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

**Answer: **A

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

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

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

error in the calculation?

**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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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:

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

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

**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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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:

**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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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?

**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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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?

**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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The population of a town increased from \(1,75,000\) to \(2,62,500 \) in a decade. The average percent increase of population per year is:

**Answer: **B

Increase in \(10\) years \(= (262500 - 175000) = 87500.\)

Increase % \(= (\frac{87500}{175000} \times 100)\) % \(= 50\) %

\(\therefore\) Required average \(= (\frac{50}{10})\) % \(= 5\) %

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A fruit seller had some apples. He sells 40 % apples and still has 420 apples. Originally, he had:

**Answer: **D

Suppose originally he had \(x\) apples.

Then, \((100 - 40)\)% of \(x = 420\)

\(\Rightarrow \frac{60}{100} \times x = 420 \)

\(\Rightarrow x = (\frac{420 \times 100}{60}) = 700\)

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