X-STANDARD MATHEMATICS

X-STANDARD MATHEMATICS

MODEL QUESTION

PREPARED BY:R.RAJENDRAN. M.A., M. Sc., M. Ed.,K.C.SANKARALINGA NADAR HR. SEC. SCHOOL,CHENNAI-21

Section – A

Note: (i) Answer all the 15 questions(ii) Choose the correct answer in each question. Each of these questions contains four options with just one correct option(iii) Each question carries One mark 15 × 1 = 15

1. Let A = { 1, 3, 4, 7, 11 }, B = {–1, 1, 2, 5, 7, 9 } and f : A B be

given by f = { (1, –1), (3, 2), (4, 1), (7, 5), (11, 9) }. Then f is

(A) one-one (B) onto

(C) bijective (D) not a function

Section – A

Section – A

,.......625

54,

125

18,

25

6,

5

2

5

2

5

3

5

4

2. The common ratio of the G.P is

(A) (B) 5

(C) (D)

3. If a1 , a2 , a3 ,……are in A.P. such that then the 13th

term of the A.P. is

(A) (B) 0

(C) 12a1 (D) 14a1

Section – A

2

3

7

4 a

a

2

3

4. The LCM of 6x2 y, 9x2 yz,12x2 y2 z is

(A) 36x2 y2 z (B) 48xy2 z2

(C) 96x2 y2 z2 (D) 72xy2 z

Section – A

5. If b = a + c , then the equation ax2 + bx + c = 0 has

(A) real roots (B) no roots

(C) equal roots (D) no real roots

Section – A

6. If then the order of A is

(A) 2 1 (B) 2 2

(C) 1 2 (D) 3 2

Section – A

2120

11

A

Section – A

Section – A

Section – A

Section – A

Section – A

12.

(A) sin + cos (B) sin cos

(C) sin – cos (D) cosec + cot

Section – A

cottan

1

13. If the total surface area of a solid hemisphere is 12 cm2 then its

curved surface area is equal to

(A) 6 cm2 (B) 24 cm2

(C) 36 cm2 (D) 8 cm2

Section – A

14. Mean and standard deviation of a data are 48 and 12 respectively. The

coefficient of variation is

(A) 42 (B) 25

(C) 28 (D) 48

Section – A

14. If A and B are mutually exclusive events and S is the sample space such

that P(A)=1/3 P(B) and S = A B, then P(A) =

(A) (B) (C) (D)

Section – A

4

1

2

1

4

3

8

3

Section – B Note:

(i) Answer 10 questions(ii) Answer any 9 questions from the first 14 questions. Question No. 30 is Compulsory.(iii) Each question carries Two marks 10 × 2 = 20

16. If A = {4,6,7,8,9}, B = {2,4,6} and C = {1,2,3,4,5,6}, then find

A (B C).

ANSWER:

B C = {2, 4, 6} {1, 2, 3, 4, 5, 6} = {2, 4, 6}

A (B C) = {4, 6, 7, 8, 9} {2, 4, 6} = {2, 4, 6, 7, 8, 9}

X = {1, 2, 3, 4}, g = {(3,1), (4,2), (2, 1)}

g is not a function. since 1 has no image.

17. Let X = { 1, 2, 3, 4 }. Examine whether the relation g = { (3, 1), (4, 2), (2, 1) } is a function from X to X or not. Explain.

1

2

3

4

1

2

3

4

18. Three numbers are in the ratio 2 : 5 : 7. If 7 is subtracted from the second, the resulting numbers form an arithmetic sequence. Determine the numbers.

The ratio of three numbers = 2 : 5 : 7

Let the numbers be 2x, 5x, 7x

2x, 5x – 7, 7x are in an AP

5x – 7 – 2x = 7x – (5x – 7)

3x – 7 = 2x – 5x + 7

3x – 7 = 2x + 7

3x – 2x = 7 + 7

x = 14

The numbers are

28, 70, 98

, are the roots of 2x2 – 3x – 1 = 0 a = 2, b = – 3, c = – 1

19. If and are the roots of the equation 2x2 – 3x – 1 = 0, find the value of – if >

2

3 roots theof Sum

a

b

2

1 roots theofProduct

a

c

4)( )( 22

2

14

2

3 )(

22

24

9

4

1

4

89

4

1

2

1

20.

