Percent

ICSE Class 6 Mathematics · Selina Solutions Concise Mathematics Class 6 ICSE · 83 solved questions

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1(i)Exercise 10(A)

Express the following statement in percentage form:

13 out of 20

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To express a statement in the percentage form, write it as a fraction and then multiply the fraction by 100 and write the percentage sign.

⇒ 13 out of 20

$\Rightarrow \frac{13}{20}$

$\Rightarrow \left(\frac{13}{20} \times 100\right)$%

$\Rightarrow \frac{1300}{20}$%

⇒ 65%

Hence, 13 out of 20 = 65%.

1(ii)Exercise 10(A)

Express the following statement in percentage form:

21 eggs out of 30 are good

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To express a statement in the percentage form, write it as a fraction and then multiply the fraction by 100 and write the percentage sign.

⇒ 21 eggs out of 30 are good

$\Rightarrow \frac{21}{30}$ eggs are good

$\Rightarrow \left(\frac{21}{30} \times 100\right)$% eggs are good

$\Rightarrow \frac{2100}{30}$% eggs are good

⇒ 70% eggs are good

Hence, 70% of the eggs are good.

2(i)Exercise 10(A)

Express the following fraction as percent:

$\frac{3}{200}$

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To convert a given fraction into percentage (percent form), multiply the given fraction by 100 and write the percentage sign.

$\Rightarrow \left(\frac{3}{200} \times 100\right)$%

$\Rightarrow \frac{300}{200}$%

$\Rightarrow \frac{3}{2}$%

⇒ 1.5%

Hence, $\mathbf{\frac{3}{200} = 1.5}$%.

2(ii)Exercise 10(A)

Express the following fraction as percent:

$\frac{5}{6}$

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To convert a given fraction into percentage (percent form), multiply the given fraction by 100 and write the percentage sign.

$\Rightarrow \left(\frac{5}{6} \times 100\right)$%

$\Rightarrow \frac{500}{6}$%

$\Rightarrow \frac{250}{3}$%

$\Rightarrow 83\frac{1}{3}$%

Hence, $\mathbf{\frac{5}{6} = 83\frac{1}{3}}$%.

2(iii)Exercise 10(A)

Express the following fraction as percent:

$\frac{65}{80}$

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To convert a given fraction into percentage (percent form), multiply the given fraction by 100 and write the percentage sign.

$\Rightarrow \left(\frac{65}{80} \times 100\right)$%

$\Rightarrow \frac{6500}{80}$%

$\Rightarrow \frac{325}{4}$%

$\Rightarrow 81\frac{1}{4}$% = 81.25%

Hence, $\mathbf{\frac{65}{80} = 81\frac{1}{4}}$% = 81.25%.

2(iv)Exercise 10(A)

Express the following fraction as percent:

$\frac{2}{3}$

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To convert a given fraction into percentage (percent form), multiply the given fraction by 100 and write the percentage sign.

$\Rightarrow \left(\frac{2}{3} \times 100\right)$%

$\Rightarrow \frac{200}{3}$%

$\Rightarrow 66\frac{2}{3}$%

Hence, $\mathbf{\frac{2}{3} = 66\frac{2}{3}}$%.

3(i)Exercise 10(A)

Express as percent:

0.10

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To convert a given decimal into percentage (percent form), multiply the given decimal by 100 and write the percentage sign.

⇒ (0.10 × 100)%

⇒ 10%

Hence, 0.10 = 10%.

3(ii)Exercise 10(A)

Express as percent:

0.02

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To convert a given decimal into percentage (percent form), multiply the given decimal by 100 and write the percentage sign.

⇒ (0.02 × 100)%

⇒ 2%

Hence, 0.02 = 2%.

3(iii)Exercise 10(A)

Express as percent:

0.7

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To convert a given decimal into percentage (percent form), multiply the given decimal by 100 and write the percentage sign.

⇒ (0.7 × 100)%

⇒ 70%

Hence, 0.7 = 70%.

3(iv)Exercise 10(A)

Express as percent:

0.15

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To convert a given decimal into percentage (percent form), multiply the given decimal by 100 and write the percentage sign.

⇒ (0.15 × 100)%

⇒ 15%

Hence, 0.15 = 15%.

