Percent — Exercise 10(C)

The price of rice rises from ₹ 30 per kg to ₹ 36 per kg. Find the percentage rise in the price of rice.

Answer

Original price of rice = ₹ 30 per kg and new price of rice = ₹ 36 per kg.

Increase in price = ₹ 36 − ₹ 30 = ₹ 6.

Increase % = $\left(\dfrac{\text{Increase in price}}{\text{Original price}} \times 100\right)$%

= $\left(\dfrac{6}{30} \times 100\right)$%

= $\dfrac{600}{30}$%

= 20%

Hence, the percentage rise in the price of rice = 20%.

The population of a small locality was 4000 in 1979 and 4500 in 1981. By what percent had the population increased?

Answer

Original population (in 1979) = 4000 and new population (in 1981) = 4500.

Increase in population = 4500 − 4000 = 500.

Increase % = $\left(\dfrac{\text{Increase in population}}{\text{Original population}} \times 100\right)$%

= $\left(\dfrac{500}{4000} \times 100\right)$%

= $\dfrac{50000}{4000}$%

= $\dfrac{25}{2}$%

= 12.5%

Hence, the population had increased by 12.5%.

The price of a scooter was ₹ 8,000 in 1975. It came down to ₹ 6,000 in 1980. By what percent had the price of the scooter came down?

Answer

Original price of the scooter = ₹ 8,000 and new price = ₹ 6,000.

Decrease in price = ₹ 8,000 − ₹ 6,000 = ₹ 2,000.

Decrease % = $\left(\dfrac{\text{Decrease in price}}{\text{Original price}} \times 100\right)$%

= $\left(\dfrac{2000}{8000} \times 100\right)$%

= $\dfrac{200000}{8000}$%

= 25%

Hence, the price of the scooter came down by 25%.

Find the resulting quantity when:

₹ 400 is decreased by 8%.

Answer

Decrease in it = 8% of ₹ 400

= $\dfrac{8}{100} \times ₹\ 400$

= $₹\ \dfrac{3{,}200}{100}$

= ₹ 32

Decreased quantity = ₹ 400 − ₹ 32 = ₹ 368.

Hence, the resulting quantity = ₹ 368.

Find the resulting quantity when:

25 km is increased by 5%.

Answer

Increase in it = 5% of 25 km

= $\dfrac{5}{100} \times 25$ km

= $\dfrac{125}{100}$ km

= 1.25 km

Increased quantity = 25 km + 1.25 km = 26.25 km.

Hence, the resulting quantity = 26.25 km.

Find the resulting quantity when:

a speed of 600 km h^-1 is increased by $12\dfrac{1}{2}$%.

Answer

Here, $12\dfrac{1}{2}$% = $\dfrac{25}{2}$%.

Increase in it = $\dfrac{25}{2}$% of 600 km h^-1

= $\dfrac{25}{2 \times 100} \times 600$ km h^-1

= $\dfrac{15000}{200}$ km h^-1

= 75 km h^-1

Increased speed = 600 km h^-1 + 75 km h^-1 = 675 km h^-1 = 675 km/h.

Hence, the resulting quantity = 675 km/h.

Find the resulting quantity when:

there is 2.5% increase in a salary of ₹ 62,500.

Answer

Increase in it = 2.5% of ₹ 62,500

= $\dfrac{2.5}{100} \times ₹\ 62{,}500$

= $₹\ \dfrac{1{,}56{,}250}{100}$

= ₹ 1,562.50

Increased salary = ₹ 62,500 + ₹ 1,562.50 = ₹ 64,062.50.

Hence, the resulting quantity = ₹ 64,062.50.

The population of a village decreased by 12%. If the original population was 25,000, find the population after the decrease?

Answer

Original population of the village = 25,000 and decrease = 12%.

Decrease in population = 12% of 25,000

= $\dfrac{12}{100} \times 25{,}000$

= $\dfrac{300000}{100}$

= 3,000

Population after the decrease = 25,000 − 3,000 = 22,000.

