obile solution of class 10 cbse sa-ii board (set...

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L.K. Gupta (Mathematic Classes) www.poineermathematics.com. MOBILE: 9815527721, 4617721 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40D, CHANDIGARH 1 Solution Of Class 10 th CBSE SA-II Board (Set-1)Mathematics 1. (2k 1) k = (2k + 1) (2k 1) 2k 1 k = 2k + 1 2k + 1 k 1 = 2 k 3 Ans. (B) 2. 0 APB 90 Ans. (D) 3. AB = 5 cm

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Page 1: OBILE Solution Of Class 10 CBSE SA-II Board (Set …jsuniltutorial.weebly.com/uploads/7/8/7/0/7870542/question_paper... · Solution Of Class 10th CBSE SA-II Board (Set-1)Mathematics

L.K. Gupta (Mathematic Classes) wwwwww..ppooiinneeeerrmmaatthheemmaattiiccss..ccoomm.. MOBILE: 9815527721, 4617721

PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH

1

Solution Of Class 10th CBSE SA-II

Board (Set-1)Mathematics

1. (2k – 1) – k = (2k + 1) – (2k – 1)

2k – 1 – k = 2k + 1 – 2k + 1

k – 1 = 2

k 3

Ans. (B)

2.

0APB 90

Ans. (D)

3.

AB = 5 cm

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L.K. Gupta (Mathematic Classes) wwwwww..ppooiinneeeerrmmaatthheemmaattiiccss..ccoomm.. MOBILE: 9815527721, 4617721

PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH

2

BC = 12 cm

AC = ?

Pythagoras theorem

= AC2 = AB2 + BC2

= 52 + 122

= 25 + 144

AC2 = 169

AC = 13 cm

BC = x + 12 – x

AB = x + 5 – x

BP = BQ ..(1) (Tangents drawn from an external point to a circle are equal)

AP = AR ..(2)

CO = CR ..(3)

AC = 13

5 – x + (12 – x) = 13

5 – x + (12 – x) = 13

–2x + 17 = 13

– 2x = – 4

x = 14

2 = x = 2

Radius = 2 cm

Ans. (C)

4. No of children = 3

Outcomes = BBB

GGG

BBG

BGB

GBB

BGG

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L.K. Gupta (Mathematic Classes) wwwwww..ppooiinneeeerrmmaatthheemmaattiiccss..ccoomm.. MOBILE: 9815527721, 4617721

PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH

3

GBG

GGB

Probability = No. of favourablecases

Total no. of cases

Total cases = 8

= 7

8 Ans. (A)

5. In ABC

oABtan30

BC

150 1

BC 3BC 150 3 m

Ans.(B)

6. Total numbers= 15

No. divisible by 4 are

4, 8, 12

Outcomes = 3 1

15 5

Ans. (C)

7. Distance of BD

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PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH

4

BD= 2 2(0 4) (0 3)

= 16 9 25 5cm

Ans. (A)

8.

AB = ?

Pythagoras theorem

AB2 = AO2 + BO2

= 102 + 102

= 100 + 100

AB2 = 200

AB = 10 2 cm

Ans.(b)

9. As given equal roots

D=0

b2 –4ac =0

a = 4, b = p, c =3

p2 – 48 =0

p2 = 48

p 48 p= 4 3

10. The numbers divisible by both 2 and 5 are

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5

110, 120 ------------------------990

d = 120–110 =10

a 110 d 10 an=110

an = a + (n –1)d.

990 = 110 + (n –1)10

990 –110 = (n –1)10

880 + 10n –10.

890n

10 n 89

11. As tangents drawn from an external point to the circle are equal in length

So EA = EC …………..(1)

EB = ED ……….(2)

Adding (1) and (2)

EA + EB =EC + ED

AB = CD.

Hence. Proved.

12.

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AB = AC (Given)

BD = BF (length of the tangent drawn from an external point to a circle are equal)…..(1)

AF = AE …………….(2)

DC = CE ………….(3)

Now, AB = AC

AF + BF = AE + CE

AE + BD = AE + DC (From equation (1) , (2) & (3))

BD DC

13. (1, 1) (2, 1) (3, 1) (4, 1) (5, 1) (6, 1)

(1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2)

(1, 3) (2, 3) (3, 3) (4, 3) (5, 3) (6, 3)

(1, 4) (2, 4) (3, 4) (4, 4) (5, 4) (6, 4)

(1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5)

(1, 6) (2, 6) (3, 6) (4, 6) (5, 6) (6, 6)

(i) p( No. of each dice is even)=9 1

36 4

favourable cases (2,2),(2,4),(2,6),(4,2),(4,4)(4,6),(6,2),(6,4),(6,6)

(ii) P( sum appearing on both dice is 5)=4 1

36 9

Favourable (1,4),(2,3),(3,2),(4,1)

14. 23 r 462

2223 r 462

7

2

2

2

462 7r

22 3

21 7r

3

r 49 r 7

Vol. of hemisphere= 32r

3

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7

=2 22

7 7 73 7

= 344 215649 718.6 cm

3 3.

