## Tamil Nadu SSLC or Class 10 Maths Question Paper 2016 with Solutions – Free Download

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### Tamil Nadu Board SSLC Class 10 Maths 2016 Question Paper with Solutions

**SECTION – I**

**1. **If A = {p, q, r, s}, B = {r, s, t u}, then A\B is:

(a) {p, q}

(b) {t, u}

(c) {r, s}

(d) {p, q, r, s}

**Solution:**

Correct answer: (a)

Given,

A = {p, q, r, s}, B = {r, s, t u}

A\B = {p, q, r, s} \ {r, s, t, u}

= {p, q}

**2. **The 8th term of the sequence 1, 1, 2, 3, 5, 8,… is:

(a) 25

(b) 24

(c) 23

(d) 21

**Solution:**

Correct answer: (d)

Given,

1, 1, 2, 3, 5, 8, â€¦.

This is a fibonacci series.

2 = 1 + 1

3 = 1 + 2

5 = 2 + 3

8 = 3 + 5

7th term = 5 + 8 = 13

8th term = 8 + 13 = 21

**3. **If 1 + 2 + 3 + … + n = k, then 1^{3} + 2^{3} + 3^{3} + … + n^{3} is equal to:

(a) k^{2}

(b) k^{3}

(c) k(k + 1)/2

(d) (k + 1)3

**Solution:**

Correct answer: (a)

Given,

1 + 2 + 3 + … + n = k

Sum of first n natural numbers = n(n + 1)/2 = k ….(i)

Sum of the cubes of first n natural numbers

1^{3} + 2^{3} + 3^{3} + … + n^{3} = [n(n + 1)/2]^{2}

= k^{2} [From (i)]

**4. **The system of equations x – 4y = 8, 3x – 12y = 24

(a) has infinitely many solutions

(b) has no solution

(c) has a unique solution

(d) may or may not have a solution

**Solution:**

Correct answer: (a)

Given,

x – 4y = 8

3x – 12y = 24

Comparing with the standard form,

a_{1} = 1, b_{1} = -4, c_{1} = -8

a_{2} = 3, b_{2} = -12, c_{2} = 24

a_{1}/a_{2} = â…“

b_{1}/b_{2} = (-4)/ (-12) = â…“

c_{1}/c_{2} = (-8)/ (-24) = â…“

a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

Hence, the given system of equations has infinitely many solutions.

**5. **If ax^{2} + bx + c = 0 has equal roots, then c is equal to

(a) b^{2}/2a

(b) b^{2}/4a

(c) -b^{2}/2a

(d) -b^{2}/4a

**Solution:**

Correct answer: (b)

Given that, ax^{2} + bx + c = 0 has equal roots.

Therefore, discriminant = 0

b^{2} – 4ac = 0

b^{2} = 4ac

â‡’ c = b^{2}/4a

**6.**

**Solution:**

Correct answer: (d)

Given,

**7. **The equation of a straight line parallel to y-axis and passing through the point (-2, 5) is:

(a) x – 2 = 0

(b) x + 2 = 0

(c) y + 5 = 0

(d) y – 5 = 0

**Solution:**

Correct answer: (b)

We know that the equation of line parallel to y-axis is x = k, where k is a constant.

Given that, the line passes through (-2, 5)

Hence, the equation of line is x = -2 or x + 2 = 0

**8. **The equation of a straight line having slope 3 and y-intercept -4 is:

(a) 3x – y – 4 = 0

(b) 3x + y – 4 = 0

(c) 3x – y + 4 = 0

(d) 3x + y + 4 = 0

**Solution:**

Correct answer: (a)

We know that the equation of a line with slope m and y-intercept c is y = mx + c

Given,

m = 3

c = – 4

Hence, the required equation of line is:

y = 3x – 4

â‡’ 3x – y – 4 = 0

**9. **In Î”ABC, DE is parallel to BC, meeting AB and AC at D and E. If AD = 3 cm, DB = 2 cm and AE = 2.7 cm, then AC is equal to:

(a) 6.5 cm

(b) 4.5 cm

(c) 3.5 cm

(d) 5.5 cm

**Solution:**

Correct answer: (b)

Given,

In Î”ABC, DE is parallel to BC, meeting AB and AC at D and E.

