## Tamil Nadu SSLC or Class 10 Maths Question Paper 2017 with Solutions – Free Download

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## Download TN SSLC 2017 Question Paper

## Download TN SSLC 2017 Question Paper With Solutions

### Tamil Nadu Board SSLC Class 10 Maths 2017 Question Paper with Solutions

**SECTION – I**

**1. **If f(x) = x^{2} + 5, then f(-4) =

(a) 26

(b) 21

(c) 20

(d) -20

**Solution:**

Correct answer: (b)

Given,

f(x) = x^{2} + 5

f(-4) = (-4)^{2} + 5

= 16 + 5

= 21

**2. **If k + 2, 4k – 6, 3k – 2 are the three consecutive terms of an AP, then the value of k is

(a) 2

(b) 3

(c) 4

(d) 5

**Solution:**

Correct answer: (b)

Given,

k + 2, 4k – 6, 3k – 2 are the three consecutive terms of an AP.

2(4k – 6) = k + 2 + 3k – 2

8k – 12 = 4k

8k – 4k = 12

4k = 12

k = 12/4

k = 3

**3. **If the product of the first four consecutive terms of a GP is 256 and if the common ratio is 4 and the first term is positive, then its 3rd term is:

(a) 8

(b) 1/16

(c) 1/32

(d) 16

**Solution:**

Correct answer: (a)

Let a be the first term and r be the common ratio of GP.

Given, r = 4

Thus, a, 4a, 16a, and 64a are the first four consecutive terms of a GP.

a Ã— 4a Ã— 16a Ã— 64a = 256

a^{4} = 256/(64 Ã— 64)

a^{4} = (4/8)^{4}

â‡’ a = 4/8

â‡’ a = 1/2

Therefore, the third term = 16a = 16 Ã— (1/2) = 8

**4. **The remainder when x^{2} – 2x + 7 is divided by x + 4 is:

(a) 28

(b) 29

(c) 30

(d) 31

**Solution:**

Correct answer: (d)

Therefore, remainder = 31

**5. **The common root of the equations x^{2} – bx + c = 0 and x^{2} + bx – a = 0 is:

(a) (c + a)/2b

(b) (c – a)/2b

(c) (c + b)/2a

(d) (a + b)/2c

**Solution:**

Correct answer: (a)

x^{2} – bx + c = 0 and x^{2} + bx – a = 0 have a common root.

â‡’ x2 – bx + c = x2 + bx – a

â‡’ c + a = bx + bx

â‡’ c + a = 2bx

â‡’ x = (c + a)/2b

**6.**

**Solution:**

Correct answer: (c)

From the given,

**7. **Slope of the straight line which is perpendicular to the straight line joining the points (-2, 6) and (4, 8) is equal to:

(a) 1/3

(b) 3

(c) -3

(d) -1/3

**Solution:**

Correct answer: (c)

Let the given points be:

(x_{1}, y_{1}) = (-2, 6)

(x_{2}, y_{2}) = (4, 8)

Slope of the line joining the given points = m = (y_{2} – y_{1})/(x_{2} – x_{1})

= (8 – 6)/(4 + 2)

= 2/6

= â…“

Slope of the straight line which is perpendicular to the straight line joining the given points = -1/m

= -1/(â…“)

= -3

**8. **If the points (2, 5), (4, 6) and (a, a) are collinear, then the value of ‘a’ is equal to:

(a) -8

(b) 4

(c) -4

(d) 8

**Solution:**

Correct answer: (d)

Given, (2, 5), (4, 6) and (a, a) are collinear.

Thus, the area of the triangle formed by these points is 0.

â‡’[2(6 – a) + 4(a – 5) + a(5 – 6)] = 0

â‡’ 12 – 2a + 4a – 20 + 5a – 6a = 0

â‡’ a – 8 = 0

â‡’ a = 8

**9. **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 cm

**Solution:**

Correct answer: (d)

Given,

Perimeters of two similar triangles are 24 cm and 18 cm respectively.

Length of side of the first triangle = 8 cm

Let x be the side of another triangle.

Ratio of perimeters = Ratio of the corresponding sides

â‡’ 24/18 = 8/x

â‡’ 4/3 = 8/x

â‡’ x = 24/4

â‡’ x = 6

Therefore, the corresponding side of the second triangle is 6 cm.

