Practice MCQ Questions and Answer on Percentage

541.

A book consist of 30 pages, 25 line on each page and 35 characters on each line. If this content is written in another note book consisting 30 lines and 28 characters per line then the required no. of pages will how much percent greater than previous pages?

(A) 4.16%
(B) 5%
(C) 6.66%
(D) 7%

Solution: $\begin{aligned} Let the required number of pages be x . \\ 30 \times 25 \times 35 = x \times 30 \times 28 \\ x = 31.25 \approx 32 \\ % increase in number of pages, \\ = \frac{2}{30} \times 100 \\ = 6.66 % \end{aligned}$

542.

In an examination, 35% of the candidates failed in Mathematics and 25% in English. If 10% failed in both Mathematics and English, then how much percent of candidates passed in both the subjects ?

(A) 50%
(B) 55%
(C) 57%
(D) 60%

543.

A shop sells floor tiles at Rs. 48 per square meter. A contractor employs a machine that polishes the tiles that damages 10% of the total number of tiles which cannot be used any more. Calculate the amount that needs to be paid by contractor to tile shop owner, if the hall is of a square shape and has a perimeter of 400 meters?

(A) Rs. 4,00,000
(B) Rs. 5,00,000
(C) Rs. 5,28,000
(D) Rs. 3,65,000

544.

When income tax is 3 paise in a rupee, a person's net income is Rs. 237650. What will it be when the income tax is raised to 7 paise ?

(A) Rs. 233000
(B) Rs. 231650
(C) Rs. 227850
(D) None of these

Solution: Let the person's gross income be Rs. $x$ Then, ⇒ (100 - 3)% of $x$ = 237650 ⇒ 97% of $x$ = 237650 ⇒ $\frac{97 x}{100}$ = 237650 ⇒ $x$ = $(\frac{237650 \times 100}{97})$ ⇒ $x$ = 245000 ∴ When income tax is raised to 7%, we have : Net income : = (100 - 7)% of Rs. 245000 = 93% of Rs. 245000 = Rs. $(\frac{93}{100} \times 245000)$ = Rs. 227850

545.

The value of which of the following fractions is less than twenty percent ?

(A) $\frac{5}{61}$
(B) $\frac{2}{3}$
(C) $\frac{2}{5}$
(D) $\frac{1}{4}$

Solution: Answer & Solution Answer: Option E Solution: $\begin{aligned} \frac{5}{6} = (\frac{5}{6} \times 100) % \\ = 83 \frac{1}{3} % \\ \frac{2}{3} = (\frac{2}{3} \times 100) % \\ = 66 \frac{2}{3} % \\ \frac{2}{5} = (\frac{2}{5} \times 100) % \\ = 40 % \\ \frac{1}{4} = (\frac{1}{4} \times 100) % \\ = 25 % \\ \frac{2}{11} = (\frac{2}{11} \times 100) % \\ = 18 \frac{2}{11} % 20 % \end{aligned}$

546.

What percent of 7.2 kg is 18 gms ?

(A) 0.025%
(B) 0.25%
(C) 2.5%
(D) 25%

Solution: Required percentage : $\begin{aligned} = (\frac{18}{7200} \times 100) % \\ = \frac{1}{4} % \\ = 0.25 % \end{aligned}$

547.

The shopkeeper increased the price of a product by 25% so that customer finds difficult to purchase the required amount. But Somehow the customer managed to purchase only 70% of the required amount. What is the net difference in the expenditure on that product ?

(A) 55 more
(B) 10% more
(C) 12.5% less
(D) 17.5% less

Solution: Let initially the quantity and rate be 100 each Quantity × rate = Expenditure 100 × 100 = 10000 Now, Increase in price is 25% and new quantity is 70% of original Quantity × rate = Expenditure 70 × 125 = 8750 Decreased expenditure, = 10000 - 8750 = 1250 % decrease $= \frac{1250 \times 100}{10000} = 12.5 %$

548.

The number which exceeds 16% of it by 42 is :

(A) 50
(B) 52
(C) 58
(D) 60

Solution: Let the number be $x$ Then, $\begin{aligned} x - 16 % of x = 42 \\ \Leftrightarrow x - \frac{16}{100} x = 42 \\ \Leftrightarrow x - \frac{4}{25 x} = 42 \\ \Leftrightarrow \frac{21}{25} x = 42 \\ \Leftrightarrow x = (\frac{42 \times 25}{21}) \\ \Leftrightarrow x = 50 \end{aligned}$

549.

A person's salary is decreased by steps of 20%, 15% and 10%. Approximately by what percent should the reduced salary be increases so as to get back the original salary ?

(A) 39%
(B) 44%
(C) 56%
(D) 63%

Solution: Let original salary = Rs. 100 New salary : 90% of 85% of 80% of Rs. 100 $\begin{aligned} = Rs . (\frac{90}{100} \times \frac{85}{100} \times \frac{80}{100} \times 100) \\ = Rs . \frac{306}{5} \end{aligned}$ Increase on $\frac{306}{5}$ : = $(100 - \frac{306}{5})$ = $\frac{194}{5}$ Increase on 100 : $\begin{aligned} = (\frac{194}{5} \times \frac{5}{306} \times 100) % \\ = \frac{9700}{153} % . \\ = 63.39 \approx 63 % \end{aligned}$

550.

The population of a town is 1771561. If it had been increasing at 10% per annum, its population 6 years ago was :

(A) 1000000
(B) 1100000
(C) 1210000
(D) 1331000

Solution: Population of 6 years ago : $\begin{aligned} = [\frac{1771561}{{(1 + \frac{10}{100})}^{6}}] \\ = 1771561 \times {(\frac{10}{11})}^{6} \\ = \frac{1771561 \times 1000000}{1771561} \\ = 1000000 \end{aligned}$

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Practice MCQ Questions and Answer on Percentage

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