Percent and Percentage — Exercise 9(C)

The salary of a man is increased from ₹ 600 per month to ₹ 850 per month. Express the increase in salary as percent.

Answer

Original salary = ₹ 600 and new salary = ₹ 850.

Increase in salary = 850 − 600 = ₹ 250.

Percentage increase = $\left(\dfrac{\text{increase in value}}{\text{original value}} \times 100\right)$%

$= \left(\dfrac{250}{600} \times 100\right)$%

$= \dfrac{25000}{600}$%

$= \dfrac{125}{3}$%

$= 41\dfrac{2}{3}$%

Hence, the increase in salary = $41\dfrac{2}{3}$%.

Increase:

60 by 5%

Answer

Increase = 5% of 60 = $\dfrac{5}{100} \times 60 = 3$.

Increased value = 60 + 3 = 63.

Hence, 60 increased by 5% = 63.

20 by 15%

Answer

Increase = 15% of 20 = $\dfrac{15}{100} \times 20 = 3$.

Increased value = 20 + 3 = 23.

Hence, 20 increased by 15% = 23.

48 by $12\dfrac{1}{2}$%

Answer

Increase = $12\dfrac{1}{2}$% of 48 = $\dfrac{25}{2 \times 100} \times 48 = 6$.

Increased value = 48 + 6 = 54.

Hence, 48 increased by $12\dfrac{1}{2}$% = 54.

80 by 140%

Answer

Increase = 140% of 80 = $\dfrac{140}{100} \times 80 = 112$.

Increased value = 80 + 112 = 192.

Hence, 80 increased by 140% = 192.

1000 by 3.5%

Answer

Increase = 3.5% of 1000 = $\dfrac{3.5}{100} \times 1000 = 35$.

Increased value = 1000 + 35 = 1035.

Hence, 1000 increased by 3.5% = 1035.

Decrease:

80 by 20%

Answer

Decrease = 20% of 80 = $\dfrac{20}{100} \times 80 = 16$.

Decreased value = 80 − 16 = 64.

Hence, 80 decreased by 20% = 64.

300 by 10%

Answer

Decrease = 10% of 300 = $\dfrac{10}{100} \times 300 = 30$.

Decreased value = 300 − 30 = 270.

Hence, 300 decreased by 10% = 270.

50 by 12.5%

Answer

Decrease = 12.5% of 50 = $\dfrac{12.5}{100} \times 50 = 6.25$.

Decreased value = 50 − 6.25 = 43.75.

Hence, 50 decreased by 12.5% = 43.75.

What number:

when increased by 10% becomes 88?

Answer

Let the number be x. When increased by 10%, it becomes 110% of x.

$\Rightarrow \dfrac{110}{100} \times x = 88\\[1em] \Rightarrow x = \dfrac{88 \times 100}{110}\\[1em] \Rightarrow x = 80$

Hence, the required number = 80.

when increased by 15% becomes 230?

Answer

Let the number be x. When increased by 15%, it becomes 115% of x.

$\Rightarrow \dfrac{115}{100} \times x = 230\\[1em] \Rightarrow x = \dfrac{230 \times 100}{115}\\[1em] \Rightarrow x = 200$

Hence, the required number = 200.

when decreased by 15% becomes 170?

Answer

Let the number be x. When decreased by 15%, it becomes 85% of x.

$\Rightarrow \dfrac{85}{100} \times x = 170\\[1em] \Rightarrow x = \dfrac{170 \times 100}{85}\\[1em] \Rightarrow x = 200$

Hence, the required number = 200.

when decreased by 40% becomes 480?

Answer

Let the number be x. When decreased by 40%, it becomes 60% of x.

$\Rightarrow \dfrac{60}{100} \times x = 480\\[1em] \Rightarrow x = \dfrac{480 \times 100}{60}\\[1em] \Rightarrow x = 800$

Hence, the required number = 800.

when increased by 100% becomes 100?

Answer

Let the number be x. When increased by 100%, it becomes 200% of x.

$\Rightarrow \dfrac{200}{100} \times x = 100\\[1em] \Rightarrow x = \dfrac{100 \times 100}{200}\\[1em] \Rightarrow x = 50$

Hence, the required number = 50.

when decreased by 50% becomes 50?

Answer

Let the number be x. When decreased by 50%, it becomes 50% of x.

$\Rightarrow \dfrac{50}{100} \times x = 50\\[1em] \Rightarrow x = \dfrac{50 \times 100}{50}\\[1em] \Rightarrow x = 100$

Hence, the required number = 100.

The price of a car is lowered by 20% to ₹ 40,000. What was the original price? Also, find the reduction in price.

Answer

Let the original price be ₹ x. When lowered by 20%, it becomes 80% of x.

$\Rightarrow \dfrac{80}{100} \times x = 40,000\\[1em] \Rightarrow x = \dfrac{40,000 \times 100}{80}\\[1em] \Rightarrow x = 50,000$

Reduction in price = 50,000 − 40,000 = ₹ 10,000.