**10. **âˆ†ABC is a right angled triangle where âˆ B = 90Â° and BD âŠ¥ AC. If BD = 8 cm, AD = 4 cm, then CD is:

(a) 24 cm

(b) 16 cm

(c) 32 cm

(d) 8 cm

**Solution:**

Correct answer: (b)

Given,

Î”DBA âˆ¼ Î”DCB

Thus,

BD/CD = AD/BD

BDÂ² = AD Ã— DC

(8)Â² = 4 Ã— DC

64 = 4DC

DC = 64/4

CD = 16 cm

**11. **In the adjoining figure, âˆ ABC =

(a) 45Â°

(b) 30Â°

(c) 60Â°

(d) 50Â°

**Solution:**

Correct answer: (c)

Let âˆ ABC = Î¸

tan Î¸ = AC/AB

tan Î¸ = 100âˆš3/100

tan Î¸ = âˆš3

tan Î¸ = tan 60Â°

â‡’ Î¸ = 60Â°

**12. **9 tan^{2}Î¸ – 9 sec^{2}Î¸ =

(a) 1

(b) 0

(c) 9

(d) -9

**Solution:**

Correct answer: (d)

9 tan^{2}Î¸ – 9 sec^{2}Î¸ = 9(tan^{2}Î¸ – sec^{2}Î¸)

= -9(sec^{2}Î¸ – tan^{2}Î¸)

= -9(1)

= -9

**13. **If the surface area of sphere is 100Ï€ cm^{2}, then its radius is equal to:

(a) 25 cm

(b) 100 cm

(c) 5 cm

(d) 10 cm

**Solution:**

Correct answer: (c)

Given,

Surface area of sphere = 100Ï€ cm^{2}

4Ï€r^{2} = 100Ï€

r^{2} = 100/4

r^{2} = 25

r = 5 cm

**14. **Standard deviation of a collection of a data is 2âˆš2. If each value is multiplied by 3, then the standard deviation of the new data is:

(a) âˆš12

(b) 4âˆš2

(c) 6âˆš2

(d) 9âˆš2

**Solution:**

Correct answer: (a)

Given,

Standard deviation = 2âˆš2

Each observation is multiplied by 3, then new standard deviation = 3 Ã— (2âˆš2) = 6âˆš2

**15. **A card is drawn from a pack of 52 cards at random. The probability of getting neither an ace nor a king card is:

(a) 2/13

(b) 11/13

(c) 4/13

(d) 8/13

**Solution:**

Correct answer: (b)

Total number of outcomes = 52

Number of ace cards = 4

Number of king cards = 4

Number of cards other than ace and king = 52 – (4 + 4) = 44

Therefore, P(neither an ace nor a king) = 44/52 = 11/13

**SECTION – II**

**16.** Given, A = {1, 2, 3, 4, 5}, B = {3, 4, 5, 6}, and C = {5, 6, 7, 8}, show that A â‹ƒ (B â‹ƒ C) = (A â‹ƒ B) â‹ƒ C.

**Solution:**

Given,

A = {1, 2, 3, 4, 5}, B = {3, 4, 5, 6}, and C = {5, 6, 7, 8}

B â‹ƒ C = {3, 4, 5, 6} â‹ƒ {5, 6, 7, 8}

= {3, 4, 5, 6, 7, 8}

A â‹ƒ (B â‹ƒ C) = {1, 2, 3, 4, 5} â‹ƒ {3, 4, 5, 6, 7, 8}

= {1, 2, 3, 4, 5, 6, 7, 8} â€¦..(i)

A â‹ƒ B = {1, 2, 3, 4, 5} â‹ƒ {3, 4, 5, 6}

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

(A â‹ƒ B) â‹ƒ C = {1, 2, 3, 4, 5, 6} â‹ƒ {5, 6, 7, 8}

= {1, 2, 3, 4, 5, 6, 7, 8} â€¦..(ii)

From (i) and (ii),

A â‹ƒ (B â‹ƒ C) = (A â‹ƒ B) â‹ƒ C

**17. **The following table represents a function from A = {5, 6, 8, 10} to B = {19, 15, 9, 11} where f(x) = 2x – 1. Find the values of a and b.

x |
5 |
6 |
8 |
10 |

f(x) |
a |
11 |
b |
19 |

**Solution:**

Given,

y = f(x) = 2x – 1

a = f(5) = 2(5) – 1 = 10 – 1 = 9

b = f(8) = 2(8) – 1 = 16 – 1 = 15

Therefore, a = 9 and b = 15.

