Questions and Answers Type: | MCQ (Multiple Choice Questions). |

Main Topic: | Quantitative Aptitude. |

Quantitative Aptitude Sub-topic: | Percentage Aptitude Questions and Answers. |

Number of Questions: | 10 Questions with Solutions. |

- An investor earns 5% return on his \(\frac{1}{4}\) capital, 10% on his \(\frac{2}{3}\)rd capital and 12% on his remaining capital. What is the average rate of return on his total capital?
- 7.61 %
- 7.95 %
- 8.15 %
- 8.97 %

Answer: (d) 8.97 %

Solution: Let the total capital of the investor is Rs \(x\).

Percentage return on parts of his capital.

5% on his \(\frac{1}{4}\)th capital. $$ = 5 \ \% \ of \ \frac{x}{4} $$ $$ = \frac{5}{100} \times \frac{x}{4} $$ $$ = \frac{x}{80} $$ 10% on his \(\frac{2}{3}\)rd capital. $$ = 10 \ \% \ of \ \frac{2x}{3} $$ $$ = \frac{10}{100} \times \frac{2x}{3} $$ $$ = \frac{x}{15} $$ 12% on his remaining capital $$ = 12 \ \% \ of \ \left[x - \{\frac{x}{4} + \frac{2x}{3}\}\right] $$ $$ = 12 \ \% \ of \ \left[x - \frac{x}{4} - \frac{2x}{3}\right] $$ $$ = 12 \ \% \ of \ \left[\frac{12x - 3x - 8x}{12}\right] $$ $$ = 12 \ \% \ of \ \frac{x}{12} $$ $$ = \frac{12}{100} \times \frac{x}{12} $$ $$ = \frac{x}{100} $$ the total return.$$ = \frac{x}{80} + \frac{x}{15} + \frac{x}{100} $$ $$ = \frac{107x}{1200} $$ Hence the average rate of return. $$ = \frac{107x}{1200} \times \frac{100}{x} $$ $$ = 8.97 \ \% $$

Solution: Let the total capital of the investor is Rs \(x\).

Percentage return on parts of his capital.

5% on his \(\frac{1}{4}\)th capital. $$ = 5 \ \% \ of \ \frac{x}{4} $$ $$ = \frac{5}{100} \times \frac{x}{4} $$ $$ = \frac{x}{80} $$ 10% on his \(\frac{2}{3}\)rd capital. $$ = 10 \ \% \ of \ \frac{2x}{3} $$ $$ = \frac{10}{100} \times \frac{2x}{3} $$ $$ = \frac{x}{15} $$ 12% on his remaining capital $$ = 12 \ \% \ of \ \left[x - \{\frac{x}{4} + \frac{2x}{3}\}\right] $$ $$ = 12 \ \% \ of \ \left[x - \frac{x}{4} - \frac{2x}{3}\right] $$ $$ = 12 \ \% \ of \ \left[\frac{12x - 3x - 8x}{12}\right] $$ $$ = 12 \ \% \ of \ \frac{x}{12} $$ $$ = \frac{12}{100} \times \frac{x}{12} $$ $$ = \frac{x}{100} $$ the total return.$$ = \frac{x}{80} + \frac{x}{15} + \frac{x}{100} $$ $$ = \frac{107x}{1200} $$ Hence the average rate of return. $$ = \frac{107x}{1200} \times \frac{100}{x} $$ $$ = 8.97 \ \% $$

**Trick:** Total percentage return from three parts of the capital = (5 + 10 + 12) % = 27 %

Hence the average rate of return = \(\frac{27}{3}\) = \(9 \ \%\)

Hence approximately average return is 9 %.

- If the salary of a man reduced by 20% then what percent should his reduced salary be raised so as to bring it, as it was before?
- 20 %
- 25 %
- 30 %
- 35 %

Answer: (b) 25 %

Solution: Let the initial salary was Rs 100, then the salary after it reduced by 20% = Rs 80.

The required percentage increase in his reduced salary so as to bring it, as it was before.$$ = \frac{20}{80} \times 100 $$ $$ = 25 \ \% $$

Solution: Let the initial salary was Rs 100, then the salary after it reduced by 20% = Rs 80.

