10 th CBSE - · PDF file · 2013-07-28Product of the zeroes of –2x2 – kx + 6 is (a) – 3 (b) 3 (c) k/2 (d) k /2 Ans. (a) 5. ... 18. If tan cot 2, find the value of 22 tan cot

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

1

General Instructions All questions are compulsory.

The question paper consists of 34 questions divided into four sections A, B, C and D. Section

A comprises of 8 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 10

questions of 4 marks each.

Question numbers 1 to 8 in Section A are multiple choice questions where you are

to select one correct option out of the given four.

There is no overall choice. However, internal choice has been provided in 1 question of two

marks each, 3 questions of three marks each and 2 questions of four marks each. You have

to attempt only one of the alternatives in all such questions,

Use of calculators is not permitted

10th CBSE Mega Test - II

Time: 3hours Max. Marks: 80

NAME OF THE CANDIDATE CONTACT NUMBER

Solution Visits: www.pioneermathematics.com/latest_updates.com

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



2

Section - A [Questions number 1 to 10 carry one mark each]

1. The product of two equal irrational numbers is always

(a) rational (b) irrational (c) rational or irrational (d) one

Ans. (a)

2. A terminating decimal when expressed in fractional form, always has denominator in

the form of

(a) 2m. 3n, m, n 0 (b) 3m5n, m, n 0 (c) 5m7n, m, n 0 (d) 2m . 5n, m, n 0

Ans. (d)

3. A polynomial of the form ax3 + bx2 + cx + d has atmost

(a) 2 zeroes (b) 3 zeroes (c) 4 zeroes (d) 5 zeroes

Ans. (b)

4. Product of the zeroes of –2x2 – kx + 6 is

(a) – 3 (b) 3 (c) k/2 (d) k /2

Ans. (a)

5. The graph of y = - 3 is a line parallel to the

(a) x-axis (b) y-axis (c) both x and y-axis (d) none of these

Ans. (a)

6. Geeta has only Re 1 and Rs 2 coins with her. If the total number of coins that she has is

50 and the amount of money with her is Rs. 75, then the number of Re 1 and Rs. 2 coins

are, respectively

(a) 35 and 15 (b) 30 and 20 (c) 10 and 40 (d) 25 and 25

Ans. (d)

7. 0 0ABC ~ DEF; A 47 , E 83 , then C is equal to

(a) 400 (b) 500 (c) 600 (d) 700

Ans. (b)

8. The area of two similar triangles is 16 cm2 and 25 cm2 respectively. The ratio of their

corresponding altitudes is



3

(a) 4 : 5 (b) 4 : 3 (c) 3 : 4 (d) 2 : 3

Ans. (a)

9. If sin θ + sin2θ = 1, then cos2 θ + cos4 θ is equal to

(a) –1 (b) 0 (c) 1 (d) 2

Ans. (c)

10. While computing mean of grouped data, we assume that the frequencies are

(a) evenly distributed over all the classes.

(b) centred at the calls marks of the classes.

(c) centred at the upper limits of the classes.

(d) centred at the lower limits of the classes.

Ans. (b)

Section - B

[Questions number 11 to 18 carry 2 marks each.]

11. Use Euclid’s division algorithm to find the HCF of 198 and 315.

198 315 1

198

117 198 1

117

81 117 1

81

36 81 2

72

9 36 4

36

X

315 > 198

315 = 198 × 1 + 117

198 = 117 × 1 + 81



4

117 = 81 × 1 + 36

81 = 36 × 2 + 9

36 = 9 × 4 + 0

H. C. F. = 9

12. In Fig. 1, the graph of some polynomial p(x) is given. Find the zeroes of the polynomial.

Sol :

The graph intersects the X-axis at x = – 3 and x = - 1

Zeroes of the polynomial are –3 and - 1

13. Find the value of k for which the following system of equations has no solutions:

2x + ky = 1; 3x – 5y = 7.