**18. **-2/7, m, -7/2(m + 2) are in GP, find the values of m.

**Solution:**

Given,

-2/7, m, -7/2(m + 2) are in GP.

If a, b, c are in GP, then b^{2} = ac.

â‡’ m^{2} = (-2/7) Ã— (-7/2)(m + 2)

â‡’ m^{2} = m + 2

â‡’ m^{2} – m – 2 = 0

â‡’ m^{2} – 2m + m – 2 = 0

â‡’ m(m – 2) + 1(m – 2) = 0

â‡’ (m – 2)(m + 1) = 0

â‡’ m = 2, -1

**19. **Solve by elimination method:

13x + 11y = 70

11x + 13y = 74

**Solution:**

Given,

13x + 11y = 70….(i)

11x + 13y = 74….(ii)

Adding (i) and (ii),

24x + 24y = 144

24(x + y) = 144

x + y = 6….(iii)

Subtracting (ii) from (i),

2x – 2y = -4

x – y = -2….(iv)

Adding (iii) and (iv),

x + y + x – y = 6 – 2

2x = 4

x = 2

Substituting x = 2 in (iii),

2 + y = 6

y = 6 – 2 = 4

Hence, the required solution is x = 2 and y = 4, i.e. (x, y) = (2, 4).

**20.** Simplify: (6x^{2} + 9x)/ (3x^{2} – 12x)

**Solution:**

(6x^{2} + 9x)/ (3x^{2} – 12x)

= [3x(2x + 3)]/ [3x(x – 4)]

= (2x + 3)/ (x – 4)

**21. **Construct a 2 x 2 matrix A = [a_{ij}] whose elements are given by a_{ij} = 2i – j.

**Solution:**

a_{ij} = 2i – j

a_{11} = 2(1) – 1 = 2 – 1 = 1

a_{12} = 2(1) – 2 = 2 – 2 = 0

a_{21} = 2(2) – 1 = 4 – 1 = 3

a_{22} = 2(2) – 2 = 4 – 2 = 2

**22.**

**Solution:**

**23. **Find the coordinates of the point which divides the line segment joining (-3, 5) and (4, -9) in the ratio 1 : 6 internally.

**Solution:**

Let P(x, y) divide the line segment joining the points A(-3, 5) and B(4, -9) internally in the ratio 1 : 6.

Here,

(x_{1}, y_{1}) = (-3, 5)

(x_{2}, y_{2}) = (4, -9)

l : m = 1 : 6

P(x, y) = [(lx_{2} + mx_{1})/ (l + m), (ly_{2} + my_{1})/ (l + m)]

= [(4 – 18)/(1 + 6), (-9 + 30)/(1 + 6)]

= (-14/7, 21/7)

= (-2, 3)

**24. **“The points (0, a), a > 0 lie on x-axis for all a”. Justify the truthness of the statement.

**Solution:**

For all a > 0, the points like (0, a) will exist on the positive side of the y-axis.

Therefore, the given statement is false.

**25. **In âˆ†PQR, AB || QR. If AB is 3 cm, PB is 2 cm and PR is 6 cm, then find the length of QR.

**Solution:**

Given that, in âˆ†PQR, AB || QR.

AB = 3 cm

PB = 2 cm

PR = 6 cm

In Î”PAB and Î”PQR,

âˆ PAB = âˆ PQR (corresponding angles)

âˆ P = âˆ P (common)

By AA similarity,

Î”PAB ~ Î”PQR

By BPT,

AB/QR = PB/PR

QR = (AB Ã— PR)/PB

= (3 Ã— 6)/2

= 9 cm

**26.** The angle of elevation of the top of a tower as seen by an observer is 30Â°. The observer is at a distance of 30âˆš3 m from the tower. If the eye level of the observer is 1.5 m above the ground level, then find the height of the tower.

**Solution:**

Let BD be the height of the tower.

AE be the distance between the eye level of the observer and the ground level.

AB = CE = 30âˆš3 m

AE = BC = 1.5 m

In right triangle DEC,

tan 30 = CD/EC

1/âˆš3 = CD/30âˆš3

â‡’ CD = 30âˆš3/âˆš3

â‡’ CD = 30 m

Therefore, height of the tower = BD

= BC + CD

= 1.5 + 30

= 31.5 m

**27. **The total surface area of a solid right circular cylinder is 1540 cm^{2}. If the height is four times the radius of the base, then find the height of the cylinder.

