Percentage

ICSE Class 7 Mathematics · Solutions for Class 7 RS Aggarwal ICSE Foundation Mathematics · 71 solved questions

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1Exercise 9(A)

Express each of the following as a per cent :

(i) $\frac{3}{5}$

(ii) $\frac{5}{16}$

(iii) $\frac{19}{40}$

(iv) $3\frac{1}{4}$

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(i) $\frac{3}{5}$

= $\Big( \frac{3}{5} \times 100 \Big)$%

= $(3 \times 20)$%

= 60%

Hence, the answer is 60%

(ii) $\frac{5}{16}$

= $\Big( \frac{5}{16} \times 100 \Big)$%

= $\Big( \frac{5 \times 25}{4} \Big)$%

= $\frac{125}{4}$%

= 31.25%

Hence, the answer is 31.25%

(iii) $\frac{19}{40}$

= $\Big( \frac{19}{40} \times 100 \Big)$%

= $\Big( \frac{19 \times 5}{2} \Big)$%

= $\frac{95}{2}$%

= 47.5%

Hence, the answer is 47.5%

(iv) $3\frac{1}{4}$

First, convert the mixed fraction to an improper fraction:

$3\frac{1}{4} = \frac{13}{4}$

We have:

$\frac{13}{4} = \Big( \frac{13}{4} \times 100 \Big)$%

= $(13 \times 25)$%

= 325%

Hence, the answer is 325%

2Exercise 9(A)

Express each of the following as a per cent :

(i) 0.45

(ii) 0.03

(iii) 0.1

(iv) 2.45

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(i) 0.45

First, convert decimal into fraction:

0.45 = $\frac{45}{100}$

We have:

$\frac{45}{100} = \Big( \frac{45}{100} \times 100 \Big)$% = 45%

Hence, the answer is 45%

(ii) 0.03

First, convert decimal into fraction:

0.03 = $\frac{3}{100}$

We have:

$\frac{3}{100} = \Big( \frac{3}{100} \times 100 \Big)$% = 3%

Hence, the answer is 3%

(iii) 0.1

First, convert decimal into fraction:

0.1 = $\frac{1}{10}$

We have:

$\frac{1}{10} = \Big( \frac{1}{10} \times 100 \Big)$% = $(1 \times 10)$% = 10%

Hence, the answer is 10%

(iv) 2.45

We have:

2.45 = $\frac{245}{100}$

First, convert decimal into fraction:

$\frac{245}{100} = \Big( \frac{245}{100} \times 100 \Big)$% = 245%

Hence, the answer is 245%

3Exercise 9(A)

Express each of the following as a per cent :

(i) 5 : 8

(ii) 11 : 18

(iii) 1 : 5

(iv) 5 : 4

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(i) 5 : 8

First, convert ratio into fraction:

5 : 8 = $\frac{5}{8}$

We have:

$\frac{5}{8} = \Big( \frac{5}{8} \times 100 \Big)$%

= $\Big( \frac{5 \times 25}{2} \Big)$%

= $\frac{125}{2}$%

= 62.5%

Hence, the answer is 62.5%

(ii) 11 : 18

First, convert ratio into fraction:

11 : 18 = $\frac{11}{18}$

We have:

$\frac{11}{18} = \Big( \frac{11}{18} \times 100 \Big)$%

= $\Big( \frac{11 \times 50}{9} \Big)$%

= $\frac{550}{9}$%

= $61\frac{1}{9}$%

Hence, the answer is $61\frac{1}{9}$%

(iii) 1 : 5

First, convert ratio into fraction:

1 : 5 = $\frac{1}{5}$

We have:

$\frac{1}{5} = \Big( \frac{1}{5} \times 100 \Big)$% = 20%

Hence, the answer is 20%

(iv) 5 : 4

First, convert ratio into fraction:

5 : 4 = $\frac{5}{4}$

We have:

$\frac{5}{4} = \Big( \frac{5}{4} \times 100 \Big)$% = (5 × 25)% = 125%

Hence, the answer is 125%

4Exercise 9(A)

Express each of the following as a fraction :

(i) 24%

(ii) $12\frac{1}{2}$%

(iii) $6\frac{2}{3}$%

(iv) 165%

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(i) 24%

To express a percentage as a fraction, divide the number by 100 and remove the % sign.

24% = $\frac{24}{100} = \frac{6}{25}$

Hence, the answer is $\frac{6}{25}$

(ii) $12\frac{1}{2}$%

First, convert the mixed fraction to an improper fraction:

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

To express a percentage as a fraction, divide the number by 100 and remove the % sign.

$\frac{25}{2}$% = $\left( \frac{25}{2} \times \frac{1}{100} \right) = \frac{25}{200} = \frac{1}{8}$

Hence, the answer is $\frac{1}{8}$

(iii) $6\frac{2}{3}$%

First, convert the mixed fraction to an improper fraction:

$6\frac{2}{3} = \frac{20}{3}$

To express a percentage as a fraction, divide the number by 100 and remove the % sign.

$\frac{20}{3}$% = $\left( \frac{20}{3} \times \frac{1}{100} \right) = \frac{20}{300} = \frac{1}{15}$

Hence, the answer is $\frac{1}{15}$

(iv) 165%

To express a percentage as a fraction, divide the number by 100 and remove the % sign.

165% = $\frac{165}{100} = \frac{33}{20} = 1\frac{13}{20}$

Hence, the answer is $1\frac{13}{20}$

5Exercise 9(A)

Express each of the following as a fraction :

(i) 0.25%

(ii) 0.05%

(iii) 2.25%

(iv) 0.006%

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(i) 0.25%

To express a percentage with a decimal as a fraction, divide the number by 100 and remove the % sign.

0.25% = $\frac{0.25}{100}$

Multiply numerator and denominator by 100 to remove the decimal:

$$\begin{aligned}\frac{0.25}{100} = \frac{0.25 \times 100}{100 \times 100} \\[1em] = \frac{25}{10000} \\[1em] = \frac{1}{400}\end{aligned}$$

Hence, the answer is $\frac{1}{400}$

(ii) 0.05%

To express a percentage with a decimal as a fraction, divide the number by 100 and remove the % sign.

