(Question paper - With Answers) STD. X - …blog.surabooks.com/wp-content/uploads/2016/12/Std_X...SURA BOOKS 2 Sura’s Mathematics - X Std. October 2016 Question Paper with Answers

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GOVT SUPPLEMENTARY EXAM OCTOBER - 2016 (Question paper - With Answers)

STD. X - MATHEMATICS[Time Allowed : 2½ Hrs.] [Maximum Marks : 100]

[1]

SECTION - INote: (i) Answer all the 15 questions. (ii) Choose the correct answer from the

given four alternatives and write the option code and the corresponding answer. [15 × 1 = 15]

1. For any two sets A and B, {(A\B)È(B\A)}Ç (AÇB) is :

(a) f (b) A È B (c) AÇB (d) A¢ Ç B¢2. The 8th term of the sequence 1, 1, 2, 3, 5, 8,

.......... is : (a) 25 (b) 24 (c) 23 (d) 213. If the third term of a G.P is 2, then the product

of first 5 term is : (a) 52 (b) 25 (c) 10 (d) 154. The sum of two zeros of the polynomial

f(x) = 2x2 + (p + 3)x + 5 is zero, then the value of p is

(a) 3 (b) 4 (c) – 3 (d) – 45. If ax2 + bx + c = 0 has equal roots, then c is

equal :

(a) ba

2

2 (b) b

a

2

4

(c) -ba

2

2 (d) -b

a

2

4

6. If (5 x 1) 213

−

= (20), then the value of x is :

(a) 7 (b) –7 (c) 1

7 (d) 0

7. Area of the quadrilateral formed by the points (1,1), (0,1), (0,0), (1, 0) is :

(a) 3 sq.units (b) 2 sq.units (c) 4 sq.units (d) 1 sq.unit

8. The equation of a straight line passing through the point (2, –7) and parallel to x - axis is :

(a) x = 2 (b) x = – 7 (c) y = – 7 (d) y = 29. In the figure if ÐPAB = 120° then ÐBPT =

P

C B

A120°

T

(a) 120° (b) 30° (c) 40° (d) 60°10. The perimeters of two similar triangles

are 24 cm and 18 cm respectively. If one side of the first triangle is 8 cm, then the corresponding side of the other triangle is :

(a) 4 cm (b) 3 cm (c) 9 cm (d) 6 cm11. In the adjoining figure, AC = (a) 25 m

(b) 25 3 m

(c) 253

m

(d) 25 2 m

12. (1 – cos2q) (1 + cot2q) = (a) sin2θ (b) 0 (c) 1 (d) tan2θ

13. If the volume of a sphere is 916

p cu.cm, then its radius is :

(a) 43

cm (b) 34

cm

(c) 32

cm (d) 23

cm

A

C

B60°

25m

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2 Sura’s Mathematics - X Std. October 2016 Question Paper with Answers

14. If t is the standard deviation of x,y,z then the standard deviation of x + 5, y + 5, z + 5 is :

(a) t3

(b) t + 5

(c) t (d) xyz

15. If p is the probability of the event A, then p satisfies:

(a) 0 < p < 1 (b) 0 £ p £ 1 (c) 0 £ p < 1 (d) 0 < p £ 1

SECTION - IINote: (i) Answer 10 questions (ii) Question number 30 is compulsory.

Select any 9 questions from the first 14 questions. [10 × 2 = 20]

16. For the given function F = {(1,3), (2, 5), (4, 7), (5, 9), (3, 1)}, write the domain and range.

17. In a flower garden, there are 23 rose plants in a first row, 21 in the second row, 19 in the third row and so on. There are 5 rose plants in the last row. How many rows are there in the flower garden?

18. Find the quotient and the remainder when 3x3 – 17x2 + 31x – 12 is divided by 3x – 2.

19. Form the quadratic equation whose roots are 3 + 7 , 3 – 7 .

20. If A =2 39 5

1 57 1−

−−

, then find the

additive inverse of A.21. Construct a 2 × 2 matrix A = [aij] whose

elements are given by aij = i ji j

−+ .

22. If the points (p2, 0) and (0, q2) and (1, 1) are

collinear, prove that 1 1 12 2p q

+ = .

23. ABCD is a quadrilateral such that all of its sides touch a circle. If AB = 6 cm, BC = 6.5 cm and CD = 7 cm, then find the length of AD.