By BPT,

AD/DB = AE/EC

3/2 = 2.7/EC

â‡’ EC = (2.7 Ã— 2)/3

â‡’ EC = 1.8 cm

AC = AE + EC

= 2.7 + 1.8

= 4.5 cm

**10. **AB and CD are two chords of a circle which when produced to meet at a point P such that AB = 5 cm, AP = 8 cm and CD = 2 cm, then PD =

(a) 12 cm

(b) 5 cm

(c) 6 cm

(d) 4 cm

**Solution:**

Correct answer: (d)

Given,

AB and CD are two chords of a circle which intersect each other externally at P.

AB = 5 cm, AP = 8 cm and CD = 2 cm

Two chords AB and CD meet at P when produced.

Let PD = x

â‡’ PA Ã— PB = PC Ã— PD

â‡’ PA Ã— (AP – AB) = (PD + CD) Ã— PD

â‡’ 8 Ã— (8 – 5) = (PD + 2) Ã— PD

â‡’ 24 = x^{2} + 2x [PD = x]

â‡’ x^{2} + 2x – 24 = 0

â‡’ (x + 6)(x – 4) = 0

â‡’ x = -6, 4

Length cannot be negative.

Therefore, x = PD = 4 cm

**11. **(1 + tan^{2}Î¸). sin^{2}Î¸ =

(a) sin^{2}Î¸

(b) cos^{2}Î¸

(c) tan^{2}Î¸

(d) cot^{2}Î¸

**Solution:**

Correct answer: (c)

(1 + tan^{2}Î¸). sin^{2}Î¸

= sec^{2}Î¸ . sin^{2}Î¸

= (1/cos^{2}Î¸) . sin^{2}Î¸

= tan^{2}Î¸

**12.** In the adjoining figure, sin Î¸ = 15/17, then BC =

(a) 85 m

(b) 65 m

(c) 95 m

(d) 75 m

**Solution:**

Correct answer: (d)

Given,

sin Î¸ = 15/17

â‡’ BC/AC = 15/17

â‡’ BC/85 = 15/17

â‡’ BC = (15 Ã— 85)/17 = 75 m

**13. **The total surface area of a solid hemisphere of diameter 2 cm is equal to:

(a) 12 cm^{2}

(b) 12Ï€ cm^{2}

(c) 4Ï€ cm^{2}

(d) 3Ï€ cm^{2}

**Solution:**

Correct answer: (d)

Given,

Diameter of hemisphere = 2 cm

Radius = r = 2/2 = 1 cm

Total surface area = 3Ï€r^{2}

= 3Ï€(1)^{2}

= 3Ï€ cm^{2}

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

(a) t/3

(b) t + 5

(c) t

(d) xyz

**Solution:**

Correct answer: (c)

We know that the standard deviation of a set of numbers will not change if every number is increased or decreased by the same amount.

Hence, the standard deviation of x + 5, y + 5, z + 5 is equal to the standard deviation of x, y, z.

i.e. t

**15. **A bag contains 5 black balls, 4 white balls and 3 red balls. If a ball is selected at random, the probability that it is not red is:

(a) 5/12

(b) 4/12

(c) 3/12

(d) 3/4

**Solution:**

Correct answer: (d)

Given,

A bag contains 5 black balls, 4 white balls and 3 red balls.

Total number of balls = 5 + 4 + 3 = 12

Number of balls other than red = 5 + 4 = 9

P(selected ball is not red) = 9/12 = Â¾

**SECTION – II**

**16. **Given A = {a, x, y, r, s}, B = {1, 3, 5, 7, -10} verify the commutative property of set union.

**Solution:**

Given,

A = {a, x, y, r, s}

B = {1, 3, 5, 7, -10}

A â‹ƒ B = {a, x, y, r, s} â‹ƒ {1, 3, 5, 7, -10}

= {a, x, y, r, s, 1, 3, 5, 7, -10}

B â‹ƒ A = {1, 3, 5, 7, -10} â‹ƒ {a, x, y, r, s}

= {1, 3, 5, 7, -10, a, x, y, r, s}

A â‹ƒ B = B â‹ƒ A

Hence, verified the commutative property of set union.

