Percentage Aptitude Questions and Answers:

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

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

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

Answer: (d) \(76043.75\)

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

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

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

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

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

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

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

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

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

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