3(v)Exercise 10(A)

Express as percent:

0.032

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To convert a given decimal into percentage (percent form), multiply the given decimal by 100 and write the percentage sign.

⇒ (0.032 × 100)%

⇒ 3.2%

Hence, 0.032 = 3.2%.

4(i)Exercise 10(A)

Convert the following into a fraction in its lowest terms:

8%

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To convert a given percentage into a fraction, remove the sign of percentage and divide by 100. Then reduce the resulting fraction to its lowest terms.

$\Rightarrow \frac{8}{100}$

$\Rightarrow \frac{2}{25}$

Hence, 8% = $\mathbf{\frac{2}{25}}$.

4(ii)Exercise 10(A)

Convert the following into a fraction in its lowest terms:

20%

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To convert a given percentage into a fraction, remove the sign of percentage and divide by 100. Then reduce the resulting fraction to its lowest terms.

$\Rightarrow \frac{20}{100}$

$\Rightarrow \frac{1}{5}$

Hence, 20% = $\mathbf{\frac{1}{5}}$.

4(iii)Exercise 10(A)

Convert the following into a fraction in its lowest terms:

85%

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To convert a given percentage into a fraction, remove the sign of percentage and divide by 100. Then reduce the resulting fraction to its lowest terms.

$\Rightarrow \frac{85}{100}$

$\Rightarrow \frac{17}{20}$

Hence, 85% = $\mathbf{\frac{17}{20}}$.

4(iv)Exercise 10(A)

Convert the following into a fraction in its lowest terms:

250%

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To convert a given percentage into a fraction, remove the sign of percentage and divide by 100. Then reduce the resulting fraction to its lowest terms.

$$\begin{aligned}\Rightarrow \frac{250}{100} \\[1em] \Rightarrow \frac{5}{2} \\[1em] \Rightarrow 2\frac{1}{2}\end{aligned}$$

Hence, 250% = $\mathbf{\frac{5}{2} = 2\frac{1}{2}}$.

4(v)Exercise 10(A)

Convert the following into a fraction in its lowest terms:

$12\frac{1}{2}$%

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To convert a given percentage into a fraction, remove the sign of percentage and divide by 100. Then reduce the resulting fraction to its lowest terms.

$\Rightarrow 12\frac{1}{2}$% = $\frac{25}{2}$%

$\Rightarrow \frac{25}{2} \times \frac{1}{100}$

$\Rightarrow \frac{25}{200}$

$\Rightarrow \frac{1}{8}$

Hence, $\mathbf{12\frac{1}{2}}$% = $\mathbf{\frac{1}{8}}$.

5(i)Exercise 10(A)

Express as a decimal fraction:

25%

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To convert a given percentage into a decimal, remove the sign of percentage and divide by 100.

$\Rightarrow \frac{25}{100}$

⇒ 0.25

Hence, 25% = 0.25.

5(ii)Exercise 10(A)

Express as a decimal fraction:

108%

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To convert a given percentage into a decimal, remove the sign of percentage and divide by 100.

$\Rightarrow \frac{108}{100}$

⇒ 1.08

Hence, 108% = 1.08.

5(iii)Exercise 10(A)

Express as a decimal fraction:

95%

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To convert a given percentage into a decimal, remove the sign of percentage and divide by 100.

$\Rightarrow \frac{95}{100}$

⇒ 0.95

Hence, 95% = 0.95.

5(iv)Exercise 10(A)

Express as a decimal fraction:

4.5%

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To convert a given percentage into a decimal, remove the sign of percentage and divide by 100.

$\Rightarrow \frac{4.5}{100}$

⇒ 0.045

Hence, 4.5% = 0.045.

5(v)Exercise 10(A)

Express as a decimal fraction:

29.2%

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To convert a given percentage into a decimal, remove the sign of percentage and divide by 100.

$\Rightarrow \frac{29.2}{100}$

⇒ 0.292

Hence, 29.2% = 0.292.

6(i)Exercise 10(A)

Express the following number as percent:

7

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To express a given number as percent, multiply the number by 100 and write the sign of percentage.

⇒ (7 × 100)%

⇒ 700%

Hence, 7 expressed as a percent is 700%.

6(ii)Exercise 10(A)

Express the following number as percent:

2

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To express a given number as percent, multiply the number by 100 and write the sign of percentage.