If then find the additive inverse of A.

17

51

59

32A

17

51

59

32A

1579

5312

616

21

Additive Inverse of A = – A

616

21

21. Find the product of the matrices, if exists

12

76

24

014

392

12

76

24

014

392

0780616

36346548

78616

3675414

122

6440

107)1(24)2(0)6()1(44

1)3(7922)2()3()6(942

22. The centre of a circle is at (-6, 4). If one end of a diameter of the circle is at the origin, then find the other end.

2

,2

Midpoint 2121 yyxx

Let the centre of the circle is (– 6, 4)One end = (0, 0), other end = (x, y)Midpoint of diameter = centre of the circle

4,62

0,

2

0

yx

4,62

,2

yx

42

,62

yx

824

1226

y

x

The other end = (– 12, 8)

23. In ABC, DE || BC and . If AE = 3.7 cm, find EC.

Given, AE = 3.7cm

3

2

DB

AD

A

B C

D E

3

2

DB

AD

EC

AE

DB

AD

In ABC, DE || BC

EC

7.3

3

2

2

37.3 EC

85.52

7.11EC

cmEC 85.5

24. A ladder leaning against a vertical wall, makes an angle of 60 with the ground. The foot of the ladder is 3.5 m away from the wall. Find the length of the ladder.

Let AC = Length of the ladder BC = distance of the wall = 3.5m

A

B C60

AC

BC60cosIn ABC, C = 60

AC

5.3

2

1

3.5m

AC = 3.5 2

= 7

Length of the ladder = 7m

25. Prove the identity:1

sec

cos

cos

sin

ec

sec

cos

cos

sin

ecLHS

cos1

cos

sin1

sin

22 cossin

RHS1

26. A right circular cylinder has radius of 14 cm and height of 8 cm. Find its curved surface area.

Radius = 14cm, height = 8cm

Curved surface area = 2rh

8147

222

2

= 2 22 2 8

= 44 16

= 704sq.cm

27. The circumference of the base of a 12 m high wooden solid cone is 44 m. Find the volume.

Height = 12cm

Circumference of the base = 44

2r = 44

447

222 r

2

r = 7cm

222

744

r

Volume = r2h

12777

22

= 22 7 12

= 154 12

= 1848cu.cm

28. Calculate the standard deviation of the first 13 natural numbers.

The standard deviation of the first ‘n’ natural numbers =12

12 n

The standard deviation of the first 13 natural numbers =12

1132

12

1169

12

168

1414. 00

3

39

56 00

.7

7469

31

7.3SD

29. Two coins are tossed together. What is the probability of getting at most one head.

When two coins are tossed the sample space

S = {HH, HT, TH. TT}

n(S) = 4

A = {atmost one head}

A = {HT, TH, TT}

n(A) = 3

)(

)()(

Sn

AnAP

4

3)( AP

30. (a) Simplify: 127

5462

2

xx

x

127

)9(6

127

5462

2

2

2

xx

x

xx

x

)4)(3(

)3)(3(6

xx

xx

)4(

)3(6

x

x

(OR)

30. (b) Show that the lines 2y = 4x + 3 and x + 2y = 10 are perpendicular.

Slope of 2y = 4x + 3 is m1 = 2

Slope of x + 2y = 10 is m2 = - ½

m1 m2 = 2 (– ½ ) = – 1

The given lines are perpendicular.

Note: (i) Answer 9 questions(ii) Answer any 8 questions from the first 14 questions. Question No. 45 is Compulsory.(iii) Each question carries Five marks 9 × 5 = 45

SECTION - C

31.Prove that A \ (B C) = (A \ B) (A \ C) using Venn diagram

A B

C

A A

A A

B B

B B

CC

C C

(1) BC (2) A – ( BC) (3) A – B

(4) A – C (5) (A – B)(A – C)

From the diagram (2)and(5)

A – (BC) = (A – B)(A – C)

32. A function f: [-7, 6)R is defined as follows

6 x 2 ; 1

2 x 5- ; 5

5- x 7- ;12

)(

2

x

x

xx

xf

Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii)