**Solution:**

Let r be the radius of the base and h be the height of the cylinder.

Given,

h = 4r

Total surface area = 1540 cm^{2}

2Ï€r(r + h) = 1540

2 Ã— (22/7) Ã— r(r + 4r) = 1540

r(5r) = (1540 Ã— 7)/ (22 Ã— 2)

5r^{2} = 35 Ã— 7

r^{2} = 7 Ã— 7

r = 7 m

Height of the cylinder = h = 4r = 4(7) = 28 cm

**28. **The smallest value of a collection of data is 12 and the range is 59. Find the largest value of the collection of data.

**Solution:**

Given,

Smallest number = 12

Range = 59

We know that,

Range = Largest number – Smallest number

59 = Largest number – 12

â‡’ Largest number = 59 + 12 = 71

**29.** In tossing a fair coin twice, find the probability of getting:

(i) Two heads

(ii) Exactly one tail

**Solution:**

Sample space = S = {HH, HT, TH, TT}

n(S) = 4

(i) Let A be the event of getting two heads.

A = {HH}

Number of outcomes favourable to A = n(A) = 1

P(A) = n(A)/n(S)

= 1/4

(ii) Let B be the event of getting exactly one tail.

B = {HT, TH}

Number of outcomes favourable to B = n(B) = 2

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

= 2/4

= 1/2

**30. (a) **If the volume of a solid sphere is 7241 1/7 cu.cm, then find its radius. (take Ï€ = 22/7)

**Solution:**

Let r be the radius of a solid sphere.

Given,

Volume of sphere = 7241 1/7 cu.cm

(4/3)Ï€r^{3} = 50688/7

(4/3) Ã— (22/7) Ã— r^{3} = 50688/7

r^{3} = (50688 Ã— 3)/ (4 Ã— 22)

r^{3} = 1728

r^{3} = 43 Ã— 33

â‡’ r = 4 Ã— 3

â‡’ r = 12

Hence, the radius of the sphere = 12 cm

**OR**

**(b) **If x = a sec Î¸ + b tan Î¸ and y = a tan Î¸ + b sec Î¸, then prove that: x^{2} – y^{2} = a^{2} – b^{2}

**Solution:**

Given,

x = a sec Î¸ + b tan Î¸

y = a tan Î¸ + b sec Î¸

LHS = x^{2} – y^{2}

= (a sec Î¸ + b tan Î¸)^{2} – (a tan Î¸ + b sec Î¸)^{2}

= a^{2} sec^{2}Î¸ + b^{2} tan^{2}Î¸ + 2ab sec Î¸ tan Î¸ – a^{2} tan^{2}Î¸ – b^{2} sec^{2}Î¸ – 2ab tan Î¸ sec Î¸

= a^{2}(sec^{2}Î¸ – tan^{2}Î¸) – b^{2}(sec^{2}Î¸ – tan^{2}Î¸)

= a^{2}(1) – b^{2}(1)

= a^{2} – b^{2}

= RHS

Hence proved.

**SECTION – III**

**31. **Let A = {a, b, c, d, e, f, g, x, y, z}, B = {1, 2, c, d, e} and C = {d, e, f, g, 2, y}. Verify A\ (B â‹ƒ C) = (A\B) â‹‚ (A\C).

**Solution:**

Given,

A = {a, b, c, d, e, f, g, x, y, z}, B = {1, 2, c, d, e} and C = {d, e, f, g, 2, y}

B â‹ƒ C = {1, 2, c, d, e} â‹ƒ {d, e, f, g, 2, y}

= {1, 2, c, d, e, f, g}

A\ (B â‹ƒ C) = {a, b, c, d, e, f, g, x, y, z} \ {1, 2, c, d, e, f, g}

= {a, b, x, z}….(i)

A\B = {a, b, c, d, e, f, g, x, y, z} \ {1, 2, c, d, e}

= {a, b, f, g, x, y, z}

A\C = {a, b, c, d, e, f, g, x, y, z} \ {d, e, f, g, 2, y}

= {a, b, c, x, z}

(A\B) â‹‚ (A\C) = {a, b, f, g, x, y, z} â‹‚ {a, b, c, x, z}

= {a, b, x, z}….(ii)

From (i) and (ii),

A\ (B â‹ƒ C) = (A\B) â‹‚ (A\C)

Hence verified.

