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 there is a couple and the income of husband is \(25 \ \%\) more than that of wife's income, then find how much percent of wife's income is less than that of husband's income?
- \(25 \ \%\)
- \(20 \ \%\)
- \(15 \ \%\)
- \(22 \ \%\)

Answer: (b) \(20 \ \%\)

Solution: Given, \(n = 25 \ \%\), then wife's income is less than that of husband's income in percent, $$ = \left(\frac{n}{100 + n} \times 100 \right) \ \% $$ $$ = \left(\frac{25}{100 + 25} \times 100 \right) \ \% $$ $$ = \left(\frac{25}{125} \times 100 \right) \ \% $$ $$ = 20 \ \% $$

Solution: Given, \(n = 25 \ \%\), then wife's income is less than that of husband's income in percent, $$ = \left(\frac{n}{100 + n} \times 100 \right) \ \% $$ $$ = \left(\frac{25}{100 + 25} \times 100 \right) \ \% $$ $$ = \left(\frac{25}{125} \times 100 \right) \ \% $$ $$ = 20 \ \% $$

- If the price of rice increases by \(10 \ \%\), then find how much percent of rice consumption be reduced so as not to increase the expenditure?
- \(9.09 \ \%\)
- \(8.25 \ \%\)
- \(9.25 \ \%\)
- \(8.09 \ \%\)

Answer: (a) \(9.09 \ \%\)

Solution: Given, \(n = 10 \ \%\), then the reduction in rice consumption, $$ = \left(\frac{n}{100 + n} \times 100 \right) \ \% $$ $$ = \left(\frac{10}{100 + 10} \times 100 \right) \ \% $$ $$ = \left(\frac{10}{110} \times 100 \right) \ \% $$ $$ = 9.09 \ \% $$

Solution: Given, \(n = 10 \ \%\), then the reduction in rice consumption, $$ = \left(\frac{n}{100 + n} \times 100 \right) \ \% $$ $$ = \left(\frac{10}{100 + 10} \times 100 \right) \ \% $$ $$ = \left(\frac{10}{110} \times 100 \right) \ \% $$ $$ = 9.09 \ \% $$

- If the price of wheat decreases by \(20 \ \%\), then find how much percent of wheat consumption be increased so as not to decrease the expenditure?
- \(20 \ \%\)
- \(28 \ \%\)
- \(25 \ \%\)
- \(30 \ \%\)

Answer: (c) \(25 \ \%\)

Solution: Given, \(n = 20 \ \%\), then the increase in wheat consumption, $$ = \left(\frac{n}{100 - n} \times 100 \right) \ \% $$ $$ = \left(\frac{20}{100 - 20} \times 100 \right) \ \% $$ $$ = \left(\frac{20}{80} \times 100 \right) \ \% $$ $$ = 25 \ \% $$

Solution: Given, \(n = 20 \ \%\), then the increase in wheat consumption, $$ = \left(\frac{n}{100 - n} \times 100 \right) \ \% $$ $$ = \left(\frac{20}{100 - 20} \times 100 \right) \ \% $$ $$ = \left(\frac{20}{80} \times 100 \right) \ \% $$ $$ = 25 \ \% $$

- If the income of Ram is \(10 \ \%\) less than that of Shyam's income, then find how much percent of Shyam's income is more than that of Ram's income?
- \(10.50 \ \%\)
- \(11.50 \ \%\)
- \(11.11 \ \%\)
- \(10.11 \ \%\)

Answer: (c) \(11.11 \ \%\)

Solution: Given, \(n = 10 \ \%\), then the Shyam's income is more than that of Ram's income in percent, $$ = \left(\frac{n}{100 - n} \times 100 \right) \ \% $$ $$ = \left(\frac{10}{100 - 10} \times 100 \right) \ \% $$ $$ = \left(\frac{10}{90} \times 100 \right) \ \% $$ $$ = 11.11 \ \% $$

Solution: Given, \(n = 10 \ \%\), then the Shyam's income is more than that of Ram's income in percent, $$ = \left(\frac{n}{100 - n} \times 100 \right) \ \% $$ $$ = \left(\frac{10}{100 - 10} \times 100 \right) \ \% $$ $$ = \left(\frac{10}{90} \times 100 \right) \ \% $$ $$ = 11.11 \ \% $$

- If the price of sugar increases \(15 \ \%\), then find how much percent of sugar comsumption be reduced so as not to increase the expenditure?
- \(15.043 \ \%\)
- \(17.042 \ \%\)
- \(18.042 \ \%\)
- \(13.043 \ \%\)

Answer: (d) \(13.043 \ \%\)

Solution: Given, \(n = 15 \ \%\), then the reduction in sugar consumption, $$ = \left(\frac{n}{100 + n} \times 100 \right) \ \% $$ $$ = \left(\frac{15}{100 + 15} \times 100 \right) \ \% $$ $$ = \left(\frac{15}{115} \times 100 \right) \ \% $$ $$ = 13.043 \ \% $$

Solution: Given, \(n = 15 \ \%\), then the reduction in sugar consumption, $$ = \left(\frac{n}{100 + n} \times 100 \right) \ \% $$ $$ = \left(\frac{15}{100 + 15} \times 100 \right) \ \% $$ $$ = \left(\frac{15}{115} \times 100 \right) \ \% $$ $$ = 13.043 \ \% $$

- If the price of milk decreases \(5 \ \%\), then find how much percent of milk consumption be increased so as not to decrease the expenditure?
- \(5.26 \ \%\)
- \(6.25 \ \%\)
- \(5.05 \ \%\)
- \(6.05 \ \%\)