Sol :

Equations are 2x + ky = 1; 3x – 5y = 7

Since these equations have no solution we have

1 1 1

2 2 2

a b c

a b c

Here a1 = 2 b1 = k c1 = 1

a1= = 3 b2= = –5 c2 = 7

2 k 1

3 5 7

10k

3



5

14. Solve for x and y:

2 313

x yx, y 0

5 42

x y

OR

Solve for x and y : ax + by – a + b = 0, bx – ay – a – b = 0.

Sol :

2 313

x y ..(i)

5 42

x y ..(ii)

Multiplying (i) by 4 and (ii) by 3, we get

8 1252

x y

15 126

x y23

46x

..(by adding)

46x = 23 1

x2

Putting the value of x in (i), we get

2 313

1 y

2

3

4 13y

3

y= 13 – 4 = 9

9y = 3 3 1

y9 3

1 1x , y

2 3

Or

ax + by – a + b = 0 ax + by = a – b ..(i)



6

bx – ay – a – b = 0 bx – ay = a + b ..(ii)

Multiplying (i) by a and (ii) by b, we get

2 2

2 2

2 2 2 2

a x aby a ab

b x aby ab b

a b x a b

…(by adding)

2 2

2 2

a bx 1

a b


a(1) + by = a – b by = a – b – a

by = – b y = –1

x = 1, y = – 1

15. A right triangle has hypotenuse of length q cm and one side of length p cm. If (q – p) = 2,

express the length of third side of the right triangle in terms of q.

Sol :

Let the third side b x

A.T.Q. x2 + p2 = q2 ..(i)

Given that q – p = 2

p = q – 2

Putting value of p in (i), we get

x2 + (q – 2)2 = q2 x2 + q2 + 4 – 4q = q2

x2 = q2 – q2 – 4 + 4q

x2 = 4q – 4 x2 = 4 (q – 1)

x 2 q 1

16. If 3 cot =2, find the value of 2 sin 3 cos

.2sin 3 cos

Sol :

23 cot 2 cot

3



7

Required expression = 2 sin 3 cos

2 sin 3 cos

=

2 sin 3 cos

2 3 cotsin sin2 sin 3cos 2 3 cotsin sin

=

22 3

2 2 03 02 2 2 42 33

17. If 3

sin5

, find the value of tan sec 2.

Sol :

3Sin

5

22 2 2 3

cos 1 sin cos 15

2 9cos 1

25

9 16

cos 125 25

4

cos5

1 5

seccos 4

3sin 3 5 35tan

4cos 5 4 45

2 2

2 3 5 8tan sec

4 4 4= (2)2 = 4



8

18. If tan cot 2,find the value of 2 2tan cot .

Sol :

tan cot 2

Squaring both sides

2 2tan 2 tan cot cot 4

2 2tan 2 cot 4 … [ tan . cot 1 ]

2 2tan cot 4 2 2

Section - C

[Question numbers 19 to 28 carry 3 marks each.]

19. Find the largest number that divides 2053 and 967 and leaves a remainder of 5 and 7

respectively.

Sol :

960 2048 2

1920

128 960 7

896

64 128 2

128

X

2053 – 5 = 2048

967 – 5 = 960

2048 > 960

2048 = 960 × 2 + 128

960 = 128 × 7 + 64

128 = 64 × 2 + 0



9

H. C. F. = 64

Required number is 64

20. Prove that 7 6 5 is an irrational number.

Sol :

Let us assume, to the contray, that 7 6 5 is rational

That is, we can find coprimes a and b b 0 , such that a

7 6 5b

a

7 6 5b

7 a 7b a

56 6b 6b

Since a and b are integers , we get 7 a

6 6b is rational, and so 5 is rational

But this contradicts the fact that 5 is irrational This contradiction has arisen because

of our incorrect assumption that 7 6 5 is rational.

So, we conclude that 7 6 5 is irrational.

21. Obtain all other zeroes 3x4 – 15x3 + 13x2 + 25x – 30, if two of its zeroes are

5 5and .

3 3

OR

On dividing p(x) by a polynomial x – 1 – x2, the quotient and remainder were x – 2 and

3 respectively. Find p(x).

