(3)

2a + 5b = 19 ..(4)

Multiply (3) & (4) by 5 & 2 respectively,

(3a – 2b = 0) × 5

15a – 10b = 0 ..(5)

& (2a + 5b = 19) × 2

4a + 10b = 38 …(6)

Adding (5) & (6)

15a 10b 0

4a 10b 38

19a 38 a 2

Put = a = 2 in (2)

3(2) – 2b = 0

6 – 2b = 0



8

–2b = – 6

b = 3

1 1

xa 2

& 1 1

yb 3

12. The following distribution give the daily income of 50 workers of a factory:

Daily income in Rs. 100–120 120–140 140–160 160–180 180–200

Number of workers 12 14 8 6 10

Write the above distribution as “more than type” cumulative frequency distribution.

Sol :

Daily income (in

Rs)

More than type

distribution

fi Cf

100–120

120–140

140–160

160–180

180–200

More than 100

More than 120

More than 140

More than 160

More than 180

12

14

8

6

10

50

38

24

16

10

Total 50

13. If 1

and are zeroes of polynomial 4x2 –2x + (k –4). Find k.

Sol :

If f(x) = 4x2 – 2x + (k – 4)

where, 1

x ,

According to relationship between zeroes & coefficients of a polynomial,

1 c

a



9

k 41

4

4 = k – 4

k = 8

Value of k = 8.

14. If tan 2A = cot (A –18o), where 2A is an acute angle, find the value of A.

Sol :

0tan 2A cot A 18

0cot 90 2A

= cot (A – 18)

90 – 2A = A – 18

90 + 18 = 3A

3A = 108

A = 360

Or

If is an acute angle and sin = cos , find the value of 3 tan2 + 2sin2 – 1.

Sol :

sin cos …(1) (Given)

2 23 tan 2sin 1

=

222 2 2

2

sin3 2 sin sin cos

cos 2 2[ sin cos 1]

=

2

2 2 2sin3 2 sin sin cos sin cos from(1)

cos

= 2 23 sin sin sin cos

= 3

Value of 2 23tan 2sin 1 3



10

Section-C

Questions number 15 to 24 carry three marks each

15. A survey regarding the height (in cm) of 50 girls of class X of a school was conducted and

the following data was obtained:

Height(in cm) 120–130 130–140 140–150 150–160 160–170 Total

Number of girls 2 8 12 20 8 50

Find the mode of the data.

Sol :

Model class = 150 – 160

f1 = 20

f0 = 12

f2 = 8

h = 10

l = 150

Mode = 1 0

1 0 2

f fl h

2f f f

= 20 12

150 1040 12 8

= 8

150 1020

= 150 + 4

= 154 cm

16. Prove that : cot A cos A cosecA 1

cot A cos A cosecA 1.

Sol :

To prove : cot A cos A cosec A 1

cot A cos A cosec A 1

L. H. S. = cot A cos A

cot A cos A



11

=

cos Acos A

sin Acos A

cos Asin A

=

cos A sin A cos A

sin Acos A sin A cos A

sin A

= cos A sin A cos A

cos A sin A cos A

= cos A 1 sin A

cos A 1 sin A

=

11

cosec A1

1cosec A

=

cosec A 1

cosec Acosec A 1

cosec A

= cosec A 1

cosec A 1 = R. H. S.

Hence proved.

17. In the given figure, oACB 90 and CD AB . Prove that 2

2

BC BD

AC AD



12

Sol:

ADC ~ ACB

AC AD CD

AB AC BC

2AC AD.AB ……….(1)

CDB ~ ACB

CD BC BD

CA AB BC

2BC AB.BD ……….(2)

Equation (2)/(1) we get

2

2

BC AB.BD

AD.ABAC

2

2

BC BD

ADAC

Or

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



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Sol :

Given : In ABC, AD BC

To Prove : AB2 + CD2 = BD2 + AC2

Proof : In ADC, AD CD (Given)

AC2 = CD2 + AD2 (By Pythagoras theorem)

AD2 = AC2 – CD2 ..(1)

In ADB, AD BD (Given)

AB2 = AD2 + BD2 (By Pythagoras theorem)

AD2 = AB2 – BD2 ..(2)

From (1) & (2)

AC2 – CD2 = AB2 – BD2

AC2 + BD2 = AB2 + CD2

AB2 + CD2 = BD2 + AC2

Hence, proved.



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18. In the given figure, the line segment XY is parallel to AC of a ABC and it divides the

triangle into two parts of equal area. FindAX

AB.

