Pioneer Education The Best Way To Success NTSE ... P, Q, R and S of the sides AB, BC, CD and DA...

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10th CBSE

Mathematics Boards Paper With Solution(Set-3) SECTION-A

1. What is the value of (cos2670 – sin2230).

Solution:

cos2670 – sin2230

cos2670 – sin2(90 – 670)

cos2670 – cos2670

= 0

2. In an A.P, if the common difference (d) = – 4, and the seventh term (a7) is 4, then find the first term.

Solution:

d = 4

a7 = 4

a + 6d = 4

a + 6 – 4 = 4

a = 4 + 24

a = 28

3. Given AB 1

ABC ~ PQR, if ,thenPQ 3

find ar ABC

ar PQR

Solution: 2 2

2 2

Area of ABC AB 1 1

Area of PQR PQ 3 9

= 1:9

4. What is the HCF of smallest prime number and the smallest composite number?

Solution:

smallest prime number = 2

smallest composite number = 4

HCF = 2

5. Find the distance of a point P(x, y) from the origin.

Solution:

2 2

x 0 y 0

2 2x y

6. If x = 3 is one root of the quadratic equation x2 – 2kx – 6 = 0, then find the value of k.

Solution:

x2 – 2kx – 6 = 0

(3)2 – 2k3 – 6 = 0

9 – 6k – 6 = 0

3–6k = 0

3 = 6k

k = 1

2




SECTION-B 7. Two different dice and tossed together. Find the probability.

(i) of getting a doublet

(ii) of getting a sum 10, of the numbers on the two dice.

Solution:

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

6P(doublet)

36

(ii) (4, 6) (6, 4) (5, 5)

3 1Probability

36 12

8. Find the ratio in which P(4, m) divides the line segment joining the points A(2, 3) and B(6, – 3). Hence

find m.

Solution:

As 6k 2

4k 1

6k + 2 = 4k + 4

2k = 2

k = 1

Hence ratio is 1:1

3 3 0

m 02 2

9. An integer is chosen at random between 1 and 100. Find the probability that it is:

(i) divisible by 8

(ii) not divisible by 8

Solution:

No. till 80 are 10

& 88, 96

No. which are divisible by 8 = 12

12P(8)

98

P(no. not divisible by 8) = 12

198

= 86

98




10. In fig, ABCD is a rectangle. Find the value of x and y.

Solution:

x + y = 30

x – y = 14

2x = 44

x 22

22 + y = 30

y = 8

11. Find the sum of first 8 multiples of 3.

Solution:

3, 6, 9,…..

8

8S [2a 7d]

2

4[2 3 7 3]

4[6 + 21]

27 4 = 108

12. Given that 2 is irrational prove that 5 3 2 is an irrational number.

Solution:

2 = given irrational number

To prove = 5 3 2 is irrational number

3 2 = irrational

5 3 2 = irrational

SECTION – C

13. If A(–2, 1), B(4, b) and D(1, 2) are the vertices of a parallelogram ABCD. Find the values of a and b. Hence

find the lengths of the sides.

Or

If A(–5, 7), B(–4, – 5), C(–1, – 6) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.

Solution:




By AC

4 2 b 1O ,

2 2

b 1O 1,

2

a 1 2 0O ,

2 2

a 11

2

,

b 11

2

2 a 1 a 1

b 1

Distance

2 2AB 3 1 10

2 2BC 4 1 1 10

2 2 22CD 4 1 1 2 3 1 10

2

AD 1 2 2 1 10

Or

Solution:

1 2 3 2 3 1 3 1 2

1A x y y x y y x y y

2

A = (–5, 7), C = (–1, – 6)

B = (–4, 5) , D = (4, 5)

ar ABC = 1

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




= 1

2[–5(1) – 4(–13) – 1(12)]

= 1

2[–5 + 52 + 12]

= 35

2sq. units

1

2[–1(–2) + 4(13) + (–5) (–11)]

1

2[2 + 52 – 55]

1

2

14. Find all zeroes of the polynomial (2x4 – 9x3 + 5x2 + 3x –1) if two of its zeroes are 2 3 and 2 3 .

Solution:

2x4 – 9x3 + 5x2 + 3x – 1

x 2 3

x 2 3

x 2 3 0

x 2 3 0

x 2 3 x 2 3 0

x2 – 4x + 1

2 4 3 2 2x 4x 1 2x 4x 5x 3x 1 2x x 1

4 3 22x 8x 2x

_________________________

–x3 + 3x2 + 3x – 1

– x3 + 4x2 – x

______________________________

(–) (–)

–x2 + 4x – 1

–x2 + 4x – 1

+ – +

________________

0

2x2 – x – 1 = 0

2x2 – 2x + x – 1 = 0

2x(x – 1) + 1(x – 1) = 0

(2x + 1) (x – 1) = 0

1x

2 , x = 1




15. Find HCF and LCM of 404 and 96 and verify that HCF LCM = Product of the two given numbers.

Solution:

404 96

101 4 25 3

H.C.F = 2 2 = 4

L.C.M = 25 3 101

L.C.M H.C.F = 25 3 101 4

= 96 404

= Product of two numbers.