15. 16 15

1x x 1

16 x 15

x x 1

(16 – x) (x +1) = 15x

16x + 16 –x2 –x = 15x –x2 +15x + 16 = 15x

–x2 + 15x + 16 -15x=0

x2 = 16

x 4

16. a5 + a9 = 30

(a + 4d) + (a + 8d) = 30 na a (n 1)d

2a + 12d = 30

a + 6d = 15 ……………(1)

a25 = 3a8.

a + 24d = 3(a + 7d)

a + 24d = 3a + 21d

24d – 21d = 3a –a.

3d = 2a.

d = 2a

3…………….(2)

Put d in the equation (1)

a + 2a

6 153

a + 4a = 15

5a =15

a 3

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8

d = 2(3) 6

23 3

d = 2

A.P. = 3, 5, 7,------------

17. Step of construction

(i) Draw a ABCwith the sides given such that AC = 5.5 cm, AB = 5 cm, BC=6.5cm

(ii) Then draw acute BAX

(iii) Make 5 arcs on line AX such that AA1=A1A2=……….=A4A5

(iv) Join BA5

(v) Draw B’ A3 parallel to BA5

(vi) Similarly drawn C’ B’ parallel to CB.

Hence C'AB' is the required triangle

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18. In PAB

o 3000 3tan60

y

3000 33

y

y 3000m

In A' B'P

o 3000 3tan30

x y

1 3000 3

x y3

x y 3000 3 3

x+3000=9000

x 6000m

From B to B’

time taken by aeroplane is 30 second

distance 6000S 200 m / sec

time 30

19. As given PA = PB

2 2 2 2(K 1 3) (2 K) (K 1 K) (2 5)

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Squaring both sides

(K –4)2 + (2 – K)2 = 1 + 9

K2 + 16 – 8K + 4 + K2 – 4K =10

2K2 –12K + 10=0

K2 –6K + 5 =0

K2– 5K –K+ 5=0

K(K–5) –1(K–5)=0

K= 1, 5.

20. Let the ratio be K:1

Let P divides AB in the ratio K : 1

2K 3 7K 3

P ,K 1 K 1

But 7K 3

0K 1

7K = 3.

K = 3

7

Ratio 3 : 7.

Co-ordinates of pt. P

3 32 3 7 3

37 7, , 0

3 3 21 17 7

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21.

Area of major sector OAYB = 2300π (42)

360

= 25π (42)

6

= 4620 cm2

Area of major sector OCZD = 2300π (21)

360

= 5

π 4416

= 1155 cm2

Area of shaded region

= Area of major section OAYB -Area of major sector OCZD

= 4620 – 1155

= 3465 cm2

22.

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a = 7 cm

Let R = radius of the sphere

2R = a

2R = 7 R = 7

2 cm

Volume of the wood left

= Volume of cube – Volume of sphere

= 3 34a π R

3

= 3

3 4 7(7) π

3 2

= 3 3

3 3 34 7 π 7 π7 π 7 7 1

3 8 6 6

=163.33cm3

23. Speed = 4 km/hr

= 4 1000 m

60 min

= 2

1003

m/min

= 200

3m/min

DS

T

200 D

3 10

200D 10

3

2000D m

3

Volume of the water flowing in a canal

= Volume of water used for irrigation

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2000 8

6 1.5 A3 100

6 15 200 100

A3 8

30 200 100

A8

15

A4

× 200 × 100

A 15 50 100

2A 75000 m

24.

Area of trapezium

= 1

2 (sum of parallel sides) × Height

1

24.5 (AD BC) AB2

1

24.5 (10 4) AB2

49 = 14 AB

49

AB14

7

AB cm2

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Area of quadrant = 21π (AB)

4

= 1 22 49

4 7 4

= 22

74 4

= 211 777 cm

8 8

Area of shaded region

= Area of trapezium – Area of quadrant

= 24.5 – 77

8

= 24.5 – 9.625

= 14.875 cm2

25. x 2 x 4 10

x 3 x 5 3 x 3,5

(x 2)(x 5) (x 4)(x 3) 10

(x 3)(x 5) 3

2 2x 5x 2x 10 x 3x 4x 12 10

(x 3)(x 5) 3

2

2

2x 5x 2x 3x 4x 10 12 10

x 5x 3x 15 3

2

2

2x 14x 22 10

x 8x 15 3

3(2x2 – 14x + 22) = 10(x2 – 8x + 15)