⇒ (2 × 100)%

⇒ 200%

Hence, 2 expressed as a percent is 200%.

6(iii)Exercise 10(A)

Express the following number as percent:

19.5

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To express a given number as percent, multiply the number by 100 and write the sign of percentage.

⇒ (19.5 × 100)%

⇒ 1950%

Hence, 19.5 expressed as a percent is 1950%.

6(iv)Exercise 10(A)

Express the following number as percent:

5.37

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To express a given number as percent, multiply the number by 100 and write the sign of percentage.

⇒ (5.37 × 100)%

⇒ 537%

Hence, 5.37 expressed as a percent is 537%.

1(i)Exercise 10(B)

Express:

₹ 5 as a percentage of ₹ 25.

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To express one quantity as a percentage of another quantity, the two quantities must have the same unit. Then, divide the first quantity by the second one, multiply the result by 100 and write the percentage sign.

$\Rightarrow \left(\frac{5}{25} \times 100\right)$%

$\Rightarrow \frac{500}{25}$%

⇒ 20%

Hence, ₹ 5 is 20% of ₹ 25.

1(ii)Exercise 10(B)

Express:

80 paise as a percent of ₹ 4.

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To express one quantity as a percentage of another quantity, the two quantities must have the same unit. Then, divide the first quantity by the second one, multiply the result by 100 and write the percentage sign.

Converting ₹ 4 into paise :

₹ 4 = 4 × 100 paise = 400 paise.

$\Rightarrow \left(\frac{80}{400} \times 100\right)$%

$\Rightarrow \frac{8000}{400}$%

⇒ 20%

Hence, 80 paise is 20% of ₹ 4.

1(iii)Exercise 10(B)

Express:

700 g as a percentage of 2.8 kg.

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To express one quantity as a percentage of another quantity, the two quantities must have the same unit. Then, divide the first quantity by the second one, multiply the result by 100 and write the percentage sign.

Converting 2.8 kg into grams :

2.8 kg = 2.8 × 1000 g = 2800 g.

$\Rightarrow \left(\frac{700}{2800} \times 100\right)$%

$\Rightarrow \frac{70000}{2800}$%

⇒ 25%

Hence, 700 g is 25% of 2.8 kg.

1(iv)Exercise 10(B)

Express:

90 cm as a percent of 4.5 m.

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To express one quantity as a percentage of another quantity, the two quantities must have the same unit. Then, divide the first quantity by the second one, multiply the result by 100 and write the percentage sign.

Converting 4.5 m into centimetres :

4.5 m = 4.5 × 100 cm = 450 cm.

$\Rightarrow \left(\frac{90}{450} \times 100\right)$%

$\Rightarrow \frac{9000}{450}$%

⇒ 20%

Hence, 90 cm is 20% of 4.5 m.

2(i)Exercise 10(B)

Express the first quantity as a percent of the second:

40 p, ₹ 2

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To express one quantity as a percentage of another quantity, the two quantities must have the same unit. Then, divide the first quantity by the second one, multiply the result by 100 and write the percentage sign.

Converting ₹ 2 into paise :

₹ 2 = 2 × 100 p = 200 p.

$\Rightarrow \left(\frac{40}{200} \times 100\right)$%

$\Rightarrow \frac{4000}{200}$%

⇒ 20%

Hence, 40 p is 20% of ₹ 2.

2(ii)Exercise 10(B)

Express the first quantity as a percent of the second:

500 g, 6 kg

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To express one quantity as a percentage of another quantity, the two quantities must have the same unit. Then, divide the first quantity by the second one, multiply the result by 100 and write the percentage sign.

Converting 6 kg into grams :

6 kg = 6 × 1000 g = 6000 g.

$\Rightarrow \left(\frac{500}{6000} \times 100\right)$%

$\Rightarrow \frac{50000}{6000}$%

$\Rightarrow \frac{25}{3}$%

$\Rightarrow 8\frac{1}{3}$%

Hence, 500 g is $\mathbf{8\frac{1}{3}}$% of 6 kg.

2(iii)Exercise 10(B)

Express the first quantity as a percent of the second:

42 seconds, 6 minutes.

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To express one quantity as a percentage of another quantity, the two quantities must have the same unit. Then, divide the first quantity by the second one, multiply the result by 100 and write the percentage sign.