Hence, the original price = ₹ 50,000 and the reduction in price = ₹ 10,000.

If the price of an article is increased by 25%, the increase is ₹ 10. Find the new price.

Answer

Let the original price be ₹ x.

Given, 25% of x = ₹ 10.

$\Rightarrow \dfrac{25}{100} \times x = 10\\[1em] \Rightarrow x = \dfrac{10 \times 100}{25}\\[1em] \Rightarrow x = 40$

New price = original price + increase = 40 + 10 = ₹ 50.

Hence, the new price = ₹ 50.

If the price of an article is reduced by 10%, the reduction is ₹ 40. What is the old price?

Answer

Let the old price be ₹ x.

Given, 10% of x = ₹ 40.

$\Rightarrow \dfrac{10}{100} \times x = 40\\[1em] \Rightarrow x = \dfrac{40 \times 100}{10}\\[1em] \Rightarrow x = 400$

Hence, the old price = ₹ 400.

The price of a chair is reduced by 25%. What is the ratio of:

(i) change in price to the old price.

(ii) old price to the new price.

Answer

Let the old price be ₹ 100.

Change in price (reduction) = 25% of 100 = ₹ 25.

New price = 100 − 25 = ₹ 75.

(i) Change in price : old price = 25 : 100 = 1 : 4.

Hence, the ratio of the change in price to the old price = 1 : 4.

(ii) Old price : new price = 100 : 75 = 4 : 3.

Hence, the ratio of the old price to the new price = 4 : 3.

If x is 20% less than y, find:

(i) $\dfrac{x}{y}$

(ii) $\dfrac{y-x}{y}$

(iii) $\dfrac{x}{y-x}$

Answer

Since x is 20% less than y,

$\Rightarrow x = y - 20$

$x = \dfrac{4y}{5}$

(i) $\dfrac{x}{y}$

$\dfrac{x}{y} = \dfrac{\dfrac{4y}{5}}{y}\\[1em] = \dfrac{4y}{5} \times \dfrac{1}{y}\\[1em] = \dfrac{4}{5}$

Hence, $\dfrac{x}{y} = \dfrac{4}{5}$.

(ii) $\dfrac{y - x}{y}$

$\dfrac{y - x}{y} = \dfrac{y - \dfrac{4y}{5}}{y}\\[1em] = \dfrac{\dfrac{y}{5}}{y}\\[1em] = \dfrac{y}{5} \times \dfrac{1}{y}\\[1em] = \dfrac{1}{5}$

Hence, $\dfrac{y - x}{y} = \dfrac{1}{5}$.

(iii) $\dfrac{x}{y - x}$

$\dfrac{x}{y - x} = \dfrac{\dfrac{4y}{5}}{y - \dfrac{4y}{5}}\\[1em] = \dfrac{\dfrac{4y}{5}}{\dfrac{y}{5}}\\[1em] = \dfrac{4y}{5} \times \dfrac{5}{y}\\[1em] = 4$

Hence, $\dfrac{x}{y - x} = 4$.

If x is 30% more than y; find:

(i) $\dfrac{x}{y}$

(ii) $\dfrac{y+x}{x}$

(iii) $\dfrac{y}{y-x}$

Answer

Since x is 30% more than y,

$\Rightarrow x = y + 30$

x = $\dfrac{13y}{10}$

(i) $\dfrac{x}{y}$

$\dfrac{x}{y} = \dfrac{\dfrac{13y}{10}}{y}\\[1em] = \dfrac{13y}{10} \times \dfrac{1}{y}\\[1em] = \dfrac{13}{10}$

Hence, $\dfrac{x}{y} = \dfrac{13}{10}$.

(ii) $\dfrac{y + x}{x}$

$\dfrac{y + x}{x} = \dfrac{y + \dfrac{13y}{10}}{\dfrac{13y}{10}}\\[1em] = \dfrac{\dfrac{23y}{10}}{\dfrac{13y}{10}}\\[1em] = \dfrac{23y}{10} \times \dfrac{10}{13y}\\[1em] = \dfrac{23}{13}$

Hence, $\dfrac{y + x}{x} = \dfrac{23}{13}$.

(iii) $\dfrac{y}{y - x}$

$\dfrac{y}{y - x} = \dfrac{y}{y - \dfrac{13y}{10}}\\[1em] = \dfrac{y}{-\dfrac{3y}{10}}\\[1em] = y \times \left(-\dfrac{10}{3y}\right)\\[1em] = -\dfrac{10}{3}$

Hence, $\dfrac{y}{y - x} = -\dfrac{10}{3}$.

The weight of a machine is 40 kg. By mistake, it was weighed as 40.8 kg. Find the error percent.

Answer

Actual weight = 40 kg and weight measured = 40.8 kg.

Error = 40.8 − 40 = 0.8 kg.

Error percent = $\left(\dfrac{\text{error}}{\text{actual weight}} \times 100\right)$%

$= \left(\dfrac{0.8}{40} \times 100\right)$%

$= \dfrac{80}{40}$%

= 2%

Hence, the error percent = 2%.

