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 salary of a man increases \(10 \ \%\) per annum then find the salary of the man after \(2\) years? present salary of the man is \(10,000 \ Rs.\)
- \(12,000 \ Rs.\)
- \(12,100 \ Rs.\)
- \(12,500 \ Rs.\)
- \(12,300 \ Rs.\)

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

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

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

- 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?
- \(2364.523 \ Rs.\)
- \(2689.567 \ Rs.\)
- \(2768.234 \ Rs.\)
- \(2893.518 \ Rs.\)

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

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

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

- 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?
- \(60,500 \ and \ 37,565.74 \)
- \(61,500 \ and \ 38,555.74 \)
- \(65,500 \ and \ 36,165.74 \)
- \(62,500 \ and \ 38,565.74 \)

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 \(x = 3\), $$ = \frac{k}{\left(1 + \frac{n}{100}\right)^x} $$ $$ = \frac{50,000}{\left(1 + \frac{10}{100}\right)^3} $$ $$ = \frac{50000 \times 1000}{1331} = 37,565.74 $$

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 \(x = 3\), $$ = \frac{k}{\left(1 + \frac{n}{100}\right)^x} $$ $$ = \frac{50,000}{\left(1 + \frac{10}{100}\right)^3} $$ $$ = \frac{50000 \times 1000}{1331} = 37,565.74 $$

- 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?
- \(30 \ \%\)
- \(20 \ \%\)
- \(35 \ \%\)
- \(25 \ \%\)

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

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

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

- If the weight of a man is \(60 \ kg\) and increases \(5 \ \%\) per annum then find the weight of the man after \(5\) years?
- \(75.57 \ kg\)
- \(76.57 \ kg\)
- \(72.57 \ kg\)
- \(78.57 \ kg\)

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

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

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

- Find out the \(10 \ \%\) of \(25 \ \%\) of \(20 \ \%\) of \(50 \ ?\)
- \(0.50\)
- \(1.25\)
- \(0.25\)
- \(0.75\)

Answer: (c) \(0.25\)

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

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

- If the population of a city is \(10,000\) and increases \(10 \ \%\) per annum, then find the population of the city \(5\) years ago?
- \(6310.216\)
- \(6209.213\)
- \(6506.328\)
- \(6150.321\)

Answer: (b) \(6209.213\)

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

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

- Find out the \(20 \ \%\) of \(25 \ \%\) of \(50 \ \%\) of \(90 \ ?\)
- \(1.25\)
- \(2.25\)
- \(1.5\)
- \(2.5\)

Answer: (b) \(2.25\)

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

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

- 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?
- \(5500 \ Rs.\)
- \(5604 \ Rs.\)
- \(5305 \ Rs.\)
- \(5202 \ Rs.\)

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

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

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

- 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?
- \(25,500 \ Rs.\)
- \(24,500 \ Rs.\)
- \(24,300 \ Rs.\)
- \(23,600 \ Rs.\)

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

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

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

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