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. |

- 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?
- \(6840 \ Rs.\)
- \(6800 \ Rs.\)
- \(6560 \ Rs.\)
- \(6720 \ Rs.\)

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, $$ \left[k \left(1 - \frac{x}{100}\right) \left(1 - \frac{y}{100}\right) \left(1 - \frac{z}{100}\right)\right]$$ $$ 10,000 \ \left(1 - \frac{5}{100}\right) \left(1 - \frac{10}{100}\right) \left(1 - \frac{20}{100}\right)$$ $$ = 10,000 \times \frac{19}{20} \times \frac{9}{10} \times \frac{4}{5} $$ $$ = 6840 \ Rs. $$

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

- 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?
- \(168,000 \ Rs.\)
- \(165,000 \ Rs.\)
- \(162,000 \ Rs.\)
- \(167,000 \ Rs.\)

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, $$ \left[k \left(1 + \frac{x}{100}\right) \left(1 + \frac{y}{100}\right) \left(1 + \frac{z}{100}\right)\right]$$ $$ 100,000 \ \left(1 + \frac{10}{100}\right) \left(1 + \frac{20}{100}\right) \left(1 + \frac{25}{100}\right)$$ $$ = 100,000 \times \frac{11}{10} \times \frac{6}{5} \times \frac{5}{4} $$ $$ = 165,000 \ Rs. $$

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

- 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?
- \(35 \ \%\)
- \(38 \ \%\)
- \(33 \ \%\)
- \(32 \ \%\)

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

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

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

- 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?
- \(-15 \ \%\)
- \(-20 \ \%\)
- \(-17 \ \%\)
- \(-19 \ \%\)

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

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

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

- 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?
- \(65.5 \ kg\)
- \(66.6 \ kg\)
- \(61.6 \ kg\)
- \(68.5 \ kg\)

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

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

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

- 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?
- \(-17.5 \ \%\)
- \(-18.5 \ \%\)
- \(-12.5 \ \%\)
- \(-14.5 \ \%\)

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.

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.

- 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?
- \(30,000 \ Rs., 50 \ \%\)
- \(35,000 \ Rs., 60 \ \%\)
- \(25,000 \ Rs., 40 \ \%\)
- \(28,000 \ Rs., 45 \ \%\)

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, $$ Percent \ change = x + y + \frac{xy}{100} $$ $$ = 20 + 25 + \frac{20 \times 25}{100} = 50 \ \% $$

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, $$ Percent \ change = x + y + \frac{xy}{100} $$ $$ = 20 + 25 + \frac{20 \times 25}{100} = 50 \ \% $$

- 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?
- \(30,300 \ Rs., -18.5 \ \%\)
- \(34,200 \ Rs., -15.5 \ \%\)
- \(35,600 \ Rs., -16.5 \ \%\)
- \(28,800 \ Rs., -20.5 \ \%\)

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, $$ Percent \ change = - x - y - \frac{xy}{100} $$ $$ = - 5 - 10 - \frac{5 \times 10}{100} $$ $$ - 5 - 10 - 0.5 = -15.5 \ \% $$

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, $$ Percent \ change = - x - y - \frac{xy}{100} $$ $$ = - 5 - 10 - \frac{5 \times 10}{100} $$ $$ - 5 - 10 - 0.5 = -15.5 \ \% $$

- 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?
- \(51.5 \ kg\)
- \(48.6 \ kg\)
- \(47.5 \ kg\)
- \(52.6 \ kg\)

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, $$ \left[k \left(1 - \frac{x}{100}\right) \left(1 - \frac{y}{100}\right) \left(1 - \frac{z}{100}\right)\right]$$ $$ \left[90 \ \left(1 - \frac{10}{100}\right) \left(1 - \frac{20}{100}\right) \left(1 - \frac{25}{100}\right)\right]$$ $$ = 90 \times \frac{9}{10} \times \frac{4}{5} \times \frac{3}{4} $$ $$ = 48.6 \ kg $$

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

- 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?
- \(590 \ Rs.\)
- \(650 \ Rs.\)
- \(630 \ Rs.\)
- \(620 \ Rs.\)

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

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

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

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