# 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. If the salary of a man increases $$10 \ \%$$ per annum then find the salary of the man after $$2$$ years? present salary of the man is $$10,000 \ Rs.$$

1. $$12,000 \ Rs.$$
2. $$12,100 \ Rs.$$
3. $$12,500 \ Rs.$$
4. $$12,300 \ Rs.$$

Answer: (b) $$12,100 \ Rs.$$

Solution: Given, $$k = 10,000 \ Rs.$$, $$n = 10$$, $$x = 2$$, then the salary of the man after $$2$$ years $$= k \ \left(1 + \frac{n}{100}\right)^x$$ $$= 10,000 \ \left(1 + \frac{10}{100}\right)^2$$ $$= 10,000 \times \frac{11}{10} \times \frac{11}{10}$$ $$= 12,100 \ Rs.$$

1. If the income of a women is $$5,000 \ Rs.$$ and increases $$20 \ \%$$ per annum then find the income of the women $$3$$ years ago?

1. $$2364.523 \ Rs.$$
2. $$2689.567 \ Rs.$$
3. $$2768.234 \ Rs.$$
4. $$2893.518 \ Rs.$$

Answer: (d) $$2893.518 \ Rs.$$

Solution: Given, $$k = 5,000 \ Rs.$$, $$n = 20$$, $$x = 3$$, then the income of the women $$3$$ years ago, $$= \frac{k}{\left(1 + \frac{n}{100}\right)^x}$$ $$= \frac{5,000}{\left(1 + \frac{20}{100}\right)^3}$$ $$= \frac{5000 \times 125}{216} = 2893.518 \ Rs.$$

1. If the population of a country is $$50,000$$ and increases $$10 \ \%$$ per annum, then find the population of the country after $$2$$ years and $$3$$ years ago?

1. $$60,500 \ and \ 37,565.74$$
2. $$61,500 \ and \ 38,555.74$$
3. $$65,500 \ and \ 36,165.74$$
4. $$62,500 \ and \ 38,565.74$$

Answer: (a) $$60,500 \ and \ 37,565.74$$

Solution: Given, $$k = 50,000 \ Rs.$$, $$n = 10$$, $$x = 2$$ then the population of the country after $$2$$ years $$= k \ \left(1 + \frac{n}{100}\right)^x$$ $$= 50,000 \ \left(1 + \frac{10}{100}\right)^2$$ $$= 50,000 \times \frac{11}{10} \times \frac{11}{10}$$ $$= 60,500$$ population of the country $$3$$ years ago $$x = 3$$, $$= \frac{k}{\left(1 + \frac{n}{100}\right)^x}$$ $$= \frac{50,000}{\left(1 + \frac{10}{100}\right)^3}$$ $$= \frac{50000 \times 1000}{1331} = 37,565.74$$

1. If the weight of John is $$20 \ \%$$ less than that of Jack's weight, then find how much of percent is jack's weight more than that of John's weight?

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

Answer: (d) $$25 \ \%$$

Solution: Given, $$n = 20 \ \%$$, then the weight of Jack is more than that of John's weight $$\left[\frac{n}{100 - n} \times 100\right] \ \%$$ $$= \left[\frac{20}{100 - 20} \times 100\right] \ \%$$ $$\frac{20}{80} \times 100 = 25 \ \%$$

1. If the weight of a man is $$60 \ kg$$ and increases $$5 \ \%$$ per annum then find the weight of the man after $$5$$ years?

1. $$75.57 \ kg$$
2. $$76.57 \ kg$$
3. $$72.57 \ kg$$
4. $$78.57 \ kg$$

Answer: (b) $$76.57 \ kg$$

Solution: Given, $$k = 60 \ kg$$, $$n = 5$$, $$x = 5$$, then the weight of the man after $$5$$ years $$= k \ \left(1 + \frac{n}{100}\right)^x$$ $$= 60 \ \left(1 + \frac{5}{100}\right)^5$$ $$= 60 \times \left(\frac{11}{10}\right)^5$$ $$= \frac{245,046,060}{3,200,000} = 76.57 \ kg$$

1. Find out the $$10 \ \%$$ of $$25 \ \%$$ of $$20 \ \%$$ of $$50 \ ?$$

1. $$0.50$$
2. $$1.25$$
3. $$0.25$$
4. $$0.75$$

Answer: (c) $$0.25$$

Solution: $$= \frac{10}{100} \times \frac{25}{100} \times \frac{20}{100} \times 50$$ $$= \frac{50}{200} = 0.25$$

1. If the population of a city is $$10,000$$ and increases $$10 \ \%$$ per annum, then find the population of the city $$5$$ years ago?

1. $$6310.216$$
2. $$6209.213$$
3. $$6506.328$$
4. $$6150.321$$

Answer: (b) $$6209.213$$

Solution: Given, $$k = 10,000$$, $$n = 10$$, $$x = 5$$, then the population of the city $$5$$ years ago, $$= \frac{k}{\left(1 + \frac{n}{100}\right)^x}$$ $$= \frac{10,000}{\left(1 + \frac{10}{100}\right)^5}$$ $$= \frac{10,000 \times 100,000}{161,051} = 6209.213$$

1. Find out the $$20 \ \%$$ of $$25 \ \%$$ of $$50 \ \%$$ of $$90 \ ?$$

1. $$1.25$$
2. $$2.25$$
3. $$1.5$$
4. $$2.5$$

Answer: (b) $$2.25$$

Solution: $$= \frac{20}{100} \times \frac{25}{100} \times \frac{50}{100} \times 90$$ $$= \frac{90}{40} = 2.25$$

1. If the expenditure of a man is $$5000 \ Rs.$$ per year and increases $$2 \ \%$$ per annum, then find the expenditure of the man after $$2$$ years?

1. $$5500 \ Rs.$$
2. $$5604 \ Rs.$$
3. $$5305 \ Rs.$$
4. $$5202 \ Rs.$$

Answer: (d) $$5202 \ Rs.$$

Solution: Given, $$k = 5000 \ Rs.$$, $$n = 2$$, $$x = 2$$, then the expenditure of the man after $$2$$ years $$= k \ \left(1 + \frac{n}{100}\right)^x$$ $$= 5000 \ \left(1 + \frac{2}{100}\right)^2$$ $$= 5000 \times \frac{51}{50} \times \frac{51}{50}$$ $$= 5202 \ Rs.$$

1. If the price of a computer is $$30,000 \ Rs.$$ and decreases $$10 \ \%$$ per annum what will be the price of computer after $$2$$ years?

1. $$25,500 \ Rs.$$
2. $$24,500 \ Rs.$$
3. $$24,300 \ Rs.$$
4. $$23,600 \ Rs.$$

Answer: (c) $$24,300 \ Rs.$$

Solution: Given, $$k = 30,000 \ Rs.$$, $$n = 10$$, $$x = 2$$, then the price of computer after $$2$$ years $$= k \ \left(1 - \frac{n}{100}\right)^x$$ $$= 30,000 \ \left(1 - \frac{10}{100}\right)^2$$ $$= 30,000 \times \frac{9}{10} \times \frac{9}{10}$$ $$= 300 \times 81 = 24,300 \ Rs.$$