Converting 6 minutes into seconds :

6 minutes = 6 × 60 seconds = 360 seconds.

$\Rightarrow \left(\frac{42}{360} \times 100\right)$%

$\Rightarrow \frac{4200}{360}$%

$\Rightarrow \frac{35}{3}$%

$\Rightarrow 11\frac{2}{3}$%

Hence, 42 seconds is $\mathbf{11\frac{2}{3}}$% of 6 minutes.

3(i)Exercise 10(B)

Find the value of the following:

20% of ₹ 150

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To find the percentage (percent) of a given quantity, express the given percent as a fraction and multiply it by the given quantity.

$$\begin{aligned}\Rightarrow \frac{20}{100} \times ₹\ 150\\[1em] \Rightarrow ₹\ \frac{3{,}000}{100}\\[1em] \Rightarrow ₹\ 30\end{aligned}$$

Hence, 20% of ₹ 150 = ₹ 30.

3(ii)Exercise 10(B)

Find the value of the following:

90% of 130

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To find the percentage (percent) of a given quantity, express the given percent as a fraction and multiply it by the given quantity.

$$\begin{aligned}\Rightarrow \frac{90}{100} \times 130\\[1em] \Rightarrow \frac{11700}{100}\\[1em] \Rightarrow 117\end{aligned}$$

Hence, 90% of 130 = 117.

3(iii)Exercise 10(B)

Find the value of the following:

15% of 2 minutes

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To find the percentage (percent) of a given quantity, express the given percent as a fraction and multiply it by the given quantity.

$$\begin{aligned}\Rightarrow \frac{15}{100} \times 2 \text{ minutes}\\[1em] \Rightarrow \frac{30}{100} \text{ minutes}\\[1em] \Rightarrow 0.3 \text{ minutes}\end{aligned}$$

Converting 0.3 minutes into seconds :

0.3 minutes = 0.3 × 60 seconds = 18 seconds.

Hence, 15% of 2 minutes = 0.3 minutes = 18 seconds.

3(iv)Exercise 10(B)

Find the value of the following:

7.5% of 500 kg.

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To find the percentage (percent) of a given quantity, express the given percent as a fraction and multiply it by the given quantity.

$$\begin{aligned}\Rightarrow \frac{7.5}{100} \times 500 \text{ kg}\\[1em] \Rightarrow \frac{3750}{100} \text{ kg}\\[1em] \Rightarrow 37.5 \text{ kg}\end{aligned}$$

Hence, 7.5% of 500 kg = 37.5 kg.

4Exercise 10(B)

If a man spends 70% of his income, what percent does he save?

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

∴ Total income of the man = 100%.

Income spent by the man = 70%.

Percent of income saved = (100 − 70)% = 30%.

Hence, the man saves 30% of his income.

5Exercise 10(B)

A girl gets 65 marks out of 80. What percentage of marks does she get?

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Marks obtained by the girl = 65 and maximum marks = 80.

Percentage of marks = $\left(\frac{65}{80} \times 100\right)$%

= $\frac{6500}{80}$%

= $\frac{325}{4}$%

= $81\frac{1}{4}$%

= 81.25%

Hence, the percentage of marks obtained by the girl is $\mathbf{81\frac{1}{4}}$%, i.e., 81.25%.

6Exercise 10(B)

A class contains 25 children, of which 6 are girls. What is the percentage of boys in the class?

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Total number of children in the class = 25 and number of girls = 6.

Number of boys = 25 − 6 = 19.

Percentage of boys = $\left(\frac{19}{25} \times 100\right)$%

= $\frac{1900}{25}$%

= 76%

Hence, the percentage of boys in the class = 76%.

7Exercise 10(B)

A tin contains 20 litres of petrol. Due to leakage, 3 litres of petrol is lost. What is the percentage of petrol left in the tin?

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Total quantity of petrol in the tin = 20 litres and petrol lost = 3 litres.

Quantity of petrol left = 20 − 3 = 17 litres.

Percentage of petrol left = $\left(\frac{17}{20} \times 100\right)$%

= $\frac{1700}{20}$%

= 85%

Hence, the percentage of petrol left in the tin = 85%.

