Percentage practice questions for competitive exams

Answer: b. 50
Solution: Let that number be x
According to the question:
$\frac{18}{100}\times x=\frac{12}{100}\times 75$
$x=\frac{12\times 75\times 100}{18\times 100}$
$x=50$

Answer: b. 20%
Solution: let the price of one article be x
Total articles 4+1=5
marked price of 5 articles = 5x
Selling price of 5 articles 4x
Discount= 5x-4x = x Discount in percentage = (x/5x)100= 20%

Answer: b. 280
Solution: Let that number be x
According to the question
$x\times \frac{52}{100}=728$
$x=728\times \frac{100}{52}=1400$
20% of that number will be
$1400\times \frac{20}{100}=280$

Answer: d. 3 litre
Solution: $15\times \frac{20}{100}=3 \mathrm{litre}$

Answer: b. 32
Solution: Let the price of the article be 100rs
Price of the article after 20% and 10% increase be:
100×120%x110%
$100\times \frac{120}{100}\times \frac{110}{100}=132$
The price of the article is increased by 32%

Answer: a. 26,620
Solution: The population of the town after 3 years will be :
$\mathrm{20,000}\times \frac{110}{100}\times \frac{110}{100}\times \frac{110}{100}=\mathrm{26,620}$

Answer: d. 14,641
Solution: There are four months between January and May.
Thus the salary of the employee after 4 months will be
$\mathrm{10,000}\times \frac{110}{100}\times \frac{110}{100}\times \frac{110}{100}\times \frac{110}{100}=\mathrm{14,641}$

Answer: b. 5600
Solution: Let the number of males in the town be x
No. of famales in the town be (9800-x)
According to the question :
$\Rightarrow x\times \frac{108}{100}+(9800–x)\times \frac{105}{100}=10458$
$\Rightarrow 108x+9800\times 105–105x=10458\times 100$
$\Rightarrow 3x+1029000=1045800$
$\Rightarrow 3x=16800$
$\Rightarrow x=5600$

Answer: b. 0.1%
Solution: $0.1\%=\frac{0.1}{100}=0.001$

Answer: a. 150
Solution: Let that number be x
According to the question:
$\Rightarrow 60\%ofx+60=x$
$\Rightarrow \frac{60}{100}\times \mathrm{x}+60=x$
$\Rightarrow \frac{3x}{5}+60=x$
$\Rightarrow \frac{3x}{5}–x=–60$
$\Rightarrow \frac{3x–5x}{5}=–60$
$\Rightarrow 2x=300\Rightarrow x=150$