**32. **Let A = {6, 9, 15, 18, 21}, B = {1, 2, 4, 5, 6} and f : A â†’ B be defined by f(x) = (x – 3)/3. Represent f by:

(i) an arrow diagram

(ii) a set of ordered pairs

(iii) a table

(iv) a graph

**Solution:**

Given,

f(x) = (x – 3)/3

f(6) = (6 – 3)/3 = 3/3 = 1

f(9) = (9 – 3)/3 = 6/3 = 2

f(15) = (15 – 3)/3 = 12/3 = 4

f(18) = (18 – 3)/3 = 15/3 = 5

f(21) = (21 – 3)/3 = 18/3 = 6

(i) An arrow diagram:

(ii) The set of ordered pairs

f = {(6, 1), (9, 2), (15, 4), (18, 5), (21, 6)}

(iii) Table

x |
6 |
9 |
15 |
18 |
21 |

f(x) |
1 |
2 |
4 |
5 |
6 |

(iv) Graph:

**33.** Find the sum of the first 2n terms of the series 1^{2} – 2^{2} + 3^{2} – 4^{2} +….

**Solution:**

Given,

1^{2} – 2^{2} + 3^{2} – 4^{2} +….

= 1 – 4 + 9 – 16 + 25 – ….. to 2n terms

= (1 – 4) + (9 – 16) + (25 – 36) + …..to n terms (after grouping)

= -3 + (-7) + (-11) + ….. n terms

This is an AP with a = -3, d = -4

Hence, the required sum = n/2 [2a + (n – 1)d]

= n/2 [2(-3) + (n – 1)(-4)]

= (n/2) [-6 – 4n + 4]

= (n/2) [-4n – 2]

= (-2n/2) (2n + 1)

= -n(2n + 1)

**34. **Find the sum of the first n terms of the series 7 + 77 + 777 +…

**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

**35. **The speed of a boat in still water is 15 km/hr. It goes 30 km upstream and returns downstream to the original point in 4 hrs.30minutes. Find the speed of the stream.

**Solution:**

Given,

Speed in still water = 15 km/hr

Total time taken = 4 hrs 30 min = 4 1/2 hr = 9/2 hrs

Let x km/hr be the speed of the stream.

Speed of the boat during upstream = 15 – x

Speed of the boat during downstream = 15 + x

Time taken for upstream = T_{1} = 30/(15 – x)

Time taken for downstream = T_{2} = 30/(15 + x)

According to the given,

T_{1} + T_{2} = 9/2

30 [(15 + x + 15- x)/ (15 – x)(15 + x)] = 9/2

30/(225 – x^{2}) = (9/ 2 Ã— 30)

30/(225 – x^{2}) = 3/20

â‡’ 600 = 675 – 3x^{2}

â‡’ 3x^{2} = 675 – 600

â‡’ x^{2} = 75/3

â‡’ x^{2} = 25

â‡’ x = 5

Therefore, speed of the stream = 5 km/hr

**36.** Find the value of a and b if 16x^{4} – 24x^{3} + (a – 1)x^{2} + (b + 1)x + 49 is a perfect square.

**Solution:**

Given that, the polynomial is a perfect square.

Therefore, remainder = 0

b + 1 + [(a – 10)/8] Ã— 6 = 0 ….(i)

49 – [(a – 10)/8]^{2} = 0

â‡’ [(a – 10)/8]^{2} = 49

â‡’ (a – 10)/8 = 7

â‡’ a = 10 = 56

â‡’ a = 56 + 10 = 66

Substituting a = 66 in (i),

b + 1 + [(66 – 10)/8] Ã— 6 = 0

b + 1 + (56/8) Ã— 6 = 0

b + 1 + 42 = 0

b = -43

Therefore, a = 66 and b = -43.

**37.**

**Solution:**

Given,

Hence, (AB)^{T} = B^{T}A^{T}

**38. **Find the area of the quadrilateral formed by the points (-4, -2), (-3, -5), (3, -2) and (2, 3).

**Solution:**

Given, vertices of the quadrilateral are (-4, -2), (-3, -5), (3, -2) and (2, 3).

Area of the quadrilateral

= (1/2) [(20 + 6 + 9 – 4) – (6 – 15 – 4 – 12)]

= (1/2) [31 + 25]

= (1/2) Ã— 56

= 28 sq.units

**39. **State and prove Pythagoras theorem.

**Solution:**

Pythagoras theorem – statement:

In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Given:

In a right triangle ABC, âˆ B = 90Â°

To prove:

AC^{2} = AB^{2} + BC^{2}

Construction:

Draw a perpendicular BD onto the side AC.

