Practice MCQ Questions and Answer on Percentage

551.

A reduction of 10% in the price of an apple enable a man to buy 10 apples more for Rs. 54. The reduced price of apples per dozen is :

(A) Rs. 6.48
(B) Rs. 12.96
(C) Rs. 10.80
(D) Rs. 14.40

Solution: Let initial expenditure = 100 units Required increment = $\frac{10}{90}$ = $\frac{1}{9}$ 1 unit = 10 apples Original consumption = 9 units = 9 × 10 = 90 apples New consumption = 10 units = 10 × 10 = 100 apples New price = $\frac{54}{100}$ × 12 = Rs. 6.48 per dozen

552.

In the price of sugar falls by $2 \frac{1}{2} %$ , a person can buy 9 kg more of sugar for Rs. 1260 than before. If the price had risen by $12 \frac{1}{2} %$ , how much sugar would he have bought for the same sum ?

(A) 288 kg
(B) 312 kg
(C) 328 kg
(D) 336 kg

Solution: Let the original price of sugar be Rs. $x$ per kg Then, reduced price : $\begin{aligned} 97 \frac{1}{2} % of Rs . x \\ = Rs . (\frac{195}{2} \times \frac{1}{100} \times x) \\ = Rs . \frac{39 x}{40} \\ ∴ \frac{1260}{(\frac{39 x}{40})} - \frac{1260}{x} = 9 \\ \Rightarrow \frac{16800}{13 x} - \frac{1260}{x} = 9 \\ \Rightarrow 13 x = \frac{420}{9} \\ \Rightarrow x = \frac{140}{39} \end{aligned}$ Increased price : $\begin{aligned} 112 \frac{1}{2} % of Rs . \frac{140}{39} \\ = Rs . (\frac{225}{2} \times \frac{1}{100} \times \frac{140}{39}) \\ = Rs . \frac{105}{26} \end{aligned}$ ∴ Quiantity of sugar bought for Rs. 1260 : $\begin{aligned} = (1260 \times \frac{26}{105}) kg \\ = 312 kg \end{aligned}$

553.

In an examination it is required to get 36% of the aggregate marks to pass. A student gets 198 marks and is declared failed by 36 marks. What is the maximum aggregate marks a student can get ?

(A) 480
(B) 550
(C) 650
(D) Cannot be determined

Solution: Let the maximum marks be x Then, 36% of x = 198 + 36 ⇒ $\frac{36 x}{100}$ = 234 ⇒ x = $\frac{234 \times 100}{36}$ ⇒ x = 650

554.

If 120 is 20% of a number, then 120% of that number will be :

(A) 20
(B) 120
(C) 480
(D) 720

Solution: Let the number = x According to the question, x × $\frac{20}{100}$ = 120 x = 600 Required answer = 600 × $\frac{120}{100}$ = 720 Alternate 20% represents → 120 1% → $\frac{120}{20}$ So, 120% = $\frac{120}{20}$ × 120 = 720

555.

The length, breadth and height of a room are in ratio 3:2:1. If breadth and height are halved while the length is doubled, then the total area of the four walls of the room will

(A) remain the same
(B) decrease by 13.64%
(C) decrease by 15%
(D) decrease by 18.75%

Solution: Answer & Solution Answer: Option E Solution: Let length, breadth and height of the room be 3, 2, 1 unit respectively. Area of walls = 2(l + b) × h = 2(3 + 2) × 1 = 10 sq. unit. Now, length, breadth and height of room will become 6, 1 and $\frac{1}{2}$ respectively. Now, area of walls = $2 (6 + 1) \times \frac{1}{2}$ = 7 sq. unit. % decrease in the area of walls = $(10 - 7) \times \frac{100}{10}$ = 30%

556.

A and B are two fixed points 5 cm apart and C is a point on AB such that AC is 3 cm. If the length of AC is increased by 6%, the length of CB is decreased by :

(A) 6%
(B) 7%
(C) 8%
(D) 9%

Solution: Original length of AB = 5 cm Original length of AC = 3 cm Original length of CB = (5 - 3) cm = 2 cm New length AC : = 106% of 3 cm = $(\frac{106}{100} \times 3)$ cm = 3.18 cm New length of CB : = (5 - 3.18) cm = 1.82 cm Decrease in length of CB : = (2 - 1.82) cm = 0.18 cm ∴ Decrease % : = $(\frac{0.18}{2} \times 100) %$ = 9%

557.

80% of a smaller number is 4 less than 40% of a larger number. The larger number is 85 greater than the smaller one. The sum of these two number is

(A) 325
(B) 425
(C) 235
(D) 500

558.

If A is equal to 20% of B and B is equal to 25% of C. Then what percentage of C equal to A ?

(A) 10%
(B) 15%
(C) 5%
(D) 20%

Solution: A = 20% of B A = $\frac{1}{5}$ B $\frac{A}{B} = \frac{1}{5}$ ----(i) B = 25% of C B = $\frac{1}{4}$ C $\frac{B}{C} = \frac{1}{4}$ ----(ii) Multiply (i) and (ii) We get $\frac{A}{B} \times \frac{B}{C} = \frac{1}{5} \times \frac{1}{4}$ $\frac{A}{C} = \frac{1}{20}$ Percent of C is equal to A $= \frac{1}{20} \times 100 = 5 %$

559.

In an examination, 35% of total students failed in Hindi, 45% failed in English and 20% failed in both. Find the percentage of those students who passed in both the subjects ?

(A) 45%
(B) 35%
(C) 20%
(D) 40%

560.

A shepherd had n goats in the year 2000. In 2001 the no. of goats increased by 40%. In 2002 the no. of goats declined to 70%. In 2003 the no. of goats grew up 30%. In 2004, he sold 10% goats and then he had only 34,398 goats. The percentage increase of the no. of goats in this duration was :

(A) 16.66%
(B) 14.66%
(C) 11.33%
(D) 20%

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Practice MCQ Questions and Answer on Percentage

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