24. Prove the ident i ty ( s in6θ+cos6θ) = 1–3sin2θcos2θ.

25. Find the angular elevation (angle of elevation from the ground level) of the Sun when the length of the shadow of a 30 m long pole is 10 3 m.

26. If the curved surface area of a solid hemisphere is 2772 sq.cm, then find its total surface area.

27. Radius and slant height of a cone are 20 cm and 29 cm respectively. Find its volume.

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

29. If A and B are two events P(A) = 12

,

P(B) = 710

, P(AÈB) = 1, find

(i) P(AÇB) (ii) P(A¢ È B¢)30. For A = {x |x - is a prime factor of 42}, B = {x |5 < x £ 12, x Î N} and C= {1, 4, 5, 6},

find A È (BÈC).[OR]

If the centroid of a triangle is at (1, 3) and two of its vertices are (–7, 6) and (8, 5) then find the third vertex of the triangle.

SECTION - IIINote: (i) Answer 9 questions. (ii) Question No. 45 is Compulsory.

Select any 8 questions from the first 14 questions. [9 × 5 = 45]

31. Use Venn diagram to verify (AÇB)¢ = A¢ÈB¢.32. A function f : [–7, 6) ® ¡ is defined as follows:

f xx x xx xx x

( );;;

=+ + − ≤ < −

+ − ≤ ≤− < <

2 2 1 7 55 5 21 2 6

Find: (i) 2f (–4) + 3f (2) (ii) f (–7) – f (–3)

(iii) 4 3 2 46 3 1

f ff f

( ) ( )( ) ( )

− +− −

33. The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find the 27th term.

SURA BOOKS

Sura’s Mathematics - X Std. October 2016 Question Paper with Answers 3

34. Find the total volume of 15 cubes whose edges are 16 cm, 17 cm, 18 cm, .......... 30 cm respectively.

35. Solve : xx

xx+

+ + =1

1 6130

36. Find the square root of x4–10x3+37x2–60x+36.

37. If a and b are the roots of the equation 3x2 – 6x + 1 = 0 form an equation whose roots are :

(i) a2b, b2a (ii) 2a+b, 2b+a

38. If A = a bc d

and I2 =

1 00 1

, then show

that A2 – (a + d)A = (bc – ad)I2.39. If the vertices of a DABC are A(2, –4), B(3,3)

and C(–1,5). Find the equation of the straight line along the altitude from the vertex B.

40. Find the area of the quadrilateral whose vertices are (–4, 5) (0, 7) (5, –5) and (–4, –2).

41. From the top and foot of 40 m high tower, the angles of elevation of the top of a lighthouse are found to be 30° and 60° respectively. Find the height of the lighthouse. Also find the distance of the top of the lighthouse from the foot of the tower.

42. The curved surface area and total surface area of solid circular cylinder are 880 sq.cm and 1188 sq.cm respectively. Find its Volume.

43. Calculate the co-efficient of variation of the following data: 18, 20, 15, 12, 25.

44. In the class, 40% of the students participated in Mathematics - quiz, 30% in Science - quiz

and 10% in both the quiz programmes. If a student is selected at random from the class, find the probability that the student participated in Mathematics or Science, or both quiz programmes.

45. (a) A circus tent is to be erected in the form of a cone surmounted on a cylinder. The total height of the tent is 49m. Diameter of the base is 42m and height of the cylinder is 21m. Find the cost of canvas needed to make te tent, if the

cost of canvas is `12.50/m2 (Take p = 227

)

[OR]

b) State and prove the converse of Angle Bisector theorem.

SECTION - IVNote: Answer both the questions choosing either

of the alternatives. [2 × 10 = 20]

46. (a) Draw a circle of diameter 10 cm. From a point P, 13 cm away from its centre, draw the two tangents PA and PB to the circle, and measure their lengths.

[OR]

(b) Construct a cyclic quadrilateral ABCD with AB = 7 cm, ÐA = 80°, AD = 4.5 cm and BC = 5 cm.

47. (a) Solve the equation x2–2x –3 = 0 graphically.

[OR]

(b) Draw the Graph of xy = 20, x, y >0. Use the graph to find y when x = 5, and to find x when y = 10.

