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 an examination conducted by a school of total \(1000\) marks. Vishal obtained \(10 \ \%\) more marks than Mohan but \(5 \ \%\) less marks than Abhi, and if marks obtained by Mohan is \(600\), then find the percentage marks obtained by Abhi?
- \(67.234 \ \%\)
- \(69.473 \ \%\)
- \(64.576 \ \%\)
- \(66.235 \ \%\)

Answer: (b) \(69.473 \ \%\)

Solution: marks ontained by Mohan = \(600\)

marks obtained by Vishal = \(\frac{110}{100} \times 600 = 660\)

Let marks obtained by Abhi is \(x\) then $$ 660 = \frac{95}{100} \times x $$ $$ x = 694.73 $$ Then marks obtained by Abhi in percent, $$ = \frac{694.73}{1000} \times 100 = 69.473 \ \% $$

Solution: marks ontained by Mohan = \(600\)

marks obtained by Vishal = \(\frac{110}{100} \times 600 = 660\)

Let marks obtained by Abhi is \(x\) then $$ 660 = \frac{95}{100} \times x $$ $$ x = 694.73 $$ Then marks obtained by Abhi in percent, $$ = \frac{694.73}{1000} \times 100 = 69.473 \ \% $$

- If a man spends \(25 \ \%\) of his income on his food, \(10 \ \%\) income on his home expenditure and saves the rest. If the man saved \(4000 \ Rs.\), then find his income?
- \(5925.9 \ Rs.\)
- \(5635.8 \ Rs.\)
- \(6325.9 \ Rs.\)
- \(5763.5 \ Rs.\)

Answer: (a) \(5925.9 \ Rs.\)

Solution: Given, \(x = 25 \ \%\), \(y = 10 \ \%\) and Let the income of the man = \(k \ Rs.\), then $$ k \ \left(1 - \frac{x}{100}\right) \left(1 - \frac{y}{100}\right) = 4000 $$ $$ k \ \left(1 - \frac{25}{100}\right) \left(1 - \frac{10}{100}\right) = 4000 $$ $$ k \times \frac{3}{4} \times \frac{9}{10} = 4000 $$ $$ k = 5925.9 \ Rs. $$

Solution: Given, \(x = 25 \ \%\), \(y = 10 \ \%\) and Let the income of the man = \(k \ Rs.\), then $$ k \ \left(1 - \frac{x}{100}\right) \left(1 - \frac{y}{100}\right) = 4000 $$ $$ k \ \left(1 - \frac{25}{100}\right) \left(1 - \frac{10}{100}\right) = 4000 $$ $$ k \times \frac{3}{4} \times \frac{9}{10} = 4000 $$ $$ k = 5925.9 \ Rs. $$

- If the current population of a city is \(50,000\), increases at the rate of \(15 \ \%\) per annum, then find the population of the city after \(3\) years?
- \(72561.23\)
- \(75235.34\)
- \(76312.65\)
- \(76043.75\)

Answer: (d) \(76043.75\)

Solution: Given, \(k = 50,000\), \(n = 15 \ \%\), \(x = 3\), then the population of the city after \(3\) years, $$ k \ \left(1 + \frac{n}{100}\right)^x $$ $$ = 50,000 \ \left(1 + \frac{15}{100}\right)^3 $$ $$ 50,000 \times \frac{115}{100} \times \frac{115}{100} \times \frac{115}{100} $$ $$ = 76043.75 $$

Solution: Given, \(k = 50,000\), \(n = 15 \ \%\), \(x = 3\), then the population of the city after \(3\) years, $$ k \ \left(1 + \frac{n}{100}\right)^x $$ $$ = 50,000 \ \left(1 + \frac{15}{100}\right)^3 $$ $$ 50,000 \times \frac{115}{100} \times \frac{115}{100} \times \frac{115}{100} $$ $$ = 76043.75 $$

- If \(500\) candidates participated in an examination, out of which \(42 \ \%\) of boys and \(40 \ \%\) of girls passed, then find the total percentage of failed candidates?
- \(60.5 \ \%\)
- \(58.8 \ \%\)
- \(56.7 \ \%\)
- \(57.5 \ \%\)

Answer: (b) \(58.8 \ \%\)

Solution: Total candidates = \(500\)

boys candidates = \(300\)

girls candidates = \(200\)

boys passed candidates $$ = 42 \ \% \ of \ 300 $$ $$ = \frac{42 \times 300}{100} = 126 $$