6x2 – 42x + 66 = 10x2 – 80x + 150

10x2 – 6x2 – 80x + 42x + 150 – 66 = 0

4x2 – 38x + 84 = 0 2x2 – 19x + 42 = 0

Quadratic formula

x=2b b 4ac

2a

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219 (19) 4 2 42

x4

19 361 336

x4

19 5

x4

24 14

x ,4 4

7

x 6,2

26. Total no. of trees planted

= 2[2 + 4 + 6 + …….24]

= 12

2 [2 2 12 1 2]2

= 12[4 + 22] [Sum of n terms of on A.P. = n

2[2A + (n – 1) D]

= 12 × 26

= 312

27. Let BD be the flagstaff

BC be the tower

In ABC

0BCtan 45

AC

BC1

120

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BC = 120 m

In ADC

0DCtan 60

AC

DC3 DC 120 3 m

120

DB = DC – BC

= 120 3 120

= 120( 3 1)

= 120(1.73 1)

= 120(0.73)

= 87.60 m

28. Total cards = 52

Cards removed = 4

52 – 4 = 48

Cards removed

2 red queens

2 blocks Jacks

(1) Probability of a king = No. of favourable cases

Total no.of cases

Outcomes = 4

= 4 1

48 12

(2) Probability of a red colour card

Outcomes = 24

=24 1

48 2

(3) Probability of a face card

Outcomes = 8

=8 1

48 6

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(4) Probability of a queen

Outcomes = 2

=2 1

48 24

29. A(–3, 5), B(–2, –7), C(1, –8) and D(6, 3)

Area = ACD

1 2 3 2 3 1 3 1 2

1x (y y ) x (y y ) x (y y )

2

1

( 3)(( 8 3)) 1(3 5) 6( 5( 8))2

1

( 3)( 11) 1( 2) 6(13)2

= 1

1092

2109cm

2

Area of ABC

1

( 3)(( 7) ( 8)) ( 2)(( 8) 5) 1(5 ( 7))2

1( 3)(1) ( 2)( 13) 1(12)

2

= 1 1

3 26 12 (35)2 2

= 235cm

2

ar of quad. ABCD

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= area of ACD + area of ABC

2109 35 14472 cm

2 2 2

30. Let the speed of the stream = x km/hr

Upstream speed = (18 – x) km/hr

Downstream speed = (18 + x) km/hr

DistanceSpeed

time

Distancetime

Speed

Upstream time = 1 + Downstream time

24 24

118 x 18 x

24 24

118 x 18 x

18 x 18 x

24 118 x 18 x

24 × 2x = 324 – x2

x2 + 48x – 324 = 0

48 2304 4 324

x2

48 3600

x2

48 60

x2

48 60 48 60

x ,2 2

x = 6 km/hr

Speed of the stream = 6 km/hr

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31.

To is the angle bisector of PTQ

OT PQ

OT bisects PQ

PR = RQ = 8 cm

Also TPO = 900

In PRO

OP2 = OR2 + PR2

(10)2 = OR2 + 82

OR = 6 cm

Let TP = x, TR = y

In PRT

PT2 = PR2 + TR2

x2 = 82 + y2 ..(1)

In PTO

OT2 = TP2 + OP2

(6 + y)2 = x2 + 102

36 + y2 + 12y = x2 + 100

x2 = y2 + 12y – 64 ..(2)

Solving (1) & (2)

y2 + 12y – 64 = 64 + y2

12y = 128

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y = 128

12

32

y3

x2 = 64 + y2

x2 = 64 + 2

32

3

x2 = 64 + 1024

9

2 64 9 1024x

9

2 576 1024x

9

2 1600x

9

40

x cm3

40

TP cm3

32.

To Prove: OP AB

Construction : Take any point Q, other than P on the tangent AB .Join OQ.

Suppose OQ meets the circle at R.

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Proof : We know that among all line segments joining the point O to a point on AB, the

shortest one is perpendicular to AB. So to prove OP AB, it is sufficient to prove that

OP is shorter than any other segment joining O to any point of AB

OP = OR (Radii of the same circle)

OQ = OR + RQ

OQ > OR

OQ > OP

OR < OQ

Thus OP is shorter than any other segment joining O to any point of AB

OP AB

33. Radius of spherical marbles = 0.7 cm

Radius of cylinder = 3.5 cm

Volume of one spherical marble 34πr

3

= 3

4 7π

3 10

Volume of 150 spherical marbles = Volume of water risen in the cylinder

3 2

4 7 7150 π π h

3 10 2

3

4 7 1150 h

3 10 4

3

4 7150 4 h

3 10

5 16 7

h100

560

h100

h 5.6 cm

34. r1 = 8 cm

r2 = 20 cm

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h = 24 cm

Volume of frustum = 2 21 2 1 2

1πh r r r r

3

= 21 22

24 8 20 8 203 7

228 64 400 160

7

= 176

6247

= 15689.142 cm3

= 15.689 litres

1 litre of milk costs Rs 21

15.689 litres ……. 21

15.6891

= Rs 329.47