0.05% = $\frac{0.05}{100}$

Multiply numerator and denominator by 100 to remove the decimal:

$$\begin{aligned}\frac{0.05}{100} = \frac{0.05 \times 100}{100 \times 100} \\[1em] = \frac{5}{10000} \\[1em] = \frac{1}{2000}\end{aligned}$$

Hence, the answer is $\frac{1}{2000}$

(iii) 2.25%

To express a percentage with a decimal as a fraction, divide the number by 100 and remove the % sign.

2.25% = $\frac{2.25}{100}$

Multiply numerator and denominator by 100 to remove the decimal:

$$\begin{aligned}\frac{2.25}{100} = \frac{2.25 \times 100}{100 \times 100} \\[1em] = \frac{225}{10000} \\[1em] = \frac{9}{400}\end{aligned}$$

Hence, the answer is $\frac{9}{400}$

(iv) 0.006%

To express a percentage with a decimal as a fraction, divide the number by 100 and remove the % sign.

0.006% = $\frac{0.006}{100}$

Multiply numerator and denominator by 1000 to remove the decimal:

$$\begin{aligned}\frac{0.006}{100} = \frac{0.006 \times 1000}{100 \times 1000} \\[1em] = \frac{6}{100000} \\[1em] = \frac{3}{50000}\end{aligned}$$

Hence, the answer is $\frac{3}{50000}$

6Exercise 9(A)

Express each of the following as a ratio :

(i) 12%

(ii) $\frac{3}{5}$%

(iii) 7.5%

(iv) 0.36%

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(i) 12%

To express a percentage as a ratio, we first write it as a fraction with a denominator of 100.

12% = $\frac{12}{100} = \frac{3}{25} = 3 : 25$

Hence, the answer is 3 : 25

(ii) $\frac{3}{5}$%

To express a percentage as a ratio, we first write it as a fraction with a denominator of 100.

$\frac{3}{5}$% = $\left( \frac{3}{5} \times \frac{1}{100} \right) = \frac{3}{500} = 3 : 500$

Hence, the answer is 3 : 500

(iii) 7.5%

To express a percentage as a ratio, we first write it as a fraction with a denominator of 100.

7.5% = $\frac{7.5}{100}$

Multiply numerator and denominator by 10 to remove the decimal:

$\frac{7.5}{100} =\frac{7.5 \times 10}{100 \times 10} = \frac{75}{1000} = \frac{3}{40} = 3 : 40$

Hence, the answer is 3 : 40

(iv) 0.36%

0.36% = $\frac{0.36}{100}$

Multiply numerator and denominator by 100 to remove the decimal:

$\frac{0.36}{100} = \frac{0.36 \times 100}{100 \times 100} = \frac{36}{10000} = \frac{9}{2500} = 9 : 2500$

Hence, the answer is 9 : 2500

7Exercise 9(A)

Find:

(i) 25% of ₹ 7

(ii) $7\frac{1}{2}$% of ₹ 40

(iii) 1% of ₹ 325

(iv) 8% of 5 kg

(v) $16\frac{2}{3}$% of 9 m

(vi) 0.6% of 5 km

(vii) $6\frac{1}{4}$% of (1 hour 20 min.)

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(i) We have:

25% of ₹ 7

$$\begin{aligned}= ₹ \frac{25}{100} \times 7 \\[1em] = ₹ \frac{1}{4} \times 7 \\[1em] = ₹ \frac{7}{4} \\[1em] = ₹ 1.75\end{aligned}$$

Hence, the answer is ₹ 1.75

(ii) We have:

$7\frac{1}{2}$% of ₹ 40

First, convert to an improper fraction:

$7\frac{1}{2}$% = $\frac{15}{2}$%.

$\frac{15}{2}$% of ₹ 40

$$\begin{aligned}= ₹ \left( 40 \times \frac{15}{2 \times 100} \right) \\[1em] = ₹ 40 \times \frac{15}{200} \\[1em] = ₹ 40 \times \frac{3}{40} \\[1em] = ₹ 3\end{aligned}$$

Hence, the answer is ₹ 3

(iii) We have:

1% of ₹ 325

$$\begin{aligned}= ₹ \frac{1}{100} \times 325 \\[1em] = ₹ 3.25\end{aligned}$$

Hence, the answer is ₹ 3.25

(iv) We have:

8% of 5 kg

$$\begin{aligned}= \frac{8}{100} \times 5 \text{ kg} \\[1em] = \frac{40}{100} \text{ kg} \\[1em] = 0.4 \text{ kg} \\[1em] = 400 \text{ g}\end{aligned}$$

Hence, the answer is 400 g

(v) We have:

$16\frac{2}{3}$% of 9 m

Convert to an improper fraction:

$16\frac{2}{3}$% = $\frac{50}{3}$%

$\frac{50}{3}$% of 9 m

$$\begin{aligned}= \left( 9 \times \frac{50}{3 \times 100} \right) \text{ m} \\[1em] = 9 \times \frac{50}{300} \text{ m} \\[1em] = 9 \times \frac{1}{6} \text{ m} \\[1em] = \frac{9}{6} \text{ m} \\[1em] = 1.5 \text{ m}\end{aligned}$$

Hence, the answer is 1.5 m

(vi) We have:

0.6% of 5 km

$$\begin{aligned}= \frac{0.6}{100} \times 5 \text{ km} \\[1em] = \frac{0.6}{20} \text{ km} \\[1em] = \frac{0.6 \times 10}{20 \times 10} \text{ km} \\[1em] = \frac{6}{200} \text{ km} \\[1em] = 0.03 \text{ km} \\[1em] = 30 \text{ m}\end{aligned}$$

Hence, the answer is 30 m

(vii) We have:

$6\frac{1}{4}$% of (1 hour 20 min.)

First, convert time to minutes:

1 hour = 60 min

∴ 1 hour 20 min = 60 min + 20 min = 80 min.