Sol :

Since two of the zeroes are 5 5

and3 3

25 5 5x x x

3 3 3is a factor of given polynomial



10

2

2 4 3 2

4 2

3 2

3

2

2

3x 15x 18

x 5/3 3x 15x 13x 25x 30

3x 5x

15x 18x 25x 30

15x 25x

18x 30

18x 30

3x4 – 15x3 + 13x2 + 25x – 30

= 2 25x 3x 15x 18

3

= 2 5x .3 x 2 x 3

3

2

2

2

3x 15x 18

3 x 5x 6

3 x 3x 2x 6

3[x x 3 2 x 3 ]

3 x 2 x 3

Remaining two zeroes are

x – 2 = 0 or x – 3 = 0

x = 2 or x = 3

Remaining two zeroes are 2 and 3

Or



11

Dividend = Divisor × Quotient + Remainder

or p(x) = g(x) × q(x) + r(x) ..(i)

Here p(x) = ? g(x) = x – 1 – x2

q(x) = x – 2, r(x) = 3

Putting these values in (i), we get

p(x) = (x – 1 – x2) (x – 2) + 3

= x2 – 2x – x + 2 – x3 + 2x2 + 3

= – x3 + 3x2 – 3x + 5

22. Draw the graph of x – y + 1 = 0 and 3x + 2y – 12 = 0. Calculate the area bounded by

these lines and X-axis.

OR

Determine the value of a and b so that the following linear equations have infinite

solutions :

(2a – 1) x + 3y – 5 = 0

3x + (b – 1) y – 2 = 0

Sol :

By plottig

In the graph, the two lines intersect at B(2, 3)

x = 2, y = 3 is the solution .

Area bounded by ABC

= 1 1

AC BD 5 32 2



12

= 7.5 sq. units

Or

(2a – 1) x + 3y = 5

3x + (b – 1) y = 2

a1 = 2a – 1 b1 = 3 c1 = 5

a2 = 3 b2 = b – 1 c2 = 2

For linear equations to have infinite solutions

1 1 1

2 2 2

a b c

a b c

2a 1 3 5

3 b 1 2

I and III

2a 1 5

3 2 2a 1 2 3 5

4a – 2 = 15 4a = 15 + 2

4a = 17 17

a4

II and III

3 5

b 1 2 3 2 5 b 1

5b = 5 – 6 5b = 6 + 5



13

5b = 11 b = 11

5

17 11

a and b4 5

23. In Fig, XY BC. Find the length of XY.

Sol:

Here AX = 1 cm, XB = 3 cm

Now AX + AB = 1 + 3 = 4 cm AB = 4 cm

In AXY and ABC

XAY = BAC ..(common)

AXY = BAC …(corresponding angles)

AXY ~ ABC (AA Similarly)

AX XY

AB BC …[sides are proportional]

1 XY

4 6

6 3XY

4 2 = 1.5 cm

24. In given figure ABC ~ PQR. Also ar ( ABC) = 4. ar ( PQR). If BC = 12 cm, find QR.

Sol :

ABC ~ PQR ..(Given)



14

2

2

ar ABC BC

ar PQR QR [ Area of two similar s is equal to the

ratio of the squares of corresponding sides]

2

2

4 ar PQR 12

ar PQR QR …[ ar ABC 4ar PQR ]

2

2

124

1 QR

2 12

1 QR [taking square-root]

2 QR = 12 QR = 6 cm

25. Prove the identity : sin 1 cos 1 sin

cos 1 sin cos

Sol :

sin 1 cosL.H.S.

cos 1 sin

Dividing numerator and denominator by cos

sin 1 cos

tan sec 1cos cos coscos 1 sin 1 sec tancos cos cos

Using formula; 2 2sec tan 1

We get

2 2tan sec sec tan

1 sec tan

= tan sec sec tan sec tan

1 sec tan

= tan sec [1 sec tan ]

1 sec tan



15

= tan sec 1 sec tan

1 sec tan

= tan sec

= sin 1 sin 1

cos cos cos = R.H.S.

