Overview:

 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.

1. 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?

1. 7.61 %
2. 7.95 %
3. 8.15 %
4. 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 %.

1. 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?

1. 20 %
2. 25 %
3. 30 %
4. 35 %

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.$$= \frac{20}{80} \times 100$$ $$= 25 \ \%$$

1. 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?

1. 2.25 kg.
2. 2.50 kg.
3. 2.78 kg.
4. 2.97 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.

1. If A's income is 25% less than that of B, then how much percent is B's income more than that of A?

1. 33.34 %
2. 37.23 %
3. 41.10 %
4. 43.34 %

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

1. 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?

1. 11.25 kg.
2. 13.34 kg.
3. 15.25 kg.
4. 16.15 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.$$\frac{x}{20} = \frac{20}{100} \times (x - 10)$$ $$5x = 20 \ (x - 10)$$ $$5x = 20x - 200$$ $$20x - 5x = 200$$ $$15x = 200$$ $$x = 13.34 \ kg.$$