16. Prove that the lengths of tangents drawn from an external point to a circle are equal.

Solution:

T.P: PA = PB

As OP = OP (common)

OAP = OBP = 900

OB = OA (radius)

OAP OBP

PA PB

17. Prove that the area of an equilateral triangle described on one side of the square is equal to half the area

of the equilateral triangle described on one of its diagonal.

Or

If the area of two similar triangles are equal, prove that they are congruent.

Solution: 2

2

Area of ABC ΑB

Area of PQR PQ

=

2 2

2 2

BC AC

QR PR

2

2

ABBut 1

PQ

ABC PQR

QB = PQ

BC = QR

AC = PR

18. A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km, away

in time, it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.

Solution:




1500 1500 1

Herex 2 x 100 2

x x 100 11500

(x(x 100) 2

2

150000 1

x 100x 2

300,000 = x2 + 100x

x2 + 10x – 300, 000.

19. The table below shows the salaries of 280 persons.

Salary (In thousand Rs) No. of persons

5-10

10-15

15-20

20-25

25-30

30-35

35-40

40-45

45-50

49

133

63

15

6

7

4

2

1

Calculate the median salary of the data.

Solution:

f 280 N 280

1402 2

c.f = 182

F = 133

i = 5

l = 10

Median =

Nc.f .

2l iF

140 4910 5

133

20. A wooden article was made by scoping out a hemisphere from each end of a solid cylinder, as shown in

fig. If the height of the cylinder is 10cm, and its base is of radius 3.5 cm. Find the total surface area of the

article.




Or

A heap of rice is in the form of a cone of base diameter 24m and height 3.5m. Find the volume of the rice.

How much canvas cloth is required to just cover the heap?

Solution:

h = 10 cm

r = 3.5 cm

T.S.A = 22πrh 2 2πr

2πr h 2r

22 3.5

2 10 2 3.57 10

22(17) = 374 cm3

Or

Solution:

d = 24cm, r = 12m

h = 3.5 cm

21V πr h

3

1 22 3.512 12

3 7 10

22 12 2

T.SA = πrl

2 2

l 12 3.5

= 12.5

22πrl 12 12.5

7

= 471 cm2

21. Find the area of the shaded region in fig. Where arcs drawn with centre A, B, C and D intersect in pairs at

mid-point P, Q, R and S of the sides AB, BC, CD and DA respectively of a square ABCD of side 42 cm.




Solution:

Side = 12cm, r = 6cm

Area of shaded = 2 21

12 4 πr4

= 144 – 3.14 36

= 30.96

22. If 4tan θ 3, evaluate 4 sin θ cos θ 1

4 sin θ cos θ 1

Or

If 0tan 2A cot A 18 , where 2A is an acute angle, find the value of A.

Solution:

4 tanθ 3

3 ptan θ

4 b

p = 3 b – 4

2 2h 5 p b

3 44 1

4 sin θ cos θ 1 5 53 44 sin θ cos θ 1 4 15 5

12 4 517 4 135

12 4 5 16 5 17

5

Or

Solution:

0tan 2A cot A 18

2A + A – 180 = 900

3A = 1080

A = 360




SECTION-D

23. As observed from the top of a 100m high light house from the sea level, the angles of depression of two

ships are 300 and 450. If one ship is exactly behind the other on the same side of the light house, find the

distance between the two ships. [use 3 1.732 ]

Solution:

ABC

0 AB 100tan 45

BC BC

1001

BC

BC = 100 m

ABD

0 AB 1tan 30

100 x 3

100 1

100 x 3

100 + x = 100 3

x 100 3 100

100 3 1

= 100 (1.732 – 1)

= 100 0.732

= 73.2 m

24. The diameters of the lower and upper ends of a bucket in the form of a frustum of a cone are 10cm and

30cm respectively. If its height is 24 cm. Find:

(i) The area of the metal sheet used to make the bucket

(ii) Why we should avoid the bucket made by ordinary plastic?

Solution:




22l h R r

= 2 224 10 576 100 676

= 26

Area = 2 2πR πr πr R r

2 21V πR r Rr

3

R = 15, h = 24

r = 5, l = 26

2 21V 3.14 15 5 15 5

3

= 3.14 2.25 + 25 + 75

1

806.5 268. 8333

25. Prove that: 3

3

sin A 2sin Atan A

2cos A cos A

Or

A train travels at a certain average speed for a distance of 63km and then travels at a distance of 72km at

an average speed of 6km/hr more than its original speed. If it takes 3hrs to complete total journey. What

is the original average speed?

Solution:

23

3 2

sin A 1 2 sin Asin A 2 sin A

cos A cos A cos A 2cos A 1

2

2

sin A 1 2 1 cos A

cos A 2cos A 1

2

2

sin A 2cos A 1

cosA 2cos A 1




sin Atan A

cos A

Or

Solution:

Given that distance = 63 km.

