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# Percentage

1.

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

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

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

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2.

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

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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3.

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:

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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4.

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

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

$$\therefore A = B$$

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5.

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

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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6.

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:

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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7.

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?

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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8.

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?

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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9.

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:

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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10.

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

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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