26. Prove that : 2 2

2 2

2 2

sin A sin Btan A tan B

cos Acos B

OR

Find the value of :

0 0 2 0 2 0

0 0 0 0 0

tan .cot 90 sec .cosec 90 sin 35 sin 55

tan10 tan20 tan30 tan70 tan80

Sol :

L. H. S. = tan2A – tan2B

= 2 2

2 2

sin A sin B

cos A cos B

= 2 2 2 2

2 2

sin A.cos B sin B.cos A

cos A.cos B

=

2 2 2 2

2 2

sin A 1 sin B sin B 1 sin A

cos A.cos B

= 2 2 2 2 2 2

2 2

sin A sin A.sin B sin B sin A. sin B

cos A.cos B

= 2 2

2 2

sin A sin B

cos A.cos B = R. H. S.

Or

0 0 2 0 2 0

0 0 0 0 0

tan .cot 90 sec .cosec 90 sin 35 sin 55

tan10 tan20 tan30 tan70 tan80

=

2 0 2 0 0

0 0 0 0 0 0

tan .tan sec . sec sin 35 sin 90 35

1tan 10 tan20 tan 90 10 .tan 90 10

3



16

= 2 2 2 0 2 0

0 0 0 0

tan sec sin 35 cos 35

1tan 10 tan20 cot 20 . cot 10

3

…

0

0

0

cot 90 tan

cosec 90 sec

1tan 30

3

= 0 0

0 0

1 1

1 1 1tan10 tan20

tan 20 tan 103

..

2 2

2 2

sec tan 1

sin cos 2

1cot

tan

= 2

2 31

3

27. Find the median from the following data :

Class interval 0–10 10–20 20–30 30–40 40-50

Frequency 3 4 6 4 3

Sol :

Median class = thnitem

2



17

= 20

2= 10th item

Median lies in class 20 – 30

I = 20, c.f. = 7, f = 6, h = 10

Median – L +

nc.f.

2 hf

= 10 7

20 106

= 20 + 3 10

6 = 20 + 5 = 25

28. Find mode from the following data :

Class interval 1–3 3–5 5–7 7–9 9-11

Frequency 4 5 8 7 6

Sol :

Maximum frequency = 8

Modal Class is 5 – 7

l = 5, f0 = 5, f1 = 8, f2 = 7, h = 2

Mode = L + 1 0

1 0 2

f fh

2f f f

= 8 5

5 22 8 5 7

= 5 + 3

16 12 × 2

= 5 + 3 2

4

= 5 + 3

5 1.5 6.52



18

Section - D

[Questions number 29 to 34 carry 4 marks each.]

29. Find the mean age in years using step deviation method from the frequency

distribution given below :

Class interval (age in years) Frequency

25-29

30–34

35–39

40–44

45–49

50–54

55–59

4

14

22

16

6

5

3

Total 70

Sol :

C.I.

Class interval

fi

Frequency

(xi)

Midpoint

ii

x au

h i if u

25–29

30–34

35–39

40–44

45–49

50–54

55–59

4

14

22

16

6

5

3

27

32

37

42

47

52

57

–3

–2

–1

0

1

2

3

12

28 62

22

0

6

10 25

9

Total in f 70 i if u 37

Here a = 42, in f 70, h = 5



19

Mean i i

i

f ux a h

f

= 42 + 5 × 37

70

= 42 – 2.64 = 39.36

30. The following table gives production yield per hectare of wheat of 100 farms of a

village :

Production yield

(in kg/hec.)

No. of farms

50–55

55–60

60–65

65–70

70–75

75–80

2

8

12

24

38

16

Total 70

Convert the above distribution to more than type cumulative frequency distribution

and draw its ogive.

Sol :

Production

yield

(in kg/hec.)

f

Number

of farms

Production

yield

more than

Cumulative

frequency

50–55

55–60

60–65

65–70

70–75

75–80

2

8

12

24

38

16

50

55

60

65

70

75

100

98

90

78

54

16



20

Now, plot the points (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) taking

production yield along X-axis and number of farms along Y-axis

31. If m = sin A cos A

, n ,sin B cos B

prove that (m2 – n2) sin2B = 1 – n2.