8Exercise 10(B)

An alloy of copper and zinc contains 45% copper and the rest is zinc. Find the weight of zinc in 20 kg of the alloy.

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

Percentage of copper in the alloy = 45%.

∴ Percentage of zinc in the alloy = (100 − 45)% = 55%.

Weight of zinc in 20 kg of the alloy = 55% of 20 kg

= $\frac{55}{100} \times 20$ kg

= $\frac{1100}{100}$ kg

= 11 kg

Hence, the weight of zinc in 20 kg of the alloy = 11 kg.

9Exercise 10(B)

A boy got 60 out of 80 in Hindi, 75 out of 100 in English, and 65 out of 70 in Arithmetic. In which subject is his percentage of marks the best? Also, find his overall percentage.

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Percentage score in Hindi = $\left(\frac{60}{80} \times 100\right)$%

= $\frac{6000}{80}$%

= 75%.

Percentage score in English = $\left(\frac{75}{100} \times 100\right)$%

= $\frac{7500}{100}$%

= 75%.

Percentage score in Arithmetic = $\left(\frac{65}{70} \times 100\right)$%

= $\frac{6500}{70}$%

= $\frac{650}{7}$%

= $92\frac{6}{7}$% (= 92.86% approximately).

Since $92\frac{6}{7}$% > 75%, his percentage of marks is the best in Arithmetic.

For the overall percentage :

Sum of the maximum marks of all the three subjects = 80 + 100 + 70 = 250

and total score in the three subjects = 60 + 75 + 65 = 200.

Overall percentage = $\left(\frac{200}{250} \times 100\right)$%

= $\frac{20000}{250}$%

= 80%

Hence, the boy's percentage of marks is the highest in Arithmetic, i.e., $92\frac{6}{7}$% (approximately 92.86%), and his overall percentage is 80%.

10Exercise 10(B)

60 more soldiers joined a camp of 500 soldiers. What percentage of the earlier soldiers do they make?

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Number of soldiers earlier in the camp = 500 and number of soldiers who joined = 60.

Required percentage = $\left(\frac{60}{500} \times 100\right)$%

= $\frac{6000}{500}$%

= 12%

Hence, the soldiers who joined make 12% of the earlier soldiers.

11Exercise 10(B)

In a plot of ground of area 6000 sq.m, only 4500 sq.m is allowed for construction. What percentage of area is to be left without construction?

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Total area of the plot = 6000 sq.m and area allowed for construction = 4500 sq.m.

Area to be left without construction = 6000 − 4500 = 1500 sq.m.

Percentage of area left without construction = $\left(\frac{1500}{6000} \times 100\right)$%

= $\frac{150000}{6000}$%

= 25%

Hence, 25% of the area is to be left without construction.

12Exercise 10(B)

Mr. Sharma has a monthly salary of ₹ 8,000. If he spends ₹ 6,400 every month, find:

(i) his monthly expenditure as percent.

(ii) his monthly savings as percent.

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Monthly salary of Mr. Sharma = ₹ 8,000 and monthly expenditure = ₹ 6,400.

(i) Monthly expenditure as percent = $\left(\frac{6400}{8000} \times 100\right)$%

= $\frac{640000}{8000}$%

= 80%

Hence, his monthly expenditure is 80% of his salary.

(ii) Monthly savings = ₹ 8,000 − ₹ 6,400 = ₹ 1,600.

Monthly savings as percent = $\left(\frac{1600}{8000} \times 100\right)$%

= $\frac{160000}{8000}$%

= 20%

Hence, his monthly savings are 20% of his salary.

13Exercise 10(B)

The monthly salary of Rohit is ₹ 24,000. If his salary increases by 12%, find his new monthly salary.

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Original monthly salary of Rohit = ₹ 24,000.

Increase in salary = 12% of ₹ 24,000

= $\frac{12}{100} \times ₹\ 24{,}000$

= $₹\ \frac{2{,}88{,}000}{100}$

= ₹ 2,880

New monthly salary = ₹ 24,000 + ₹ 2,880 = ₹ 26,880.

Hence, the new monthly salary of Rohit = ₹ 26,880.

14Exercise 10(B)

In a sale, the price of an article is reduced by 30%. If the original price of the article is ₹ 1,800, find:

(i) the reduction in the price of the article

(ii) the reduced price of the article.

