# Ratio and Proportion Aptitude Questions and Answers:

#### Overview:

 Questions and Answers Type: MCQ (Multiple Choice Questions). Main Topic: Quantitative Aptitude. Quantitative Aptitude Sub-topic: Ratio and Proportion Aptitude Questions and Answers. Number of Questions: 10 Questions with Solutions.

1. Find the number that must be added to each of the numbers $$5, \ 16, \ 11, \ 33$$ to make the ratio of first two numbers equal to the ratio of last two numbers?

1. $$1$$
2. $$2$$
3. $$3$$
4. $$4$$

Answer: (a) $$1$$

Solution: Let the number = $$k$$, then $$\frac{5 + k}{16 + k} = \frac{11 + k}{33 + k}$$ from the options $$k = 1$$ is satisfied the equation.

1. If ratio of two numbers is $$4 : 5$$. If $$10$$ be subtracted from each, then the ratio become $$15 : 20$$. Find the lower number?

1. $$20$$
2. $$30$$
3. $$40$$
4. $$50$$

Answer: (c) $$40$$

Solution: Let the numbers = $$4k : 5k$$, then $$\frac{4k - 10}{5k - 10} = \frac{15}{20}$$ $$k = 10$$ now the numbers $$= 4k : 5k$$ $$= 40 : 50$$ Hence, the lower number is $$40$$.

1. Three friends shares an amount $$Rs.500$$ in the ratio of $$2 : 3 : 5$$. Find the share of first friend?

1. $$Rs.100$$
2. $$Rs.150$$
3. $$Rs.200$$
4. $$Rs.250$$

Answer: (a) $$Rs.100$$

Solution: Let the shares = $$2k : 3k : 5k$$, then $$2k + 3k + 5k = 500$$ $$k = 50$$ now the shares of friends, $$= 2k : 3k : 5k$$ $$= 100 : 150 : 250$$ Hence, the share of friend is $$Rs.100$$.

1. The ratio of two numbers is $$1 : 2$$. If each number is increased by $$10$$, then ratio becomes $$3 : 5$$. Find the numbers?

1. $$20 \ and \ 30$$
2. $$30 \ and \ 40$$
3. $$40 \ and \ 50$$
4. $$20 \ and \ 40$$

Answer: (d) $$20 \ and \ 40$$

Solution: Let the numbers = $$k : 2k$$, then $$\frac{k + 10}{2k + 10} = \frac{3}{5}$$ $$k = 20$$ Hence, the numbers, $$= k : 2k$$ $$= 20 : 40$$

1. A mixture of $$25$$ litres of wine and water in the proportion of $$3 : 2$$. How much water must be added to it so that the ratio of wine and water will become $$5 : 4$$?

1. $$1 \ litre$$
2. $$2 \ litres$$
3. $$3 \ litres$$
4. $$4 \ litres$$

Answer: (b) $$2 \ litres$$

Solution: Let the present ratio of wine and water = $$3k : 2k$$, then $$3k + 2k = 25$$ $$k = 5$$ now the ratio = $$15 : 10$$

on adding $$x$$ litres of water, $$15 : (10 + x) = 5 : 4$$ $$\frac{15}{10 + x} = \frac{5}{4}$$ $$x = 2 \ litres$$

1. The incomes of $$M$$ and $$N$$ are in the ratio of $$5 : 4$$ and their expenditures in the ratio of $$6 : 5$$. If each saves $$Rs.500$$, then find the income of $$M$$?

1. $$Rs.2500$$
2. $$Rs.3500$$
3. $$Rs.4000$$
4. $$Rs.4500$$

Answer: (b) $$Rs.3500$$

Solution: Let income of $$M$$ and $$N$$ = $$5k : 4k$$

expenditures of $$M$$ and $$N$$ = $$6P : 5P$$, then $$5k - 6P = 500.....(1)$$ $$4k - 5P = 500.....(2)$$ on solving equations (1) and (2) we get, $$P = 500 \ and \ k = 700$$ Hence, the income of $$M$$ = $$5k$$ = $$Rs.3500$$

1. The incomes of $$x$$ and $$y$$ are in the ratio of $$2 : 3$$ and their expenditures in the ratio of $$1 : 2$$. If each saves $$Rs.100$$, then find the income of $$y$$?

1. $$Rs.500$$
2. $$Rs.450$$
3. $$Rs.400$$
4. $$Rs.300$$

Answer: (d) $$Rs.300$$

Solution: Let income of $$x$$ and $$y$$ = $$2k : 3k$$

expenditures of $$x$$ and $$y$$ = $$P : 2P$$, then $$2k - P = 100.....(1)$$ $$3k - 2P = 100.....(2)$$ on solving equations (1) and (2) we get, $$P = 100 \ and \ k = 100$$ Hence, the income of $$y$$ = $$3k$$ = $$Rs.300$$

1. Two numbers are in the ratio of $$3 : 2$$. If $$15$$ be subtracted from each, the ratio will become $$25 : 32$$. Find the difference between two numbers?

1. $$2.28$$
2. $$3.36$$
3. $$4.24$$
4. $$5.12$$

Answer: (a) $$2.28$$

Solution: Let the numbers = $$3k : 2k$$

on subtracting $$15$$ from both the numbers we get, $$\frac{3k - 15}{2k - 15} = \frac{25}{32}$$ $$k = 2.28$$ Hence difference between two numbers, $$= 3k - 2k$$ $$k = 2.28$$

1. The present ratio of ages of $$x$$ and $$y$$ is $$6 : 7$$. If $$10$$ years ago, this ratio was $$3 : 5$$, then find the present ages of $$x$$ and $$y$$?

1. $$13.2 \ and \ 14.6 \ years$$
2. $$13.2 \ and \ 15.4 \ years$$
3. $$12.2 \ and \ 16.2 \ years$$
4. $$15.2 \ and \ 18.6 \ years$$

Answer: (b) $$13.2 \ and \ 15.4 \ years$$

Solution: Let present ages of $$x$$ and $$y$$ = $$6k : 7k$$

ten years ago the ages of $$x$$ and $$y$$, $$\frac{6k - 10}{7k - 10} = \frac{3}{5}$$ $$k = 2.2$$ Hence, present ages of $$x$$ and $$y$$, $$= 6k : 7k$$ $$= 13.2 : 15.4$$

1. If $$50$$ Apples were distributed among two children in the ratio of $$4 : 3$$, then find how many Apples did the first child get?

1. $$24.5 \ Apples$$
2. $$25.6 \ Apples$$
3. $$28.4 \ Apples$$
4. $$29.2 \ Apples$$

Answer: (c) $$28.4 \ Apples$$

Solution: Let ratio of distribution = $$4k : 3k$$, then $$4k + 3k = 50$$ $$k = 7.1$$ Hence, first child did get the Apples = $$4k$$ = $$28.4 \ Apples$$