**17.** Let A = {5, 6, 7, 8}, B = {-11, 4, 7, -10, -7, -9, -13} and f = {(x, y) : y = 3 – 2x, x âˆˆ A, y âˆˆ B}

(i) Write down the elements of f.

(ii) What is the range?

**Solution:**

Given,

y = 3 – 2x

When x = 5

y = 3 – 2(5) = 3 – 10 = -7

When x = 6

y = 3 – 2(6) = 3 – 12 = -9

When x = 7

y = 3 – 2(7) = 3 – 14 = -11

When x = 8

y = 3 – 2(8) = 3 – 16 = -13

(i) f = {(5, -7), (6, -9), (7, -11), (8, -13)}

(ii) Range of f = {-7, -9, -11, -13}

**18. **Find the 12th term of the AP âˆš2, 3âˆš2, 5âˆš2,…..

**Solution:**

Given AP is: âˆš2, 3âˆš2, 5âˆš2,…..

First term = a = âˆš2

Common difference = d = 3âˆš2 – âˆš2 = âˆš2(3 – 1) = 2âˆš2

nth term an = a + (n – 1)d

12th term of the AP = âˆš2 + (12 – 1)2âˆš2

= âˆš2 + 11(2âˆš2)

= âˆš2 + 22âˆš2

= 23âˆš2

**19. **Simplify: [x^{3}/(x – 2)] + [8/ (2 – x)]

**Solution:**

^{3}/(x – 2)] + [8/ (2 – x)]

= [x^{3}/(x – 2)] – [8/ (x – 2)]

= (x^{3} – 8)/ (x – 2)

= (x^{3} – 2^{3})/ (x – 2)

= [(x – 2)(x^{2} + 2x + 2^{2})/ (x – 2)

= x^{2} + 2x + 4

**20. **If Î± and Î² are the roots of the equation 3x^{2} – 6x + 4 = 0, find the value of Î±^{2} + Î²^{2}.

**Solution:**

Given,

Î± and Î² are the roots of the equation 3x^{2} – 6x + 4 = 0.

Comparing with the standard form ax^{2} + bx + c = 0,

a = 3, b = -6, c = 4

Î± + Î² = -b/a = -(-6)/3 = 2

Î±Î² = c/a = 4/3

Î±^{2} + Î²^{2} = (Î± + Î²)^{2} – 2Î±Î²

= (2)^{2} – 2(4/3)

= 4 – (8/3)

= (12 – 8)/3

= 4/3

**21. **

**Solution:**

**22. **

**Solution:**

From the given,

x + y = 7 ….(i)

y + z = 9 ….(ii)

z – 5 = 0

z = 5

Substituting z = 5 in (ii),

y + 5 = 9

y = 9 – 5 = 4

Substituting y = 4 in (i),

x + 4 = 7

x = 7 – 4 = 3

Therefore, x = 3, y = 4 and z = 5.

**23.** 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.

**Solution:**

Given,

Two vertices of the triangle are (-7, 6) and (8, 5).

Centroid of the triangle is (1, 3).

Let (x, y) be the third vertex of the triangle.

Using centroid formula,

Centroid = [(x_{1} + x_{2} + x_{3})/3, (y_{1} + y_{2} + y_{3})/3]

(1, 3) = [(-7 + 8 + x)/3, (6 + 5 + y)/3]

(1, 3) = [(1 + x)/3, (11 + y)/3]

â‡’ 1 + x = 3

â‡’ x = 2

And

11 + y = 9

â‡’ y = -2

Hence, the third vertex is (2, -2).

**24. **Find the value of ‘a’, if the straight lines 5x – 2y – 9 = 0 and ay + 2x – 11 = 0 are perpendicular to each other.

**Solution:**

Given equations of straight lines are:

5x – 2y – 9 = 0

â‡’ 2y = 5x – 9

â‡’ y = (5/2)x – (9/2)

Slope = m_{1} = 5/2

And

ay + 2x – 11 = 0

ay = -2x + 11

y = (-2/a)x + (11/a)

Slope = m_{2} = -2/a

Given that, the straight lines are perpendicular to each other.