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SECTION – I1. (a) 2. (d) 3. (b) 4. (c) 5. (b) 6. (b) 7. (d) 8. (c) 9. (d) 10. (d)11. (b) 12. (c) 13. (b) 14. (c) 15. (b)

SECTION – II

16. Solution: Domain = {1, 2, 4, 5, 3} Range = {3, 5, 7, 9, 1}17. Solution: Let n be the number of rows in the flower garden. The number of rose plants in the 1st, 2nd, 3rd,...,

nth rows are 23, 21, 19, ..., 5 respectively. t1 = 23 , t2 = 21, t3 = 19 tn = 5

Here d = t2 – t1 = 21 – 23 = – 2

In an A.P. tn = a+(n –1)d

tn Þ 23 + (n – 1)(–2) = 5

Þ 23 –2n + 2 = 5

25–2n = 5

–2n = 5 – 25

– 2n = –20

n = 10

So, there are 10 rows in the flower garden.

18. Solution: Let P(x) = 3x3 – 17x2 + 31x –12 Given that the divisor is 3x – 2

When 3x – 2 = 0 Þ x = 23

23

13 15 21 2

3 17 31 12 2 0 14

− −↓ −

−

So, 3x3–17x2+31x–12= x −

23

(3x2–15x+21)+2

Þ (3x – 2) 13 (3x2 – 15x + 21) + 2

Þ (3x – 2) (x2 – 5x + 7) + 2

\ Quotient = x2 – 5x + 7, Remainder = 2

19. Solution : Sum of the roots = (3 + 7 )+(3 – 7 ) = 6

Product of the roots = (3 + 7 )(3 – 7 ) = 32 – 7 2

= 9 – 7 = 2

\ Required quadratic equation

= x2– (Sum of the roots)x

+ (Product of the roots) = x2 – 6x + 2 = 0.20. Solution :

A =2 39 5

1 57 1−

−−

= 1 216 6

−−

Additive inverse of A = (–A) = −

−

1 216 621. Solution :

aij = i ji j

−+

A

=

a a

a a11 12

21 22

a =1 11+1

= 0, a =1 21+ 2

=1

3, 11 12

− − −

a =2 12 +1

=13

, a =2 22 + 2

=04

= 021 22− −

Required Matrix = A = 0

13

13

0

−

22. Solution: Given points are (p2, 0) (0, q2), (1, 1)

ANSWERS

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p 0 1

0

2 p

q

2

2{ {0 1 p2 q2 + 0 + 0 = 0 + q2 + p2

Þ p2q2 = q2+p2

Dividing by p2q2 we get,

p qp q

2 2

2 2 =

qp q

pp q

2

2 2

2

2 2+

1 = 1 1

2 2p q+

Þ 1 12 2p q

+ = 1. Hence it is proved.

23. Solution: Let P, Q, R and S be the points where the circle

touches the quadrilateral. We know that the lengths of the two tangents

drawn from an exterior point to a circle are equal.

\ AB = AS .....(1) BP = BQ .....(2) CR = CQ .....(3) DR = DS .....(4) Adding the equations (1), (2), (3) and (4) we get, AP + BP + CR + DR = AS + BQ + CQ + DS Þ AB + CD = AD + BC. AD = AB + CD – BC = 6 + 7 – 6.5 = 6.5 Thus, AD = 6.5 cm.

24. Solution: sin6 θ+ cos6 θ

= (sin2 θ)3 + (cos2 θ)3

= (sin2θ+cos2θ)3–3sin2θcos2 θ (sin2 θ + cos2 θ) ( a3 + b3 = (a + b)3 – 3ab(a + b)) = 1 – 3 sin2 θ cos2 θ. ( sin2 θ+ cos2 θ= 1)25. Solution: Let S be the position of the Sun and BC be the

pole. Let AB denote the length of the shadow of the

pole.

Let the angular elevation of the Sun be q.

A B

CD

6 cm

6.5

cm

7 cm

P

Q

R

S

Given that AB = 10 3 m and BC = 30 m

In the Right D CAB,

tan q = BCAB

= 3010 3

33

=

tan q = 3

q = 60° Thus, the angular elevation of the Sun from the

ground level is 60°.

26. Solution: Curved surface area of a solid hemisphere = 2pr2 sq.units. 2pr2 = 2772

pr2 = 27722

= 1386 Its total surface area 3pr2 = 3×1386 Total surface area of a hemisphere = 4158 sq.cm27. Solution: Radius of a cone r = 20 cm

Slant height l = 29 cm

\ Height h = l r2 2−

= 29 202 2− = 841 400−

= 441

= 21 cm

Volume of the cone V = 13

pr2h cu.units

= 13

× 227

×20×20×21

Volume of the solid cone = 8800 cu.cm.

