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. |

- In a mixture of \(35\) litres, the ratio of milk and water is \(5 : 2\). find how much water must be added to this mixture so that the ratio of milk and water becomes \(3 : 4\)?
- \(24.2 \ litres\)
- \(22.2 \ litres\)
- \(23.3 \ litres\)
- \(23.8 \ litres\)

Answer: (c) \(23.3 \ litres\)

Solution: Let the ratio of milk and water = \(5k : 2k\)

the given mixture is \(35\) litres, so $$ 25 : 10 = 5k : 2k $$ now the amount of water k must be added to the given ratio become \(3 : 4\), $$ 25 : (10 + k) = 3 : 4 $$ $$ \frac{25}{10 + k} = \frac{3}{4} $$ $$ 10 + k = \frac{100}{3} $$ $$ k = 23.3 \ litres $$

Solution: Let the ratio of milk and water = \(5k : 2k\)

the given mixture is \(35\) litres, so $$ 25 : 10 = 5k : 2k $$ now the amount of water k must be added to the given ratio become \(3 : 4\), $$ 25 : (10 + k) = 3 : 4 $$ $$ \frac{25}{10 + k} = \frac{3}{4} $$ $$ 10 + k = \frac{100}{3} $$ $$ k = 23.3 \ litres $$

- Find the ratio of squares of the numbers, if three numbers are in the ratio of \(2 : 3 : 5\) and half the sum of the numbers is \(10\)?
- \(16 : 36 : 81\)
- \(16 : 36 : 100\)
- \(36 : 49 : 81\)
- \(36 : 81 : 100\)

Answer: (b) \(16 : 36 : 100\)

Solution: Let ratio is \(2k : 3k : 5k\), then $$ \frac{2k + 3k + 5k}{2} = 10 $$ $$ k = 2 $$ then the numbers are, \(4 : 6 : 10\) and square of numbers are \(16 : 36 : 100\)

Solution: Let ratio is \(2k : 3k : 5k\), then $$ \frac{2k + 3k + 5k}{2} = 10 $$ $$ k = 2 $$ then the numbers are, \(4 : 6 : 10\) and square of numbers are \(16 : 36 : 100\)

- The ratio of ages of \(x\) and \(y\) is \(3 : 2\) at present and after \(10\) years, the ratio will become \(5 : 4\), find the present age of \(x\)?
- \(10 \ years\)
- \(12 \ years\)
- \(15 \ years\)
- \(20 \ years\)

Answer: (c) \(15 \ years\)

Solution: Let their present ages = \(3k : 2k\)

then after \(10\) years, $$ \frac{3k + 10}{2k + 10} = \frac{5}{4} $$ $$ k = 5 $$ then the present ages of \(x\) and \(y\), $$ x : y = 3k : 2k $$ $$ x : y = 15 : 10 $$ Hence, present age of \(x\) is \(15\) years.

Solution: Let their present ages = \(3k : 2k\)

then after \(10\) years, $$ \frac{3k + 10}{2k + 10} = \frac{5}{4} $$ $$ k = 5 $$ then the present ages of \(x\) and \(y\), $$ x : y = 3k : 2k $$ $$ x : y = 15 : 10 $$ Hence, present age of \(x\) is \(15\) years.

- If \(115\) books was distributed among three students in the ratio of \(2/3 : 3/4 : 1/2\), then find how many books the first student got?
- \(25 \ books\)
- \(30 \ books\)
- \(40 \ books\)
- \(45 \ books\)

Answer: (c) \(40 \ books\)

Solution: The ratio of books given, $$ \frac{2}{3} : \frac{3}{4} : \frac{1}{2} $$ $$ = \frac{8 : 9 : 6}{12} $$ so the books in the ratio \(8k : 9k : 6k\) are distributed, then $$ 8k + 9k + 6k = 115 $$ $$ k = 5 $$ now the ratio $$ = 8k : 9k : 6k $$ $$ = 40 : 45 : 30 $$ Hence, the first student got \(40\) books.

Solution: The ratio of books given, $$ \frac{2}{3} : \frac{3}{4} : \frac{1}{2} $$ $$ = \frac{8 : 9 : 6}{12} $$ so the books in the ratio \(8k : 9k : 6k\) are distributed, then $$ 8k + 9k + 6k = 115 $$ $$ k = 5 $$ now the ratio $$ = 8k : 9k : 6k $$ $$ = 40 : 45 : 30 $$ Hence, the first student got \(40\) books.