Convert mixed to an improper fraction:

$6\frac{1}{4}$% = $\frac{25}{4}$%

$\frac{25}{4}$% of 80 min

$$\begin{aligned}= \left( 80 \times \frac{25}{4 \times 100} \right) \text{ min} \\[1em] = 80 \times \frac{25}{400} \text{ min} \\[1em] = 80 \times \frac{1}{16} \text{ min} \\[1em] = \frac{80}{16} \text{ min} \\[1em] = 5 \text{ min}\end{aligned}$$

Hence, the answer is 5 min

8Exercise 9(A)

Express:

(i) 56 p as a per cent of ₹ 2.80

(ii) 45 cm as a per cent of 3 m

(iii) 630 g as a per cent of 3.5 kg

(iv) 18 hours as a per cent of 2 days

(v) 2.8 mm as a per cent of 2.24 cm

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(i) 56 p as a per cent of ₹ 2.80

First quantity = 56 p

Convert rupee into paise:

1 rupee = 100 paise

∴ Second quantity = ₹ 2.80 = 2.80 x 100 p = 280 p

Percentage =

$\left( \frac{\text{First Quantity}}{\text{Second Quantity}} \times 100 \right)$%

= $\left( \frac{56}{280} \times 100 \right)$%

= $\left( \frac{1}{5} \times 100 \right)$%

= 20%

Hence, the answer is 20%.

(ii) 45 cm as a per cent of 3 m

First quantity = 45 cm

Convert meter into centimeter:

1 m = 100 cm

∴ Second quantity = 3 m = 3 x 100 cm = 300 cm

Percentage =

$\left( \frac{\text{First Quantity}}{\text{Second Quantity}} \times 100 \right)$%

= $\left( \frac{45}{300} \times 100 \right)$%

= $\frac{45}{3}$%

= 15%

Hence, the answer is 15%

(iii) 630 g as a per cent of 3.5 kg

First quantity = 630 g

Convert kg into grams:

1 kg = 1000 g

∴ Second quantity = 3.5 kg = 3.5 x 1000 g = 3500 g

Percentage =

$\left( \frac{\text{First Quantity}}{\text{Second Quantity}} \times 100 \right)$%

= $\left( \frac{630}{3500} \times 100 \right)$%

= $\frac{630}{35}$%

= 18%

Hence, the answer is 18%

(iv) 18 hours as a per cent of 2 days

First quantity = 18 hours

Convert days into hours:

1 day = 24 hours

∴ Second quantity = 2 days = 2 x 24 hours = 48 hours

Percentage =

$\left( \frac{\text{First Quantity}}{\text{Second Quantity}} \times 100 \right)$%

= $\left( \frac{18}{48} \times 100 \right)$%

= $\left( \frac{3}{8} \times 100 \right)$%

= 37.5%

Hence, the answer is 37.5%

(v) 2.8 mm as a per cent of 2.24 cm

First quantity = 2.8 mm

Convert cm into mm:

1 cm = 10 mm

∴ Second quantity = 2.24 cm = 2.24 x 10 mm = 22.4 mm

Percentage =

$\left( \frac{\text{First Quantity}}{\text{Second Quantity}} \times 100 \right)$%

= $\left( \frac{2.8}{22.4} \times 100 \right)$%

= $\left( \frac{1}{8} \times 100 \right)$%

= 12.5%

Hence, the answer is 12.5%

9Exercise 9(A)

(i) What per cent of 1.5 m is 60 cm?

(ii) What per cent of 2 kg is 125 g?

(iii) What per cent of ₹ 6 is 40 p?

(iv) What per cent of a day is 10 h?

(v) What per cent of $7\frac{1}{3}$m is $5\frac{1}{2}$m?

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(i) What per cent of 1.5 m is 60 cm?

Convert m into cm:

1 m = 100 cm

∴ Total = 1.5 x 100 cm = 150 cm

Part = 60 cm

Percentage =

$\left( \frac{\text{Part}}{\text{Total}} \times 100 \right)$%

= $\left( \frac{60}{150} \times 100 \right)$%

= $\left( \frac{60}{30} \times 20 \right)$%

= $\left( \frac{2}{1} \times 20 \right)$%

= 40%

Hence, the answer is 40%

(ii) What per cent of 2 kg is 125 g?

Convert kg into grams:

1 kg = 1000 g

∴ Total = 2 kg = 2 x 1000 g = 2000 g

Part = 125 g

Percentage =

$\left( \frac{\text{Part}}{\text{Total}} \times 100 \right)$%

= $\left( \frac{125}{2000} \times 100 \right)$%

= $\frac{125}{20}$%

= $6\frac{1}{4}$%

Hence, the answer is $6\frac{1}{4}$%

(iii) What per cent of ₹ 6 is 40 p?

Convert rupee into paise:

₹ 1 = 100 paise

∴ Total = ₹ 6 = 6 x 100 p = 600 p

Part = 40 p

Percentage =

$\left( \frac{\text{Part}}{\text{Total}} \times 100 \right)$%

= $\left( \frac{40}{600} \times 100 \right)$%

= $\left( \frac{40}{6} \times 1 \right)$%

= $\frac{40}{6}$%

= $6\frac{2}{3}$%

Hence, the answer is $6\frac{2}{3}$%

(iv) What per cent of a day is 10 h?

Convert day into hours:

1 day = 24 hours

∴ Total = 24 hours

Part = 10 hours

Percentage =

$\left( \frac{\text{Part}}{\text{Total}} \times 100 \right)$%

= $\left( \frac{10}{24} \times 100 \right)$%

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

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

Hence, the answer is $41\frac{2}{3}$%

(v) What per cent of $7\frac{1}{3}$m is $5\frac{1}{2}$m?