We know that,

â–³ADB ~ â–³ABC

Therefore, AD/AB = AB/AC (by similarity)

AB^{2} = AD Ã— AC …..(i)

Also, â–³BDC ~â–³ABC

Therefore, CD/BC = BC/AC (by similarity)

BC^{2} = CD Ã— AC ….(ii)

Adding (i) and (ii),

AB^{2} + BC^{2} = AD Ã— AC + CD Ã— AC

AB^{2} + BC^{2} = AC (AD + CD)

Since, AD + CD = AC

Therefore, AC^{2} = AB^{2} + BC^{2}

Hence, the Pythagorean theorem is proved.

**40. **A flag post stands on the top of a building. From a point on the ground, the angles of elevation of the top and bottom of the flag post are 60Â° and 45Â° respectively. If the height of the flag post is 10 m, find the height of the building. (âˆš3 = 1.732)

**Solution:**

Let AB be the flag post and BC be the building.

In right triangle BCD,

tan 45Â° = BC/CD

1 = h/x

â‡’ h = x ….(i)

In right triangle ACD,

tan 60Â° = AC/CD

âˆš3 = (10 + h)/x

âˆš3x = 10 + h

âˆš3h = 10 + h [From (i)]

(âˆš3 – 1)h = 10

h = 10/(âˆš3 – 1)

h = [10/(âˆš3 – 1)] [(âˆš3 + )/ (âˆš3 + 1)]

h = 10(âˆš3 + 1)/ (3 – 1)

h = 10(1.732 + 1)/2

= 5 (2.732)

= 13.66 m

Therefore, the height of the building is 13.66 m.

**41. **The perimeter of the ends of a frustum of a cone are 44 cm and 8.4Ï€ cm. If the depth is 14 cm, then find its volume.

**Solution:**

Given,

Depth of the frustum of a cone = h = 14 cm

Perimeter of circular base with radius R = 44 cm

2Ï€R = 44

(22/7) R = 22

R = 7 cm

Perimeter of circular end with radius r = 8.4Ï€ cm

2Ï€r = 8.4Ï€

r = 8.4/2

r = 4.2 cm

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

= (1/3) Ã— (22/7) Ã— 14 Ã— [7^{2} + 4.2^{2} + (7)(4.2)]

= (44/3) Ã— [49 + 17.64 + 29.4]

= (44/3) Ã— 96.04

= 1408.58 cm^{3}

**42.** The length, breadth and height of a solid metallic cuboid are 44 cm, 21 cm and 12 cm respectively. It is melted and a solid cone is made out of it. If the height of the cone is 24 cm, then find the diameter of its base.

**Solution:**

Given, dimensions of the cuboid are:

Length = l = 44 cm

Breadth = b = 21 cm

Height = h = 12 cm

Let r be the radius of the cone.

Height of the cone = H = 24 cm (given)

Volume of cone = Volume of cuboid

(1/3)Ï€r^{2}H = lbh

(1/3) Ã— (22/7) Ã— r^{2} Ã— 24 = 44 Ã— 21 Ã— 21

r^{2} = (44 Ã— 21 Ã— 12 Ã— 7 Ã— 3)/ (22 Ã— 24)

r^{2} = 7 Ã— 3 Ã— 7 Ã— 3

r^{2} = (7 Ã— 3)^{2}

r = 21 cm

Diameter of the base of cone = 2r = 2(21) = 42 cm

**43. **Find the coefficient of variation of the following data.

18, 20, 15, 12, 25

**Solution:**

Given,

18, 20, 15, 12, 25

Mean = (18 + 20 + 15 + 12 + 25)/5

= 90/5

= 18

x |
d = x – 18 |
d |

18 |
0 |
0 |

20 |
2 |
4 |

15 |
-3 |
9 |

12 |
-6 |
36 |

25 |
7 |
49 |

âˆ‘d = 0 |
âˆ‘d |

Standard deviation (Ïƒ) = âˆš(âˆ‘d^{2}/n)

= âˆš(98/5)

= âˆš19.6

= 4.428

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

= (4.428/18) Ã— 100

= 442.8/18

= 24.6

**44. **If a die is rolled twice, find the probability of getting an even number in the first time or a total of 8.

**Solution:**

Total number of outcomes = n(S) = 6^{2} = 36

Let A be the event of getting an even number for the first time.