28. Solution:

Standard deviation of the first n natural numbers

= n2 112

−

B

30 m

A

C

10 3 m

S

�

CSA 2772sq.cm

20 cm

29 cm

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Thus, the standard deviation of the first 13 natural numbers

=−

=−

=

=

13 112

169 112

16812

143 74

2

� .

3

67

744

14 0000

500

3 74..

9

469

3100 2976

124

29. Solution: (i) P(A Ç B) = P(A) + P(B) – P(A È B)

By addition theorem,

= 12

710

1+ −

= 5 7 1010

210

15

+ − = =

(ii) P(A¢È B¢) = P(A ÇB)¢ = 1 – P(A ÇB)

= 1– 15

= 45

30. Solution:

A = {x / x - is a prime factor of 42 } = {2, 3, 7} B = {x / 5 < x ≤ 12, x Î N} = {6, 7, 8, 9, 10, 11, 12} C = {1, 4, 5, 6}

B ÈC ={6,7,8,9,10,11,12}È{1,4, 5, 6} = { 1, 4, 5, 6, 7, 8, 9, 10, 11, 12} \ AÈ(BÈC) = {2,3,7} È {1,4,5,6,7,8,9,10,11,12} = {1,2,3,4,5,6,7,8,9,10,11,12}

(OR) Solution: Centroid of triangle G (x, y) = (1, 3), Vertices of triangles are A (x1, y1) = (–7, 6), B (x2, y2) = (8, 5) C (x3, y3) = ?

\ Centroid of triangle

G (x, y) = G x x x y y y1 2 3 1 2 3

3 3+ + + +

,

G(1, 3) = G− + + + +

7 83

6 53

3 3x y,

= G 1

311

33 3+ +

x y,

Equating the x-co-ordinates, we get

1

33+ x

= 1 Þ x3 = 3 –1 = 2

Equating the y-co-ordinates, we get

11

33+ y

= 3 Þ y3 = 9 –11 = – 2

\ Third vertex C(x3, y3) = (2, – 2)

SECTION – III

31. Solution:

(AÇB)¢ = A¢ÈB¢

(A B) ∩

U

(A B)� ∩

U

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A�

U

B�

U

A��B�

U

From (2) & (5) it is clear that (AÇB)¢ = A¢ÈB¢32. Solution:

−∞ −7 −5 2 6 ∞x x2+2 +1 x + 5 x – 1

(i) f (–4) = (–4) + 5 = 1 (Since –4 lies in the interval –7£ x <–5 use x+5)

f (2) = 2 + 5 = 7 (Since 2 lies in the interval –5£ x £ 2 use x + 5) 2f (–4) + 3f (2) = 2(1) + 3(7) = 2+21 = 23 (ii) f (–7) = (–7)2 + 2(–7) +1 = 49 – 14 + 1 = 36 ((Since –7 lies in the interval –7£ x <–5

use x2+2x + 1)) f (–3) = (–3) + 5 = 2 (Since –3 lies in the interval –5£ x £ 2

use x + 5)

f (–7) – f(–3) = 36 – 2 = 34 (iii) f (–3) = 2 from (ii) f (4) = 4 – 1 = 3 f (–6) = (–6)2 + 2(–6) + 1 = 36 – 12 + 1 = 25 (Since –6 lies in the interval –7£ x <–5

use x2+2x+1) f (1) = 1 + 5 = 6 (Since 1 lies in the interval –5£ x £ 2 use x + 5)

\ 4 3 2 46 3 1

f ff f

( ) ( )( ) ( )

− +− −

= 4 2 2 325 3 6( ) ( )

( )+

−

= 8 6

25 18+−

= 147

= 2

33. Solution: Given, t10 =41 Þ a + (10 – 1)d = 41 Þ a + 9d = 41 ... (1) and t18 = 73 Þ a + (18 – 1)d = 73 Þ a + 17d = 73 ... (2)

Solving (1) and (2) ⇒a + 9d = 41 a + 17d = 73

(–) (–) (–) –8d = –32

⇒ d = 4

Substituting in (1)

Þ a + 9 × 4 = 41 Þ a + 36 = 41 a = 5 \ t27 = a + (27 – 1)d = a + 26d = 5 + 26 (4) = 5 + 104 = 109

\ The 27th term =109.