Sol :

Let Area of BXY = k

Area of BAC = 2k

In XBY and ABC

XBY = ABC (Common angle)

BXY = BAC (Corresponding angles)

BXY ~ BAC by AA similarity criteria

2Area of BXY BX

Area of BAC AB

2K BX

2k AB

BX 1

AB 2



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AB AX 1

AB 2

2AB 2AX AB

AX 2 1

AB 2

2 2 1

2 2=

AX

AB

AX 2 2

AB 2

19. Prove that: 3

3

sin 2sintan

2cos cos

Sol :

To prove : 3

3

sin 2 sintan

2 cos cos

Proof : = 3

3

sin 2 sin

2 cos cos

= 2

2

sin 1 2 sin

cos 2cos cos

= 2

2

1 2 sintan

2 1 sin 1

= 2

2

1 2 sintan

2 2 sin 1

= 2

2

1 2 sintan

1 2 sin

= tan = R. H. S.

Hence, proved.

20. Find the mean of the following data:



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Or

Find the missing frequency for the given data if mean of distribution is 52.

Wages (in Rs.) 10–20 20–30 30–40 40–50 50–60 60–70 70–80

No. of workers 5 3 4 f 2 6 13

Sol :

Class interval Frequency (fi) xi fixi

100-120

120-140

140-160

160-180

180-200

12

14

8

6

10

110

130

150

170

190

1320

1820

1200

1020

1900

if 50 i if x 7260

Mean = i i

i

x f

f

= 7260

50

= 145.2

Class- interval 100–120 120–140 140–160 160–180 180–200

Frequency 12 14 8 6 10



17

Or

Sol :

Wages in Rs. No. of workers xi di = xi – 45 fidi

10-20 20-30 30-40 40-50 50-60 60-70 70-80

5 3 4 f 2 6

13

15 25 35 45 55 65 75

–30 –20 –10

0 10 20 30

–150 –60 –40

0 20

120 390

if 33 f i if d 280

Mean = A + i i

i

f d

f

52 = 45 + 280

33 f

7 = 280

33 f

231 + 7f = 280

7f = 280 – 231

f = 7

21. Prove that 2 – 5 is an irrational number.

Sol :

Let us assume the 2 5 is a rational number. Then, according to Euclid’s division

lemma, use can find two co-prime integers, such them,

a2 5

b

a5 2

b

2b a5

b



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As, a & b are integers, this implies, (2b – a)/b is a rational no.

5 is also a rational number. But it contradicts the fact the 5 is irrational. This

contradiction has arisen due to our incorrect assumption that 2 5 is a rational

number.

2 5 is an irrational number.

22. If one zeros of polynomial p(x) = 3x2 – 8x + 2k + 1 is seven times of the other, then find

the zeroes and the value of k.

Sol :

p(x) = 3x2 – 8x + (2k + 1)

x = , 7

b

7a

8 88

3 3

1

3

1

x3

..(1)

7x 7

3 ..(2)

And, c 2k 1

7a 3

1 7 2k 1

3 3 3 (From (1) & (2))

7 = 6k + 3

6k = 4

k = 2

3

Value of k = 2

3



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23. A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40

km upstream and 55 km downstream. Find the speed of the stream and that of the boat in

still water.

Sol :

Speed of stream = y km/hr

Speed of boat = x km/hr

5 = D

T

Case I

30 4410

x y x y

Case II

40 5513

x y x y

Let 1 1

a, bx y x y

1 1

a , b5 11

on solving x = 8 km/hr y = 3 km/hr

Or

Solve by cross multiplication method: ax + by =a –b; bx – ay = a + b

Sol :

ax by a b

bx ay a b



20

2 2 2 2

x y

ab b a ab ab b a ab

= 2 2

1

a b

2 2 2 2 2 2

x y 1

a b a b a b

I II III

Equating I & III

2 2 2 2

x 1

a b a b

x = 1

Equating II & III

2 22 2

y 1

a ba b

y = – 1

Values of x = 1 & y = – 1

24. Evaluate: sinA cosA –sin A cos(90 A)cos A cos A sin(90 A)sin A

sec(90 A) cosec(90 A).

Sol :

= sin A sin A cos A cos A cos A sin A

sin A cos Acosec A sec A

= sin A cos A - sin3A cos A – cos3A sin A

= sin A cos A – sin A cos A (sin2A + cos2A)

= sin A cos A – sin A cos A = 0

= 0



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Section-D

Questions number 25to 34 carry Four marks each

25. Draw ‘more than ogive’ and less than ogive’ for the following distribution and hence

obtain the median.