â‡’ m1m2 = -1

â‡’ (5/2) Ã— (-2/a) = -1

â‡’ a = 5

**25.** In the figure TP is a tangent to the circle. A and B are two points on the circle. If âˆ BTP = 72Â° and âˆ ATB = 43Â° find âˆ ABT.

**Solution:**

Given,

âˆ BTP = 72Â° and âˆ ATB = 43Â°

âˆ BAT = âˆ BTP = 72Â° (alternate angles)

Sum of all angles of a triangle = 180Â°

In triangle ABT,

âˆ ABT + âˆ BTA + âˆ TAB = 180Â°

â‡’ âˆ ABT + 43Â° + 72Â° = 180Â°

â‡’ âˆ ABT + 115Â° = 180Â°

â‡’ âˆ ABT = 180Â° – 115Â°

â‡’ âˆ ABT = 65Â°

**26. **If Î”ABC is right angled at ‘C’, then find the values of cos (A + B) and sin (A + B).

**Solution:**

Given that, in Î”ABC, C is the right angle.

Thus, A + B = 90Â°

cos (A + B) = cos 90Â° = 0

sin (A + B) = sin 90Â° = 1

**27.** Curved surface area and circumference at the base of a solid right circular cylinder are 4400 sq. cm and 110 cm respectively. Find its height and diameter.

**Solution:**

Let r be the radius and h be the height of the right circular cylinder.

Given,

Circumference of the base = 110 cm

2Ï€r = 110

2 Ã— (22/7) Ã— r = 110

r = (110 Ã— 7)/ (22 Ã— 2)

r = 17.5 cm

Curved surface area = 4400 sq.cm (given)

2Ï€rh = 4400

2 Ã— (22/7) Ã— 17.5 Ã— h = 4400

h = 4400/(22 Ã— 5)

h = 40 cm

Diameter = 2r = 2 Ã— 17.5 = 35 cm

Therefore, height is 40 cm and diameter is 35 cm.

**28. **If the coefficient of variation of a collection of data is 57 and its S.D. is 6.84, then find the mean.

**Solution:**

Given,

Coefficient of variation = 57

Standard deviation = 6.84

We know that,

Coefficient of variation = (standard deviation/ mean) Ã— 100

57 = (6.84 Ã— 100)/mean

Mean = 684/57

= 12

**29. **20 cards are numbered from 1 to 20. One card is drawn at random. What is the probability that the number on the card is

(i) A multiple of 4

(ii) Not a multiple of 6

**Solution:**

Total number of cards = n(S) = 20

(i) Let A be the event of getting a card with a number as a multiple of 4.

A = {4, 8, 12, 16, 20}

n(A) = 5

P(A) = n(A)/n(S) = 5/20 = Â¼

(ii) Let B be the event of getting a card with numbers which is not a multiple of 6.

B = {1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 19, 20}

n(B) = 17

P(B) = n(B)/n(S) = 17/20

**30. (a) **The central angle and radius of a sector of a circular disc are 180Â° and 21 cm respectively. If the edges of the sector are joined together to make a hollow cone, then find the radius of the cone.

**Solution:**

Given,

Radius of the sector = R = 21 cm

Angle of the sector = Î¸ = 180Â°

The cone is obtained by joining the radius and hence, the radius of the sector is equal to the slant height of the cone.

Slant height = l = 21 cm

Let r be the radius of the cone.

Arc length of the sector = Circumference of the base of the cone

(Î¸/360Â°) Ã— 2Ï€R = 2Ï€r

(180Â°/360Â°) Ã— (21) = r

â‡’ r = 21/2

â‡’ r = 10.5 cm

**OR**

**(b) **Prove that: (1 + sec Î¸)/sec Î¸ = sin^{2}Î¸/ (1 – cos Î¸)

**Solution:**

LHS = (1 + sec Î¸)/sec Î¸

= (1/sec Î¸) + 1

= cos Î¸ + 1

= [1 + cos Î¸] Ã— [(1 – cos Î¸)/ (1 – cos Î¸)]

= (1 – cos^{2}Î¸)/ (1 – cos Î¸)

= sin^{2}Î¸/(1 – cos Î¸)

= RHS

Hence proved.

