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. |

- If \(x : y = 3 : 2\) then find the value of \(x^2 - y^2 : x^2 + y^2\)?
- \(13 : 5\)
- \(5 : 13\)
- \(7 : 12\)
- \(12 : 7\)

Answer: (b) \(5 : 13\)

Solution: Given, $$ \frac{x}{y} = \frac{3}{2} $$ then $$ \frac{x^2}{y^2} = \frac{3^2}{2^2} = \frac{9}{4} $$ Hence, $$ \frac{x^2 - y^2}{x^2 + y^2} = \frac{9 - 4}{9 + 4} $$ $$ = \frac{5}{13} $$

Solution: Given, $$ \frac{x}{y} = \frac{3}{2} $$ then $$ \frac{x^2}{y^2} = \frac{3^2}{2^2} = \frac{9}{4} $$ Hence, $$ \frac{x^2 - y^2}{x^2 + y^2} = \frac{9 - 4}{9 + 4} $$ $$ = \frac{5}{13} $$

- The ratio between two numbers is \(7 : 8\). If each number is reduced by \(15\), then the ratio becomes \(4 : 5\). Find the numbers?
- \(28 \ and \ 32\)
- \(32 \ and \ 28\)
- \(24 \ and \ 30\)
- \(30 \ and \ 24\)

Answer: (a) \(28 \ and \ 32\)

Solution: Let ratio of two numbers = \(7k : 8k\)

by reducing 12 from each number, $$ \frac{7k - 12}{8k - 12} = \frac{4}{5} $$ $$ k = 4 $$ Hence the numbers = \(28 : 32\)

Solution: Let ratio of two numbers = \(7k : 8k\)

by reducing 12 from each number, $$ \frac{7k - 12}{8k - 12} = \frac{4}{5} $$ $$ k = 4 $$ Hence the numbers = \(28 : 32\)

- A mixture of wine and water in the ratio of \(5 : 4\). If two litres of water is added to the mixture the ratio becomes \(7 : 8\). Find the quantity of wine and water in the mixture?
- \(6.2 \ and \ 5.4 \ litres\)
- \(5.4 \ and \ 6.2 \ litres\)
- \(5.8 \ and \ 4.6 \ litres\)
- \(4.6 \ and \ 5.8 \ litres\)

Answer: (c) \(5.8 \ and \ 4.6 \ litres\)

Solution: Let the ratio of wine and water in the mixture = \(5k : 4k\)

on adding two litres of water, $$ \frac{5k}{4k + 2} = \frac{7}{8} $$ $$ k = 1.16 $$ Hence, quantity of wine and water $$ = 5k : 4k $$ $$ = 5.8 : 4.6 $$

Solution: Let the ratio of wine and water in the mixture = \(5k : 4k\)

on adding two litres of water, $$ \frac{5k}{4k + 2} = \frac{7}{8} $$ $$ k = 1.16 $$ Hence, quantity of wine and water $$ = 5k : 4k $$ $$ = 5.8 : 4.6 $$

- If \(x : y = 5 : 4\) then find the value of \(x - y : x + y\)?
- \(1 : 5\)
- \(5 : 1\)
- \(9 : 1\)
- \(1 : 9\)

Answer: (d) \(1 : 9\)

Solution: Given, $$ \frac{x}{y} = \frac{5}{4} $$ then, $$ \frac{x - y}{x + y} = \frac{5 - 4}{5 + 4} $$ $$ = \frac{1}{9} $$

Solution: Given, $$ \frac{x}{y} = \frac{5}{4} $$ then, $$ \frac{x - y}{x + y} = \frac{5 - 4}{5 + 4} $$ $$ = \frac{1}{9} $$

- A \(500 \ ml\) mixtute of tea and water in the ratio of \(5 : 3\). Find how much more water to be included to get a new mixture of tea and water in the ratio of \(5 : 4\)?
- \(60.2 \ ml\)
- \(62.5 \ ml\)
- \(65.3 \ ml\)
- \(68.5 \ ml\)

Answer: (b) \(62.5 \ ml\)

Solution: quantity of tea in the mixture $$ \frac{500}{8} \times 5 $$ $$ = 312.5 \ ml $$ quantity of water in the mixture $$ \frac{500}{8} \times 3 $$ $$ = 187.5 \ ml $$ Let \(k \ ml\) water to be added then $$ \frac{312.5}{187.5 + k} = \frac{5}{4} $$ $$ k = 62.5 \ ml $$

Solution: quantity of tea in the mixture $$ \frac{500}{8} \times 5 $$ $$ = 312.5 \ ml $$ quantity of water in the mixture $$ \frac{500}{8} \times 3 $$ $$ = 187.5 \ ml $$ Let \(k \ ml\) water to be added then $$ \frac{312.5}{187.5 + k} = \frac{5}{4} $$ $$ k = 62.5 \ ml $$

- Number of students in math and biology in an institute are in the ratio of \(3 : 5\) respectively. If \(100\) more students join math and \(50\) more students join biology, then the ratio becomes \(2 : 3\). Find the total students in both subjects?
- \(100\)
- \(125\)
- \(175\)
- \(200\)