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Original price of the article = ₹ 1,800 and reduction = 30%.

(i) Reduction in the price = 30% of ₹ 1,800

= $\frac{30}{100} \times ₹\ 1{,}800$

= $₹\ \frac{54{,}000}{100}$

= ₹ 540

Hence, the reduction in the price of the article = ₹ 540.

(ii) Reduced price of the article = Original price − Reduction in price

= ₹ 1,800 − ₹ 540

= ₹ 1,260

Hence, the reduced price of the article = ₹ 1,260.

15(i)Exercise 10(B)

Evaluate:

30% of 200 + 20% of 450 - 25% of 600

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Solving,

$\Rightarrow \frac{30}{100} \times 200 + \frac{20}{100} \times 450 - \frac{25}{100} \times 600$

⇒ 60 + 90 - 150

⇒ 150 - 150

⇒ 0

Hence, 30% of 200 + 20% of 450 − 25% of 600 = 0.

15(ii)Exercise 10(B)

Evaluate:

10% of ₹ 450 - 12% of ₹ 500 + 8% of ₹ 500.

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Solving,

$\Rightarrow \frac{10}{100} \times ₹\ 450 - \frac{12}{100} \times ₹\ 500 + \frac{8}{100} \times ₹\ 500$

⇒ ₹ 45 - ₹ 60 + ₹ 40

⇒ ₹ 85 - ₹ 60

⇒ ₹ 25

Hence, 10% of ₹ 450 − 12% of ₹ 500 + 8% of ₹ 500 = ₹ 25.

1Exercise 10(C)

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

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

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

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

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

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

= 20%

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

2Exercise 10(C)

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

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

Increase in population = 4500 − 4000 = 500.

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

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

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

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

= 12.5%

Hence, the population had increased by 12.5%.

3Exercise 10(C)

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?

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

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

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

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

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

= 25%

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

4(i)Exercise 10(C)

Find the resulting quantity when:

₹ 400 is decreased by 8%.

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Decrease in it = 8% of ₹ 400

= $\frac{8}{100} \times ₹\ 400$

= $₹\ \frac{3{,}200}{100}$

= ₹ 32

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

Hence, the resulting quantity = ₹ 368.

4(ii)Exercise 10(C)

Find the resulting quantity when:

25 km is increased by 5%.

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Increase in it = 5% of 25 km

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

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

= 1.25 km

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

Hence, the resulting quantity = 26.25 km.

4(iii)Exercise 10(C)

Find the resulting quantity when:

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

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

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

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

= $\frac{15000}{200}$ 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.

4(iv)Exercise 10(C)

Find the resulting quantity when:

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

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Increase in it = 2.5% of ₹ 62,500

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

= $₹\ \frac{1{,}56{,}250}{100}$

= ₹ 1,562.50

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

Hence, the resulting quantity = ₹ 64,062.50.

5Exercise 10(C)

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

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

Decrease in population = 12% of 25,000

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

= $\frac{300000}{100}$

= 3,000

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

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

6Exercise 10(C)

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?

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Total salary = ₹ 13,500.

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

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

Money spent on food = 50% of ₹ 9,000

= $\frac{50}{100} \times ₹\ 9{,}000$

= $₹\ \frac{4{,}50{,}000}{100}$

= ₹ 4,500

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

= $\frac{20}{100} \times ₹\ 9{,}000$

= $₹\ \frac{1{,}80{,}000}{100}$

= ₹ 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.

7Exercise 10(C)

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?

Show SolutionHide Solution↓

Total capacity of the tank = 50 litres.

Water present in the tank = 30% of 50 litres

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

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

= 15 litres

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

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

= $\frac{2500}{100}$ 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.

8Exercise 10(C)

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?

Show SolutionHide Solution↓

Total number of voters = 80,000.

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

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

= $\frac{6400000}{100}$

= 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

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

= $\frac{2560000}{100}$

= 25,600

Hence, B gets 25,600 votes.

9Exercise 10(C)

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.

Show SolutionHide Solution↓

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

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

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

= 39.2 kg

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

10Exercise 10(C)

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.

Show SolutionHide Solution↓

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

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

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

= 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%.

11Exercise 10(C)

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.

