Percentages Quiz Set 001

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Question 1

If 82 is m% of p, and B is p% of m, then what is B?

 A

82.

 B

92.

 C

72.

 D

102.

Soln.
Ans: a

Let 82 = A, so A = $m/100$ × p = $p/100$ × m. = B, so A and B are same.


Question 2

A number is first increased by 60%, and thereafter it is reduced by 60% again. What is the overall reduction?

 A

36 %.

 B

38 %.

 C

34 %.

 D

40 %.

Soln.
Ans: a

Let us derive the shortcut formula first, so that you can remember it and use it when the need arises. Suppose the number is 100, and let the increase be x%. The number becomes 100 + x. When it is reduced by x%, we get 100 + x - $((100 + x) × x)/100$ = 100 + x - $(x + x^2/100)$ = 100 - $x^2/100$, hence a reduction of $x^2/100$ = 36%


Question 3

If 20% of X is Y, then Y% of 20 is?

 A

4 % of X.

 B

14 % of X.

 C

34 % of X.

 D

24 % of X.

Soln.
Ans: a

We have Y = $(20 × X)/100$, so Y% of 20 is same as $(20 × X)/100$% × 20, which can be rearranged to $(20 × 20)/100$% × X = 4% of X.


Question 4

A carton contains 3000 led bulbs, some of them are red, some green and others are blue. How many are green if 9% are red and 5% are blue?

 A

2580.

 B

2680.

 C

2480.

 D

2780.

Soln.
Ans: a

The percentage of green is 100 - (9 + 5) = 86. So number of green bulbs = $86/100$ × 3000 = 2580.


Question 5

A number is first increased by 80%. By what percent must it be reduced so as to restore it to its previous value?

 A

$44{4/9}$ %.

 B

$51{1/8}$ %.

 C

$35{6/11}$ %.

 D

$38{9/11}$ %.

Soln.
Ans: a

Let us derive the shortcut formula first, so that you can remember it and use it when the need arises. Suppose the number is 100, and let the increase be x%. The number becomes 100 + x. Let us suppose that it has to be reduced by p% to restore it to 100. Then, (100 + x) - p × $(100 + x)/100$ = 100. We can cancel away 100, and easily simplify it to p = ${100 × x}/{100 + x}$ = ${100 × 80}/{100 + 80}$ = ${400/9}$, which is same as: $44{4/9}$%.


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Creative Commons License
This Blog Post/Article "Percentages Quiz Set 001" by Parveen (Hoven) is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
Updated on 2017-06-24.

Posted by Parveen(Hoven),
Aptitude Trainer


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