34. Solution:

The volume of the cubes from the series = 163 + 173 + 183 ............ + 303 = (13 + 23 + 33 + ...........+ 303) – (13 + 23 + 33 ...... + 153) = ∑ 303 – ∑ 153

= 30 30 12

15 15 12

2 2( ) ( )+

− +

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= 30 312

15 162

2 2×

− ×

= (15×31)2 – (15×8)2

= 4652 – 1202

= (465 + 120) (465 – 120) = 585 × 345 = 201825 cu.cm The total volume of the 15 cubes= 201825 cu.cm

35. Solution:

Given x

xx

x++ +

11 = 61

30

Þ x xx x

2 211

+ ++

( )( )

= 6130

Þ x x xx x

2 2

22 1+ + +

+ = 61

30

Þ 2 2 12

2x x

x x+ +

+ = 61

30

Þ 30(2x2+2x + 1) = 61(x2+x) Þ 60x2+60x + 30 = 61x2+ 61x Þ 61x2+61x –60x2 –60x – 30 = 0 Þ x2 + x – 30 = 0 Þ (x + 6) (x – 5) = 0 Þ x = –6, 5 \ Solution set is = {–6, 5}36. Solution:

Given polynomial x4 – 6x3 + 37x2 – 60x + 36 is already in descending powers of x.

– 5 + 6

– 10 + 37 – 60 + 36

2 – 5 – 10 + 37

–10 + 25

2 – 10 + 6 12 – 60 + 36

12 – 60 + 36

0

x xx x x x x

xx x x x

x xx x x x

x x

2

2 4 2

4

2 3 2

3 2

2 2

2

3

\ x x x x x x4 3 210 37 60 36 2 5 6− + + =− − +( )

–30

61

–5

37. Solution: a and b are the roots of the equation

3x2 – 6x + 1 = 0

a + b = − −( )6

3 = 2

ab = 13

(i) a2b, b2a

Sum of the roots of required equation

= a2b + b2a = ab(a + b) =

13

×2 = 23

Product of the roots = a2b × b2a = a3b3 = (ab)3

= 13

3

=

127

\ Required equation = x2– 23 x +

127 = 0

Þ 27x2 – 18x + 1 = 0

(ii) 2a+b, 2b+a Sum of the roots = (2a+b) + (2b+a) = 3a + 3b = 3×2 = 6 Product of the roots = (2a+b)× (2b+a) = 4ab+ 2a2 + 2b2 + ab = 5ab+ 2(a2 + b2) = 5ab+ 2((a + b)2 –2 ab)

= 5×13

+ 2((2)2 –2 ×13

)

= 53 + 8 – 4

3

= 13

+ 8 =1 24

3+

= 253

\ Required equation = x2– 6x + 253

= 0

Þ 3x2 – 18x + 25 = 0

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38. Solution:

A = a bc d

and I2

= 1 00 1

Consider A2 = A × A

A2 = a bc d

×

a bc d

= a bc ab bdac cd bc d

2

2

+ ++ +

...(1)

(a + d)A = (a + d) a bc d

= a ad ab bdac cd ad d

2

2

+ ++ +

...(2)

From equation (1) and (2)

A2 – (a + d)A = a bc ab bdac cd bc d

2

2

+ ++ +

– a ad ab bdac cd ad d

2

2

+ ++ +

= bc ad

bc ad−

−

00

= (bc – ad) 1 00 1

= (bc – ad) I2 \ A2 – (a + d)A = (bc – ad)I2

Hence it is proved39. Solution: Let (x1, y1) = A (2, –4) (x2, y2) = B (–1, 5).

\ Slope of AC

= y yx x

2 1

2 1

−−

= 5 41 2

− −− −

( )

C(–1,5)B(3,3)

D

A(2, –4)

m1 = 93−

= – 3

Also BD passes through B to AC \ Slope of BD is m2

\ m1 × m2 = –1 ( BD ^ AC) –3 × m2 = –1

m2 = 13

The equation of BD at slope 13 , and passes

through B(3, 3) is

Þ y – 3 = 13 (x – 3)

3y – 9 = x – 3

\ The required equation is x – 3y + 6 = 0.

40. Solution:

The vertices of the quadrilateral are (–4, 5) (0, 7), (5, –5) and (–4, –2)

-4 0 5 -4 -4

5 7 -5 -2 512

Area = 12

28 0 10 20 0 35 20 8( ) ( )− − − − − + + +{ }

= 12

58 63( ) ( )− −{ }

= 12

121−{ }

= 1212

= 60.5 Sq.units

41. Solution:

Required to find CD and AD. Let the distance of the top of the light house

ED be ‘x’ m CD = (x +40) m.

