# 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 income of a women decreases $$5 \ \%$$, $$10 \ \%$$, and $$20 \ \%$$, successively in three years and the present income of the women is $$10,000 \ Rs.$$, then find the income of the women after three years?

1. $$6840 \ Rs.$$
2. $$6800 \ Rs.$$
3. $$6560 \ Rs.$$
4. $$6720 \ Rs.$$

Answer: (a) $$6840 \ Rs.$$

Solution: Given, $$k = 10,000 \ Rs.$$, $$x = 5$$, $$y = 10$$, and $$z = 20$$, then the income of the women after three years, $$\left[k \left(1 - \frac{x}{100}\right) \left(1 - \frac{y}{100}\right) \left(1 - \frac{z}{100}\right)\right]$$ $$10,000 \ \left(1 - \frac{5}{100}\right) \left(1 - \frac{10}{100}\right) \left(1 - \frac{20}{100}\right)$$ $$= 10,000 \times \frac{19}{20} \times \frac{9}{10} \times \frac{4}{5}$$ $$= 6840 \ Rs.$$

1. If the current salary of a person is $$100,000 \ Rs.$$, increases $$10 \ \%$$, $$20 \ \%$$, and $$25 \ \%$$, successively in three years, then find the salary of the person after three years?

1. $$168,000 \ Rs.$$
2. $$165,000 \ Rs.$$
3. $$162,000 \ Rs.$$
4. $$167,000 \ Rs.$$

Answer: (b) $$165,000 \ Rs.$$

Solution: Given, $$k = 100,000 \ Rs.$$, $$x = 10$$, $$y = 20$$, and $$z = 25$$, then the salary of the person after three years, $$\left[k \left(1 + \frac{x}{100}\right) \left(1 + \frac{y}{100}\right) \left(1 + \frac{z}{100}\right)\right]$$ $$100,000 \ \left(1 + \frac{10}{100}\right) \left(1 + \frac{20}{100}\right) \left(1 + \frac{25}{100}\right)$$ $$= 100,000 \times \frac{11}{10} \times \frac{6}{5} \times \frac{5}{4}$$ $$= 165,000 \ Rs.$$

1. If the current price of a mobile is $$7000 \ Rs.$$, increases $$10 \ \%$$ in 1st years and $$20 \ \%$$ in second year then find the percent change of the price of mobile after two years?

1. $$35 \ \%$$
2. $$38 \ \%$$
3. $$33 \ \%$$
4. $$32 \ \%$$

Answer: (d) $$32 \ \%$$

Solution: Given, $$k = 7000 \ Rs.$$, $$x = 10$$, and $$y = 20$$, then $$Percent \ change = x + y + \frac{xy}{100}$$ $$10 + 20 + \frac{10 \times 20}{100} = 32 \ \%$$

1. If the current price of wheat is $$120 \ Rs/kg$$, decreases $$5 \ \%$$ and $$10 \ \%$$ successively in two years, then find the net percent change in the price of wheat after two years?

1. $$-15 \ \%$$
2. $$-20 \ \%$$
3. $$-17 \ \%$$
4. $$-19 \ \%$$

Answer: (c) $$-17 \ \%$$

Solution: Given, $$k = 120$$, $$x = 5$$, and $$y = 10$$, then $$Percent \ change = -x - y - \frac{xy}{100}$$ $$= -5 - 10 - \frac{5 \times 10}{100} = -17 \ \%$$

1. If the current weight of a man is $$70 \ kg$$, increases $$10 \ \%$$ in 1st year and decreases $$20 \ \%$$ in second year, then find the weight of the man after two years?

1. $$65.5 \ kg$$
2. $$66.6 \ kg$$
3. $$61.6 \ kg$$
4. $$68.5 \ kg$$

Answer: (c) $$61.6 \ kg$$

Solution: Given, $$k = 70 \ kg$$, $$x = 10$$, $$y = 20$$, then the weight of the man after two years, $$k \left(1 + \frac{x}{100}\right) \left(1 - \frac{y}{100}\right)$$ $$= 70 \ \left(1 + \frac{10}{100}\right) \left(1 - \frac{20}{100}\right)$$ $$= 70 \times \frac{11}{10} \times \frac{4}{5} = 61.6 \ kg$$

1. If the current rate of Gold is $$30,000 \ Rs.$$ per ten gram, increases $$10 \ \%$$ in first year and decreases $$25 \ \%$$ in second year, then find the net percent change in the rate of the Gold?

