Percentage Questions and Answers

Q: The present population of a village is 11,000. If the numbers of males increase by 11% and the numbers of females increase by 20%, then the population will become 12,660. What is the present population of females in the village ?

The present female population is 5,000.Let the numbers of males and females be M and F. Then \(M + F = 11000\), and after the increases:\[ \begin{aligned} 1.11M + 1.2F &= 12660 \\ 1.11(11000) + 0.09F &= 12660 \\ 12210 + 0.09F &= 12660 \\ F &= 5000 \end{aligned} \]Check: 6,000 males rise by 660 and 5,000 females by 1,000, a total rise of 1,660, from 11,000 to 12,660. 6,000 is the number of males, which is the trap.

Q: Find the greatest from the following options :14.28% of 700, 11.11% of 819, 9.09% of 704, 6.25% of 960

The greatest is 14.28% of 700.Each percentage is close to a simple fraction:\[ \begin{aligned} 14.28\% \times 700 &\approx \tfrac{1}{7} \times 700 = 100 \\ 11.11\% \times 819 &\approx \tfrac{1}{9} \times 819 = 91 \\ 9.09\% \times 704 &\approx \tfrac{1}{11} \times 704 = 64 \\ 6.25\% \times 960 &= \tfrac{1}{16} \times 960 = 60 \end{aligned} \]Knowing these fraction equivalents makes the comparison quick: 100 is clearly the largest.

Q: In a college election between two candidates, one candidate got 45% of the total valid votes. 20% of the votes were invalid. If the total votes were 14,500, then what is the number of valid votes which the other candidate got ?

The other candidate got 6,380 valid votes.20% of the votes were invalid, so 80% were valid. The first candidate got 45% of the valid votes, so the other got 55%:\[ \begin{aligned} \text{valid} &= 0.8 \times 14500 = 11600 \\ \text{other} &= 0.55 \times 11600 = 6380 \end{aligned} \]5,220 is the first candidate's share, and 6,525 and 7,975 come from taking 45% or 55% of all 14,500 votes instead of the valid ones.

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61.

The present population of a village is 11,000. If the numbers of males increase by 11% and the numbers of females increase by 20%, then the population will become 12,660. What is the present population of females in the village ?

A.6,000
B.7,000
C.5,000
D.5,500

Exam Year(s)

RSMSSB - 2024, RSMSSB Mahila Supervisor - 2024,

Answer: C

The present female population is 5,000.

Let the numbers of males and females be M and F. Then \(M + F = 11000\), and after the increases:

\[ \begin{aligned} 1.11M + 1.2F &= 12660 \\ 1.11(11000) + 0.09F &= 12660 \\ 12210 + 0.09F &= 12660 \\ F &= 5000 \end{aligned} \]

Check: 6,000 males rise by 660 and 5,000 females by 1,000, a total rise of 1,660, from 11,000 to 12,660. 6,000 is the number of males, which is the trap.

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62.

Find the greatest from the following options :

14.28% of 700, 11.11% of 819, 9.09% of 704, 6.25% of 960

A.14.28% of 700
B.11.11% of 819
C.9.09% of 704
D.6.25% of 960

Exam Year(s)

RSMSSB - 2024, RSMSSB Mahila Supervisor - 2024,

Answer: A

The greatest is 14.28% of 700.

Each percentage is close to a simple fraction:

\[ \begin{aligned} 14.28\% \times 700 &\approx \tfrac{1}{7} \times 700 = 100 \\ 11.11\% \times 819 &\approx \tfrac{1}{9} \times 819 = 91 \\ 9.09\% \times 704 &\approx \tfrac{1}{11} \times 704 = 64 \\ 6.25\% \times 960 &= \tfrac{1}{16} \times 960 = 60 \end{aligned} \]

Knowing these fraction equivalents makes the comparison quick: 100 is clearly the largest.

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63.

In a college election between two candidates, one candidate got 45% of the total valid votes. 20% of the votes were invalid. If the total votes were 14,500, then what is the number of valid votes which the other candidate got ?