**SECTION – III**

**31. **A radio station surveyed 190 students to determine the types of music they liked. The survey revealed that 114 liked rock music, 50 liked folk music and 41 liked classical music, 14 liked rock music and folk music, 15 liked rock music and classical music, 11 liked classical music and folk music, 5 liked all three of music.

Find:

(i) How many did not like any of the 3 types?

(ii) How many liked any two types only?

(iii) How many liked folk music but not rock music?

**Solution:**

Total number of students = 190

R = Rock music

F = Folk music

C = Classical music

From the given,

n(R â‹‚ F) = 14

n(R â‹‚ C) = 15

n(C â‹‚ F) = 11

n(R â‹‚ F â‹‚ C) = 5

n(R â‹‚ F â‹‚ C’) = 14 – 5 = 9

n(R â‹‚ C â‹‚ F’) = 15 – 5 = 10

n(F â‹‚ C â‹‚ R’) = 11 – 5 = 6

The number of students who liked any one of the music = 90 + 9 + 30 + 10 + 5 + 6 + 20

= 170

(a) The number of students who didn’t like any 3 types of music = 190 – 170 = 20

(b) Number of students who liked any two types ony = 9 + 6 + 10 = 25

(c) Number of students who liked folk music but not rock music = 30 + 6 = 36

**32.** A function f : [-3, 7) â†’ R is defined as follows:

Find:

(i) f(-2) – f(4)

(ii) [f(3) + f(-1)]/ [2f(6) – f(1)]

**Solution:**

(i) f(-2) = 4x^{2} – 1 = 4(-2)^{2} – 1 = 16 – 1 = 15

f(4) = 3x – 2 = 3(4) – 2 = 12 – 2 = 10

f(-2) – f(4) = 15 – 10 = 5

(ii) f(3) = 3x – 2 = 3(3) – 2 = 9 – 2 = 7

f(-1) = 4x^{2} – 1 = 4(-1)^{2} – 1 = 4 – 1 = 3

f(6) = 2x – 3 = 2(6) – 3 = 12 – 3 = 9

f(1) = 4x^{2} – 1 = 4(1)^{2} – 1 = 4 – 1 = 3

= 10/(18 – 3)

= 10/15

= 2/3

**33. **Find the sum to n terms of the series 7 + 77 + 777 + â€¦

**Solution:**

Given,

7 + 77 + 777 +…

Sum of first n terms

Sn = 7 + 77 + 777 + ….. + (n terms)

Sn = 7(1 + 11 + 111 + …. + n terms)

= (7/9) [9 + 99 + 999 + … + n terms]

= (7/9) [(10 – 1) + (100 – 1) + (1000 – 1) + …. + (10^{n} – 1)]

= (7/9) [(10 + 10^{2} + 10^{3} + … + 10^{n}) – (1 + 1 + 1 + … + n terms)]

= (7/9) {[10(10^{n} – 1)/ (10 – 1)] – n}

= (7/9) {[10(10^{n} – 1)/9] – n}

= (70/81) (10^{n} – 1) – 7n/9

**34.** If 7 times the 7th term of an Arithmetic Progression is equal to 11 times its 11th term, show that its 18th term is zero. Can you find the first term and the common difference? Justify your answer.

**Solution:**

Let a be the first term and d be the common difference an AP.

Given,

7a_{7} = 11a_{11}

7(a + 6d) = 11(a + 10d)

7a + 42d = 11a + 110d

â‡’ 11a + 110d – 7a – 42d = 0

â‡’ 4a + 68d = 0

â‡’ 4(a + 17) = 0

â‡’ a + 17 = 0

â‡’ a_{18} = 0

It is not possible here to find both the first term and common difference. Since only one equation obtained from the given data.