Answer: (d) \(200\)

Solution: Let students in math and biology = \(3k : 5k\)

on adding \(100\) students in math and \(50\) students in biology, $$ \frac{3k + 100}{5k + 50} = \frac{2}{3} $$ $$ k = 200 $$

Solution: Let students in math and biology = \(3k : 5k\)

on adding \(100\) students in math and \(50\) students in biology, $$ \frac{3k + 100}{5k + 50} = \frac{2}{3} $$ $$ k = 200 $$

- The ratio of length and breadth of a rectangle is \(2 : 3\) and by raising length and breadth the ratio becomes \(3 : 4\). Find the ratio of previous area and present area of rectangle?
- \(1 : 2\)
- \(2 : 1\)
- \(3 : 2\)
- \(2 : 3\)

Answer: (a) \(1 : 2\)

Solution: let ratio of previous length and breadth of the rectangle = \(2A : 3A\)

then previous area of rectangle = \(2A \times 3A\) = \(6A^2\)

now let the ratio of present length and breadth of the rectangle = \(3A : 4A\)

then present area of the rectangle = \(3A \times 4A\) = \(12A^2\)

Hence, the ratio of previous area and present area of rectangle, $$ = 6A^2 : 12A^2 $$ $$ = 1 : 2 $$

Solution: let ratio of previous length and breadth of the rectangle = \(2A : 3A\)

then previous area of rectangle = \(2A \times 3A\) = \(6A^2\)

now let the ratio of present length and breadth of the rectangle = \(3A : 4A\)

then present area of the rectangle = \(3A \times 4A\) = \(12A^2\)

Hence, the ratio of previous area and present area of rectangle, $$ = 6A^2 : 12A^2 $$ $$ = 1 : 2 $$

- \(250\) bananas are divided among five boys, \(10\) girls and \(15\) women. If the ratio of one boy, one girl and one women is \(5 : 7 : 9\), then find the share of one girl?
- \(6.7\)
- \(7.6\)
- \(8.2\)
- \(9.6\)

Answer: (b) \(7.6\)

Solution: Given, the ratio of one boy, one girl and one women = \(5 : 7 : 9\)

then overall ratio of \(5\) boys, \(10\) girls and \(15\) women $$ = 5 \times 5 : 7 \times 10 : 9 \times 15 $$ $$ = 25 : 70 : 135 $$ now the share of \(10\) girls, $$ = \frac{250}{230} \times 70 $$ $$ = 76 \ bananas $$ Hence, the share of one girl $$ = \frac{76}{10} $$ $$ = 7.6 \ bananas $$

Solution: Given, the ratio of one boy, one girl and one women = \(5 : 7 : 9\)

then overall ratio of \(5\) boys, \(10\) girls and \(15\) women $$ = 5 \times 5 : 7 \times 10 : 9 \times 15 $$ $$ = 25 : 70 : 135 $$ now the share of \(10\) girls, $$ = \frac{250}{230} \times 70 $$ $$ = 76 \ bananas $$ Hence, the share of one girl $$ = \frac{76}{10} $$ $$ = 7.6 \ bananas $$

- The ratio between two numbers is \(5 : 6\). If each number is reduced by \(25\), then the ratio becomes \(7 : 8\). Find the numbers?
- \(62.5 \ and \ 75\)
- \(75 \ and \ 62.5\)
- \(68.5 \ and \ 72\)
- \(72 \ and \ 68.5\)

Answer: (a) \(62.5 \ and \ 75\)

Solution: Let ratio of two numbers = \(5k : 6k\)

each number is increased by \(25\), then $$ \frac{5k + 25}{6k + 25} = \frac{7}{8} $$ $$ k = 12.5 $$ Hence the numbers = \(62.5 : 75\)

Solution: Let ratio of two numbers = \(5k : 6k\)

each number is increased by \(25\), then $$ \frac{5k + 25}{6k + 25} = \frac{7}{8} $$ $$ k = 12.5 $$ Hence the numbers = \(62.5 : 75\)

- A mixture of milk and water in the ratio of \(4 : 3\). If \(5\) litres of water is added to the mixture the ratio becomes \(5 : 4\). Find the quantity of water in the mixture?
- \(50 \ litres\)
- \(75 \ litres\)
- \(80 \ litres\)
- \(85 \ litres\)

Answer: (b) \(75 \ litres\)

Solution: Let the ratio of milk and water in the mixture = \(4k : 3k\)

on adding \(5\) litres of water, $$ \frac{4k}{3k + 5} = \frac{5}{4} $$ $$ k = 25 $$ Hence, quantity of water $$ = 3k = 75 \ litres $$

Solution: Let the ratio of milk and water in the mixture = \(4k : 3k\)

on adding \(5\) litres of water, $$ \frac{4k}{3k + 5} = \frac{5}{4} $$ $$ k = 25 $$ Hence, quantity of water $$ = 3k = 75 \ litres $$

Lec 1: Introduction
Exercise-1
Lec 2: Rules of Componendo and Dividendo
Exercise-2
Lec 3: Rules of Partnership
Exercise-3
Exercise-4
Exercise-5