A.6,380
B.5,220
C.6,525
D.7,975

Exam Year(s)

RSMSSB - 2024, RSMSSB Mahila Supervisor - 2024,

Answer: A

The other candidate got 6,380 valid votes.

20% of the votes were invalid, so 80% were valid. The first candidate got 45% of the valid votes, so the other got 55%:

\[ \begin{aligned} \text{valid} &= 0.8 \times 14500 = 11600 \\ \text{other} &= 0.55 \times 11600 = 6380 \end{aligned} \]

5,220 is the first candidate's share, and 6,525 and 7,975 come from taking 45% or 55% of all 14,500 votes instead of the valid ones.

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64.

Three candidates contested an election and received \(1136, 7636\) and \(11628\) votes respectively. What percentage of the total votes did the winning candidate get?

A.57 %
B.60 %
C.65 %
D.90 %

Answer: A

Total number of votes polled \(= (1136 + 7636 + 11628) = 20400.\)

\(\therefore\) Requried percentage \(= (\frac{11628}{20400} \times 100)\)% \(= 57\)%

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65.

A student multiplied a number by \(\frac{5}{6}\) instead of \(\frac{6}{5}\). What is the percentage
error in the calculation?

A.44 %
B.30.56 %
C.15.28 %
D.22 %

Answer: B

The number be 30

When the student multiplied it by \(\frac{5}{6}\), ⇒ original result \(= \frac{5}{6} \times 30 = 25\)

When the student multiply it by \(\frac{6}{5}\), ⇒ New result \(= \frac{6}{5} \times 30 = 36\)

⇒ Percentage error in calculation \(= \frac{(36-25)}{36} \times 100 \)

\(\Rightarrow \frac{1100}{36} = 30.56\)

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66.

In an election between two candidates, one got \(55\) % of the total valid votes, \(55\)% 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

Answer: A

Number of valid votes \(= 80\) % of \(7500 = 6000.\)

\(\therefore \) Valid votes polled by other candidate \(= 45\)% of \(6000\)

\(= (\frac{45}{100} \times 6000) = 2700.\)

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67.

If \(A = x\)% of \(y\) and \(B = y\)% of \(x\), then which of the following is true?

A.A is smaller than B.
B.A is greater than B
C.Relationship between A and B cannot be determined.
D.None of these

Answer: D

\(x\)% of \(y\) \(= (\frac{x}{100} \times y) = (\frac{y}{100} \times x) = y\)% of \(x\)

\(\therefore A = B\)

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68.

A grain dealer cheats to the extent of 10% while buying as well as selling by using false weights. His total profit percentage is :

A.20 %
B.22.22 %
C.23.23 %
D.25 %

Answer: B

The retailer buys 110 gram in Rs 100 because of 10 % cheating so cost price = 1 gram = Rs \(\frac{100}{110}\) = Rs 0.909 and retailer sells 90 gram in Rs 100

and hence selling price = Rs 100 / 90 = Rs 1.111

Profit % \(= \frac{1.111-0.909}{0.909} = 22.22\)

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69.

Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56 % of the sum of their marks. The marks obtained by them are:

A.39, 30
B.41, 32
C.42, 33
D.43, 34

Answer: C

Their marks be \((x + 9)\) and \(x\)

Then, \(x + 9 = \frac{56}{100}(x + 9 + x)\)

\(\Rightarrow 25(x + 9) = 14(2x + 9) \)

\(\Rightarrow 3x = 99 \)

\(\Rightarrow x = 33\)

So, their marks are 42 and 33.

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70.

Two tailors \(X\) and \(Y\) are paid a total of \(Rs. 550\) per week by their employer. If \(X\) is paid \(120\) % of the sum paid to \(Y\), how much is \(Y\) paid per week?

A.Rs. 200
B.Rs. 250
C.Rs. 300

Answer: B

The sum paid to \(Y\) per week be \(Rs. Z.\)

Then, \(Z + 120\) % of \(Z = 550.\)

\(\Rightarrow Z + \frac{120}{100}Z =550\)

\(\Rightarrow \frac{11}{5}Z = 550\)

\(\Rightarrow Z = (\frac{550 \times 5}{11}) = 250.\)

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