# 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 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?

1. $$67.234 \ \%$$
2. $$69.473 \ \%$$
3. $$64.576 \ \%$$
4. $$66.235 \ \%$$

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, $$= \frac{694.73}{1000} \times 100 = 69.473 \ \%$$

1. 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?

1. $$5925.9 \ Rs.$$
2. $$5635.8 \ Rs.$$
3. $$6325.9 \ Rs.$$
4. $$5763.5 \ Rs.$$

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

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

1. 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?

1. $$72561.23$$
2. $$75235.34$$
3. $$76312.65$$
4. $$76043.75$$

Answer: (d) $$76043.75$$

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

1. 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?

1. $$60.5 \ \%$$
2. $$58.8 \ \%$$
3. $$56.7 \ \%$$
4. $$57.5 \ \%$$

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 $$= \left(\frac{294}{500} \times 100 \right) \ \% = 58.8 \ \%$$

1. 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?

1. $$80 \ matches$$
2. $$75 \ matches$$
3. $$78 \ matches$$
4. $$77 \ matches$$

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

1. 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?

1. $$320 \ marks$$
2. $$350 \ marks$$
3. $$360 \ marks$$
4. $$375 \ marks$$

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

1. 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?

1. $$21.5 \ \%$$
2. $$22.2 \ \%$$
3. $$23.3 \ \%$$
4. $$24.2 \ \%$$

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 $$\frac{222.224}{1000} \times 100 = 22.2 \ \%$$

1. 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?

1. $$550 \ pieces$$
2. $$620 \ pieces$$
3. $$600 \ pieces$$
4. $$630 \ pieces$$

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 $$= 5 \times 100 + 10 \times 10$$ $$= 500 + 100 = 600 \ pieces$$

1. If the current expenditure of a man is $$8000 \ Rs.$$, increases $$25 \ \%$$ per annum, then find the expenditure of the man after three years?

1. $$16,525 \ Rs.$$
2. $$15,250 \ Rs.$$
3. $$15,350 \ Rs.$$
4. $$15,625 \ Rs.$$

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

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

1. If current fee of a course is $$70,000 \ Rs.$$, decreases $$7 \ \%$$ per annum, then find the fee of the course after $$2$$ years?

1. $$62,350 \ Rs.$$
2. $$60,543 \ Rs.$$
3. $$59,342 \ Rs.$$
4. $$58,445 \ Rs.$$

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

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