# Percentage Aptitude Questions and Answers:

#### 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. A woman buys a house for Rs 40 Lac and rents it. She puts 15% of each month's rent aside for repairs and pays Rs 3000 as annual tax. If she realizes 20% on her investment thereafter then find the monthly rent of her house?

1. Rs 70135.5
2. Rs 78725.5
3. Rs 79566.5
4. Rs 80221.5

Solution: Let the annual rent of the house is Rs $$x$$ then.

$$x - (15 \ \% \ of \ x + 3000)$$ = $$20 \ \% \ of \ 4,000,000$$

$$x - \frac{15x}{100} - 3000$$ = 800,000

$$x - \frac{15x}{100} = 803,000$$

$$\frac{85x}{100} = 803,000$$

$$x = 944,705.89$$

$$\approx 944,706$$

Hence monthly rent of the house.

$$= \frac{944,706}{12}$$

$$= Rs \ 78,725.5$$

1. If $$z = \frac{x^2}{y}$$ and $$x$$, $$y$$ are both increased in value by 10% then find the percentage change in the value of $$z$$?

1. 10 %
2. 11 %
3. 12 %
4. 13 %

Solution: After increasing the value of $$x$$ and $$y$$ by 10%. $$x = \frac{110x}{100}$$ $$x = \frac{11x}{10}$$ $$y = \frac{110y}{100}$$ $$y = \frac{11y}{10}$$ $$z = \frac{x^2}{y}$$ $$z = \frac{\left[\frac{11x}{10}\right]^2}{\frac{11y}{10}}$$ $$z = \frac{11}{10} \times \frac{x^2}{y}$$ $$z = \frac{11z}{10}$$ The increase in the value of z. $$= \frac{11z}{10} - z$$ $$= \frac{z}{10}$$ Hence the percentage increase in the value of z. $$= \frac{z}{10} \times \frac{100}{z}$$ $$= 10 \ \%$$

1. If the wage of an employee reduced by 25% and again increased by 50% then the final loss or profit of that employee in percentage?

1. 12.5 % Profit
2. 12.5 % Loss
3. 10.5 % Profit
4. 10.5 % Loss

Solution: Let the initial wage of the employee was \$100 then.

Wage after reducing 25% = 75 % of 100. $$= \frac{75}{100} \times 100$$ $$= 75$$ Final wage after increasing 50% = 150 % of 75. $$= \frac{150}{100} \times 75$$ $$= 112.5$$ Hence the employee is in profit of 12.5 %.

1. If the price of petrol increased by 20% then how much percent, a car owner should use less petrol so as not to increase his expenditure on petrol?

1. 10.89 %
2. 12.25 %
3. 15.50 %
4. 16.67 %

Solution: The reduction in petrol consumption. $$= \left[\frac{n}{100 + n} \times 100\right]$$ $$= \left[\frac{20}{100 + 20} \times 100\right]$$ $$= \left[\frac{20}{120} \times 100\right]$$ $$= 16.67 \ \%$$

1. The population of a town is 40,000. If it increases at the rate of 10% per annum then what will be the population of the town after 2 years?

1. 48,250
2. 48,400
3. 48,500
4. 48,650

Solution: Given $$k = 40,000$$, $$n = 10$$, and $$x = 2$$ then the population of the town after 2 years. $$= k \times \left[1 + \frac{n}{100}\right]^x$$ $$= 40,000 \times \left[1 + \frac{10}{100}\right]^2$$ $$= 40,000 \times \frac{11}{10} \times \frac{11}{10}$$ $$= 48,400$$ Hence the population of the town after 2 years will be 48,400.