)1(3)6(

)4(2)3(4

ff

ff

32. A function f: [-7, 6)R is defined as follows

6 x 2 ; 1

2 x 5- ; 5

5- x 7- ;12

)(

2

x

x

xx

xf


(i) Since -4 lies in the interval -5 x 2 f(x) = x + 5

f(-4) = – 4 + 5

= 1

Since 2 lies in the interval -5 x 2 f(x) = x + 5

f(2) = 2 + 5

= 7

)1(3)6(

)4(2)3(4

ff

ff

2f(-4) + 3f(2)

= 2(1) + 3(7)

= 2 + 21

= 23

A function f: [-7, 6)R is defined as follows

f(x) =

6 x 2 ; 1

2 x 5- ; 5

5- x 7- ;122

x

x

xx


(ii) Since -7 lies in the interval -7 x < -5

f(x) = x2 + 2x + 1

f(-7) = (–7)2 + 2(–7) + 1

= 49 – 14 + 1 = 36

Since -3 lies in the interval -5 x 2 f(x) = x + 5

f(2) = -3 + 5

= 2

)1(3)6(

)4(2)3(4

ff

ff

f(-7) – f(–3)

= 36 – 2

= 34

A function f: [-7, 6)R is defined as follows

f(x) =

6 x 2 ; 1

2 x 5- ; 5

5- x 7- ;122

x

x

xx


(iii) Since -3 lies in the interval -5 x 2 f(x) = x + 5 f(-3) = -3 + 5 = 2

Since -6 lies in the interval -7 x < -5

f(x) = x2 + 2x + 1

f(-6) = (–6)2 + 2(–6) + 1

= 36 – 12 + 1 = 25

)1(3)6(

)4(2)3(4

ff

ff

4 lies in the interval 2 < x < 6

f(x) = x – 1

f(4) = 4 – 1

= 3

Since 1 lies in the interval -5 x 2 f(x) = x + 5

f(1) = 1 + 5

= 6

)6(3)25(

)3(2)2(4

)1(3)6(

)4(2)3(4

ff

ff

1825

68

7

14

2

33. Find the sum of the first 40 terms of the series 12 – 22 + 32 – 42 + ……. .

12 – 22 + 32 – 42 + ……. . = 1 - 4 + 9 - 16 + 25 ….. to 40 terms

= (1 – 4) + (9 – 16) + (25 – 36) + …… to 20 terms.

= (-3) + (-7) + (-11) +………20 terms

It is an AP with a = –3, d = –4, n = 20

dn )1(2a2

nSn

)4)(120(2(-3)2

20S20

)4(19)62

20

7662

20

822

20

10

= 10(-82)

= – 820

40

34. Factorize the following x3 – 5x2 – 2x + 24

1 – 5 – 2 + 24

– 2

0

1

– 2

– 7

14

12

– 24

0 x + 2 is a factor

Other factor = x2 – 7x + 12

= x2 – 4x – 3x + 12

= x(x – 4) – 3(x – 4)

= (x – 4)(x – 3)

x3 – 5x2 – 2x + 24 = (x + 2)(x – 4)(x – 3)

9x4 + 12x3 + 28x2 – nx + m3x2

3x2

9x2(-)

6x2 + 12x3 + 28x2+2x

+ 2x

+ 12x3 + 4x2(–) (-)

24x2– nx + m6x2 + 4x + 4

+ 4

24x2 + 16x + 16

(-) (–) (–)

0 n = -16, m = - 16

35. If m – nx + 28x2 + 12x3 + 9x4 is a perfect square, then find the values of m, n

37. If verify that (AB)T = BTAT .

11

12 ,

37

25BA

LHS = RHS (AB)T = BTAT .