First, convert mixed into improper fraction:

$7\frac{1}{3} \text{ m} = \frac{22}{3}\text{ m}$

$5\frac{1}{2} \text{ m} = \frac{11}{2}\text{ m}$

Total = $\frac{22}{3}\text{ m}$

Part = $\frac{11}{2}\text{ m}$

Percentage =

$\left( \frac{\text{Part}}{\text{Total}} \times 100 \right)$%

= $\left( \frac{\frac{11}{2}}{\frac{22}{3}} \times 100 \right)$%

= $\left( \frac{11}{2} \times \frac{3}{22} \times 100 \right)$%

= $\left( \frac{1}{2} \times \frac{3}{2} \times 100 \right)$%

= $\left( \frac{3}{4} \times 100 \right)$%

= 75%

Hence, the answer is 75%

1Exercise 9(B)

If 16% of a number is 60, find the number.

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Given:

16% of a number is 60

Let the required number be x.

16% of x = 60

$$\begin{aligned}\Rightarrow x \times \frac{16}{100} = 60 \\[1em] \Rightarrow x = 60 \times \frac{100}{16} \\[1em] \Rightarrow x = 15 \times \frac{100}{4} \\[1em] \Rightarrow x = 15 \times 25 \\[1em] \Rightarrow x = 375\end{aligned}$$

Hence, the number is 375

2Exercise 9(B)

If $13\frac{1}{3}$% of a number is 90, find the number.

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Given:

$13\frac{1}{3}$% of a number is 90

Convert mixed fraction to improper:

$13\frac{1}{3}$% = $\frac{40}{3}$%.

Let the number be x.

$\frac{40}{3}$% of x = 90

$$\begin{aligned}\Rightarrow x \times \frac{40}{3 \times 100} = 90 \\[1em] \Rightarrow x \times \frac{40}{300} = 90 \\[1em] \Rightarrow x = 90 \times \frac{300}{40} \\[1em] \Rightarrow x = 90 \times \frac{15}{2} \\[1em] \Rightarrow x = 45 \times 15 \\[1em] \Rightarrow x = 675\end{aligned}$$

Hence, the number is 675.

3Exercise 9(B)

If 0.6% of a number is 15, find the number.

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Given:

0.6% of a number is 15

Let the number be x.

0.6% of x = 15

$$\begin{aligned}\Rightarrow x \times \frac{0.6}{100} = 15 \\[1em] \Rightarrow x = 15 \times \frac{100}{0.6} \\[1em] \Rightarrow x = 15 \times \frac{1000}{6} \\[1em] \Rightarrow x = \frac{5000}{2} \\[1em] \Rightarrow x = 2500\end{aligned}$$

Hence, the number is 2500.

4Exercise 9(B)

If $\frac{3}{4}$% of a number is 9, find the number.

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Given:

$\frac{3}{4}$% of a number is 9

Let the number be x.

$\frac{3}{4}$% of x = 9

$$\begin{aligned}\Rightarrow x \times \frac{3}{4 \times 100} = 9 \\[1em] \Rightarrow x \times \frac{3}{400} = 9 \\[1em] \Rightarrow x = 9 \times \frac{400}{3} \\[1em] \Rightarrow x = 3 \times 400 \\[1em] \Rightarrow x = 1200 \\[1em]\end{aligned}$$

Hence, the number is 1200.

5Exercise 9(B)

There are 42 boys and 18 girls in a class. What is the percentage of boys in the class?

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Given:

Boys = 42

Girls = 18

Total students = 42 + 18 = 60.

Percentage of boys =

$\left( \frac{\text{Number of boys}}{\text{Total students}} \times 100 \right)$%

= $\left( \frac{42}{60} \times 100 \right)$%

= $\left( \frac{42}{3} \times 5 \right)$%

= $\left( \frac{14}{1} \times 5 \right)$%

= 70%

Percentage of boys in the class = 70%

6Exercise 9(B)

A team won 6 hockey matches and lost 9 matches. What per cent of the matches did the team win?

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Given:

Won matches = 6

Lost matches = 9

Total matches played = 6 + 9 = 15.

Percentage of matches won =

$\left( \frac{\text{Number of won matches}}{\text{Total matches played}} \times 100 \right)$%

= $\left( \frac{6}{15} \times 100 \right)$%

= $\left( \frac{2}{5} \times 100 \right)$%

= $2 \times 20$%

= 40%

Percentage of matches won = 40%

7Exercise 9(B)

A batsman scored 75 runs which included 3 boundaries and 8 sixes. What per cent of his total score did he make by running between the wickets?

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Given:

Total runs = 75

Boundaries = 3

Sixes = 8

Runs from boundaries = 3 x 4 = 12 runs.

Runs from sixes = 8 x 6 = 48 runs.

Total runs from hits = 12 + 48 = 60 runs.

Runs made by running = 75 - 60 = 15 runs.

Percentage of runs by running =

$\left( \frac{\text{Runs made by running}}{\text{Total runs}} \times 100 \right)$%

= $\left( \frac{15}{75} \times 100 \right)$%

= $\left( \frac{1}{5} \times 100 \right)$%

= 20%

Percentage of runs by running = 20%

8Exercise 9(B)

12% of a sum of money is ₹ 42. What is 20% of the same sum?

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Given:

12% of a sum of money is ₹ 42

20% of the same sum = ?

First, find the sum:

Let the sum be x.

$$\begin{aligned}\phantom{=} x \times \frac{12}{100} = ₹ 42 \\[1em] \Rightarrow x = ₹ 42 \times \frac{100}{12} \\[1em] \Rightarrow x = ₹ 7 \times \frac{100}{2} \\[1em] \Rightarrow x = ₹ \frac{700}{2} \\[1em] \Rightarrow x = ₹ 350\end{aligned}$$

Now, find 20% of ₹ 350:

20% of ₹ 350 = $\frac{20}{100} \times ₹ 350$

$$\begin{aligned}= ₹ \frac{1}{5} \times 350 \\[1em] = ₹ 70\end{aligned}$$

20% of ₹ 350 = ₹ 70

9Exercise 9(B)

$6\frac{1}{4}$% of a weight is 0.25 kg. What is 45% of this weight?

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Given:

$6\frac{1}{4}$% of a weight is 0.25 kg

45% of the weight = ?