The required percentage increase in his reduced salary so as to bring it, as it was before.$$ = \frac{20}{80} \times 100 $$ $$ = 25 \ \% $$

- How many kg. of pure salt must be added to 50 kg. of 5% solution of salt and water to increase it to a 10% solution?
- 2.25 kg.
- 2.50 kg.
- 2.78 kg.
- 2.97 kg.

Answer: (c) 2.78 kg.

Solution: The current amount of salt in 50 kg solution. $$ = \frac{5}{100} \times 50 $$ $$ = 2.5 \ kg. $$ If \(x\) kg. of pure salt be added then. $$ \frac{2.5 + x}{50 + x} = \frac{10}{100} $$ $$ 25 + 10x = 50 + x $$ $$ 10x - x = 50 - 25 $$ $$ 9x = 25 $$ $$ x = 2.78 \ kg. $$ Hence 2.78 kg. pure salt must be added to increase it to 10% of the solution.

Solution: The current amount of salt in 50 kg solution. $$ = \frac{5}{100} \times 50 $$ $$ = 2.5 \ kg. $$ If \(x\) kg. of pure salt be added then. $$ \frac{2.5 + x}{50 + x} = \frac{10}{100} $$ $$ 25 + 10x = 50 + x $$ $$ 10x - x = 50 - 25 $$ $$ 9x = 25 $$ $$ x = 2.78 \ kg. $$ Hence 2.78 kg. pure salt must be added to increase it to 10% of the solution.

- If A's income is 25% less than that of B, then how much percent is B's income more than that of A?
- 33.34 %
- 37.23 %
- 41.10 %
- 43.34 %

Answer: (a) 33.34 %

Solution: Required percentage $$ = \left[\frac{25}{100 - 25} \times 100\right] $$ $$ = \frac{25}{75} \times 100 $$ $$ = 33.34 \ \% $$

Solution: Required percentage $$ = \left[\frac{25}{100 - 25} \times 100\right] $$ $$ = \frac{25}{75} \times 100 $$ $$ = 33.34 \ \% $$

- If 10 kg. of water has been evaporated from a solution of salt and water, which had 5% salt, the remaining solution has 20% salt then find the weight of the original solution?
- 11.25 kg.
- 13.34 kg.
- 15.25 kg.
- 16.15 kg.

Answer: (b) 13.34 kg.

Solution: Let the weight of the original solution is \(x\) kg. then.

The initial weight of salt in \(x\) kg. solution.$$ = 5 \ \% \ of \ x $$ $$ = \frac{5x}{100} $$ $$ = \frac{x}{20} \ kg. $$ The weight of solution after evaporation \(= (x - 10) \ kg.\)

The weight of the original solution.$$ \frac{x}{20} = \frac{20}{100} \times (x - 10) $$ $$ 5x = 20 \ (x - 10) $$ $$ 5x = 20x - 200 $$ $$ 20x - 5x = 200 $$ $$ 15x = 200 $$ $$ x = 13.34 \ kg. $$

Solution: Let the weight of the original solution is \(x\) kg. then.

The initial weight of salt in \(x\) kg. solution.$$ = 5 \ \% \ of \ x $$ $$ = \frac{5x}{100} $$ $$ = \frac{x}{20} \ kg. $$ The weight of solution after evaporation \(= (x - 10) \ kg.\)

The weight of the original solution.$$ \frac{x}{20} = \frac{20}{100} \times (x - 10) $$ $$ 5x = 20 \ (x - 10) $$ $$ 5x = 20x - 200 $$ $$ 20x - 5x = 200 $$ $$ 15x = 200 $$ $$ x = 13.34 \ kg. $$

Lec 1: Introduction to Percentage
Exercise-1
Lec 2: Percentage Case-1
Exercise-2
Lec 2: Percentage Case-2
Exercise-3
Lec 2: Percentage Case-3 & Case-4
Exercise-4
Exercise-5
Exercise-6
Exercise-7
Exercise-8
Exercise-9
Exercise-10