11

12 ,

37

25BA

11

12

37

25AB

13)1(7)1(327

12)1(5)1(225

37314

25210

411

38

LHSAB T ........43

118)(

11

12 ,

32

75 TT BA

32

75

11

12TT AB

317)1(215)1(

3)1(722)1(52

3725

314210

RHSAB TT ........43

118

38. Find the area of the quadrilateral whose vertices are(-4, -2), (-3,-5), (3,-2) and (2, 3)

Area of the quadrilateral ABCD

= ½ {(x1y2 + x2 y3 + x3y4 + x4y1)

– (x2y1 + x3 y2 + x4y3 + x1y4)}

-4

-2

-3

-5

3

-2

2

3

-4

-2

= ½ {(-4)(-5) + (-3)(-2) + (3)(3) + (2)(-2)} –

{(-4)(3) + (2)(-2) + (3)(-5) + (-3)(-2)}

= ½ {(20 + 6 + 9 – 4) – (–12 – 4 – 15 + 6)}

= ½ {31 – (–1)}

= ½ (31 + 1)

= ½ (32)

= 16 sq. units

A(-4,-2)

D(2,3)

B(-3,-5)

C(3,-2)

39. The vertices of ABC are A(2, 1), B(6, –1) and C(4, 11). Find the equation of the straight line along the altitude from the vertex A.

The vertices of the triangle are A(2, 1), B(6, –1), C(4, 11)

Let AD be the altitude.

A

B CD

12

12

xx

yySlope

46

111

BCofSlope 62

12

6

1 ADofSlope

A(2, 1), Slope m = 1/6

Required Equation of the line is

y – y1 = m(x – x1)

)2(6

11 xy

6(y – 1) = x – 2

6y – 6 = x – 2

x – 2 – 6y + 6 = 0

x – 6y + 4 = 0

40. A boy is designing a diamond shaped kite, as shown in the figure where AE = 16cm, EC = 81 cm. He wants to use a straight cross bar BD. How long should it be?

The vertices of the triangle are A(2, 1), B(6, –1), C(4, 11)

Let AD be the altitude.

A

B CD

12

12

xx

yySlope

46

111

BCofSlope 62

12

6

1 ADofSlope

A(2, 1), Slope m = 1/6

Required Equation of the line is

y – y1 = m(x – x1)

)2(6

11 xy

6(y – 1) = x – 2

6y – 6 = x – 2

x – 2 – 6y + 6 = 0

x – 6y + 4 = 0

41. A vertical tree is broken by the wind. The top of the tree touches the ground and makes an angle 30with it. If the top of the tree touches the ground 30 m away from its foot, then find the actual height of the tree.

B C30BC

AC30tan

303

1 AC

A

30m

Let A be the point at which the tree is broken and let the top of the tree touch the ground at B.

In ABC, BC = 30m, mB = 30

41. A vertical tree is broken by the wind. The top of the tree touches the ground and makes an angle 30with it. If the top of the tree touches the ground 30 m away from its foot, then find the actual height of the tree.

B C30

3

30AC

3

330

3

3

3

30

= 10 1.732

= 17.32

A

30m

AB

BC30cos

AB

30

2

3

3

230AB

3

360

3

3

3

60

= 20 1.732

= 34.64

Actual height of the tree

= 17.32 + 34.64

= 51.96m

42. Using clay, a student made a right circular cone of height 48 cm and base radius 12 cm. Another student reshapes it in the form of a sphere. Find the radius of the sphere.

Cone: Radius = 12cm,

Height = 48cm

Volume of the sphere = volume of the cone

hrr 23 3

1

3

4

r3 = 12 12 12

r = 12cm

4812123

1

3

4 3 r12

43.Calculate the standard deviation of the following data.

3 7

8 10

13 15

18 10

23 8

50

x f d = x – 3 d2 fd

fd2 3 – 3 = 0

8 – 3 = 5

13 – 3 = 10

18 – 3 = 15

23 – 3 = 20

0

25

100

225

400

0

50

150

150

160

510

0

250

1500

2250

3200

7200

X 3 8 13 18 23

f 7 10 15 10 8

Standard deviation

22

f

fd

f

fd

2

50

510

50

7200

22.10144

04.104144

96.39

39.966

6

36

396123369

27

.3

3.6

44. The probability that a new car will get an award for its design is 0.25, the probability that it will get an award for efficient use of fuel is 0.35 and the probability that it will get both the awards is 0.15.

Find the probability that (i) it will get atleast one of the two awards (ii) it will get only one of the awards.

Let A be the event that a new car will get an award for its design

B be the event that a new car will get an award for efficient use of fuel.