A = {(2, 1), (2, 2), (2, 3) (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

n(A) = 18

P(A) = n(A)/n(S) = 18/36

Let B be the event of getting a sum of numbers on dice 8.

B = {(2, 6), (6, 2), (3, 5), (5, 3), (4, 4)}

n(B) = 5

P(B) = n(B)/n(S) = 5/36

A â‹‚ B = {(2, 6), (6, 2), (4, 4)}

n(A â‹‚ B) = 3

P(A â‹‚ B) = n(A â‹‚ B)/n(S) = 3/36

P(A â‹ƒ B) = P(A) + P(B) – P(A â‹‚ B)

= (18/36) + (5/36) – (3/36)

= (18 + 5 – 3)/36

= 20/36

= 5/9

Hence, the required probability is 5/9.

**45. (a)** Find the GCD of the following polynomials 3x^{4} + 6x^{3} – 12x^{2} – 24x and 4x^{4} + 14x^{3} + 8x^{2} – 8x.

**Solution:**

Let f(x) = 3x^{4} + 6x^{3} – 12x^{2} – 24x

= 3x(x3 + 2×2 – 4x – 8)

g(x) = 4x^{4} + 14x^{3} + 8x^{2} – 8x

= 2x(2x^{3} + 7x^{2} + 4x – 4)

Now, let us find the GCD of the below polynomials.

x^{3} + 2x^{2} – 4x – 8 and 2x^{3} + 7x^{2} + 4x – 4

Thus, the common factor of x^{3} + 2x^{2} – 4x – 8 and 2x^{3} + 7x^{2} + 4x – 4 is x^{2} + 4x + 4.

Also, the common factor of 3x and 2x is x.

Therefore, the required GCD the given polynomials = x(x^{2} + 4x + 4).

**OR**

**(b) **A straight line cuts the coordinate axes at A and B. If the midpoint of AB is (3, 2), then find the equation of AB.

**Solution:**

Let A(a, 0) and B(0, b) the points on the x-axis and y-axis respectively.

Given that, (3, 2) is the midpoint of AB.

M(3, 2) = Midpoint of AB

(3, 2) = [(a + 0)/2, (0 + b)/2]

(3, 2) = (a/2, b/2)

â‡’ a = 6 and b = 4

The equation of line passing through A and B is (x/a) + (y/b) = 1

â‡’ (x/6) + (y/4) = 1

â‡’ (2x + 3y)/12 = 1

â‡’ 2x + 3y = 12

Hence, the required equation is 2x + 3y – 12 = 0

**SECTION – IV**

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

**Solution:**

Measure of length of the tangents = PA = PB = 8 cm

**OR**

**(b) **Construct a cyclic quadrilateral ABCD, given AB = 6 cm, âˆ ABC = 70Â°, BC = 5 cm and âˆ ACD = 30Â°.

**Solution:**

**47. (a) **Solve graphically 2x^{2} + x – 6 = 0

**Solution:**

Given equation is:

2x^{2} + x – 6 = 0

Let y = 2x^{2} + x – 6

Plot the ordered pair of points (â€“3, 9), (â€“2, 0), (â€“1, â€“5), (0, â€“6), (1, â€“3), (2, 4) and (3, 15) on the graph.

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

x2 |
9 |
4 |
1 |
0 |
1 |
4 |
9 |

2×2 |
18 |
8 |
2 |
0 |
2 |
8 |
18 |

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

-6 |
-6 |
-6 |
-6 |
-6 |
-6 |
-6 |
-6 |

y |
9 |
0 |
-5 |
-6 |
-3 |
4 |
15 |

The curve in the graph, cuts the x-axis at the points (â€“2, 0) and (1.5, 0).

Hence, the solution set is x = -2 and x = 1.5.

**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.

**Solution:**

Given, xy = 20

x |
1 |
2 |
4 |
5 |

y |
20 |
10 |
5 |
4 |

From the above table, it is clear that as x increases y decreases. This type of variation is called indirect variation.

y Î± 1/x

or

xy = k, where k = proportionality constant

Now,

1 Ã— 20 = 2 Ã— 10 = 4 Ã— 5 = 20 = k

Thus, k = 20

Plot the points (1, 20), (2, 10), (4, 5) and (5, 4) in the graph.

When x = 5, y = 4

When y = 10, x = 2