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. |

- A certain amount of money is to be divided between x, y and z in the ratio of \(3 : 5 : 7\) respectively. If total share of x and z is \(Rs.1200\) more than the share of y, then find what will be the share of x?
- \(Rs.680\)
- \(Rs.700\)
- \(Rs.720\)
- \(Rs.750\)

Answer: (c) \(Rs.720\)

Solution: Let the ratio of x, y and z = \(3k : 5k : 7k\), then $$ \frac{3k}{15} + \frac{7k}{15} - \frac{3k}{15} = 1200 $$ $$ k = Rs.3600 $$ Hence, the share of x, $$ = 3600 \times \frac{3}{15} $$ $$ = Rs.720 $$

Solution: Let the ratio of x, y and z = \(3k : 5k : 7k\), then $$ \frac{3k}{15} + \frac{7k}{15} - \frac{3k}{15} = 1200 $$ $$ k = Rs.3600 $$ Hence, the share of x, $$ = 3600 \times \frac{3}{15} $$ $$ = Rs.720 $$

- In a college, seats for Arts, Science and Maths are divided in the ratio of \(2 : 3 : 5\) respectively. If the number of students studying Arts, Science and Maths are increased by \(5 \ \%\), \(10 \ \%\) and \(20 \ \%\) respectively, then find the new ratio of seats in Arts, Science and Maths respectively.
- \(3.2 : 2.2 : 6\)
- \(2.1 : 3.3 : 6\)
- \(2.2 : 3.3 : 4\)
- \(3.3 : 2.2 : 4\)

Answer: (b) \(2.1 : 3.3 : 6\)

Solution: The new ratio of seats in Arts, Science and Maths will be, $$ = 2 \times \frac{105}{100} : 3 \times \frac{110}{100} : 5 \times \frac{120}{100} $$ $$ = 2.1 : 3.3 : 6 $$

Solution: The new ratio of seats in Arts, Science and Maths will be, $$ = 2 \times \frac{105}{100} : 3 \times \frac{110}{100} : 5 \times \frac{120}{100} $$ $$ = 2.1 : 3.3 : 6 $$

- \(10\) boys and \(15\) girls from a class. During the admission, the same number of boys and girls taken the admission in the same class, Find how many students does the group have now, if the ratio of girls and boys is \(5 : 4\)?
- \(50 \ Students\)
- \(45 \ Students\)
- \(40 \ Students\)
- \(35 \ Students\)

Answer: (b) \(45 \ Students\)

Solution: Let k boys and girls are taken the admission, then $$ \frac{10 + k}{15 + k} = \frac{4}{5} $$ $$ k = 10 $$ Hence, total students $$ = (10 + k) + (15 + k) $$ $$ = (10 + 10) + (15 + 10) $$ $$ = 45 \ Students $$

Solution: Let k boys and girls are taken the admission, then $$ \frac{10 + k}{15 + k} = \frac{4}{5} $$ $$ k = 10 $$ Hence, total students $$ = (10 + k) + (15 + k) $$ $$ = (10 + 10) + (15 + 10) $$ $$ = 45 \ Students $$

- Find the ratio of squares of the numbers, if three numbers are in the ratio of \(5 : 7 : 8\) and half the sum of the numbers is \(5\)?
- \(6.25 : 12.25 : 16\)
- \(12.25 : 6.25 : 18\)
- \(6.25 : 12.25 : 12\)
- \(12.25 : 6.25 : 14\)

Answer: (a) \(6.25 : 12.25 : 16\)

Solution: Let ratio is \(5k : 7k : 8k\), then $$ \frac{5k + 7k + 8k}{2} = 5 $$ $$ k = \frac{1}{2} $$ then the numbers are, \(2.5 : 63.5 : 4\) and square of numbers are \(6.25 : 12.25 : 16\)

Solution: Let ratio is \(5k : 7k : 8k\), then $$ \frac{5k + 7k + 8k}{2} = 5 $$ $$ k = \frac{1}{2} $$ then the numbers are, \(2.5 : 63.5 : 4\) and square of numbers are \(6.25 : 12.25 : 16\)

- If \(150\) fruits was distributed among three persons in the ratio of \(1/2 : 2/3 : 5/6\), then find how many fruits the second person got?
- \(25 \ Fruits\)
- \(35 \ Fruits\)
- \(45 \ Fruits\)
- \(50 \ Fruits\)

Answer: (d) \(50 \ Fruits\)

Solution: The ratio of fruits given, $$ \frac{1}{2} : \frac{2}{3} : \frac{5}{6} $$ $$ = \frac{3 : 4 : 5}{6} $$ so the fruits in the ratio \(3k : 4k : 5k\) are distributed, then $$ 3k + 4k + 5k = 150 $$ $$ k = 12.5 $$ now the ratio $$ = 3k : 4k : 5k $$ $$ = 37.5 : 50 : 62.5 $$ Hence, the second person got \(50\) fruits.