P(A) = 0.25, P(B) = 0.35 P(AB) = 0.15

P(atleast one of the two awards) = P(AB)



Let A be the event that a new car will get an award for its design

B be the event that a new car will get an award for efficient use of fuel.

P(A) = 0.25, P(B) = 0.35 P(AB) = 0.15

P(atleast one of the two awards) = P(AB)

P(AB) = P(A) + P(B) – P(AB)

= 0.25 + 0.35 – 0.15

= 0.60 – 0.15

= 0.45



P(A) = 0.25, P(B) = 0.35 P(AB) = 0.15

P(getting only one of the awards)

= P (only A or only B)

B)AP()BP(A

= [P(A) – P(A B)] + [P(B) – P(A B)]

= (0.25 – 0.15) + (0.35 – 0.15)

= 0.10 + 0.20

= 0.30

45. The sum of three consecutive terms in an AP is – 6 and their product is 90. Find the numbers.

Let a – d, a, a + d be three numbers in an AP

Sum = –6

a – d + a + a + d = – 6

3a = –6

a = – 6/3

a = –2

Product = 90

(a – d) a (a + d) = 90

(a2 – d2)a = 90

{(–2)2 – d2 }(–2) = 90

4 – d2 = – 45

– d2 = –45 –4

d2 = 49

d = 7The given numbers

are

– 2, 5, 12

(or) –2, –9, –16

45. A Cylindrical jar of diameter 14cm and depth 20cm is half-full of water . 300 lead shots of same size are dropped into the jar and the level of water raises by 2.8cm. Find the diameter of each lead shots.

Cylindrical jar:

Diameter = 14cm

Radius = 7cm

Height of water level raises = 2.8cm

Volume of 300 lead shots = volume of raised water

45. A Cylindrical jar of diameter 14cm and depth 20cm is half-full of water . 300 lead shots of same size are dropped into the jar and the level of water raises by 2.8cm. Find the diameter of each lead shots.

hrr 23 3

4300

8.277 400 3 r

100

10 400

2877 3

r

7

100

10 1010

7773

r

cmr 7.010

7

Note: 2 ×10 = 20 marks

(i) This section contains Two questions, each with two alternatives.

(ii) Answer both the questions choosing either of the alternatives.

(iii) Each question carries Ten marks

SECTION - D

46. Draw two tangents from a point which is 10cm away from the centre of a circle of radius 6cm. Also measure the lengths of the tangents.

O P10

6

A

B

8

8

Length of the tangent segment

22 610

36100

864

46. Draw two tangents from a point which is 10cm away from the centre of a circle of radius 6cm. Also measure the lengths of the tangents.

M

47. (a) Draw the graph of x2 – x – 8 = 0 and hence find the roots of x2 – 2x – 15 = 0.

x2 – 2x – 15 = 0 (x + 3)(x – 5) = 0 x = –3, 5

In x axis 1cm = 1 unit

y axis 1cm = 2 units.

x -4 -3 -2 -1 0 1 2 3 4 5

x2 16 9 4 1 0 1 4 9 16 25

-x 4 3 2 1 0 -1 -2 - 3 -4 -5

-8 -8 –8 –8 –8 –8 –8 –8 –8 –8 –8

y 12 4 -2 -6 -8 -8 -6 -2 4 12

-10

-8

-6

-4

-2

0

2

4

6

8

10

12

14

16

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

Solution set = {– 3, 5}

47. (b) A cyclist travels from a place A to a place B along the same route at a uniform speed on different days. The following table gives the speed of his travel and the

corresponding time he took to cover the distance.

Speed in km/hr

2 4 6 10 12

Time in hrs

60 30 20 12 10

Draw the speed-time graph and use it to find

(i) the number of hours he will take if he travels at a speed of 5 km / hr

(ii) the speed with which he should travel if he has to cover the distance in 40 hrs.

Speed in km/hr 2 4 6 10 12

Time in hrs 60 30 20 12 10

0

10

20

30

40

50

60

70

0 1 2 3 4 5 6 7 8 9 10 11 12 13

If x = 5,

y = 24

If y = 40,

y = 2.4Speed in km/hr

Tim

e in

hr

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X-STANDARD MATHEMATICS