**35.** Factorise: x^{3} – 5x^{2} – 2x + 24

**Solution:**

Let the given polynomial be p(x) = x^{3} – 5x^{2} – 2x + 24

Let us substitute various values of x until we get p(x) = 0.

p(0) = (0)^{3} – 5(0)^{2} – 2(0) + 24 = 24 â‰ 0

p(1) = (1)^{3} – 5(1)^{2} – 2(1) + 24 = 1 – 5 – 2 + 24 = 18 â‰ =

p(-2) = (-2)^{3} – 5(-2)^{2} – 2(-2) + 24 = -8 – 10 + 4 + 24 = 0

Therefore, x + 2 is a factor of p(x).

p(x) = (x + 2)(x^{2} – 7x + 12)

= (x + 2)[x^{2} – 3x – 4x + 12]

= (x + 2) [x(x – 3) – 4(x – 3)]

= (x + 2)(x – 3)(x – 4)

Hence, the factors of the given polynomial are (x + 2), (x – 3) and (x – 4).

**36. **Simplify: [1/(x^{2} + 3x + 2)] + [1/(x^{2} + 5x + 6)] – [2/(x^{2} + 4x + 3)]

**Solution:**

x^{2} + 3x + 2 = x^{2} + 2x + x + 2 = (x + 1)(x + 2)

x^{2} + 5x + 6 = x^{2} + 2x + 3x + 6 = (x + 2)(x + 3)

x^{2} + 4x + 3 = x^{2} + 3x + x + 3 = (x + 1)(x + 3)

Now,

[1/(x^{2}+ 3x + 2)] + [1/(x

^{2}+ 5x + 6)] – [2/(x

^{2}+ 4x + 3)]

= [1/ (x + 1)(x + 2)] + [1/ (x + 2)(x + 3)] – [2/ (x + 1)(x + 3)]

= [(x + 3) + (x + 1) – 2(x + 2)]/ [(x + 1)(x + 2)(x + 3)]

= [2x + 4 – 2x – 4]/ [(x + 1)(x + 2)(x + 3)]

= 0

**37.**

**Solution:**

Therefore, A(B + C) = AB + AC

**38. **Find the area of the quadrilateral whose vertices are (-4, 5), (0, 7), (5, -5) and (-4, -2).

**Solution:**

Given, vertices of the quadrilateral are (-4, 5), (0, 7), (5, -5) and (-4, -2).

Area of the quadrilateral

= 1/2 |(-28 – 0 – 10 – 20) – (0 + 35 + 20 + 8)|

= (1/2) |-58 – 63]|

= (1/2) Ã— 121

= 60.5 sq.units

**39.** State and prove Thales theorem.

**Solution:**

Statement:

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Proof:

In triangle ABC, a line parallel to side BC intersects other two sides namely AB and AC at D and E respectively.

Join BE and CD.

Also, draw DM âŠ¥ AC and EN âŠ¥ AB.

area of âˆ† ADE = 1/2 (AD Ã— EN)

ar(âˆ†ADE) = 1/2 (AD Ã— EN)

Similarly,

ar(âˆ†BDE) = 1/2 (DB Ã— EN)

ar(âˆ†ADE) = 1/2 (AE Ã— DM)

ar(âˆ†DEC) = 1/2 (EC Ã— DM)

Now,

ar(âˆ†ADE)/ar(âˆ†BDE) = [1/2 AD Ã— EN)]/ [1/2 (DB Ã— EN)] = AD/DB ….(i)

ar(âˆ†ADE)/ar(âˆ†DEC) = [1/2 (AE Ã— DM)]/ [1/2 (EC Ã— DM)] = AE/EC ….(i)

Triangle BDE and DEC are on the same base DE and between the same parallels.

Therefore, ar(âˆ†BDE) = ar(âˆ†DEC) ….(iii)

From (i), (ii) and (iii),

AD/DB = AE/EC

Hence proved.

BC and DE.

**40. **A person in a helicopter flying at a height of 700 m, observes two objects lying opposite to each other on either banks of a river. The angles of depression of the objects are 30Â° and 45Â°; find the width of the river. (âˆš3 = 1.732)

**Solution:**

Let A be the position of flight.

B and C be the positions of two objects.