Solution: The ratio of fruits given, $$ \frac{1}{2} : \frac{2}{3} : \frac{5}{6} $$ $$ = \frac{3 : 4 : 5}{6} $$ so the fruits in the ratio \(3k : 4k : 5k\) are distributed, then $$ 3k + 4k + 5k = 150 $$ $$ k = 12.5 $$ now the ratio $$ = 3k : 4k : 5k $$ $$ = 37.5 : 50 : 62.5 $$ Hence, the second person got \(50\) fruits.

- Find the fourth proportional of the numbers \(20\), \(25\) and \(30\)?
- \(28.5\)
- \(36.5\)
- \(37.5\)
- \(39.5\)

Answer: (c) \(37.5\)

Solution: Let k is the fourth proportional, then $$ = 20 : 25 :: 30 : k $$ $$ \frac{20}{25} = \frac{30}{k} $$ $$ k = 37.5 $$

Solution: Let k is the fourth proportional, then $$ = 20 : 25 :: 30 : k $$ $$ \frac{20}{25} = \frac{30}{k} $$ $$ k = 37.5 $$

- Find the mean proportional of \(9\) and \(4\)?
- \(4\)
- \(6\)
- \(7\)
- \(8\)

Answer: (b) \(6\)

Solution: Let mean proportional = \(k\), then $$ 9 : k :: k : 4 $$ $$ \frac{9}{k} = \frac{k}{4} $$ $$ k^2 = 36 $$ $$ k = \sqrt{36} $$ $$ k = 6 $$

Solution: Let mean proportional = \(k\), then $$ 9 : k :: k : 4 $$ $$ \frac{9}{k} = \frac{k}{4} $$ $$ k^2 = 36 $$ $$ k = \sqrt{36} $$ $$ k = 6 $$

- Find the third proportional of \(3\) and \(5\)?
- \(3.33\)
- \(5.22\)
- \(8.33\)
- \(9.11\)

Answer: (c) \(8.33\)

Solution: Let third proportional = \(k\), then $$ 3 : 5 :: 5 : k $$ $$ \frac{3}{5} = \frac{5}{k} $$ $$ k = 8.33 $$

Solution: Let third proportional = \(k\), then $$ 3 : 5 :: 5 : k $$ $$ \frac{3}{5} = \frac{5}{k} $$ $$ k = 8.33 $$

- If the ratio of two numbers is \(6 : 4\) and their sum is \(250\), then find the greater number?
- \(100\)
- \(120\)
- \(150\)
- \(180\)

Answer: (c) \(150\)

Solution: Let numbers = \(6k : 4k\), then $$ 6k + 4k = 250 $$ $$ k = 25 $$ then numbers are,$$ 6k : 4k = 150 : 100 $$ Hence, the greater number is \(150\)

Solution: Let numbers = \(6k : 4k\), then $$ 6k + 4k = 250 $$ $$ k = 25 $$ then numbers are,$$ 6k : 4k = 150 : 100 $$ Hence, the greater number is \(150\)

- The ratio of two numbers is \(3 : 4\). If each number is increased by \(15\), then ratio becomes \(6 : 7\). Find the numbers?
- \(15 \ and \ 20\)
- \(20 \ and \ 30\)
- \(12 \ and \ 24\)
- \(18 \ and \ 40\)

Answer: (a) \(15 \ and \ 20\)

Solution: Let the numbers = \(3k : 4k\), then $$ \frac{3k + 15}{4k + 15} = \frac{6}{7} $$ $$ k = 5 $$ Hence, the numbers, $$ = 3k : 4k $$ $$ = 15 : 20 $$

Solution: Let the numbers = \(3k : 4k\), then $$ \frac{3k + 15}{4k + 15} = \frac{6}{7} $$ $$ k = 5 $$ Hence, the numbers, $$ = 3k : 4k $$ $$ = 15 : 20 $$

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