Aptitude - Ratio and Proportion
Exercise : Ratio and Proportion - General Questions
- Ratio and Proportion - Formulas
- Ratio and Proportion - General Questions
1.
A and B together have Rs. 1210. If
of A's amount is equal to
of B's amount, how much amount does B have?


Answer: Option
Explanation:
4 | A | = | 2 | B | |
15 | 5 |
![]() |
![]() |
2 | x | 15 | ![]() |
5 | 4 |
![]() |
3 | B |
2 |
![]() |
A | = | 3 |
B | 2 |
A : B = 3 : 2.
![]() |
![]() |
1210 x | 2 | ![]() |
= Rs. 484. |
5 |
2.
Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is:
Answer: Option
Explanation:
Let the third number be x.
Then, first number = 120% of x = | 120x | = | 6x |
100 | 5 |
Second number = 150% of x = | 150x | = | 3x |
100 | 2 |
![]() |
![]() |
6x | : | 3x | ![]() |
= 12x : 15x = 4 : 5. |
5 | 2 |
3.
A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Rs. 1000 more than D, what is B's share?
Answer: Option
Explanation:
Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.
Then, 4x - 3x = 1000
x = 1000.
B's share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000.
4.
Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats?
Answer: Option
Explanation:
Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively.
Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x).
![]() |
![]() |
140 | x 5x | ![]() |
, | ![]() |
150 | x 7x | ![]() |
and | ![]() |
175 | x 8x | ![]() |
100 | 100 | 100 |
![]() |
21x | and 14x. |
2 |
![]() |
21x | : 14x |
2 |
14x : 21x : 28x
2 : 3 : 4.
5.
In a mixture 60 litres, the ratio of milk and water 2 : 1. If this ratio is to be 1 : 2, then the quantity of water to be further added is:
Answer: Option
Explanation:
Quantity of milk = | ![]() |
60 x | 2 | ![]() |
3 |
Quantity of water in it = (60- 40) litres = 20 litres.
New ratio = 1 : 2
Let quantity of water to be added further be x litres.
Then, milk : water = | ![]() |
40 | ![]() |
. |
20 + x |
Now, | ![]() |
40 | ![]() |
= | 1 |
20 + x | 2 |
20 + x = 80
x = 60.
Quantity of water to be added = 60 litres.
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