Convert mixed to improper fraction:

$6\frac{1}{4}$% = $\frac{25}{4}$%

Let the total weight be x.

$$\begin{aligned}\phantom{=} x \times \frac{25}{4 \times 100} = 0.25 \text{ kg} \\[1em] x \times \frac{25}{400} = 0.25 \text{ kg} \\[1em] \Rightarrow x = 0.25 \times \frac{400}{25} \text{ kg} \\[1em] \Rightarrow x = 0.25 \times 16 \text{ kg} \\[1em] \Rightarrow x = 4 \text{ kg}\end{aligned}$$

Now, find 45% of 4 kg:

45% of 4 = $\frac{45}{100} \times 4$ kg

$$\begin{aligned}= \frac{180}{100} \text{ kg} \\[1em] = \frac{9}{5} \text{ kg} \\[1em] = 1.8 \text{ kg}\end{aligned}$$

45% of 4 kg = 1.8 kg

10Exercise 9(B)

In a class of 60 pupils, 15% remained absent on a rainy day. How many pupils were present in the class on that day?

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Given:

Total pupils in class = 60

Percentage of pupils absent = 15%

Number of absent pupils = 15% of 60 = $\frac{15}{100} \times 60$ = 9

Number of present pupils = Total pupils - Absent pupils

Substituting the values in above, we get:

Number of present pupils = 60 - 9 = 51

51 pupils were present in the class on that day.

11Exercise 9(B)

The monthly income of Mr. Amit Goel is ₹ 30400. He saves 12.5% of his income and the rest he spends. How much does he spend each month?

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Given:

Monthly income = ₹ 30400

Percentage saved = 12.5%

Percentage spent = (100 - 12.5)% = 87.5%

Amount spent = 87.5% of ₹ 30400

= ₹ $\frac{87.5}{100} \times 30400$

= ₹ 87.5 x 304

= ₹ 26600

He spends ₹ 26600 each month.

12Exercise 9(B)

An ore contains 15% iron. How much ore will be required to get 18 kg of iron?

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Given:

Percentage of iron in ore = 15%

Required quantity of iron = 18 kg

Let the total weight of the ore be x kg.

Then,

15% of x = 18 kg

$$\begin{aligned}\Rightarrow x \times \frac{15}{100} = 18 \text{ kg} \\[1em] \Rightarrow x = 18 \times \frac{100}{15} \text{ kg} \\[1em] \Rightarrow x = 6 \times \frac{100}{5} \text{ kg} \\[1em] \Rightarrow x = 6 \times \frac{20}{1} \text{ kg} \\[1em] \Rightarrow x = 6 \times 20 \text{ kg} \\[1em] \Rightarrow x = 120 \text{ kg}\end{aligned}$$

120 kg of ore will be required to get 18 kg of iron.

13Exercise 9(B)

A property dealer charges a commission of 2% on the first ₹ 25000 and 1.5% on the remainder. What commission does he charge for selling a plot of land for ₹ 130000?

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Given:

Selling price of plot = ₹ 130000

Commission on first ₹ 25000 = 2%

Commission on remainder = 1.5%

Commission on first part = 2% of ₹ 25000

= ₹ $\frac{2}{100} \times 25000$

= ₹ 500

Remainder amount = ₹ 130000 - ₹ 25000 = ₹ 105000

Commission on remainder = 1.5% of ₹ 105000

= ₹ $\frac{1.5}{100} \times 105000$

= ₹ 1575

Total Commission = ₹ 500 + ₹ 1575 = ₹ 2075

Charge for selling a plot of land for ₹ 130000 is ₹ 2075.

14Exercise 9(B)

In an examination, the maximum marks are 850. Rohit gets 34% marks and fails by 17 marks. Find

(i) the passing marks and

(ii) the minimum percentage for passing the examination.

Show SolutionHide Solution↓

Given:

Maximum marks = 850

Rohit's percentage = 34%

Failing margin = 17 marks

Rohit's marks = 34% of 850

= $\frac{34}{100} \times 850$

= 289

(i) Passing marks = Rohit's marks + Failing margin

Substituting the values in above, we get:

Passing marks = 289 + 17 = 306

Passing marks = 306

(ii) Minimum passing percentage =

$\left( \frac{\text{Passing marks}}{\text{Maximum marks}} \times 100 \right)$%

$$\begin{aligned}= \frac{306}{850} \times 100 \\[1em] = \frac{3060}{85} \\[1em]\end{aligned}$$

= 36%

The minimum percentage for passing the examination = 36%

15Exercise 9(B)

A student secures 90%, 60% and 54% marks in three test papers with maximum marks 100, 150 and 200 respectively. Find his aggregate percentage.

Show SolutionHide Solution↓

Given:

Test 1: Max 100, Scored 90%

Test 2: Max 150, Scored 60%

Test 3: Max 200, Scored 54%

Marks in Test 1 = 90% of 100

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

= 90

Marks in Test 2 = 60% of 150

$$\begin{aligned}= \frac{60}{100} \times 150 \\[1em] = \frac{60}{2} \times 3 \\[1em] = \frac{180}{2} \\[1em] = 90\end{aligned}$$

Marks in Test 3 = 54% of 200

$$\begin{aligned}\phantom{=} \frac{54}{100} \times 200 \\[1em] = \frac{54}{1} \times 2 \\[1em] = 108\end{aligned}$$

Total marks scored = Marks in Test 1 + Marks in Test 2 + Marks in Test 3

Total marks scored = 90 + 90 + 108 = 288

Total maximum marks = 100 + 150 + 200 = 450

Aggregate percentage =

$\frac{\text{Total marks scored}}{\text{Total maximum marks}} \times 100$%

= $\left( \frac{288}{450} \times 100 \right)$%

= $\frac{2880}{45}$%

= 64%

Aggregate percentage = 64%

16Exercise 9(B)

In an examination, 5% of the applicants were found ineligible and 85% of the eligible candidates belonged to the general category. If 4275 candidates belonged to other categories, then how many candidates applied for the examination?

