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. |

- 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?
- \(1\)
- \(2\)
- \(3\)
- \(4\)

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.

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.

- If ratio of two numbers is \(4 : 5\). If \(10\) be subtracted from each, then the ratio become \(15 : 20\). Find the lower number?
- \(20\)
- \(30\)
- \(40\)
- \(50\)

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 \(40\).

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 \(40\).

- Three friends shares an amount \(Rs.500\) in the ratio of \(2 : 3 : 5\). Find the share of first friend?
- \(Rs.100\)
- \(Rs.150\)
- \(Rs.200\)
- \(Rs.250\)

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 \(Rs.100\).

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 \(Rs.100\).

- The ratio of two numbers is \(1 : 2\). If each number is increased by \(10\), then ratio becomes \(3 : 5\). Find the numbers?
- \(20 \ and \ 30\)
- \(30 \ and \ 40\)
- \(40 \ and \ 50\)
- \(20 \ and \ 40\)

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, $$ = k : 2k $$ $$ = 20 : 40 $$

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

- 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 \(5 : 4\)?
- \(1 \ litre\)
- \(2 \ litres\)
- \(3 \ litres\)
- \(4 \ litres\)

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

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

- 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 \(M\)?
- \(Rs.2500\)
- \(Rs.3500\)
- \(Rs.4000\)
- \(Rs.4500\)

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 \(M\) = \(5k\) = \(Rs.3500\)

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 \(M\) = \(5k\) = \(Rs.3500\)

- 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 \(y\)?
- \(Rs.500\)
- \(Rs.450\)
- \(Rs.400\)
- \(Rs.300\)

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 \(y\) = \(3k\) = \(Rs.300\)

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 \(y\) = \(3k\) = \(Rs.300\)

- 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?
- \(2.28\)
- \(3.36\)
- \(4.24\)
- \(5.12\)

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, $$ = 3k - 2k $$ $$ k = 2.28 $$

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, $$ = 3k - 2k $$ $$ k = 2.28 $$

- 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 \(y\)?
- \(13.2 \ and \ 14.6 \ years\)
- \(13.2 \ and \ 15.4 \ years\)
- \(12.2 \ and \ 16.2 \ years\)
- \(15.2 \ and \ 18.6 \ years\)

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 \(y\), $$ = 6k : 7k $$ $$ = 13.2 : 15.4 $$

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 \(y\), $$ = 6k : 7k $$ $$ = 13.2 : 15.4 $$

- If \(50\) Apples were distributed among two children in the ratio of \(4 : 3\), then find how many Apples did the first child get?
- \(24.5 \ Apples\)
- \(25.6 \ Apples\)
- \(28.4 \ Apples\)
- \(29.2 \ Apples\)

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 = \(4k\) = \(28.4 \ Apples\)

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

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