**Problem 1 :**

A rubber ball dropped from a height of 50 m rebounds at every impact from the floor to a height half of that from which is has fallen. Find the distance described by the time it comes to rest.

**Solution :**

Distance described in the 1^{st} impact = 50 m

Distance described in the 2^{nd} impact = 25

Distance described in the 3^{rd} impact = 25/2

Distance described in the 4^{rth }impact = 25/4

Distance described by the time it comes to rest

= 50 + [25 + (25/2) + (25/4) + ............]

Sum of infinite geometric series = a/(1 - r)

a = 25 r = t_{2}/t_{1 }==> 1/2

S_{∞} = a/(1 - r)

= 25/(1/2)

= 50

Distance described by the time it comes to rest

= 50 + 2(50)

= 50 + 100

= 150

So, the correct answer for this question is 150 m.

**Problem 2 :**

A work was assigned to me on a Tuesday. I completed to work after 72 days. On what day, i completed the work.

**Solution :**

Tuesday corresponds to 2 in the clock.

Then, 2 + 72 = 74

When we divide 74 by 7, we get 4 as remainder.

4 corresponds to Thursday.

So, i completed the work on Thursday.

**Problem 3 :**

The radius of the top of a bucket is 18 cm and that of the bottom is 6 cm.Its depth is 24 cm. Determine the capacity of the bucket.

**Solution :**

Volume of frustum cone = (1/3) π h (R^{2} + r ^{2} + R r)

= (1/3) π (24) (18^{2} + 6 ^{2} + (18)(6))

= 8π(324 + 36 + 108)

= 8π(468)

= 3744 π cm^{3}

So, the correct answer is 3744 π cm^{3}

**Problem 4 :**

A solid metal cylinder is 20 cm in height and has a radius of 1.5 cm. This is melted down and cast into spheres each of radius 1.5 cm.How many spheres each of radius 1.5 cm can be cast from the cylinder ?

**Solution :**

Height of the cylinder(h) = 20 cm

Radius of cylinder ( R) = 1.5 cm

Radius of sphere (r) = 1.5 cm

Volume of cylinder = n ⋅ Volume of sphere

π R^{2} h = n x (4/3) ⋅ π r^{3}

(1.5)^{2} ⋅ 20 = n ⋅ (4/3) ⋅ (1.5)^{3}

(1/1.5) ⋅ (3/4) ⋅ 20 = n

n = (1/1.5) ⋅ 3 ⋅ 5

n = (5/0.5)

n = 10

**Problem 5 :**

In a higher secondary class, 66 play football,56 play hockey, 63 play cricket, 27 play both foot ball and hockey, 25 play hockey and cricket, 23 play cricket and foot ball and 5 do not play any game. if the strength of the class is 130. Calculate the number who play all the three games.

**Solution :**

Let A, B and C be the number of students who play foot ball, hockey and cricket respectively.

Number of students who play foot ball = 66

Number of students who play hockey = 56

Number of students who play cricket = 63

Number of students who play football and hockey = 27

Number of students who play hockey and cricket = 25

Number of students who play cricket and football = 23

Number of students who do not play any game = 5

Let x be the number of students who play all the three game.

Number of students who play any one of the games =

16+x+27-x+x+23-x+25-x+4+x+15+x

125 = 16 + 27 + 23 + 25 + 4 + 15 + x

x = 125 - 110

x = 15

Number of students who play all the three games = 15.

**Problem 6 :**

Determine the value of m if x+1 is a factor of

x^{3} + mx^{2} + 19x + 12

**Solution :**

Let p(x) = x^{3} + mx^{2} + 19 x + 12

x + 1 = 0

x = -1

Since(x + 1) is a factor p(-1) = 0

p(-1) = (-1)^{3} + m (-1)^{2} + 19 (-1) + 12

0 = -1 + m - 19 + 12

0 = -20 + 12 + m

0 = -8 + m

m = 8

So, the value of m is 8.

**Problem 7 :**

Find the square root of

(x^{2} - 4) (x^{2} + x - 6) (x^{2 }+ 5x + 6)

**Solution :**

= √(x^{2} - 4) (x^{2} + x - 6) (x^{2} + 5 x + 6)

= √(x^{2} - 2^{2}) (x + 3) (x - 2) (x + 2)(x + 3)

= √(x + 2) (x - 2)(x + 3) (x - 2) (x + 2)(x + 3)

= (x + 2) (x - 2)(x + 3)

So the correct answer is (x + 2) (x - 2)(x + 3).

**Problem 8 :**

The outer dimension of a bordered table are 72 cm and 108 cm. If the area of a table, excluding the border is 6400 cm^{2}, how wide is the border ?

**Solution :**

Let x be the width of the border.

Outer length of table = 108 cm

Width of the table = 72 cm

Area of the table including the border

= 108(72) - (108-2x)(72-2x)

6400 = 108(72) - [108(72) - 108(2x) - 2x(72) + 2x^{2})]

6400 = 108(72) - 108(72) - 216x - 144x + 2x^{2}

2x^{2} - 360x = 6400

x^{2} - 180x - 3200 = 0

(x - 160) (x - 20) = 0

x = 160 and x = 20

**Problem 9 :**

Find the area of the triangle whose vertices are

(4, 5) (4, 2) (-2, 2).

**Solution :**

Now we should take anticlockwise direction. So, we have to take the points in the following order.

A (4, 5) C (-2, 2) and B (4, 2)

Area of the triangle

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

x_{1} = 4 x_{2} = -2 x_{3} = 4

y_{1} = 5 y_{2} = 2 y_{3} = 2

Area of the triangle ACB

= 1/2 [4(2-2) + (-2)(2-5) + 4(5-2)]

= 1/2 [(-2)(-3) + 4(3)]

= 1/2 {0 + 6 + 12}

= 18/2

= 9 square units

So, required area of the triangle is 9 square units.

**Problem 10 :**

Find the equation of the straight line which joins the points A (5,1) and B (-2,2)

**Solution :**

Equation of the line joining two points

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

x_{1} = 5, x_{2} = -2, y_{1} = 1 and y_{2} = 2

(y - 1)/(2 - 1) = (x - 5)/(-2 - 5)

(y - 1)/1 = (x - 5)/(-7)

-7 (y - 1) = 1 (x - 5)

-7 y + 7 = x - 5

x + 7 y - 5 - 7 = 0

x + 7 y - 12 = 0

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