Show SolutionHide Solution↓

Given:

Ineligible applicants = 5%

General category (of eligible) = 85%

Other categories (of eligible) = 100% - 85% = 15%

Number of "Other category" candidates = 4275

Let the total applicants be x

Eligible candidates = 95% of x

= $\frac{95}{100} \times x$

= 0.95x

15% of these eligible candidates = 4275

$$\begin{aligned}\phantom{=} \frac{15}{100} \times 0.95x = 4275 \\[1em] 0.15 \times 0.95x = 4275 \\[1em] 0.1425x = 4275 \\[1em] \Rightarrow x = \frac{4275}{0.1425} \\[1em] \Rightarrow x = 30000\end{aligned}$$

30000 candidates applied for the examination.

17Exercise 9(B)

Gun-powder contains 75% nitre, 10% sulphur and the rest of it is charcoal. Find the quantity of charcoal in 8 kg of gun-powder.

Show SolutionHide Solution↓

Given:

Nitre = 75%

Sulphur = 10%

Charcoal = Rest of the powder

Total quantity = 8 kg

Percentage of charcoal = 100% - (75% + 10%) = 15%

Quantity of charcoal = 15% of 8 kg

$$\begin{aligned}= \frac{15}{100} \times 8 \text{ kg} \\[1em] = \frac{15}{25} \times 2 \text{ kg} \\[1em] = \frac{3}{5} \times 2 \text{ kg} \\[1em] = \frac{6}{5} \text{ kg} \\[1em] = 1.2 \text{ kg}\end{aligned}$$

8 kg of gun-powder contains 1.2 kg of charcoal.

18Exercise 9(B)

An alloy consists of 13 parts of copper, 7 parts of zinc and 5 parts of nickel. Find the percentage of copper in the alloy.

Show SolutionHide Solution↓

Given:

Copper = 13 parts

Zinc = 7 parts

Nickel = 5 parts

Total parts = 13 + 7 + 5 = 25 parts.

Percentage of copper =

$\left( \frac{\text{Parts of copper}}{\text{Total parts}} \times 100 \right)$%

= $\frac{13}{25} \times 100$%

= $\frac{13}{1} \times 4$%

= $13 \times 4$%

= 52%

Percentage of copper in the alloy = 52%.

19Exercise 9(B)

Two candidates A and B contested an election. The total votes polled were 9650. If A got 54% of the votes, find the number of votes received by each.

Show SolutionHide Solution↓

Given:

Total votes = 9650

Candidate A's share = 54%

Votes for A = 54% of 9650

$$\begin{aligned}= \frac{54}{100} \times 9650 \\[1em] = \frac{27}{50} \times 9650 \\[1em] = \frac{27}{1} \times 193 \\[1em] = 5211 \text{ votes}\end{aligned}$$

Votes for B = Total votes - A's share

Votes for B = 9650 - 5211 = 4439 votes

Votes for A = 5211, Votes for B = 4439

20Exercise 9(B)

At an election between two candidates, 68 votes were declared invalid. The winning candidate secures 52% of the valid votes and wins by 354 votes. Find the total number of votes polled.

Show SolutionHide Solution↓

Given:

Invalid votes = 68

Winner's share of valid votes = 52%

Winner's margin = 354 votes

Let valid votes be x.

Loser's share = 100% - 52% = 48%

Difference (Margin) = 52% - 48% = 4%

4% of x = 354

$$\begin{aligned}= \frac{4}{100} \times x = 354 \\[1em] \Rightarrow x = \frac{354 \times 100}{4} \\[1em] \Rightarrow x = \frac{354 \times 25}{1} \\[1em] \Rightarrow x = 354 \times 25 \\[1em] \Rightarrow x = 8850 \text{ (valid votes)}\end{aligned}$$

Total votes polled = Valid + Invalid

= 8850 + 68 = 8918 votes.

Total number of votes polled = 8918

21Exercise 9(B)

The salary of Mrs Sarita is ₹ 32000 per month. 10% of it is deducted by the employer as provident fund. Of the remaining money, she spends 20% on house rent, 46% on food, 14% on the education of children and 10% on other expenses. Rest she saves. Find :

(i) how much is credited each month to her Provident Fund Account.

(ii) how much is spent on food.

(iii) how much is paid as house rent.

(iv) how much is spent on the education of children.

(v) how much does she save every month.

Show SolutionHide Solution↓

Given:

Total Salary = ₹ 32000

Provident Fund Deduction = 10% of total Salary

Remaining Money = Total Salary - Provident Fund Deduction

House Rent = 20% of Remaining Money

Food = 46% of Remaining Money

Education = 14% of Remaining Money

Other Expenses = 10% of Remaining Money

(i) Monthly Credit to Provident Fund Account

Provident Fund = 10% of ₹ 32000

$$\begin{aligned}= \frac{10}{100} \times ₹ 32000 \\[1em] = ₹ \frac{10}{1} \times 320 \\[1em] = ₹ 3200\end{aligned}$$

₹ 3200 is credited each month to her Provident Fund Account.

(ii) Amount spent on food:

Food = 46% of Remaining Money

Let us calculate Remaining Money:

Remaining Money = Total Salary - Provident Fund Deduction

Remaining Money = ₹ 32000 - ₹ 3200 = ₹ 28800

Food = 46% of Remaining Money

Food = 46% of ₹ 28800

$$\begin{aligned}= ₹ \frac{46}{100} \times 28800 \\[1em] = ₹ \frac{46}{1} \times 288 \\[1em] = ₹ 46 \times 288 \\[1em] = ₹ 13248\end{aligned}$$

Amount spent on food = ₹ 13248.

(iii) Amount paid as House Rent

House Rent = 20% of Remaining Money

House Rent = 20% of ₹ 28800

$$\begin{aligned}= ₹ \frac{20}{100} \times 28800 \\[1em] = ₹ \frac{20}{1} \times 288 \\[1em] = ₹ 20 \times 288 \\[1em] = ₹ 5760\end{aligned}$$

Amount paid as House Rent = ₹ 5760.

(iv) Amount spent on Education of children:

Education = 14% of Remaining Money

Education = 14% of ₹ 28800

$$\begin{aligned}= ₹ \frac{14}{100} \times 28800 \\[1em] = ₹ \frac{14}{1} \times 288 \\[1em] = ₹ 14 \times 288 \\[1em] = ₹ 4032\end{aligned}$$

Amount spent on Education of children = ₹ 4032.