- If a mixture contains milk and water in the ratio of \(3 : 1\) and on adding two litres of water, the ratio becomes \(5 : 3\), then find the quantity of milk in the mixture?
- \(10.2 \ litres\)
- \(12.5 \ litres\)
- \(15.5 \ litres\)
- \(16.8 \ litres\)

Answer: (b) \(12.5 \ litres\)

Solution: Let the present ratio of milk and water = \(3k : k\)

then on adding two litres of water, $$ 3k : (k + 2) = 5 : 3 $$ $$ \frac{3k}{k + 2} = \frac{5}{3} $$ $$ k = 2.5 $$ so the ratio after adding two litres of water, $$ = 5k : 3k $$ $$ = 5 \times 2.5 : 3 \times 2.5 $$ $$ = 12.5 : 7.5 $$ Hence, the quantity of milk = \(12.5 \ litres\)

Solution: Let the present ratio of milk and water = \(3k : k\)

then on adding two litres of water, $$ 3k : (k + 2) = 5 : 3 $$ $$ \frac{3k}{k + 2} = \frac{5}{3} $$ $$ k = 2.5 $$ so the ratio after adding two litres of water, $$ = 5k : 3k $$ $$ = 5 \times 2.5 : 3 \times 2.5 $$ $$ = 12.5 : 7.5 $$ Hence, the quantity of milk = \(12.5 \ litres\)

- If by adding \(5\) on the ratio of the numbers \(5 : 7\), it becomes \(7 : 9\), then find the lower number?
- \(20.5\)
- \(16.5\)
- \(15.5\)
- \(12.5\)

Answer: (d) \(12.5\)

Solution: Let the ratio = \(5k : 7k\)

on adding \(5\) ratio becomes \(7 : 9\), then $$ \frac{5k + 5}{7k + 5} =\frac{7}{9} $$ $$ k = 2.5 $$ then ratio $$ = 5k : 7k $$ $$ = 5 \times 2.5 : 7 \times 2.5 $$ $$ = 12.5 : 17.5 $$ Hence, the lower number = \(12.5\)

Solution: Let the ratio = \(5k : 7k\)

on adding \(5\) ratio becomes \(7 : 9\), then $$ \frac{5k + 5}{7k + 5} =\frac{7}{9} $$ $$ k = 2.5 $$ then ratio $$ = 5k : 7k $$ $$ = 5 \times 2.5 : 7 \times 2.5 $$ $$ = 12.5 : 17.5 $$ Hence, the lower number = \(12.5\)

- Find the fourth proportional of the numbers \(10\), \(15\) and \(20\)?
- \(20\)
- \(30\)
- \(40\)
- \(50\)

Answer: (b) \(30\)

Solution: Let k is the fourth proportional, then $$ = 10 : 15 :: 20 : k $$ $$ \frac{10}{15} = \frac{20}{k} $$ $$ k = 30 $$

Solution: Let k is the fourth proportional, then $$ = 10 : 15 :: 20 : k $$ $$ \frac{10}{15} = \frac{20}{k} $$ $$ k = 30 $$

- Find the mean proportional of \(8\) and \(2\)?
- \(4\)
- \(3\)
- \(2\)
- \(1\)

Answer: (a) \(4\)

Solution: Let mean proportional = \(k\), then $$ 8 : k :: k : 2 $$ $$ \frac{8}{k} = \frac{k}{2} $$ $$ k^2 = 16 $$ $$ k = \sqrt{16} $$ $$ k = 4 $$

Solution: Let mean proportional = \(k\), then $$ 8 : k :: k : 2 $$ $$ \frac{8}{k} = \frac{k}{2} $$ $$ k^2 = 16 $$ $$ k = \sqrt{16} $$ $$ k = 4 $$

- Find the third proportional of \(2\) and \(4\)?
- \(4\)
- \(6\)
- \(8\)
- \(10\)

Answer: (c) \(8\)

Solution: Let third proportional = \(k\), then $$ 2 : 4 :: 4 : k $$ $$ \frac{2}{4} = \frac{4}{k} $$ $$ k = 8 $$

Solution: Let third proportional = \(k\), then $$ 2 : 4 :: 4 : k $$ $$ \frac{2}{4} = \frac{4}{k} $$ $$ k = 8 $$

- If the ratio of two numbers is \(9 : 6\) and their sum is \(150\), then find the greater number?
- \(50\)
- \(70\)
- \(80\)
- \(90\)

Answer: (d) \(90\)

Solution: Let numbers = \(9k : 6k\), then $$ 9k + 6k = 150 $$ $$ k = 10 $$ then numbers are,$$ 9k : 6k = 90 : 60 $$ Hence, the greater number is \(90\)

Solution: Let numbers = \(9k : 6k\), then $$ 9k + 6k = 150 $$ $$ k = 10 $$ then numbers are,$$ 9k : 6k = 90 : 60 $$ Hence, the greater number is \(90\)

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