Class-interval 5–10 10–15 15–20 20–25 25–30 30–35 35–40

Frequency 2 12 2 4 3 4 3

Sol:

Less then type of distribution

Class interval Less than type distribution

f.i c.f

5–10 Less than 10 2 2

10–15 Less than15 12 14

15–20 Less than20 2 16

20–25 Less than 25 4 20

25–30 Less than30 3 23

30–35 Less than35 4 27

35–40 Less than40 3 30

We will plot the points (10, 2); (15, 14); (20, 16); (25, 20); (30, 23) (35, 27); (40, 30)

More than type of distribution

Class interval More than type distribution

f.i c.f

5–10 More than 5 2 30

10–15 More than10 12 28

15–20 More than15 2 16

20–25 More than 20 4 14

25–30 More than25 3 10

30–35 More than30 4 7



22

35–40 Less than35 3 3

Total 30

We will plot the points (5, 30), (10, 28), (15, 16), (20, 14), (25, 10), (30, 7), (35, 3)

26. Evaluate:2 2 2 o 2 o 2 o

2 o 2 o 2 o 2 o

sin sin (90 ) 3cot 30 sin 54 sec 36

3 sec 61 cot 29 2 cosec 65 tan 25.

Or

Prove that :1 cos A sin A 1 sin A

sin A cos A 1 cos A

Sol:

2 2 2

2 2 2 2 2

sin cos 3 3 sin 54

3 sec 61 tan 61 cos 36 2 sec 25 tan 25

= 2

2

9 sin 90 361

3 1 cos 36 2 1



23

= 2

2

1 9 cos 36

3 2 cos 36

= 1 9

3 2

= 2 27

6

= 25

6

Or

To prove : 1 cos A sin A 1 sin A

sin A cos A 1 cos A

L.H.S. = 1 sin A cos A 1 sin A cos A

sin A cos A sin A cos A 1

2 2

2 2

1 sin A cos A

sin A cos A 1

2 2

2 2

1 sin 2sin A cos A

sin A cos A 2sin Acos A 1

2 2sin A sin A 2sin A

1 1 2sin Acos A

2sin A sin A 1

2sin Acos A

sin A 1R.H.S.

cos A Hence proved

27. If tan A + sin A = m and tan A – sin A = n, Show that m2 –n2 = 4 mn .

Sol:

Given : sin A tan A m & tan A sin A n

To prove :

2 2m n 4 mn

Proof:



24

2 2m n L.H.S.

=2 2

sin A tan A tan A sin A

= 2 2 2 2sin A tan A 2sin A tan A tan A sin A 2sin A tan A

= 4sinA tanA

R.H.S. = 4 mn

= 4 tan A sin A tan A sin A

= 2 24 tan A sin A

=2

2

2

sin A4 sin A

cos A

=2 2 2

2

sin A sin Acos A4

cos A

=2

2

2

1 cos A4 sin A

cos A

=2

2

2

sin A4 sin A

cos A

= 2 24 sin A tan A

= 4sin A tan A L.H.S. Hence proved

28. If the median of the distribution given below is 32.5. find x and y.

Class interval 0–10 10–20 20–30 30–40 40–50 50–60 60–70 Total

Frequency x 5 9 12 y 3 2 40



25

Sol:

Class interval f.i c.f

0–10 x x

10–20 5 x+5

20–30 9 x+14

30–40 12 x+26

40–50 y x+ y+26

50–60 3 x+ y+29

60–70 2 x+ y+31

40

x+ y + 31 = 40

x + y = 9

Median = 32.5 = 30 + 20 x 14

1012

x = 3, x + y = 9

3 + y = 9

y = 6

29. In the given figure, AB||PQ||CD, AB = x units, CD = y units and PQ = z units,

prove that, 1 1 1

x y z



26

Or

In an equilateral triangle ABC, D is a point on BC, such that BD=1

BC3

. Prove that 9 AD2 = 7

AB2.

Sol :

ABD ~ PQD

x z

l m m

xm = z (l + m)

x m = zl + zm

xm – zm = zl

m (x – z) = zl



27

m z

l x z ..(1)

BPQ ~ BCP

z l

y l m

z(l + m) = yl

zl + mz = yl

mz = yl – zl

mz = l(y – z)

m y z

l z ..(2)

From (1) & (2)

z y z

x 2 z 2z y z x z

z2 = xy – yz – xz + z2

xy = yz +zx xy yz zx

xyz xyz xyz

1 1 1

2 x y



28

Or

Sol:

Given : Equilateral ABC , BD=1

BC3

To prove : 9AD2 = 7AB2

Construction: Draw AM BC.