(v) Monthly Savings:

First, let's find the total percentage spent from the remaining money:

Total Spent % = 20% + 46% + 14% + 10% = 90%

Savings % = 100% - 90% = 10%

Savings = 10% of 28800

$$\begin{aligned}= ₹ \frac{10}{100} \times 28800 \\[1em] = ₹ \frac{10}{1} \times 288 \\[1em] = ₹ 10 \times 288 \\[1em] = ₹ 2880\end{aligned}$$

Monthly Savings = ₹ 2880.

1Exercise 9(C)

Increase :

(i) 375 by 4%

(ii) 500 by 3.4%

(iii) 70 by 140%

(iv) 48 by $12\frac{1}{2}$%

(v) 90 by 2%

Show SolutionHide Solution↓

(i) 375 by 4%

Given:

Increase = 4% of 375

$$\begin{aligned}= \left( 375 \times \frac{4}{100} \right) \\[1em] = \left( 15 \times \frac{4}{4} \right) \quad \text{[Dividing 375 and 100 by 25]} \\[1em] = \left( 15 \times \frac{1}{1} \right) \\[1em] = 15\end{aligned}$$

Increased value = 375 + 15 = 390

∴ Increased value is 390

(ii) 500 by 3.4%

Increase = 3.4% of 500

$$\begin{aligned}= \left( 500 \times \frac{3.4}{100} \right) \\[1em] = \left( 5 \times \frac{3.4}{1} \right) \quad \text{[Dividing 500 and 100 by 100]} \\[1em] = 5 \times 3.4 \\[1em] = 17\end{aligned}$$

Increased value = 500 + 17 = 517

∴ Increased value is 517

(iii) 70 by 140%

Increase = 140% of 70

$$\begin{aligned}= \left( 70 \times \frac{140}{100} \right) \\[1em] = \left( 7 \times \frac{140}{10} \right) \quad \text{[Dividing 70 and 100 by 10]} \\[1em] = \left( 7 \times \frac{14}{1} \right) \quad \text{[Dividing 140 and 10 by 10]} \\[1em] = 7 \times 14 \\[1em] = 98\end{aligned}$$

Increased value = 70 + 98 = 168

∴ Increased value is 168

(iv) 48 by $12\frac{1}{2}$%

First, convert to an improper fraction: $12\frac{1}{2}$% = $\frac{25}{2}$%.

Increase = $\frac{25}{2}$% $\text{ of } 48$

$$\begin{aligned}= \left( 48 \times \frac{25}{2 \times 100} \right) \\[1em] = \left( 48 \times \frac{1}{2 \times 4} \right) \quad \text{[Dividing 25 and 100 by 25]} \\[1em] = 48 \times \frac{1}{8} \\[1em] = 6 \times \frac{1}{1} \quad \text{[Dividing 48 and 8 by 8]} \\[1em] = 6\end{aligned}$$

Increased value = 48 + 6 = 54

∴ Increased value is 54

(v) 90 by 2%

Increase = 2% of 90

$$\begin{aligned}= \left( 90 \times \frac{2}{100} \right) \\[1em] = \left( 9 \times \frac{2}{10} \right) \quad \text{[Dividing 90 and 100 by 10]} \\[1em] = \frac{18}{10} \\[1em] = 1.8\end{aligned}$$

Increased value = 90 + 1.8 = 91.8

∴ Increased value is 91.8

2Exercise 9(C)

Decrease :

(i) 70 by 40%

(ii) 340 by 35%

(iii) 65 by 4%

(iv) 36 by $16\frac{2}{3}$%

(v) 260 by 1.5%

Show SolutionHide Solution↓

(i) 70 by 40%

Decrease = 40% of 70

$$\begin{aligned}= \left( 70 \times \frac{40}{100} \right) \\[1em] = \left( 7 \times \frac{40}{10} \right) \quad \text{[Dividing 70 and 100 by 10]} \\[1em] = \left( 7 \times \frac{4}{1} \right) \quad \text{[Dividing 40 and 10 by 10]} \\[1em] = 28\end{aligned}$$

Decreased value = 70 - 28 = 42

∴ Decreased value is 42

(ii) 340 by 35%

Decrease = 35% of 340

$$\begin{aligned}= \left( 340 \times \frac{35}{100} \right) \\[1em] = \left( 34 \times \frac{35}{10} \right) \quad \text{[Dividing 340 and 100 by 10]} \\[1em] = \left( 34 \times \frac{7}{2} \right) \quad \text{[Dividing 35 and 10 by 5]} \\[1em] = \left( 17 \times \frac{7}{1} \right) \quad \text{[Dividing 34 and 2 by 2]} \\[1em] = 17 \times 7 \\[1em] = 119\end{aligned}$$

Decreased value = 340 - 119 = 221

∴ Decreased value is 221

(iii) 65 by 4%

Decrease = 4% of 65

$$\begin{aligned}= \left( 65 \times \frac{4}{100} \right) \\[1em] = \left( 13 \times \frac{4}{20} \right) \quad \text{[Dividing 65 and 100 by 5]} \\[1em] = \left( 13 \times \frac{1}{5} \right) \quad \text{[Dividing 4 and 20 by 5]} \\[1em] = \frac{13}{5} \\[1em] = 2.6\end{aligned}$$

Decreased value = 65 - 2.6 = 62.4

∴ Decreased value is 62.4

(iv) 36 by $16\frac{2}{3}$%

Convert to an improper fraction:

$16\frac{2}{3}$% = $\frac{50}{3}$%

Decrease = $\frac{50}{3}$% of 36

$$\begin{aligned}= \left( 36 \times \frac{50}{3 \times 100} \right) \\[1em] = \left( 36 \times \frac{1}{3 \times 2} \right) \quad \text{[Dividing 50 and 100 by 50]} \\[1em] = 36 \times \frac{1}{6} \quad \text{[Dividing 36 and 6 by 6]} \\[1em] = 6\end{aligned}$$

Decreased value = 36 - 6 = 30

∴ Decreased value is 30

(v) 260 by 1.5%

Decrease = 1.5% of 260

$$\begin{aligned}= \left( 260 \times \frac{1.5}{100} \right) \\[1em] = 2.6 \times 1.5 \quad \text{[Dividing 260 and 100 by 100]} \\[1em] = 3.9\end{aligned}$$

Decreased value = 260 - 3.9 = 256.1

∴ Decreased value is 256.1

3Exercise 9(C)

Find a number :

(i) Which when increased by 10% becomes 66.

