Percentage Questions

Percentage Questions

by Amit Prabhu | Updated on 08 August 2023

by Amit Prabhu | Updated on 08 August 2023


Percentage Questions With Answers

Percentage is one of the topics from Quantitative aptitude on which fixed questions are asked. 


1) A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?

A) 45%

B) 45(5/11)%

C) 54(6/11)%

D) 54(6/11)%

Solution:


2) Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. What are the marks obtained by them?

A) 42, 36

B) 42, 33

C) 44, 36

D) 44, 33

Solution :


3) In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, the number of valid votes that the other candidate got, was:

A) 2700

B) 2900

C) 3000

D) 3100

Solution :


4) Two employees X and Y are paid a total of Rs. 550 per week by their employer. If X is paid 120 percent of the sum paid to Y, how much is Y paid per week?

A) Rs. 250

B) Rs. 150

C) Rs. 300

D) Rs. 200

Solution :


5) A student has to obtain 33% of the total marks to pass. He got 125  marks and failed by 40 marks. The maximum marks are:  

A. 500  

B. 600  

C. 800  

D. 1000

Solution:    

Given that the student got 125 marks and still he failed by 40 marks  
=> The minimum pass mark = 125 + 40 = 165  
Given that minimum pass mark = 33% of the total mark  
=> total mark =33/100 =165  
=> total mark = 16500/33 = 500  

6) A man spends 35% of his income on food, 25% on children's  education and 80% of the remaining on house rent. What percent of his income he is left with?  

A) 6%  

B) 8%  

C) 10%  

D) 12%  

Solution:  

Let the total income be x. Then, income left = (100 -80)% of [100 - (35 + 
25)] % of x  
= 20% of 40% of x = 20/100 * 40/100 * 100) % of x = 8 % of x.   

7) A shopkeeper bought 600 oranges and 400 bananas. He found 15% of oranges and 8% of bananas were rotten. Find the percentage of fruits in good condition.  

A) 23.4%  

B) 54.8%  

C) 64.5%  

D) 87.8%  

Solution:  

Total number of fruits shopkeeper bought = 600 + 400 = 1000  
Number of rotten oranges = 15% of 600  
= 15/100 × 600  
= 9000/100  
= 90  
Number of rotten bananas = 8% of 400  
= 8/100 × 400  
= 3200/100  
= 32  
Therefore, total number of rotten fruits = 90 + 32 = 122  
Therefore Number of fruits in good condition = 1000 - 122 = 878  
Therefore Percentage of fruits in good condition = (878/1000 × 100)%  
= (87800/1000)%  
= 87.8%  

8) A reduction of 10% in the price of wheat enables a man to buy 50g of wheat   more for a rupee. How much wheat could originally be had for a  rupee?  

A) 400g  

B) 500g  

C) 450g  

D) 350g  

Solution:  

Original prices per kg= Rs/(100-p)y per kg  
{(1 x 10)/(100 – 10)0.05} = Rs. 20/9 /kg  
It means the man can buy 1 kg or 1000 gm in Rs 20/9 or he can buy  
(1000 x 9)/20 = 450g in one rupee.

9) In a certain school, 20% of students are below 8 years of age. The  number of students above 8 years of age is 2/3 of the number of  students of 8 years of age which is 48. What is the total number of  students in the school?  

A) 72  

B) 80  

C) 120  

D) 100  

Solution:  

Let the number of students be x.  
Then, Number of students above 8 years of age = (100 - 20)% of x = 
80% of x.  
80% of x = 48 + 2/3 of 48  
80/100x = 80  
x = 100.  

10) If A's height is 40% less than that of B, how much percent B's  height is more than that of A?  

A) 66.66%  

B) 76.66%  

C) 96.66%  

D) 86.66%  

Solution:

Excess of B's height over A's = [(40/(100 - 40)] x 100% = 66.66%.

 

Relates Topics

 Time and Work  Questions
Time Speed and Distance Questions  
Pipe and Cistern Questions  
Ratio and Proportion Questions  
Permutation and Combination Questions  
Probability Questions
Percentage Questions  
Alligation and Mixture Questions  
Age Related Questions

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