In right triangle ADC,

tan 45Â° = AD/CD

1 = 700/CD

â‡’ CD = 700 m

In right triangle ADB,

tan 30Â° = AD/BD

1/âˆš3 = 700/BD

â‡’ BD = 700âˆš3 m

Width of the river = BD + CD = 700âˆš3 + 700

= 700(âˆš3 + 1)

= 700(1.732 + 1)

= 700 Ã— 2.732

= 1912.40 m

**41. **The radii of two circular ends of a frustum shaped bucket are 15 cm and 8 cm. If its depth is 63 cm, find the capacity of the bucket in litres. (Take Ï€ = 22/7)

**Solution:**

Given,

The radii of two circular ends of a frustum shaped bucket are 15 cm and 8 cm.

R = 15 cm and r = 8 cm

Depth = h = 63 cm

Volume of frustum shaped bucket = = (1/3)Ï€h[R^{2} + r^{2} + Rr]

= (1/3) Ã— (22/7) Ã— 63 Ã— [15^{2} + 8^{2} + (15)(8)]

= (66) Ã— [225 + 64 + 120]

= 66 Ã— 409

= 26994 cm^{3}

= 26.994 lit

Hence, the capacity of the bucket is 26.994 lit.

**42. **A circus tent is to be erected in the form of a cone surmounted on a cylinder. The total height of the tent is 49 m. Diameter of the base is 42 m and the height of the cylinder is 21 m. Find the cost of canvas needed to make the tent, if the cost of canvas is Rs. 12.50/m^{2}. (Take Ï€ = 22/7)

**Solution:**

Given,

Diameter of cylinder = 42 m

Radius = r = 21 m

Height of cylinder = h = 21 m

Curved surface area of cylinder = 2Ï€rh

= 2 Ã— (22/7) Ã— 21 Ã— 21

= 44 Ã— 62

= 2772 m^{2}

Radius of the cone = R = 21 m (given)

Height = H = 28 m

Slant height = l = âˆš(R^{2} + H^{2})

= âˆš(21^{2} + 28^{2})

= âˆš(441 + 784)

= âˆš1225

= 35 m

Curved surface area of cone = Ï€Rl

= (22/7) Ã— 21 Ã— 35

= 2310 m^{2}

Total surface area of circus tent = 2772 + 2310 = 5082 m^{2}

Cost of canvas 1 m^{2} = Rs. 12.50

Total cost of canvas = 5082 Ã— Rs. 12.50 = Rs. 63525

43. Calculate the standard deviation of the following data.

x | 3 | 8 | 13 | 18 | 23 |

f | 7 | 10 | 15 | 10 | 8 |

**Solution:**

x | f | d = x – 13 | d^{2} |
fd^{2} |

3 | 7 | -10 | 100 | 700 |

8 | 10 | -5 | 25 | 250 |

13 | 15 | 0 | 0 | 0 |

18 | 10 | 5 | 25 | 250 |

23 | 8 | 10 | 100 | 800 |

âˆ‘d = 50 | âˆ‘d^{2} = 2000 |

Ïƒ = âˆš(Î£fdÂ²/Î£f)

= âˆš(2000/50)

= âˆš40

= 6.325

Therefore, the standard deviation is 6.325.

**44.** A box contains 4 white balls, 6 red balls, 7 black balls and 3 blue balls. One ball is drawn at random from the bag. Find the probability that the ball drawn is :

(i) Â·neither white nor black

(ii) red or white, and

(iii) either white or red or black or blue

**Solution:**

Given,

4 white balls, 6 red balls, 7 black balls and 3 blue balls

n(S) = 4 + 6 + 7 + 3 = 20

(i) Number of balls other than white and black = 6 (red) + 3 (blue) = 9

P(neither white nor black) = 9/20

(ii) Probability of getting red ball = P(R) = 6/20

Probability of getting white ball = P(W) = 4/20

P(red or white ball) = P(R) + P(W)

= (6/20) + (4/20)

= 10/20

= Â½

(iii) Probability of getting black ball = 7/20

Probability of getting blue ball = 3/20

P(either white or red or black or blue) = (6/20) + (4/20) + (7/20) + (3/20)

= 20/20

= 1

**45. (a) **If the equation (1 + m^{2})x^{2} + 2mcx + c^{2} – a^{2} = 0 has equal roots, then prove that c^{2} = a^{2}(1 + m^{2})

**Solution:**

Given that the equation (1 + m^{2})x^{2} + 2mcx + c^{2} – a^{2} = 0 has equal roots.