From the rt.Dle BED;

tan 30° = xy

Þ 13

= xy

Þ y = 3 x .....(1)

( 4, 5)–

( 4, 2)– – (5, 5)–

(0, 7)AB

CD

y

x

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From the rt Dle ACD;

\ tan 60° = 40 + xy

Þ 3 = 40 + xy

Þ 3 y = 40 + x .....(2)

Sub.(1) in (2) 3 ( 3 x) = 40 + x

Þ 3x – x = 40

\ 2x = 40 or x = 20

\ Height of the light house CD = 20 + 40

= 60 m.

From the rt D le ACD;

sin 60° = x + 40AD

Þ 32

= x + 40AD

\ AD = 2 40

3( )x +

= × =2 60

333

2 60 33

20

( ) ( )

= 40 3 m

600

300

D

EB

A C

x my m

y m

Light house40 mTower

40 m

Distance of the top of the lighthouse from the foot of the tower = 40 3 m

42. Solution: The curved surface area of Solid Cylinder

(CSA) = 880 sq.cm

Þ 2prh = 880 ..... (1) Total surface area (TSA) = 1188 sq.cm Þ 2pr(h+r) = 1188 ..... (2) Substituting (1) in (2) we get Þ 2prh+2pr2 = 1188 Þ 880 + 2pr2 = 1188 Þ 2pr2 = 1188 – 880 Þ 2pr2 = 308 Þ pr2 = 154

Þ 227

× r2 = 154

Þ r2 = 154 722

7

× = 49

Þ r = 7 cm Substituting r = 7 cm in equation (1) we get

2×227

× 7/×h = 880 h = 880

2 22× = 880

44 = 20 cm

volume of the cylinder = pr2h

= 227

×7× 7/× 20 = 3080 cm3

43. Solution:

Average x– = 12 15 18 20 25

5+ + + +

= 905

= 18

x d = x – x– d2

12 – 6 3615 – 3 918 0 0

20 2 425 7 49

Sd = 0 Sd2 =98

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s = Σdn

2 985

=

= 19 6 4 428. .�

Co-efficeient of variation (C.V) = σx

× 100

= 4 428

18.

× 100

= 442 8

18.

\ Co-efficeient of variation = 24.6

44. Solution: n(S) = 100

No. of students participated in Maths Quiz = 40%

No. of students participated in Science Quiz = 30%

No. of students participated in both quiz programmes = 10%

Let A denote the event of taking part in Math Quiz.

\ n(A) = 40

Þ P(A) = nn( )( )AS

= 40100

Let B denote the event of taking part in Science quiz.

\ n(B) = 30

Þ P(B) = nn( )( )BS

= 30100

and (AÇB) denotes the event of taking part in both Maths and Science Quiz

\ n(AÇB) = 10

Þ P(AÇB) = nn

( )( )

A BS∩ = 10

100

Using addition theorem on probabilities P(AÈB) = P(A) + P(B) – P(AÇB)

= 40100

30100

10100

+ −

= 60

10035

=

45. (a) Solution:

Cylindrical Part Conical Part Diameter, 2r = 42 m Radius, r = 21 m

Radius, r = 21 m Height, h1=49–21=28 m

Height, h = 21 m Slant height,l= h r2 2+

= 28 212 2+

= 7 4 32 2+ = 7 16 9+

= 7 25

= 7× 5 = 35 m

Total area of the canvas needed

= CSA of the cylindrical part

+ CSA of the conical part

= 2prh + prl = pr (2h + l)

= 227

× 21 (2 ×21 + 35) = 5082

Therefore, area of the canvas = 5082 m2

Now, the cost of the canvas per sq.m = `12.50 Thus, the total cost of the canvas

= 5082 × 12.5 = `63525.

(OR)(b) Theorem: If a straight line through one

vertex of a triangle divides the opposite side internally (externally) in the ratio of the other two sides, then the line bisects the angle internally (externally) at the vertex.

Case : (Internally)

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Given : In DABC, the line AD divides the opposite side BC internally such that

BDDC

= ABAC

..... (1)

To Prove : AD is the internal bisector of

ÐBAC

i.e., to prove ÐBAD = ÐDAC

Construction : Through C draw CE || AD meeting BA produced at E.

