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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 = πr2 and new area = πr/22

 


decrease in area =  3πr2/4*1πr2 *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: 8C4= 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 = a2 

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

New side = 125a/100 = 5a/4 

 

New area = 5a42  

 

Increased area= 25a216-a2 

 

Increase %=9a2/16a2*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, AC2=AD2+DC2 

x2=82+16-x2 x=10 

BC = 32-2x = 32-20 = 12 cm 

Hence, required area = 12*BC*AD12*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:

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:

In how many ways can the letters of the word EDUCATION be rearranged so that the relative position of the vowels and consonants remain the same as in the word EDUCATION?

A) 4! x 4! B) 5! x 5!
C) 4! x 5! D) 3! x 4!
 
Answer & Explanation Answer: C) 4! x 5!

Explanation:

The word EDUCATION is a 9 letter word, with none of the letters repeating.

The vowels occupy 3rd,5th,7th and 8th position in the word and the remaining 5 positions are occupied by consonants

As the relative position of the vowels and consonants in any arrangement should remain the same as in the word EDUCATION, the vowels can occupy only the afore mentioned 4 places and the consonants can occupy1st,2nd,4th,6th and 9th positions.

The 4 vowels can be arranged in the 3rd,5th,7th and 8th position in 4! Ways.

Similarly, the 5 consonants can be arranged in1st,2nd,4th,6th and 9th position in5! Ways.

Hence, the total number of ways = 4! × 5!

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Q:

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

A) 55% B) 65%
C) 75% D) 85%
 
Answer & Explanation Answer: C) 75%

Explanation:

Let original radius = R. 

 

New radius = 50100R50100R 

 

Original area =R2  and new area = πR2

 

3πR24*1πR2*100 

 

Decrease in area = πR22=πR24 = 75%

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