Hence, the population after the decrease = 22,000.

Out of a salary of ₹ 13,500, I keep one-third as savings. Of the remaining money, I spend 50% on food and 20% on house rent. How much do I spend on food and house rent?

Answer

Total salary = ₹ 13,500.

Savings = $\dfrac{1}{3}$ of ₹ 13,500 $= ₹\ \dfrac{13{,}500}{3} = ₹\ 4{,}500$.

Remaining money = ₹ 13,500 − ₹ 4,500 = ₹ 9,000.

Money spent on food = 50% of ₹ 9,000

= $\dfrac{50}{100} \times ₹\ 9{,}000$

= $₹\ \dfrac{4{,}50{,}000}{100}$

= ₹ 4,500

Money spent on house rent = 20% of ₹ 9,000

= $\dfrac{20}{100} \times ₹\ 9{,}000$

= $₹\ \dfrac{1{,}80{,}000}{100}$

= ₹ 1,800

Money spent on both = ₹ 4,500 + ₹ 1,800 = ₹ 6,300.

Hence, ₹ 4,500 is spent on food, ₹ 1,800 is spent on house rent and ₹ 6,300 is spent in total on both food and rent.

A tank can hold 50 litres of water. At present it is only 30% full. How many litres of water should I put into the tank so that it becomes 50% full?

Answer

Total capacity of the tank = 50 litres.

Water present in the tank = 30% of 50 litres

= $\dfrac{30}{100} \times 50$ litres

= $\dfrac{1500}{100}$ litres

= 15 litres

Water required to make the tank 50% full = 50% of 50 litres

= $\dfrac{50}{100} \times 50$ litres

= $\dfrac{2500}{100}$ litres

= 25 litres

Water to be put into the tank = 25 litres − 15 litres = 10 litres.

Hence, 10 litres of water should be put into the tank.

In an election, there are a total of 80,000 voters and two candidates, A and B. 80% of the voters go to the polls, out of which 60% vote for A. How many votes does B get?

Answer

Total number of voters = 80,000.

Number of voters who went to the polls = 80% of 80,000

= $\dfrac{80}{100} \times 80{,}000$

= $\dfrac{6400000}{100}$

= 64,000

Since 60% of these voters vote for A, the remaining (100 − 60)% = 40% vote for B.

Number of votes B gets = 40% of 64,000

= $\dfrac{40}{100} \times 64{,}000$

= $\dfrac{2560000}{100}$

= 25,600

Hence, B gets 25,600 votes.

70% of our body weight is made up of water. Find the weight of water in the body of a person whose body weight is 56 kg.

Answer

Body weight of the person = 56 kg and water content = 70% of the body weight.

Weight of water in the body = 70% of 56 kg

= $\dfrac{70}{100} \times 56$ kg

= $\dfrac{3920}{100}$ kg

= 39.2 kg

Hence, the weight of water in the body of the person = 39.2 kg.

Only one-fifth of the available water is in liquid form. This limited amount of water is replenished and used by man recurrently. Express this information as percent, showing:

(i) water available in liquid form.

(ii) water available in frozen form.

Answer

(i) Water available in liquid form = $\dfrac{1}{5}$ of the available water.

Percentage of water available in liquid form = $\left(\dfrac{1}{5} \times 100\right)$%

= $\dfrac{100}{5}$%

= 20%

Hence, the water available in liquid form = 20%.

(ii) In percentage, the whole quantity is taken as 100%.

Percentage of water available in frozen form = (100 − 20)% = 80%.

Hence, the water available in frozen form = 80%.

By weight, 90% of tomato and 78% of potato is water. Find:

(i) the weight of water in 25 kg of tomato.

(ii) the total quantity, by weight, of water in 90 kg of potato and 30 kg of tomato.