Answer: (a) \(5.26 \ \%\)

Solution: Given, \(n = 5 \ \%\), then the increase in milk consumption, $$ = \left(\frac{n}{100 - n} \times 100 \right) \ \% $$ $$ = \left(\frac{5}{100 - 5} \times 100 \right) \ \% $$ $$ = \left(\frac{5}{95} \times 100 \right) \ \% $$ $$ = 5.26 \ \% $$

Solution: Given, \(n = 5 \ \%\), then the increase in milk consumption, $$ = \left(\frac{n}{100 - n} \times 100 \right) \ \% $$ $$ = \left(\frac{5}{100 - 5} \times 100 \right) \ \% $$ $$ = \left(\frac{5}{95} \times 100 \right) \ \% $$ $$ = 5.26 \ \% $$

- If the current price of a x-ray machine is \(500,000 \ Rs.\), increases \(2 \ \%\) and \(10 \ \%\) successively in two years, then find the price of x-ray machine after two years?
- \(565,000 \ Rs.\)
- \(560,000 \ Rs.\)
- \(561,000 \ Rs.\)
- \(568,000 \ Rs.\)

Answer: (c) \(561,000 \ Rs.\)

Solution: Given, \(k = 500,000 \ Rs.\), \(x = 2\), \(y = 10\), then the price of the x-ray machine after two years, $$ k \left(1 + \frac{x}{100}\right) \left(1 + \frac{y}{100}\right)$$ $$ = 500,000 \ \left(1 + \frac{2}{100}\right) \left(1 + \frac{10}{100}\right)$$ $$ = 500,000 \times \frac{51}{50} \times \frac{11}{10} = 561,000 \ Rs. $$

Solution: Given, \(k = 500,000 \ Rs.\), \(x = 2\), \(y = 10\), then the price of the x-ray machine after two years, $$ k \left(1 + \frac{x}{100}\right) \left(1 + \frac{y}{100}\right)$$ $$ = 500,000 \ \left(1 + \frac{2}{100}\right) \left(1 + \frac{10}{100}\right)$$ $$ = 500,000 \times \frac{51}{50} \times \frac{11}{10} = 561,000 \ Rs. $$

- If the weight of mohan is \(25 \ \%\) less than that of Rohan's weight, then find how much percent of Rohan's weight is more than that of Mohan's weight in percent?
- \(35.50 \ \%\)
- \(33.33 \ \%\)
- \(31.33 \ \%\)
- \(32.25 \ \%\)

Answer: (b) \(33.33 \ \%\)

Solution: Given, \(n = 25 \ \%\), then Rohan's weight is more than that of Mohan's weight in percent, $$ = \left(\frac{n}{100 - n} \times 100 \right) \ \% $$ $$ = \left(\frac{25}{100 - 25} \times 100 \right) \ \% $$ $$ = \left(\frac{25}{75} \times 100 \right) \ \% $$ $$ = 33.33 \ \% $$

Solution: Given, \(n = 25 \ \%\), then Rohan's weight is more than that of Mohan's weight in percent, $$ = \left(\frac{n}{100 - n} \times 100 \right) \ \% $$ $$ = \left(\frac{25}{100 - 25} \times 100 \right) \ \% $$ $$ = \left(\frac{25}{75} \times 100 \right) \ \% $$ $$ = 33.33 \ \% $$

- If the price of vegetables increases by \(2 \ \%\), then find how much percent of vegetable consumption be reduced so as not to increase the expenditure?
- \(1.50 \ \%\)
- \(2.25 \ \%\)
- \(2.82 \ \%\)
- \(1.96 \ \%\)

Answer: (d) \(1.96 \ \%\)

Solution: Given, \(n = 2 \ \%\), then the reduction in vegetable consumption, $$ = \left(\frac{n}{100 + n} \times 100 \right) \ \% $$ $$ = \left(\frac{2}{100 + 2} \times 100 \right) \ \% $$ $$ = \left(\frac{2}{102} \times 100 \right) \ \% $$ $$ = 1.96 \ \% $$

Solution: Given, \(n = 2 \ \%\), then the reduction in vegetable consumption, $$ = \left(\frac{n}{100 + n} \times 100 \right) \ \% $$ $$ = \left(\frac{2}{100 + 2} \times 100 \right) \ \% $$ $$ = \left(\frac{2}{102} \times 100 \right) \ \% $$ $$ = 1.96 \ \% $$

- If the price of water decreases by \(7 \ \%\), then find how much percent of water consumption be increased so as not to decrease the expenditure?
- \(8.52 \ \%\)
- \(7.52 \ \%\)
- \(7.25 \ \%\)
- \(8.25 \ \%\)

Answer: (b) \(7.52 \ \%\)

Solution: Given, \(n = 7 \ \%\), then the increase in water consumption, $$ = \left(\frac{n}{100 - n} \times 100 \right) \ \% $$ $$ = \left(\frac{7}{100 - 7} \times 100 \right) \ \% $$ $$ = \left(\frac{7}{93} \times 100 \right) \ \% $$ $$ = 7.52 \ \% $$

Solution: Given, \(n = 7 \ \%\), then the increase in water consumption, $$ = \left(\frac{n}{100 - n} \times 100 \right) \ \% $$ $$ = \left(\frac{7}{100 - 7} \times 100 \right) \ \% $$ $$ = \left(\frac{7}{93} \times 100 \right) \ \% $$ $$ = 7.52 \ \% $$

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