Proof: BD = 1

3BC ……(1)(Given)

BM = 1

2BC …(2) [Altitude of an equilateral is also its median]

DM = BM –BD

DM = BC BC

2 3 [From (1) and (2)]

DM = 3BC 2BC BC

6 6 ….(5)

In ADM,AM DM

AD2 = AM2 + DM2 (By Pythagoras theorem)

AM2 = AD2 –DM2 …..(3)

Similarly, in ABM,



29

AM2 = AB2 –BM2 …(4)

From (3) and (4)

AD2 –DM2 = AB2–BM2

AD2 –AB2 = DM2 –BM2

AD2 –AB2 = 2 2

BC BC

6 2 (from (2) and (5)

AD2 –AB2 = 2 2BC BC

36 4

=2 2BC 9BC

36

=28BC

36

AD2 –AB2 = 22BC

9

AD2 = 2 22BC 9AB

9

9AD2 = –2AB2 +9AB2

9AD2 = 7AB2 Hence proved

30. Prove that in a right angle triangle, the square of the hypotenuse is equal to the sum of the

squares of other two sides.

Sol:



30

Given : 0ABC, BAC 90 ; AD BC

To prove : BC2 = AB2 +AC2

Proof: According to a theorem, if a is drawn from the right of a to its hypotenuse,

then the two s formed are similar to each other & to the whole .

BAD ACD BCA ……(1)

BAD BCA From (1)

AB BC

BD AB (by CPST)

AB2 = BC . BD ……..(2)

ACD BCA (from (1))

AC BC

CD AC

AC2 = BC . CD …….(2)

Adding (1) & (2)

AC2 +AB2 = BC .BD +BC .CD

= BC(BD + CD)

= BC × BC

= BC2

AC2 +AB2 = BC2 Hence proved

31. Use Euclid's division lemma to show that the cube of any positive integer is of the form

9m, 9m +1 or 9m + 8.

Sol:

Let a be any +ve integer and b = 3, the according to Euclid’s division lemma, a can be



31

written as,

a = 3q +r, whose, 0 r<3.

a is of the form 3q, 3q+ 1&3q +2.

Case I: a = 3q

cubing both sides

a3 = (3q)3

= 27 q3

= 9(3q3)

=9m, where m = 9q3

a3 is of the of 9m.

Case II: a = 3q + 1

cubing both sides

a3 = (3q +1)3

= 27 q3 + 1 + 27 q2 +9q

= 9(3q3 + 3 q2 +q) +1

= 9m +1, where m = 3q3 +3q2 +q

a3 is of the form 9m +1.

Case III: a = 3q +2

cubing both sides

a3 = (3q+2)3

= 27q3 +8 +54q2 + 36q

= 9(3q3 +6q2 +4q) + 8

= 9m +8, where m = 3q3 + 6q2 +4q

a3 is of the form 9m +8

32. Draw the graph of 2x + y = 6 and 2x – y + 2 = 0. Shade the region bounded by

these lines with x axis. Find the area of the shaded region.

Sol:

2x + y = 6

2x – y =–2



32

Area of 1

b h2

= 214 4 8 cm

2

33. Find other zeroes of the polynomial p(x) = 2x4– 21x3 + 49x2 – 10x –20, if two of its zeroes

are 5 ± 5 .

Sol:

P(x) = 2x4 –213 + 49x2 – 10x –20

x = 5 5

x –5 5 = 0

( x –5) – 5 =0 …….(1)

x = 5 5

x –5 + 5 = 0

(x –5) + 5 = 0 ……….(2)

Multiplying (1) & (2)



33

(x – 5) 2 – 2

5 =0

x2 + 25 –10x – 5 = 0

f(x) = x2 –10x +20 = 0

2

2 4 3 2

4 3 2

3 2

3 2

2

2

2x x 1

x –10x 20 2x 21x 49x 10x 20

2x 20x 40x

x 9x 10x

x 10x 20x

x 10x 20

x 10x 20

0

g(x) = 2x2 – x – 1 = 0

= 2x2 – 2x + x + 1 = 0

= 2x ( x –1) + 1(x – 1)

= (2x + 1) ( x–1) = 0

x = 1

,1& 5 5 , 5 52

Ans.

34. Let days taken by 1 women = x

Let days taken by 1 man = y

2 5 1

x y 4 ……(1)

&

3 6 1

x y 3 ……(2)

Put 1

x = a &

1b

y in (1) & (2)



34

2a +5b = 1

4 ……(3)

3a + 6b = 1

3

a 2b 1/9 2

2a + 4b = 2/9 …..(4)

Subtracting (3) from (4) we get

2a + 5b = ¼

2a + 4b = 2/9

Put b = 1/36 in (3)

2a + 51 1

36 4

72a + 5 = 36

94

72a = 4

a = 4 1

72 18

x = 1

18 daysa

1

y 36 daysb

All The Best

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