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Q:

If the radius of a circle is decreased by 50% , find the percentage decrease in its area.

A) 75%	B) 65%
C) 35%	D) 25%

Answer & Explanation Answer: A) 75%

Explanation:

let original radius = r and new radius = (50/100) r = r/2

original area = ${πr}^{2}$ and new area = $π {(r / 2)}^{2}$

decrease in area = $3 {πr}^{2} / 4$ * $\frac{1}{{πr}^{2}}$ *100 = 75%

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Q:

There are eight boxes of chocolates, each box containing distinct number of chocolates from 1 to 8. In how many ways four of these boxes can be given to four persons (one boxes to each) such that the first person gets more chocolates than each of the three, the second person gets more chocolates than the third as well as the fourth persons and the third person gets more chocolates than fourth person?

A) 70	B) 40
C) 72	D) 80

Answer & Explanation Answer: A) 70

Explanation:

All the boxes contain distinct number of chocolates.
For each combination of 4 out of 8 boxes, the box with the greatest number has to be given to the first person, the box with the second highest to the second person and so on.

The number of ways of giving 4 boxes to the 4 person is: 8 $C_{4}$ = 70

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Q:

If each side of a square is increased by 25%, find the percentage change in its area?

A) 65.25	B) 56.25
C) 65	D) 56

Answer & Explanation Answer: B) 56.25

Explanation:

let each side of the square be a , then area = $a^{2}$

As given that The side is increased by 25%, then

New side = 125a/100 = 5a/4

New area = ${(\frac{5 a}{4})}^{2}$

Increased area= $\frac{25 a^{2}}{16} - a^{2}$

Increase %= $\frac{[9 a^{2} / 16]}{a^{2}} * 100$ % = 56.25%

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Q:

The altitude drawn to the base of an isosceles triangle is 8cm and the perimeter is 32cm. Find the area of the triangle?

A) 50	B) 60
C) 70	D) 80

Answer & Explanation Answer: B) 60

Explanation:

let ABC be the isosceles triangle, the AD be the altitude

Let AB = AC = x then BC= 32-2x [because parameter = 2 (side) + Base]

since in an isoceles triange the altitude bisects the base so

BD = DC = 16-x

In a triangle ADC, ${(A C)}^{2} = {(A D)}^{2} + {(D C)}^{2}$

$x^{2} = 8^{2} + {(16 - x)}^{2}$ $\Rightarrow x = 10$

BC = 32-2x = 32-20 = 12 cm

Hence, required area = $\frac{1}{2} * B C * A D$ = $\frac{1}{2} * 12 * 10$ = 60 sq cm

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Q:

A tea expert claims that he can easily find out whether milk or tea leaves were added first to water just by tasting the cup of tea. In order to check this claims 10 cups of tea are prepared, 5 in one way and 5 in other. Find the different possible ways of presenting these 10 cups to the expert.

A) 340	B) 210
C) 290	D) 252

Answer & Explanation Answer: D) 252

Explanation:

Since there are 5 cups of each kind,prepared with milk or tea leaves added first,are identical hence,total number of different people ways of presenting the cups to the expert is 10!/(5! x 5!)= 252

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Q:

In measuring the sides of a rectangle, one side is taken 5% in excess and the other 4% in deficit. Find the error percent in the area, calculate from the those measurements.

A) .7%	B) 0.8%
C) 0.9%	D) 0.3%

Answer & Explanation Answer: B) 0.8%

Explanation:

let x and y be the sides of the rectangle then

correct area = $\frac{105}{100} x \times \frac{96}{100} y$

Error% = $\frac{504}{500} x y - x y = \frac{4}{500} x y$ %

$\frac{4}{500} x y \times \frac{1}{x y} \times 100 = \frac{4}{5} = 0.8$

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Q:

The length of a rectangle is twice its breadth if its length is decreased by 5cm and breadth is increased by 5cm, the area of the rectangle is increased by 75 sq cm. Find the length of the rectangle.

A) 30	B) 40
C) 50	D) 60

Answer & Explanation Answer: B) 40

Explanation:

let length = 2x and breadth = x then
(2x-5) (x+5) = (2x * x)+75
5x-25 = 75 => x=20
length of the rectangle = 40 cm

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Q:

A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4. In how many ways can they travel?

A) 126	B) 120
C) 146	D) 156

Answer & Explanation Answer: A) 126

Explanation:

There are 8 students and the maximum capacity of the cars together is 9.

We may divide the 8 students as follows

Case I: 5 students in the first car and 3 in the second
Case II: 4 students in the first car and 4 in the second

Hence, in Case I: 8 students are divided into groups of 5 and 3 in8C3 ways.

Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4ways.

Therefore, the total number of ways in which 8 students can travel is:
$8 C_{3} + 8 C_{4}$ =56 + 70= 126

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