Girls passed candidates $$ = 40 \ \% \ of \ 200 $$ $$ = \frac{40 \times 200}{100} = 80 $$ Then total passed candidates = \(126 + 80 = 206\), and

total failed candidates = \(500 - 206 = 294\)

Then failed candidates in percent $$ = \left(\frac{294}{500} \times 100 \right) \ \% = 58.8 \ \% $$

Solution: Total candidates = \(500\)

boys candidates = \(300\)

girls candidates = \(200\)

boys passed candidates $$ = 42 \ \% \ of \ 300 $$ $$ = \frac{42 \times 300}{100} = 126 $$

Girls passed candidates $$ = 40 \ \% \ of \ 200 $$ $$ = \frac{40 \times 200}{100} = 80 $$ Then total passed candidates = \(126 + 80 = 206\), and

total failed candidates = \(500 - 206 = 294\)

Then failed candidates in percent $$ = \left(\frac{294}{500} \times 100 \right) \ \% = 58.8 \ \% $$

- If a football team won \(42 \ \%\) of total number of matches the team played in a year, lost \(45 \ \%\) of matches and \(10\) were drawn, then find total number of matches the team played in the year?
- \(80 \ matches\)
- \(75 \ matches\)
- \(78 \ matches\)
- \(77 \ matches\)

Answer: (d) \(77 \ matches\)

Solution: Let total number of matches = \(k\)

matches won by the team = \(42 \ \% \ of \ k = \frac{42 \ k}{100}\)

matches lost by the team = \(45 \ \% \ of \ k = \frac{45 \ k}{100}\)

matches drew by the team = \(10\)

Then total number of matches $$ 42 \ \% \ of \ k + 45 \ \% \ of \ k + 10 = k $$ $$ \frac{42 \ k}{100} + \frac{45 \ k}{100} + 10 = k $$ $$ 42 \ k + 45 \ k + 1000 = 100 \ k $$ $$ k = 76.92 \approx 77 \ matches $$

Solution: Let total number of matches = \(k\)

matches won by the team = \(42 \ \% \ of \ k = \frac{42 \ k}{100}\)

matches lost by the team = \(45 \ \% \ of \ k = \frac{45 \ k}{100}\)

matches drew by the team = \(10\)

Then total number of matches $$ 42 \ \% \ of \ k + 45 \ \% \ of \ k + 10 = k $$ $$ \frac{42 \ k}{100} + \frac{45 \ k}{100} + 10 = k $$ $$ 42 \ k + 45 \ k + 1000 = 100 \ k $$ $$ k = 76.92 \approx 77 \ matches $$

- In an examination a student has to secure \(50 \ \%\) marks to pass. If he obtained \(150\) marks and fails by \(30\) marks, then find the maximum marks of the examination?
- \(320 \ marks\)
- \(350 \ marks\)
- \(360 \ marks\)
- \(375 \ marks\)

Answer: (c) \(360 \ marks\)

Solution: Let maximum marks = \(k\)

the student has to secure marks = \(50 \ \% \ of \ k\)

marks obtained by student = \(150\) marks

the student fails by marks = \(30\) marks, then

maximum marks, $$ 50 \ \% \ of \ k = 150 + 30 $$ $$ \frac{50 \ k}{100} = 180 $$ $$ k = 360 \ marks $$

Solution: Let maximum marks = \(k\)

the student has to secure marks = \(50 \ \% \ of \ k\)

marks obtained by student = \(150\) marks

the student fails by marks = \(30\) marks, then

maximum marks, $$ 50 \ \% \ of \ k = 150 + 30 $$ $$ \frac{50 \ k}{100} = 180 $$ $$ k = 360 \ marks $$

- If there is aexamination and maximum marks of an examination were \(1000\). Four students A, B, C and D participated in the examination. A obtained \(20 \ \%\) less than B, B obtained \(10 \ \%\) less than C, C obtained \(25 \ \%\) more than D. If A obtained \(200\) marks, then find percentage marks obtained by D?
- \(21.5 \ \%\)
- \(22.2 \ \%\)
- \(23.3 \ \%\)
- \(24.2 \ \%\)

Answer: (b) \(22.2 \ \%\)

Solution: marks obtained by A = \(200\)

marks obtained by B = \(\frac{200}{80} \times 100 = 250\)

marks obtained by C = \(\frac{250}{90} \times 100 = 277.78\)

marks obtained by D = \(\frac{277.78}{125} \times 100 = 222.224\)

Hence percentage marks obtained by D $$ \frac{222.224}{1000} \times 100 = 22.2 \ \% $$