Answer

(i) Weight of water in 25 kg of tomato = 90% of 25 kg

= $\dfrac{90}{100} \times 25$ kg

= $\dfrac{2250}{100}$ kg

= 22.5 kg

Hence, the weight of water in 25 kg of tomato = 22.5 kg.

(ii) Weight of water in 90 kg of potato = 78% of 90 kg

= $\dfrac{78}{100} \times 90$ kg

= $\dfrac{7020}{100}$ kg

= 70.2 kg

Weight of water in 30 kg of tomato = 90% of 30 kg

= $\dfrac{90}{100} \times 30$ kg

= $\dfrac{2700}{100}$ kg

= 27 kg

Total quantity of water = 70.2 kg + 27 kg = 97.2 kg.

Hence, the total quantity, by weight, of water = 97.2 kg.

Find the number that is:

50% more than 48

Answer

Increase in it = 50% of 48

= $\dfrac{50}{100} \times 48$

= $\dfrac{2400}{100}$

= 24

Required number = 48 + 24 = 72.

Hence, the number that is 50% more than 48 = 72.

Find the number that is:

30% less than 70.

Answer

Decrease in it = 30% of 70

= $\dfrac{30}{100} \times 70$

= $\dfrac{2100}{100}$

= 21

Required number = 70 − 21 = 49.

Hence, the number that is 30% less than 70 = 49.

Evaluate:

8% of 900 - 12% of 750 + 20% of 165.

Answer

$\Rightarrow \dfrac{8}{100} \times 900 - \dfrac{12}{100} \times 750 + \dfrac{20}{100} \times 165$

⇒ 72 - 90 + 33

⇒ 105 - 90

⇒ 15

Hence, 8% of 900 − 12% of 750 + 20% of 165 = 15.

Evaluate:

70% of 70 + 90% of 90 - 120% of 120.

Answer

$\Rightarrow \dfrac{70}{100} \times 70 + \dfrac{90}{100} \times 90 - \dfrac{120}{100} \times 120$

⇒ 49 + 81 - 144

⇒ 130 - 144

⇒ -14

Hence, 70% of 70 + 90% of 90 − 120% of 120 = −14.

Approximately 97.3% water on the earth is not fit for drinking. Find:

(i) the percentage of water on the earth that is fit for drinking.

(ii) the total volume of water available in a certain part of the earth where there is 21,600 m³ of drinking water.

Answer

(i) In percentage, the whole quantity is taken as 100%.

Percentage of water not fit for drinking = 97.3%.

Percentage of water fit for drinking = (100 − 97.3)% = 2.7%.

Hence, the percentage of water on the earth that is fit for drinking = 2.7%.

(ii) Let the total volume of water available be $x$ m³.

Since 2.7% of the water is fit for drinking,

$\Rightarrow \dfrac{2.7}{100} \times x = 21{,}600\\[1em] \Rightarrow x = \dfrac{21{,}600 \times 100}{2.7}\\[1em] \Rightarrow x = \dfrac{21{,}60{,}000}{2.7}\\[1em] \Rightarrow x = 8{,}00{,}000$

Hence, the total volume of water available = 8,00,000 m³.

Air is an important inexhaustible natural resource. It is essential for the survival of human beings, microbes, plants and animals. The following table shows the percentage of various gases in air.

In 800 m³ of air, calculate the approximate quantities of nitrogen, oxygen and other gases.

Answer

Total quantity of air = 800 m³.

Quantity of nitrogen = 78% of 800 m³

= $\dfrac{78}{100} \times 800$ m³

= $\dfrac{62400}{100}$ m³

= 624 m³

Quantity of oxygen = 21% of 800 m³

= $\dfrac{21}{100} \times 800$ m³

= $\dfrac{16800}{100}$ m³

= 168 m³

Quantity of other gases = 1% of 800 m³

= $\dfrac{1}{100} \times 800$ m³

= $\dfrac{800}{100}$ m³

= 8 m³

Hence, nitrogen = 624 m³, oxygen = 168 m³ and other gases = 8 m³.