Practice MCQ Questions and Answer on Percentage

41.

The price of sugar per kg increased from Rs. 16 to Rs. 20. The percentage reduction in the use of sugar so that the expenditure does not increase, should be :

(A) 15%
(B) 20%
(C) 25%
(D) 40%

Solution: Let original consumption = 100 kg and new consumption = $x$ kg So, ⇔ 100 × 16 = $x$ × 20 ⇔ $x$ = 80 ∴ Reduction in consumption = 20%

42.

Last year, the population of a town was x and if it increases at the same rate, next year it will be y. the present population of the town is

(A) $\frac{x + y}{2}$
(B) $\frac{y - x}{2}$
(C) $\frac{2 x y}{x + y}$
(D) $\sqrt{x y}$

Solution: $\begin{aligned} Let the present population of the town be P \\ Using compound interest formula \\ Then, \\ P = x [1 + (\frac{R}{100})] - - - (i) \\ And y = P [1 + (\frac{R}{100})] \\ = P \times \frac{P}{x} - - - - (i i) \\ P^{2} = x y; \\ Hence, P = \sqrt{x y} \end{aligned}$

43.

If x% of a is the same as y% of b, then z% of b is :

(A) $\frac{x y}{z} % of a$
(B) $\frac{y z}{x} % of a$
(C) $\frac{x z}{y} % of a$
(D) None of these

Solution: $\begin{aligned} x % of a = y % of a \\ \Rightarrow \frac{x}{100} a = \frac{y}{100} b \\ \Rightarrow b = (\frac{x}{y}) a \\ ∴ z % of b : \\ = (z % of \frac{x}{y}) a \\ = (\frac{x z}{y \times 100}) a \\ = (\frac{x z}{y}) % of a \end{aligned}$

44.

If 25% of 400 + 35% of 1260 + 27% of 1800 = 1020 + x, then the value of x lies between:

(A) 6 to 10
(B) 0 to 5
(C) 11 to 15
(D) 16 to 20

45.

A man spends 75% of his income, if his income increases by 28% and his expenditure increases by 20%, then what is the increase or decrease percentage in his savings?

(A) 52% increase
(B) 13% decrease
(C) 13% increase
(D) 52% decrease

Solution: $\begin{array}{l} 75 \to \frac{3}{4} \\ \begin{array}{c} I n c o m e & E x p e n d i t u r e & S a v i n g \\ 400 & 300 & 100 \\ ↓ + 28 % & ↓ + 20 % & ↓ \\ 112 & - & 60 & = 52 % \end{array} \\ S a v i n g = 52 % \end{array}$

46.

Mr. More spent 20% of his monthly income on food and 15% on children's education. 40% of the remaining he spent on entertainment and transport together and 30% on medical. He is left with an amount of Rs. 8775 after all these expenditures. What is Mr. More's monthly income ?

(A) Rs. 35000
(B) Rs. 38000
(C) Rs. 40000
(D) Rs. 42000

Solution: Answer & Solution Answer: Option E Solution: Let Mr. More's monthly income be Rs. $x$ Then, [100 - (40 + 30)]% of [100 - 20 + 15]% of $x$ = 8775 ⇒ 30% of 65% of $x$ = 8775 ⇒ $(\frac{30}{100} \times \frac{65}{100} \times x)$ = 8775 ⇒ $\frac{39}{200}$ $x$ = 8775 ⇒ $x$ = $(\frac{8775 \times 200}{39})$ ⇒ $x$ = 45000

47.

Each side of a rectangular field is diminished by 40%. By how much percent is the area of the field diminished ?

(A) 32%
(B) 64%
(C) 25%
(D) 16%

Solution: By using successive formula Net decrease in area $\begin{aligned} = 40 + 40 - \frac{40 \times 40}{100} \\ = 64 % \end{aligned}$

48.

One-fourth of sixty percent of a number is equal to two-fifth of twenty percent of another number. What is the respective ratio of the first number to the second number ?

(A) 4 : 7
(B) 5 : 9
(C) 8 : 13
(D) Cannot be determined

Solution: Answer & Solution Answer: Option E Solution: Let the first number be x and the second number be y $\frac{1}{4} of 60% of x = \frac{2}{5} of 20% of y$ $\begin{aligned} \Rightarrow \frac{1}{4} \times \frac{60}{100} \times x = \frac{2}{5} \times \frac{20}{100} \times y \\ \Rightarrow \frac{3 x}{20} = \frac{2 y}{25} \\ \Rightarrow \frac{x}{y} = \frac{2}{25} \times \frac{20}{3} \\ \Rightarrow \frac{x}{y} = \frac{8}{15} \end{aligned}$

49.

In a class, the number of girls is 20% more than that of the boys. The strength of the class is 66. If 4 more girls are admitted to the class, the ratio of the number of boys to that of the girls is :

(A) 1 : 2
(B) 3 : 4
(C) 1 : 4
(D) 3 : 5

Solution: 20% = $\frac{1}{5}$ = $\frac{6 \to Girls}{5 \to Boys}$ Boys : Girls 5 : 6 According to the question, (5 + 6) units = 66 11 units = 66 1 unit = 6 Hence Boys = 6 × 5 = 30 Girls = 6 × 6 = 36 The number of girls when 4 is admitted = (36 + 4) = 40 Required ratio : = 30 : 40 = 3 : 4

50.

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%

«
1
2
3
4
5
6
7
8
9
10
...
57
58
Next ›
Last »

Indian Polity Mcq IAS GK Question India GK World GK GK questions GK Quiz Biology GK Science GK Physics GK Chemistry GK General Awareness Computer GK Political GK Indian Political GK Economics GK History GK Sports GK Geography GK Teaching Aptitude GK CTET/TET GK B Ed Entrance GK Banking GK SSC GK UPSC GK Indian Army GK Hindi Grammar GK Hindi Grammar MCQ Reasoning questions Bihar GK Rajasthan GK Jharkhand GK MP GK UP GK Chhattisgarh GK Haryana GK HP GK Maharashtra GK Electronics GK Agriculture GK States/Capitals GK Books Name & Author Arthimetic Verbal Reasoning Non Verbal Reasoning English

Practice MCQ Questions and Answer on Percentage

General Knowledge