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 $$ = k \ \left(1 - \frac{n}{100}\right)^x $$ $$ = 30,000 \ \left(1 - \frac{10}{100}\right)^2 $$ $$ = 30,000 \times \frac{9}{10} \times \frac{9}{10} $$ $$ = 300 \times 81 = 24,300 \ Rs. $$