Show SolutionHide Solution↓

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

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

= $\frac{2250}{100}$ 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

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

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

= 70.2 kg

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

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

= $\frac{2700}{100}$ 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.

12(i)Exercise 10(C)

Find the number that is:

50% more than 48

Show SolutionHide Solution↓

Increase in it = 50% of 48

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

= $\frac{2400}{100}$

= 24

Required number = 48 + 24 = 72.

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

12(ii)Exercise 10(C)

Find the number that is:

30% less than 70.

Show SolutionHide Solution↓

Decrease in it = 30% of 70

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

= $\frac{2100}{100}$

= 21

Required number = 70 − 21 = 49.

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

13(i)Exercise 10(C)

Evaluate:

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

Show SolutionHide Solution↓

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

⇒ 72 - 90 + 33

⇒ 105 - 90

⇒ 15

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

13(ii)Exercise 10(C)

Evaluate:

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

Show SolutionHide Solution↓

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

⇒ 49 + 81 - 144

⇒ 130 - 144

⇒ -14

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

14Exercise 10(C)

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.

Show SolutionHide Solution↓

(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,

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

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

15Exercise 10(C)

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.

Contents of airPercentage (by volume)
Nitrogen78
Oxygen21
Others (carbon dioxide, inert gases, water vapour, etc.)1

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

Show SolutionHide Solution↓

Total quantity of air = 800 m3.

Quantity of nitrogen = 78% of 800 m3

= $\frac{78}{100} \times 800$ m3

= $\frac{62400}{100}$ m3

= 624 m3

Quantity of oxygen = 21% of 800 m3

= $\frac{21}{100} \times 800$ m3

= $\frac{16800}{100}$ m3

= 168 m3

Quantity of other gases = 1% of 800 m3

= $\frac{1}{100} \times 800$ m3

= $\frac{800}{100}$ m3

= 8 m3

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

11 Mark · Multiple Choice Questions

0.75 as percentage is:

Show SolutionHide Solution↓

To convert a decimal into percentage, multiply it by 100 and write the sign of percentage.

⇒ (0.75 × 100)%

⇒ 75%

Hence, option 2 is the correct option.

21 Mark · Multiple Choice Questions

4 : 5 as percentage is:

Show SolutionHide Solution↓

Writing the ratio 4 : 5 as a fraction and then multiplying the fraction by 100 and writing the percentage sign :

⇒ 4 : 5

$\Rightarrow \frac{4}{5}$

$\Rightarrow \left(\frac{4}{5} \times 100\right)$%

$\Rightarrow \frac{400}{5}$%

⇒ 80%

Hence, option 1 is the correct option.

31 Mark · Multiple Choice Questions

80% of 2 kg is:

Show SolutionHide Solution↓

The given quantity is in kilogram, whereas all the options are in gram. So, 2 kg is first converted into gram, as the answer and the options must be in the same unit.

Converting 2 kg into grams :

2 kg = 2 × 1000 g = 2000 g.

80% of 2000 g

$$\begin{aligned}\Rightarrow \frac{80}{100} \times 2000 \text{ g}\\[1em] \Rightarrow \frac{160000}{100} \text{ g}\\[1em] \Rightarrow 1600 \text{ g}\end{aligned}$$

Hence, option 1 is the correct option.

41 Mark · Multiple Choice Questions

400 - 40% of 400 is:

Show SolutionHide Solution↓

⇒ 40% of 400

$$\begin{aligned}\Rightarrow \frac{40}{100} \times 400\\[1em] \Rightarrow \frac{16000}{100}\\[1em] \Rightarrow 160\end{aligned}$$

∴ 400 − 40% of 400 = 400 − 160 = 240.

Hence, option 2 is the correct option.

51 Mark · Multiple Choice Questions

80% of 500 - 30% of 500 is:

Show SolutionHide Solution↓

⇒ 80% of 500 − 30% of 500

$\Rightarrow \frac{80}{100} \times 500 - \frac{30}{100} \times 500$

⇒ 400 - 150

⇒ 250

Hence, option 3 is the correct option.

61 Mark · Multiple Choice Questions

800 is increased by 50%, the increase is:

Show SolutionHide Solution↓

The increase is 50% of 800.

$$\begin{aligned}\Rightarrow \frac{50}{100} \times 800\\[1em] \Rightarrow \frac{40000}{100}\\[1em] \Rightarrow 400\end{aligned}$$

Hence, option 2 is the correct option.