1. $$-17.5 \ \%$$
2. $$-18.5 \ \%$$
3. $$-12.5 \ \%$$
4. $$-14.5 \ \%$$

Answer: (a) $$-17.5 \ \%$$

Solution: Given, $$k = 30,000 \ Rs.$$, $$x = 10$$, and $$y = 25$$, then $$Percent \ change = + x - y + \frac{(+x)(-y)}{100}$$ $$= 10 - 25 + \frac{10 \times (-25)}{100}$$ $$10 - 25 - 2.5 = -17.5 \ \%$$ Rate of Gold will be $$17.5 \ \%$$ less than that of current price.

1. If the current income of a man is $$20,000 \ Rs.$$, increases $$20 \ \%$$ and $$25 \ \%$$ successively in two years, then find salary of the man and net percent change in the salary after two years?

1. $$30,000 \ Rs., 50 \ \%$$
2. $$35,000 \ Rs., 60 \ \%$$
3. $$25,000 \ Rs., 40 \ \%$$
4. $$28,000 \ Rs., 45 \ \%$$

Answer: (a) $$30,000, 50 \ \%$$

Solution: Given, $$k = 20,000 \ Rs.$$, $$x = 20$$, and $$y = 25$$, then salary of the man after two years $$k \left(1 + \frac{x}{100}\right) \left(1 + \frac{y}{100}\right)$$ $$= 20,000 \left(1 + \frac{20}{100}\right) \left(1 + \frac{25}{100}\right)$$ $$= 20,000 \times \frac{6}{5} \times \frac{5}{4} = 30,000 \ Rs.$$ and net percent change in the salary of the man is, $$Percent \ change = x + y + \frac{xy}{100}$$ $$= 20 + 25 + \frac{20 \times 25}{100} = 50 \ \%$$

1. If the current price of a laptop is $$40,000 \ Rs.$$, decreases $$5 \ \%$$ and $$10 \ \%$$ successively in two years, then find the price and net percent change in the price after two years?

1. $$30,300 \ Rs., -18.5 \ \%$$
2. $$34,200 \ Rs., -15.5 \ \%$$
3. $$35,600 \ Rs., -16.5 \ \%$$
4. $$28,800 \ Rs., -20.5 \ \%$$

Answer: (b) $$34,200 \ Rs., -15.5 \ \%$$

Solution: Given, $$k = 40,000 \ Rs.$$, $$x = 5$$, and $$y = 10$$, then price of the laptop after two years $$k \left(1 - \frac{x}{100}\right) \left(1 - \frac{y}{100}\right)$$ $$= 40,000 \left(1 - \frac{5}{100}\right) \left(1 - \frac{10}{100}\right)$$ $$= 40,000 \times \frac{19}{20} \times \frac{9}{10} = 34,200 \ Rs.$$ and net percent change in the price of the laptop is, $$Percent \ change = - x - y - \frac{xy}{100}$$ $$= - 5 - 10 - \frac{5 \times 10}{100}$$ $$- 5 - 10 - 0.5 = -15.5 \ \%$$

1. if the current weight of a boy is $$90 \ kg$$, decreases $$10 \ \%$$, $$20 \ \%$$, and $$25 \ \%$$ successively in three years, then find the weight of the boy after three years?

1. $$51.5 \ kg$$
2. $$48.6 \ kg$$
3. $$47.5 \ kg$$
4. $$52.6 \ kg$$

Answer: (b) $$48.6 \ kg$$

Solution: Given, $$k = 90 \ kg$$, $$x = 10$$, $$y = 20$$, and $$z = 25$$, then the weight of the boy after three years, $$\left[k \left(1 - \frac{x}{100}\right) \left(1 - \frac{y}{100}\right) \left(1 - \frac{z}{100}\right)\right]$$ $$\left[90 \ \left(1 - \frac{10}{100}\right) \left(1 - \frac{20}{100}\right) \left(1 - \frac{25}{100}\right)\right]$$ $$= 90 \times \frac{9}{10} \times \frac{4}{5} \times \frac{3}{4}$$ $$= 48.6 \ kg$$

1. If the current price of a book is $$500 \ Rs.$$, increases $$5 \ \%$$ and $$20 \ \%$$, successively in two years, find the price of the book after two years?

1. $$590 \ Rs.$$
2. $$650 \ Rs.$$
3. $$630 \ Rs.$$
4. $$620 \ Rs.$$

Answer: (c) $$630 \ Rs.$$

Solution: Given, $$k = 500 \ kg$$, $$x = 5$$, $$y = 20$$, then the price of the book after two years, $$k \left(1 + \frac{x}{100}\right) \left(1 + \frac{y}{100}\right)$$ $$= 500 \ \left(1 + \frac{5}{100}\right) \left(1 + \frac{20}{100}\right)$$ $$= 500 \times \frac{21}{20} \times \frac{6}{5} = 630 \ Rs.$$