Sol :

L. H. S. = (m2 – n2) sin2B

= 2 2

2

2 2

sin A cos Asin B

sin B cos B

= 2 2 2 2

2

2 2

sin Acos B cos Asin Bsin B

sin B.cos B

=

2 2 2 2

2

1 cos A .cos B cos A 1 cos B

cos B

= 2 2 2 2 2 2

2

cos B cos Acos B cos A cos A cos B

cos B ..(i)

= 2 2

2

cos B cos A

cos B

R. H. S. = 1 – n2

= 2

2

cos A1

cos B

= 2 2

2

cos B cos A

cos B ..(ii)

L.H.S. = R.H.S.



21

32. If 3 3x sin ycos sin cos , and x sin = y cos , then prove that x2 + y2 = 1.

OR

Prove that: 2 2

2

2 2 2

tan cos1 2 cos

tan 1 sin cos

Sol :

3 3x sin y cos sin cos

x sin y cos

xsiny ...(i)

cos

3 3sinx sin x cos sin cos

cos [From (i)]

3 2 2x sin xsin cos sin cos

2 2x sin sin cos sin cos

x sin .1 sin cos

sin cos

xsin

x cos

From (i), y = sin

cos .cos

y sin

L. H. S. = x2 + y2

= 2 2sin cos

= 1= R. H. S.

Or

L. H. S = 2 2

2 2 2

tan cos

tan 1 sin cos

=

2

22

2 2 2

2

sincoscos

sin 1 sin cos

1cos



22

=

2

22

2 2 2 2

2

sincoscos

sin cos sin cos

cos

= 2 2

2 2 2 2

sin cos

sin cos sin cos

= 2 2

2 2 2 2

sin cos 1

sin cos sin cos

= 2 2

1

1 cos cos

= 2

1

1 2 cos = R. H. S.

33. The ratio of the areas of similar triangles is equal to the ratio squares on the

corresponding sides–Prove This.

Using the above theorem, prove that of the area of the equilateral triangle described on

the side of a square is half the area of the equilateral triangle described on its diagonal.

Sol :

Part I. See theorem 1, Page 96

Part II.

Given : ACP and BCQ are two similar s, drawn on the diagonal and side of a

square ABCD

To Prove : 2 BCQ = ACP

Proof : ACP ~ BCQ



23

2

2

ar ACP AC

ar BCQ BC

But AC2 = BC2 + AB2

= 2BC2 [ AB = BC]

2

2

ar ACP 2BC2

ar BCQ BC

ar ACP 2 ar BCQ O.E.D.

34. Solve the system of equations:bx ay

a b + a + b = 0 and bx – ay + 2ab = 0.

OR

The sum of the digits of a two-digit number is 12. The number obtained by

interchanging the two digits exceeds the given number by 18. Find the number.

Sol:

bx aya b 0

a b

bx ay

a ba b

2 2b x a y

a bab

b2x – a2y = –a2b – ab2 ..(i)

Again, bx – ay = 2ab = 0

bx – ay = – 2ab ..(ii)

Multiplying (i) by 1 and (ii) by a, we get

2 2 2 2

2 2

2 2 2

b x a y a b ab

abx a y 2a b

b x abx a b ab ....(By subtracting)

bx (b – a) = ab(a – b)



24

ab a b

x ab b a

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

b(–a) – ay = –2ab

–ab – ay = – 2ab

–ay = –2ab + ab

ab

y ba

x =–a, y = b

Or

Let the unit’s place digit be x and the ten’s place digit be y

Original number = x + 10y and Interchanged number = y + 10x

Given, x + y = 12

y = 12 – x ..(i)

A.T.Q.,

(10x + y) – (x + 10y) = 18

10x + y –x –10y = 18

9x – 9y = 18

x – y = 2 …(Dividing by 9)

x – (12 – x) = 2 ..[From (i)]

x – 12 + x = 2

2x = 2 + 12 = 14 x = 7


y = 12 – 7 = 5

The original number = x + 10y

= 7 + 10(5) = 57

Documents

10 th CBSE - · PDF file · 2013-07-28Product of the zeroes of –2x2 – kx + 6 is (a) – 3 (b) 3 (c) k/2 (d) k /2 Ans. (a) 5. ... 18. If tan cot 2, find the value of 22 tan cot