Practice MCQ Questions and Answer on Percentage

491.

In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, the number of valid votes that the other candidate got, was:

(A) 2700
(B) 2900
(C) 3000
(D) 3100

Solution: Number of valid votes = 80% of 7500 = 6000 ∴ Valid votes polled by other candidate = 45% of 6000 = $\frac{45}{100} \times 6000$ = 2700

492.

What will come in the place of (?) in the expression below :
x% of y is y% of (?)

(A) x
(B) 100x
(C) $\frac{x}{100}$
(D) $\frac{y}{100}$

Solution: x% of y = $\frac{x y}{100}$ ..... (i) And y% of x = $\frac{x y}{100}$ ..... (ii) From (i) and (ii) x% of y = y% of x

493.

Solve this : 15% of 578 + 22.5% of 644 = ?

(A) 231.4
(B) 231.6
(C) 231.8
(D) 233.6

Solution: 15% of 578 + 22.5% of 644 $= (\frac{15}{100} \times 578) +$ $(\frac{225}{10} \times \frac{1}{100} \times 644)$ $\begin{aligned} = 86.7 + 144.9 \\ = 231.6 \end{aligned}$

494.

The ratio of the number of boys to that of girls in a village is 3 : 2. If 30% of boys and 70% of girls appeared in an examination, the ratio of the number of students, appeared in the examination to that not appeared in the same examination is :

(A) 1 : 1
(B) 27 : 23
(C) 9 : 14
(D) 23 : 27

Solution: According to the question, Let the total number of students = 100 Ratio of Boys and Girls = $\frac{3}{2}$ Total number of Boy = $\frac{3}{5} \times 100 = 60$ Total number of Girl = $\frac{2}{5} \times 100 = 40$ 30% of boy appeared in exam = 30% of 60 = 18 70% of girl appeared in exam = 70% of 40 = 28 Total number of student appeared in exam = 18 + 28 = 46 Students not appeared in exam = 100 - 46 = 54 $∴ \frac{Ratio of students appeared in exam}{Not appeared in exam}$ = $\frac{46}{54}$ = $\frac{23}{27}$ = 23 : 27

495.

At an election there were two candidate got 38% of votes and lost by 7200 votes. The total numbers of valid votes were :

(A) 13000
(B) 13800
(C) 16200
(D) 30000

496.

A population of variety of tiny bush in an experiment field increased by 10% in the first year, increased by 8% in the second year but decreased by 10% in third year. If the present number of bushes in the experiment field is 26730, then the number of variety of bushes in beginning was:

(A) 35000
(B) 27000
(C) 25000
(D) 36000

Solution: Let the number of bushes originally be 100 Number of bushes after one year 100 ==10% (↑) ==> 110 After second year it becomes 110 ==8%(↑) ==> 118.8 After third year, 118.8 ==8%(↓)==> 109.3 Now, according to the question 109.3 = 26730 1 = $\frac{26730}{109.3}$ So, 100 = $\frac{26730}{109.3} \times 100$ = 25000 Thus, number of bushes originally was 25000 NOTE:You can take number of bushes originally as x then solve for the x

497. Solve this :

(12% of 555) + (15% of 666) = ?

(A) 166.5
(B) 167.5
(C) 168.5
(D) None of these

Solution: Given 12% of 555 + 15% of 666 $\begin{aligned} \Leftrightarrow (?) = (\frac{12}{100} \times 555) + (\frac{15}{100} \times 666) \\ = (66.6 + 99.9) \\ = 166.5 \end{aligned}$

498.

Raman's salary is increased by 5% this year. If his present salary is Rs. 1806, the last year's salary was :

(A) Rs. 1720
(B) Rs. 1620
(C) Rs. 1520
(D) Rs. 1801

Solution: Required last year salary = $\frac{1806}{100 + 5}$ × 100 = Rs. 1720

499.

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

500.

A person saves $33 \frac{1}{3} %$ ÃÂÃÂ of his income. If the saving increases by 22% and the expenditure increases by 10%, then the percentage increase in his income is:

(A) 14%
(B) 16%
(C) 18%
(D) 22%

Solution: $\begin{aligned} 33 \frac{1}{3} % = \frac{1}{3} \\ Let income = 300 \\ Saving = 100 \\ Expenditure = 300 - 100 = 200 \end{aligned}$ $\begin{matrix} Income & = & Expenditure & + & Saving \\ Old \to & 300 & = & 200 & + & 100 \\ ↓^{+ 10 %} & ↓^{+ 22 %} \\ New \to & x & = & 220 & + & 122 \end{matrix}$ $\begin{aligned} x = 220 + 122 = 342 \\ Increase = 342 - 300 = 42 \\ Increase % = \frac{42}{300} \times 100 = 14 % \end{aligned}$

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Practice MCQ Questions and Answer on Percentage

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