Percentage Questions and Answers

Q: Mr. Manish Das is a fruit vendor, and he sells 60% of Mangoes he had and throws away 15% of the remainder. On the next day, he sells 50% of the remainder and throws away the rest. What percentage (%) of his mangoes does Manish throw in total ?

Manish throws away 23 per cent of his mangoes.Start with 100 mangoes and follow them through both days:\[ \begin{aligned} \text{after selling 60\%} &: 40\ \text{left} \\ \text{thrown on day 1} &= 15\% \times 40 = 6 \\ \text{left} &= 34 \\ \text{sold on day 2} &= 50\% \times 34 = 17 \\ \text{thrown on day 2} &= 17 \end{aligned} \]In total he throws away \(6 + 17 = 23\) of every 100 mangoes. A common slip is to take 15% of the original 100 instead of the 40 left, which gives 15 thrown on the first day and a different total; since 23 is offered, 'None of these' is wrong.

Q: 12.5% of a positive number is 62.5% of the other positive number. If the sum of squares of both numbers is 416, smallest number is what percentage of sum of both the numbers ?

The smaller number is \(16\frac{2}{3}\)% of the sum.Let the numbers be x and y. Since 12.5% of x equals 62.5% of y, x = 5y. Then use the sum of squares:\[ \begin{aligned} (5y)^{2} + y^{2} &= 416 \\ 26y^{2} &= 416 \\ y^{2} = 16, \ y &= 4, \ x = 20 \end{aligned} \]The smaller number is 4 and the sum is 24:\[ \frac{4}{24} \times 100 = 16\tfrac{2}{3}\% \]\(83\frac{1}{3}\)% is the larger number's share (20 out of 24), and 20% and 25% compare the smaller number with the larger (4/20) or use a wrong ratio, not with the sum.

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91.

Mr. Manish Das is a fruit vendor, and he sells 60% of Mangoes he had and throws away 15% of the remainder. On the next day, he sells 50% of the remainder and throws away the rest. What percentage (%) of his mangoes does Manish throw in total ?

A.19
B.23
C.27
D.None of these.

Exam Year(s)

Tripura Public Service Commission - 2013,

Answer: B

Manish throws away 23 per cent of his mangoes.

Start with 100 mangoes and follow them through both days:

\[ \begin{aligned} \text{after selling 60\%} &: 40\ \text{left} \\ \text{thrown on day 1} &= 15\% \times 40 = 6 \\ \text{left} &= 34 \\ \text{sold on day 2} &= 50\% \times 34 = 17 \\ \text{thrown on day 2} &= 17 \end{aligned} \]

In total he throws away \(6 + 17 = 23\) of every 100 mangoes. A common slip is to take 15% of the original 100 instead of the 40 left, which gives 15 thrown on the first day and a different total; since 23 is offered, 'None of these' is wrong.

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92.

12.5% of a positive number is 62.5% of the other positive number. If the sum of squares of both numbers is 416, smallest number is what percentage of sum of both the numbers ?

A.20%
B.25%
C.\(83\frac{1}{3}\)%
D.\(16\frac{2}{3}\)%

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: D

The smaller number is \(16\frac{2}{3}\)% of the sum.

Let the numbers be x and y. Since 12.5% of x equals 62.5% of y, x = 5y. Then use the sum of squares:

\[ \begin{aligned} (5y)^{2} + y^{2} &= 416 \\ 26y^{2} &= 416 \\ y^{2} = 16, \ y &= 4, \ x = 20 \end{aligned} \]

The smaller number is 4 and the sum is 24:

\[ \frac{4}{24} \times 100 = 16\tfrac{2}{3}\% \]

\(83\frac{1}{3}\)% is the larger number's share (20 out of 24), and 20% and 25% compare the smaller number with the larger (4/20) or use a wrong ratio, not with the sum.

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93.

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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94.

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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95.

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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96.

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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97.

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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98.

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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99.

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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100.

In a certain school, 20 % of students are below 8 years of age. The number of students above 8 years of age is\(\frac{2}{3}\) of the number of students of 8 years of age which is 48. What is the total number of students in the school?

A.72
B.80
C.120
D.100

Answer: D

Let the number of students be \(x\).

Then, Number of students above 8 years of age \(= (100 -20)\)% of \(x = 80\)% of \(x\)

\(\therefore 80\)% of \(x = 40 + \frac{2}{3} \) of 48

\(\Rightarrow \frac{80}{100}x = 80\)

\(\Rightarrow x = 100.\)

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