Proof : Since CE || AD, by Thales theorem,

we have BDDC

= BAAE

..... (2)

Thus, from (1) and (2) we have, ABAE

ABAC

=

\ AE = AC

Now, in D ACE, we have

Ð ACE = Ð AEC ( AE = AC ) ..... (3)

SECTION – IV46. Solution:(a) Given: Diameter of the circle = 10 cm

\ radius = 102

5= cm

OP = 13 cmRough Diagram

5 cm

13 cm PO

A

B

5 cm

E

CDB

A

Since AC is a transversal of the parallel lines AD and CE,

we get, ÐDAC = ÐACE (alternate interior angles are equal) ..... (4) Also BE is a transversal of the parallel lines

AD and CE. we get ÐBAD = ÐAEC ( corresponding angles are equal) ..... (5) From (3), (4) and (5), we get ÐBAD = ÐDAC \ AD is the angle bisector of ÐBAC. Hence the theorem.

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5 cm

12 cm

12 cm

13 cm PO

A

M

B

Steps of Construction:(i) Let the two circles intersect at A and B(ii) With O as the centre draw a circle of radius 5 cm(iii) Draw the perpendicular bisector of OP. Let it meet OP at M.(iv) With M as centre and MO as radius, draw another circle.(v) Let the two circles intersect at A and B(vi) Join PA and PB. They are the required tangents. Length of the tangent, PA = 12 cmVerification:

In the right angled D OPA, PA = − = − = − =OP OA2 2 2 213 5 169 25 144 = 12 cm \ PA = PB = 12 cm

(OR)

Fair Diagram

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(b) Given: In the cyclic quadrilateral ABCD, AB = 7 cm, ∠A = 80º, AD = 4.5 cm and BC = 5 cm. Rough Diagram

5 cm

O

CD

X

BA 7 cm

4.5

cm

80°

Steps of Construction:

(i) Draw a rough diagram and mark the measurements. Draw a line segment AB = 7 cm.(ii) Through A draw AX such that ∠BAX = 80º.(iii) With A as centre and radius 4.5 cm, draw an arc intersecting AX at D and join AD.(iv) Draw the perpendicular bisectors of AB and AD intersecting each other at O.(v) With O as centre and OA = OD as radius, draw the circumcircle of DABD(vi) With B as centre and radius 5 cm, draw an arc intersecting the circle at C.(vii) Join BC and CD. Now, ABCD is the required cyclic quadrilateral.

Fair Diagram

5 cm

BA 7 cm

4.5

cm

80°

CD

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47. (a) Solution:

x2–2x –3 = 0 Þ y = x2–2x –3 Table

x –3 –2 –1 0 1 2 3 4x2 9 4 1 0 1 4 9 16

–2x 6 4 2 0 –2 –4 –6 –8–3 –3 –3 –3 –3 –3 –3 –3 –3

y 12 5 0 –3 –4 –3 0 5

Plot the points (–3, 12), (–2, 5), (–1, 0), (0, –3), (1, –4), (2, –3), (3, 0), (4, 5) Draw the graph by joining the points by a smooth curve.

O

y

y�

xx� y =

x2 –2

x–3

The curve intersects the x-axis at the points (–1, 0) and (3, 0).

The x-coordinates of the above points are –1 and 3.

Thus the solution set is {–1, 3}.

Scale: x axis 1 cm = 1 unit y axis 1 cm = 2 units

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47. (b) Solution:

We form the table for Rectangular hyperbola xy = 20

Here y = 20x

x 1 2 4 5 10 20

y = 20x

20 10 5 4 2 1

Plot the points (1, 20), (2, 10), (4, 5) and (5, 4) and join them

2018161412108642

Y

1

(1,20)

(2,10)

(4,5)(5,4)

2 3 4 5 6xx′

y′

0 8 10 12

(10,2)

y

y = 20x

\ The relation xy = 20 is a rectangular hyperbola as exhibited in the graph.

From the graph we find,

(i) When x = 5, y = 4

(ii) When y =10, x =2.

Scale: x axis 2 cm = 1 unit y axis 1 cm = 2 units

(Question paper - With Answers) STD. X - …blog.surabooks.com/wp-content/uploads/2016/12/Std_X...SURA BOOKS 2 Sura’s Mathematics - X Std. October 2016 Question Paper with Answers

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