Solution: marks obtained by A = \(200\)

marks obtained by B = \(\frac{200}{80} \times 100 = 250\)

marks obtained by C = \(\frac{250}{90} \times 100 = 277.78\)

marks obtained by D = \(\frac{277.78}{125} \times 100 = 222.224\)

Hence percentage marks obtained by D $$ \frac{222.224}{1000} \times 100 = 22.2 \ \% $$

- There are \(100\) students and \(10\) teachers in a school. In an event organised in the school, sweet pieces were distributed, each student got sweet pieces which are \(5 \ \%\) of the total number of students and each teacher got sweet pieces which are \(10 \ \%\) of the total number of students, then find the total number of sweet pieces distributed in the school?
- \(550 \ pieces\)
- \(620 \ pieces\)
- \(600 \ pieces\)
- \(630 \ pieces\)

Answer: (c) \(600 \ pieces\)

Solution: Total number of students = \(100\)

total number of teachers = \(10\)

each student got the sweet pieces $$ = 5 \ \% \ of \ 100 $$ $$ = \frac{5 \times 100}{100} = 5 \ pieces $$ each teacher got the sweet pieces $$ = 10 \ \% \ of \ 100 $$ $$ = \frac{10 \times 100}{100} = 10 \ pieces $$ Hence total number of sweet pieces distributed in the school $$ = 5 \times 100 + 10 \times 10 $$ $$ = 500 + 100 = 600 \ pieces $$

Solution: Total number of students = \(100\)

total number of teachers = \(10\)

each student got the sweet pieces $$ = 5 \ \% \ of \ 100 $$ $$ = \frac{5 \times 100}{100} = 5 \ pieces $$ each teacher got the sweet pieces $$ = 10 \ \% \ of \ 100 $$ $$ = \frac{10 \times 100}{100} = 10 \ pieces $$ Hence total number of sweet pieces distributed in the school $$ = 5 \times 100 + 10 \times 10 $$ $$ = 500 + 100 = 600 \ pieces $$

- If the current expenditure of a man is \(8000 \ Rs.\), increases \(25 \ \%\) per annum, then find the expenditure of the man after three years?
- \(16,525 \ Rs.\)
- \(15,250 \ Rs.\)
- \(15,350 \ Rs.\)
- \(15,625 \ Rs.\)

Answer: (d) \(15,625 \ Rs.\)

Solution: Given, \(k = 8000 \ Rs.\), \(n = 25\), \(x = 3\), then the expenditure of the man after \(3\) years, $$ = k \ \left(1 + \frac{n}{100}\right)^x $$ $$ = 8000 \ \left(1 + \frac{25}{100}\right)^3 $$ $$ = 8000 \times \frac{5}{4} \times \frac{5}{4} \times \frac{5}{4} $$ $$ = 125 \times 125 = 15,625 \ Rs. $$

Solution: Given, \(k = 8000 \ Rs.\), \(n = 25\), \(x = 3\), then the expenditure of the man after \(3\) years, $$ = k \ \left(1 + \frac{n}{100}\right)^x $$ $$ = 8000 \ \left(1 + \frac{25}{100}\right)^3 $$ $$ = 8000 \times \frac{5}{4} \times \frac{5}{4} \times \frac{5}{4} $$ $$ = 125 \times 125 = 15,625 \ Rs. $$

- If current fee of a course is \(70,000 \ Rs.\), decreases \(7 \ \%\) per annum, then find the fee of the course after \(2\) years?
- \(62,350 \ Rs.\)
- \(60,543 \ Rs.\)
- \(59,342 \ Rs.\)
- \(58,445 \ Rs.\)

Answer: (b) \(60,543 \ Rs.\)

Solution: Given, \(k = 70,000 \ Rs.\), \(n = 7 \ \%\), \(x = 2\), then the fee of course after \(2\) years, $$ = k \ \left(1 - \frac{n}{100}\right)^x $$ $$ = 70,000 \ \left(1 - \frac{7}{100}\right)^2 $$ $$ = 70,000 \times \frac{93}{100} \times \frac{93}{100} $$ $$ = 60,543 \ Rs. $$

Solution: Given, \(k = 70,000 \ Rs.\), \(n = 7 \ \%\), \(x = 2\), then the fee of course after \(2\) years, $$ = k \ \left(1 - \frac{n}{100}\right)^x $$ $$ = 70,000 \ \left(1 - \frac{7}{100}\right)^2 $$ $$ = 70,000 \times \frac{93}{100} \times \frac{93}{100} $$ $$ = 60,543 \ Rs. $$

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