Therefore, discriminant = 0

â‡’ (2mc)^{2} – 4(1 + m^{2})(c^{2} – a^{2}) = 0

â‡’ 4m^{2}c^{2} – 4[c^{2} – a^{2} + m^{2}c^{2} – m^{2}a^{2}] = 0

â‡’ 4m^{2}c^{2} – 4c^{2} + 4a^{2} – 4m^{2}c^{2} + 4m^{2}a^{2} = 0

â‡’ -4c^{2} + 4a^{2}(1 + m^{2}) = 0

â‡’ 4c^{2} = 4a^{2}(1 + m^{2})

â‡’ c^{2} = a^{2}(1 + m^{2})

Hence proved.

**OR**

**(b) **Find the equation of the straight line segment whose endpoints are the points of intersection of the straight lines 2x – 3y + 4 = 0, x – 2y + 3 = 0 and the midpoint of the line joining the points (3, -2) and (-5, 8).

**Solution:**

Given equations of lines are:

2x – 3y + 4 = 0â€¦.(i)

x – 2y + 3 = 0â€¦.(ii)

(ii) Ã— 2 – (i),

2x – 4y + 6 – (2x – 3y + 4) = 0

-y + 2 = 0

y = 2

Substituting y = 2 in (ii)

x – 2(2) + 3 = 0

x – 4 + 3 = 0

x = 1

Therefore, the point of intersection of lines = A(1, 2)

Midpoint of the line joining the points (3, -2) and (-5, 8)= B = [(3 – 5)/2, (-2 + 8)/2]

= (-2/2, 6/2)

= (-1, 3)

Equation of the line segment joining A and B is

(y – y_{1})/ (y_{2} – y_{1}) = (x – x_{1})/ (x_{2} – x_{1})

(y – 2)/ (3 – 2) = (x – 1)/ (-1 – 1)

(y – 2)/1 = (x – 1)/ (-2)

(-2)(y – 2) = x – 1

-2y + 4 = x – 1

â‡’ x + 2y – 1 – 4 = 0

â‡’ x + 2y – 5 = 0

**SECTION – IV**

**46. (a)** Draw the two tangents from a point which is 9 cm away from the centre-of a circle of radius 3 cm. Also, measure the lengths of the tangents.

**Solution:**

Length of the tangents = 8.4 cm

**OR**

**(b) **Construct a cyclic quadrilateral PQRS given PQ = 5 cm, QR = 4 cm, âˆ QPR = 35Â° and âˆ PRS = 70Â°.

**Solution:**

**47. (a) **Draw the graph of y = x^{2} + 2x – 3 and hence find the roots of x^{2} – x – 6 = 0.

**Solution:**

Given,

y = x^{2} + 2x – 3

x | -3 | -2 | -1 | 0 | 1 | 2 | 3 |

y | 0 | -3 | -4 | -3 | 0 | 5 | 12 |

Now, subtracting x^{2} – x – 6 = 0 from y = x^{2} + 2x – 3,

y – 0 = x^{2} + 2x – 3 – (x^{2} – x – 6)

y = 3x + 3

x | -2 | -1 | 0 | 1 | 2 | 3 |

y | -3 | 0 | 3 | 6 | 9 | 12 |

The graph of given equations intersect each other at (-2, -3) and (3, 12).

Hence, the solution set is (-2, -3) and (3, 12).

**OR**

**(b) **

Number of workers | 3 | 4 | 6 | 8 | 9 | 16 |

Number of days | 96 | 72 | 48 | 36 | 32 | 18 |

Draw the graph for the data given in the table. Hence find the number of days taken by 12 workers to complete the work.

**Solution:**

Hence, the number of days taken by 12 workers to complete the work is 24.