Let original speed of train = x km/hr.

time = distance / time = 63/x hrs.

And it travels a distance of 72 km at a average speed of 6 km/hr more than the original speed.

distance = 72 km ; speed = (x + 6) km/hr .

time= 72/(x+6) hrs.

If it takes 3 hours to complete the whole journey 63/x + 72/(x + 6) = 3 hrs

⇒ 63(x + 6) + 72x = 3x(x + 6)

⇒ 21(x + 6) + 24x = x(x+6)

⇒ 45x + 21×6 = x2 + 6x

⇒ x2 – 39x – 126 = 0

⇒ x2 – 39x – 126 = 0

⇒ (x – 42)(x + 3) = 0

∴ x = 42 km/hr

∴ the original average speed = 42 km/hr

26. The mean of the following distribution is 18. Find the frequency f of the class 19-21.

Class 11-13 13-15 15-17 17-19 19-21 21-23 23-25

Frequency 3 6 9 13 f 5 4

Or

The following distribution given the daily income of 50 workers of a factory.

Daily Income (in rs) 100-120 120-140 140-160 160-180 180-200

No. of workers 12 14 8 6 10

Convert the distribution above to less than type cumulative frequency distribution and draw its ogive.

Solution:

f 40 f x f 524 20f

Mean = x f

f

524 20f18

40 f

720 + 18f = 524 + 20f

720 – 524 = 2f

196 = 2f

f 98

Or

Solution:

We can find frequency distribution table of less than type as following: -

Daily income (in Rs) Cumulative frequency




(upper class limits)

Less than 120 12

Less than 140 12 + 14 = 26

Less than 160 26 + 8 = 34

Less than 180 34 + 6 = 40

Less than 200 40 + 10 = 50

Now taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis we

can draw its ogive as following-

27. A motor boat whose speed is 18 km/hr in still water takes 1hr more to go 24km upstream than to return

downstream to the same spot. Find the speed of the stream.

Solution:

24 24Here 1

18 x 18 x

2

24[18 x (18 x)]1

182 x

2

24 18 x 18 x1

324 x

48x = 324 – x2

x2 + 48x – 324 = 0

x2 + 54x – 6x – 324 = 0

x(x + 54) – 6(x + 54) = 0

x + 54 = 0

x 54

x – 6 = 0

x 6km / h

28. The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last

term to the product of two middle terms is 7.15. Find the numbers

Solution:

Let the four consecutive numbers in A.P be (a - 3d), (a - d), (a + d) and (a + 3d)

So, according to the question.




a – 3d + a – d + a + d + a + 3d = 32

4a = 32

a = 32/4

a = 8 ......(1)

Now, (a – 3d)(a + 3d)/(a – d)(a + d) = 7/15

15(a² – 9d²) = 7(a² – d²)

15a² – 135d² = 7a² – 7d²

15a² – 7a² = 135d² –7d²

8a² = 128d²

Putting the value of a = 8 in above we get.

8(8)² = 128d²

128 d² = 512

d² = 512/128

d² = 4

d = 2

So, the four consecutive numbers are

8 – (3x2)

[8 –6 = 2]

[8 – 2 = 6]

[8 + 2 = 10]

8 + (3x2)

[8 + 6 = 14]

Four consecutive numbers are 2, 6, 10 and 14.

29. Draw a triangle ABC with BC = 6cm. AB = 5 cm and 0ABC 60 . Then construct a triangle where sides are

3

4of the corresponding sides of the ABC.

Solution:




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

3BC. Prove that

2 29 AD 7 AB .

Solution:

Given: Equilateral triangle ABC D is a point on BC

Such that BD = 1

3BC

To prove: 9AD2 = 7AB2

Construction: Let draw AE BC

Proof:

All sides of equilateral triangle is equal,

AB = BC = AC

Let AB = BC = AC = x

Given

BD = 1

3BC

BD = x

3

In AEB and AEC

AE = AE (Common)

AB = AC (Both as it is equilateral triangle)

AEB AEC (Both 900 as AE BC)

Hence by RHS congruency

AEB AEC

BE = EC (CPT)

So, BE = EC = 1

2BC

BE = EC = x

2

So,

BE = x

2

BD + DE = x

2




x xDE

3 2

x xDE

2 3

3x 2xDE

2 3

xDE

6

Using Pythagoras theorem

(Hypotenuse)2 = (Height)2 + (Base)2

Now in right AEB

AB2 = AE2 + (BE)2

x2 = (AE)2 + 2

x

2

2

22 xx AE

4

22 2x

x AE4

22 3x

AE4

…(1)

Similarly

In right AED

AD2 = AE2 + DE2 22

2 3x xAD

4 6

(from (1))

2 22 3x x

AD4 36

2 2

23x 9 x

AD36

2 22 27x x

AD36

22 28x

AD36

22 7x

AD9

9AD2 = 7x2

9 AD2 = 7 AB2

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Pioneer Education The Best Way To Success NTSE ... P, Q, R and S of the sides AB, BC, CD and DA...

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