- Aptitude Test Preparation
- Aptitude - Home
- Aptitude - Overview
- Quantitative Aptitude

- Aptitude Useful Resources
- Aptitude - Questions & Answers

# Aptitude - Percentages Online Quiz

Following quiz provides Multiple Choice Questions (MCQs) related to **Percentages**. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using **Show Answer** button. You can use **Next Quiz** button to check new set of questions in the quiz.

### Answer : D

### Explanation

60% of 264 =^{60}⁄_{100}x 264 = 158.4 50% of 234 =^{50}⁄_{100}x 234 = 117 70% of 164 =^{70}⁄_{100}x 164 = 114.8 20% of 1500 =^{20}⁄_{100}x 1500 = 300 15% of 1056 =^{15}⁄_{100}x 1056 = 158.4

Q 2 - The sum of two numbers is 2490. If 6.5% of one number is equal to 8.5% of the other, then the numbers are?

### Answer : A

### Explanation

Let p and q be the numbers. Then, 6.5% of p = 8.5% of q p =^{85}⁄_{65}q i.e p =^{17}⁄_{13}q Now, p + q = 2490 i.e.^{17}⁄_{13}q + q = 2490 Therefore, q = 1079 This shows the two numbers are : q = 1079 and^{17}⁄_{13}q = 1411 = p

Q 3 - 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?

### Answer : C

### Explanation

Number of runs made by running = 110 - (3 x 4 + 8 x 6) = 50 Therefore, percentage = (^{50}⁄_{100}x 100)% = 45^{5}⁄_{11}%

Q 4 - Rahul gave 40% of the amount he had to Pankaj. Pankaj in turn gave one-fourth of what he received from Rahul to Sanyam. After paying Rs. 200 to the taxi driver out of the amount he got from Pankaj, Sanyam now has Rs. 600 left with him. How much amount did Rahul have?

### Answer : B

### Explanation

Let p be the amount with Rahul. ↠ amount received by Sanyam =^{1}⁄_{4}of 40% of Rs. p = 10% of Rs. p. Therefore, 10% of p = 600 + 200 → p = 8000 i.e. Amount with Rahul

Q 5 - Of the 1000 inhabitants of a town, 60% are males of whom 20% are literate. If, of all the inhabitants, 25% are literate, then what percent of the females of the town are literate?

### Answer : D

### Explanation

^{60}⁄_{100}x 1000 = 600 = number of males → 1000 - 600 = 400 = number of females Literates = 25% of 1000 = 250 Literate males = 20% of 600 = 120 Literate females = 250 - 120 = 130 Therefore, the percent of the females literate =^{130}⁄_{400}x 100 = 32.5%

Q 6 - A's salary is 50% more than that of B. How much percent is B's salary less than that of A?

### Answer : B

### Explanation

Required percentage = {50/(100+50)}*100% = 50/150 * 100 = 100/3

Q 7 - Water tax is increased by 30% but its consumption is decreased by 20%. Then, the increase or decreased in the expenditure of the money is

### Answer : D

### Explanation

Let tax =Rs.100 and Consumption = 100 units Original expenditure = Rs.100*100=Rs.10000 New expenditure=Rs.(130*80) =Rs. 10400 Increase in expenditure by 4 %

Q 8 - The estimation of a machine deteriorates at the rate of 10% for every annum. On the off chance that its present worth is Rs. 8100, what will be its worth following 3 years?

### Answer : B

### Explanation

Value of the machine after three years = Rs{81000*(1-10/100)^{3}} = 81000* 9/10*9/10*9/10 = Rs 59049

Q 9 - In an examination, 60% of the candidates passed in English and 70% of the candidates passed in Mathematics but 20% failed in both the subjects. If 2500 candidates passed in both the subjects, the number of candidates that appeared in the examination was:

### Answer : D

### Explanation

Let, the total number of candidates = x Hence, number of candidates passed in English = 0.6x Similarly, number of candidates passed in Mathematics = 0.7x Number of candidates failed in both the subjects = 0.2x Number of candidates passed in at least one of the subjects =x-0.2x = 0.8x ∴ 0.6x+0.7x-2500=0.8x Or,1.3x–0.8x=2500 Or,0.5x=2500 Or,x=5000

Q 10 - If the radius of a circle is increased by 4 times, what will be the percentage increase in its area?

### Answer : D

### Explanation

Area of circle = Πr2 Area of new circle = Π(4r^{2}) = 16Πr^{2}= 16 times the area of the original circle ∴ Required percentage = (16 – 1)/1 * 100 = 1500%