Practice MCQ Questions and Answer on Percentage

461.

the price of sugar increases by 32%. A family reduces its consumption so that the expenditure of the sugar is up only by 10%. If the total consumption of the sugar before the price rise was 10 kg per month, then the consumption of sugar per month at present (in kg) is :

(A) $8 \frac{1}{3}$
(B) $8 \frac{1}{2}$
(C) $8 \frac{3}{4}$
(D) 9

Solution: Let the original price of sugar be Rs. 10 per kg Then, original expenditure = Rs. (10 × 10) = Rs. 100 New expenditure = 110% of Rs. 100 = Rs. 110 New price of sugar = 132% of Rs. 10 = Rs. 13.20 New consumption : = $\frac{110}{13.20}$ kg = $\frac{25}{3}$ kg = $8 \frac{1}{3}$ kg

462.

x * 12 = 75% of 336 Find x.

(A) 27
(B) 25
(C) 21
(D) 19

Solution: $\begin{aligned} 12 x = \frac{75 \times 336}{100} \\ x = \frac{75 \times 336}{100 \times 12} \\ x = 21 \end{aligned}$

463.

In an examination, 96% of students passed and 500 students failed. How many students did appear at the examination ?

(A) 14000
(B) 12500
(C) 12000
(D) 13500

Solution: Let number of students who appeared at the examination be $x$ Passed percentage = 96% Failed percentage = 4% According to given information we get Unsuccessful students = 4% ∴ 4% of $x$ = 500 ⇒ $(\frac{x \times 4}{100})$ = 500 ⇒ $x$ = $(\frac{500 \times 100}{4})$ ⇒ $x$ = 12500

464.

The numbers are respectively $12 \frac{1}{2} %$ ÃÂÃÂ and $25 %$ more than a third number. The as a percentage of the second number is :

(A) 50%
(B) 60%
(C) 75%
(D) 90%

Solution: Let the third number be $x$ Then, first number : = $112 \frac{1}{2} %$ of $x$ = $\frac{9 x}{8}$ Second number : = 125% of $x$ = $\frac{5 x}{4}$ ∴ Required percentage : = $(\frac{9 x}{8} \times \frac{4}{5 x} \times 100)$ % = 90%

465.

A saves 35% of his income. If his income increase by 20.1% and his expenditure increase by 20%, then by what percentage do his saving increase or decrease? (correct to one decimal place)

(A) 20.3% of increase
(B) 18.5% of decrease
(C) 19.75% of decrease
(D) 21.9% of increase

Solution: $\begin{array}{l} 35 % = \frac{35}{100} \\ I n c o m e = 100 \\ S a v i n g = 35 \\ E x p e n d i t u r e = 65 \\ \begin{array}{c} I n c o m e & = & E x p e n d i t u r e & + & S a v i n g \\ 100 & = & 65 & + & 35 \\ ↓ + 20.1 & ↓ + 20 \\ 120.1 & = & 78 & + & x \end{array} \\ x = 120.1 - 78 = 42.1 \\ I n c r e a s e i n s a v i n g = 42.1 - 35 = 7.1 \\ I n c r e a s e % = \frac{7.1}{35} \times 100 = 20.3 % \end{array}$

466.

Reena goes to a shop to buy a radio costing Rs. 2568. The rate of sales tax is 7% and the final value is rounded off to the next higher integer. She tells the shopkeeper to reduce the price of the radio so that she has to pay Rs. 2568 inclusive of sales tax. Find the reduction needed in the price of the radio.

(A) Rs. 180
(B) Rs. 210
(C) Rs. 168
(D) Rs. 170

Solution: $\begin{aligned} Reduction \\ = \frac{7}{107} \times 2568 \\ = 168 \end{aligned}$

467.

A student took five papers in an examination, where the full marks were the same for each paper. His marks in these papers were in the proportion of 6:7:8:9:10. In all papers together, the candidate obtained 60% of the total marks then, the number of papers in which he got more than 50% marks is

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

Solution: Let the marks obtained in five subjects be 6x, 7x, 8x, 9x and 10x. Total marks obtained = 40x Max. Marks of the five subjects = $\frac{40 x}{0.6}$ [40x is 60% of total marks] Max. Marks in each subject = $\frac{40 x}{0.6 \times 5}$ = 13.33x Hence, % of each subject = $\frac{6 x \times 100}{13.33}$ = 45.01% Or, $\frac{7 x \times 100}{13.33}$ = 52.51 In same way other percentage are 60.01%, 67.52%, 75.01%. Hence, number of subjects in which he gets more than 50% marks = 4

468.

The average weight of a class of students is 67.5 kg. The weight of the class teacher is 25% more than the average weight of the class. The average weight of the class is less than the class teacher by x%. The value of x is:

(A) 33.335
(B) 25%
(C) 20%
(D) 22%

Solution: 100 == 25% ↑ ==> 125 == x%↓ ==> 100 $x % = \frac{25 \times 100}{100} = 25 %$ We don't need the actual value to calculate the answer

469.

In an election between two candidates, 75% of the voters cast their votes, out of which 2% votes were declared invalid. A candidate got 9261 votes which were 75% of the valid votes. The total number of voters enrolled in that election was :

(A) 16000
(B) 16400
(C) 16800
(D) 18000

Solution: Total valid votes got candidates = $\frac{9261}{75}$ × 100 = 12348 Let total number of votes = x Total votes polled = x × $\frac{75}{100}$ = $\frac{75 x}{100}$ Valid votes = $\frac{75 x}{100}$ × $\frac{98}{100}$ According to the question, ⇔ $\frac{75 x}{100}$ × $\frac{98}{100}$ = 12348 ⇔ x = 16800 Hence, total votes = 16800

470.

Every day a mango seller sells half his stock, 10% of the stock overnight gets spoiled. If 1983 mangoes rotted over 3 nights then how many did hi start with on the first day ?

(A) 25,000
(B) 24,000
(C) 30,000
(D) 32,000

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Practice MCQ Questions and Answer on Percentage

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