71 Mark · Multiple Choice Questions

1600 is decreased by 25%, the decreased number is:

Show SolutionHide Solution↓

Decrease in it = 25% of 1600

= $\frac{25}{100} \times 1600$

= $\frac{40000}{100}$

= 400

Decreased number = 1600 − 400 = 1200.

Hence, option 1 is the correct option.

81 Mark · Multiple Choice Questions

80 is increased by 50% of it, the increased number is:

Show SolutionHide Solution↓

Increase in it = 50% of 80

= $\frac{50}{100} \times 80$

= $\frac{4000}{100}$

= 40

Increased number = 80 + 40 = 120.

Hence, option 4 is the correct option.

91 Mark · Multiple Choice Questions

500 is decreased by 20% of it, the decreased number is:

Show SolutionHide Solution↓

Decrease in it = 20% of 500

= $\frac{20}{100} \times 500$

= $\frac{10000}{100}$

= 100

Decreased number = 500 − 100 = 400.

Hence, option 3 is the correct option.

101 Mark · Multiple Choice Questions

50% of 20 - 20% of 50 is equal to:

Show SolutionHide Solution↓

⇒ 50% of 20 − 20% of 50

$\Rightarrow \frac{50}{100} \times 20 - \frac{20}{100} \times 50$

⇒ 10 - 10

⇒ 0

Hence, option 3 is the correct option.

11Statement I-II Type Questions

Statement 1: Out of 1000 students 60% are boys and the rest are girls. The number of girls is 600.

Statement 2: The number of boys is 600.

Which of the following options is correct?

Show SolutionHide Solution↓

Total number of students = 1000 and percentage of boys = 60%.

Number of boys = 60% of 1000

= $\frac{60}{100} \times 1000$

= $\frac{60000}{100}$

= 600

Number of girls = 1000 − 600 = 400.

The number of girls is 400 and not 600.

Thus, statement 1 is false.

The number of boys is 600, as calculated above.

Thus, statement 2 is true.

Hence, option 4 is the correct option.

12Statement I-II Type Questions

Statement 1: 10% of 600 + 15% of 600 = 150.

Statement 2: 10% of 600 + 15% of 600 = 25% of 600 = $\frac{25}{100}$ × 600 = 150

Which of the following options is correct?

Show SolutionHide Solution↓

Solving,

⇒ 10% of 600 + 15% of 600

$\Rightarrow \frac{10}{100} \times 600 + \frac{15}{100} \times 600$

⇒ 60 + 90

⇒ 150

So, 10% of 600 + 15% of 600 = 150.

Thus, statement 1 is true.

Since both the percentages are taken of the same quantity 600, they can be added.

⇒ (10% + 15%) of 600

⇒ 25% of 600

$\Rightarrow \frac{25}{100} \times 600$

⇒ 150

Thus, statement 2 is true.

Hence, option 1 is the correct option.

13Assertion Reason Type Questions

Assertion (A): 800 - 40% of 800 is 240.

Reason (R): 800 - 40% of 800 = 60% of 800.

Show SolutionHide Solution↓

40% of 800 $= \frac{40}{100} \times 800 = 320$.

∴ 800 − 40% of 800 = 800 − 320 = 480, and not 240.

Thus, Assertion (A) is false.

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

⇒ 800 - 40% of 800

⇒ (100 - 40)% of 800

⇒ 60% of 800

Thus, Reason (R) is true.

Hence, option 2 is the correct option.

14Assertion Reason Type Questions

Assertion (A): 300 increases to 390. Increase% = 30%

Reason (R): Increase % = $\frac{\text{increase in value}}{\text{original value}}$ × 100%

Show SolutionHide Solution↓

Original value = 300 and new value = 390.

Increase in value = 390 − 300 = 90.

⇒ Increase% = $\left(\frac{90}{300} \times 100\right)$%

⇒ Increase% = $\frac{9000}{300}$%

⇒ Increase% = 30%

Thus, Assertion (A) is true.

The formula to find the increase percent is

Increase% = $\frac{\text{Increase in value}}{\text{Original value}} \times 100$%, which is correct.

Thus, Reason (R) is true.

Hence, option 3 is the correct option.

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