From a cask, containing 450 litres of petrol, 8% of the petrol was lost by leakage and evaporation. How many litres of petrol were left in the cask?

Answer

Total petrol = 450 litres.

Petrol lost = 8% of 450 = $\dfrac{8}{100} \times 450 = 36$ litres.

Petrol left = 450 − 36 = 414 litres.

Hence, 414 litres of petrol were left in the cask.

An alloy consists of 13 parts of copper, 7 parts of zinc and 5 parts of nickel. What is the percentage of each metal in the alloy?

Answer

Let the weight corresponding to each part be x kg.

So, copper = 13x, zinc = 7x and nickel = 5x.

Total weight of the alloy = 13x + 7x + 5x = 25x.

Percentage of copper = $\left(\dfrac{13x}{25x} \times 100\right)$%

$= \dfrac{1300}{25}$%

= 52%

Percentage of zinc = $\left(\dfrac{7x}{25x} \times 100\right)$%

$= \dfrac{700}{25}$%

= 28%

Percentage of nickel = $\left(\dfrac{5x}{25x} \times 100\right)$%

$= \dfrac{500}{25}$%

= 20%

Hence, copper = 52%, zinc = 28% and nickel = 20%.

In an examination, first division marks are 60%. A student secures 538 marks and misses the first division by 2 marks. Find the total marks of the examination.

Answer

Marks secured by the student = 538.

Since the student misses the first division by 2 marks,

Marks needed for first division = 538 + 2 = 540.

Let the total marks be x. Since 60% of the total marks are needed for first division,

$\Rightarrow \dfrac{60}{100} \times x = 540\\[1em] \Rightarrow x = \dfrac{540 \times 100}{60}\\[1em] \Rightarrow x = 900$

Hence, the total marks of the examination = 900.

Out of 1200 pupils in a school, 900 are boys and the rest are girls. If 20% of the boys and 30% of the girls wear spectacles, find:

(i) how many pupils in all wear spectacles.

(ii) what percent of the total number of pupils wear spectacles.

Answer

Total number of pupils = 1200.

Number of boys = 900 and number of girls = 1200 − 900 = 300.

Boys wearing spectacles = 20% of 900 = $\dfrac{20}{100} \times 900 = 180$.

Girls wearing spectacles = 30% of 300 = $\dfrac{30}{100} \times 300 = 90$.

(i) Total pupils wearing spectacles = 180 + 90 = 270.

Hence, 270 pupils wear spectacles.

(ii) Percentage wearing spectacles = $\left(\dfrac{270}{1200} \times 100\right)$%

$= \dfrac{27000}{1200}$%

= 22.5%

Hence, 22.5% of the total number of pupils wear spectacles.

Out of 25 identical bulbs, 17 are red, 3 are black and the remaining are yellow. Find the difference between the numbers of red and yellow bulbs and express this difference as percent.

Answer

Total number of bulbs = 25.

Number of red bulbs = 17 and number of black bulbs = 3.

Number of yellow bulbs = 25 − (17 + 3) = 25 − 20 = 5.

Difference between red and yellow bulbs = 17 − 5 = 12.

Difference as a percent of total = $\left(\dfrac{12}{25} \times 100\right)$%

$= \dfrac{1200}{25}$%

= 48%

Hence, the difference is 12 bulbs, which is 48% of the total.

A number first increases by 20% and then decreases by 20%. Find the percentage increase or decrease on the whole.

Answer

Let the number be 100.

After an increase of 20%:

$\Rightarrow 100 + \dfrac{20}{100} \times 100$

⇒ 100 + 20

⇒ 120

After a decrease of 20% on 120:

$\Rightarrow 120 - \dfrac{20}{100} \times 120$

⇒ 120 - 24

⇒ 96

Net change = 96 − 100 = −4, i.e. a decrease of 4.

Percentage change = $\left(\dfrac{4}{100} \times 100\right)$% = 4% decrease.

Hence, there is a 4% decrease on the whole.

A number is first decreased by 40% and then again decreased by 60%. Find the percentage increase or decrease on the whole.

Answer

Let the number be 100.

After a decrease of 40%:

$\Rightarrow 100 - \dfrac{40}{100} \times 100$

⇒ 100 - 40

⇒ 60

After a decrease of 60% on 60:

$\Rightarrow 60 - \dfrac{60}{100} \times 60$

⇒ 60 - 36

⇒ 24

Net change = 24 − 100 = −76, i.e. a decrease of 76.

Percentage change = $\left(\dfrac{76}{100} \times 100\right)$% = 76% decrease.

Hence, there is a 76% decrease on the whole.

If 150% of a number is 750, find 60% of this number.

Answer

Let the number be x.

$\Rightarrow \dfrac{150}{100} \times x = 750\\[1em] \Rightarrow x = \dfrac{750 \times 100}{150}\\[1em] \Rightarrow x = 500$

Now, 60% of 500 = $\dfrac{60}{100} \times 500 = 300$.

Hence, 60% of the number = 300.