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. |

- 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?
- Rs 70135.5
- Rs 78725.5
- Rs 79566.5
- Rs 80221.5

Answer: (b) Rs 78725.5

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.

\(= \frac{944,706}{12}\)

\(= Rs \ 78,725.5\)

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.

\(= \frac{944,706}{12}\)

\(= Rs \ 78,725.5\)

- 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 \(z\)?
- 10 %
- 11 %
- 12 %
- 13 %

Answer: (a) 10 %

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. $$ = \frac{z}{10} \times \frac{100}{z} $$ $$ = 10 \ \% $$

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. $$ = \frac{z}{10} \times \frac{100}{z} $$ $$ = 10 \ \% $$

- 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?
- 12.5 % Profit
- 12.5 % Loss
- 10.5 % Profit
- 10.5 % Loss

Answer: (a) 12.5 % Profit

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 %.

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 %.

- 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?
- 10.89 %
- 12.25 %
- 15.50 %
- 16.67 %

Answer: (d) 16.67 %

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

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

- 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?
- 48,250
- 48,400
- 48,500
- 48,650

Answer: (b) 48,400

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.

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.

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