(ii) Which when increased by 120% becomes 77.

(iii) Which when increased by 2.5% becomes 246.

Show SolutionHide Solution↓

(i) Which when increased by 10% becomes 66.

Let the required number be x.

Increase = 10% of x

$$\begin{aligned}= x \times \frac{10}{100} \\[1em] =\frac{10}{100}x \\[1em] = \frac{x}{10}\end{aligned}$$

Increased number = $x + \frac{x}{10} = \frac{11}{10}x$

$$\begin{aligned}\therefore \frac{11}{10}x = 66 \\[1em] \Rightarrow x = \frac{66 \times 10}{11} \\[1em] \Rightarrow x = \frac{6 \times 10}{1} \\[1em] \Rightarrow x = 60 \\[1em]\end{aligned}$$

Hence, the required number is 60

(ii) Which when increased by 120% becomes 77.

Let the required number be x.

Increase = 120% of x

$$\begin{aligned}= x \times \frac{120}{100} \\[1em] = \frac{6}{5}x \\[1em]\end{aligned}$$

Increased number = $x + \frac{6}{5}x = \frac{11}{5}x$

$$\begin{aligned}\therefore \frac{11}{5}x = 77 \\[1em] \Rightarrow x = \frac{77 \times 5}{11} \\[1em] \Rightarrow x = \frac{7 \times 5}{1} \\[1em] \Rightarrow x = 35 \\[1em]\end{aligned}$$

Hence, the required number is 35

(iii) Which when increased by 2.5% becomes 246.

Let the required number be x.

Increase = 2.5% of x

$$\begin{aligned}= x \times \frac{2.5}{100} \\[1em] = x \times \frac{2.5 \times 10}{100 \times 10} \\[1em] = \frac{25}{1000}x\end{aligned}$$

Increased number = $x + \frac{25}{1000}x = \frac{1025}{1000}x$

$$\begin{aligned}\therefore \frac{1025}{1000}x = 246 \\[1em] \Rightarrow x = \frac{246 \times 1000}{1025} \\[1em] \Rightarrow x = \frac{246 \times 40}{41} \\[1em] \Rightarrow x = \frac{6 \times 40}{1} \\[1em] \Rightarrow x = 240 \\[1em]\end{aligned}$$

Hence, the required number is 240.

4Exercise 9(C)

Find a number :

(i) Which when decreased by 35% becomes 52.

(ii) Which when decreased by 8% becomes 115.

(iii) Which when decreased by 3.4% becomes 483.

Show SolutionHide Solution↓

(i) Which when decreased by 35% becomes 52.

Let the required number be x.

Decrease = 35% of x

$$\begin{aligned}= x \times \frac{35}{100} \\[1em] = x \times \frac{7}{20} \\[1em] = \frac{7}{20}x\end{aligned}$$

Decreased number = $x - \frac{7}{20}x = \frac{13}{20}x$

$$\begin{aligned}\therefore \frac{13}{20}x = 52 \\[1em] \Rightarrow x = \frac{52 \times 20}{13} \\[1em] \Rightarrow x = \frac{4 \times 20}{1} \\[1em] \Rightarrow x = 80\end{aligned}$$

Hence, the required number is 80.

(ii) Which when decreased by 8% becomes 115.

Let the required number be x.

Decrease = 8% of x

$$\begin{aligned}= x \times \frac{8}{100} \\[1em] = x \times \frac{2}{25} \\[1em] = \frac{2}{25}x\end{aligned}$$

Decreased number = $x - \frac{2}{25}x = \frac{23}{25}x$

$$\begin{aligned}\therefore \frac{23}{25}x = 115 \\[1em] \Rightarrow x = \frac{115 \times 25}{23} \\[1em] \Rightarrow x = \frac{5 \times 25}{1} \\[1em] \Rightarrow x = 125\end{aligned}$$

Hence, the required number is 125.

(iii) Which when decreased by 3.4% becomes 483.

Let the required number be x.

Decrease = 3.4% of x

$$\begin{aligned}= x \times \frac{3.4}{100} \\[1em] = x \times \frac{3.4 \times 10}{100 \times 10} \\[1em] = x \times \frac{34}{1000} \\[1em] = \frac{17}{500}x\end{aligned}$$

Decreased number = $x - \frac{17}{500}x = \frac{483}{500}x$

$$\begin{aligned}\therefore \frac{483}{500}x = 483 \\[1em] \Rightarrow x = \frac{483 \times 500}{483} \\[1em] \Rightarrow x = \frac{1 \times 500}{1} \\[1em] \Rightarrow x = 500\end{aligned}$$

Hence, the required number is 500.

5Exercise 9(C)

By what number must a given number be multiplied to increase it by 12%?

Show SolutionHide Solution↓

Let the number be x.

Increase in its value = 12% of x

$$\begin{aligned}= x \times \frac{12}{100} \\[1em] = \frac{12}{100}x \\[1em] = \frac{3}{25}x\end{aligned}$$

∴ Increased value = $\Big(x + \frac{3}{25}x\Big) = \frac{28}{25}x$

Hence, for an increase of 12%, the given number should be multiplied by $\frac{28}{25}$.

6Exercise 9(C)

By what number must a given number be multiplied to decrease it by 30%?

Show SolutionHide Solution↓

Let the number be x.

Decrease in its value = 30% of x

$$\begin{aligned}= x \times \frac{30}{100} \\[1em] = \frac{3}{10}x\end{aligned}$$

∴ Decreased value = $\Big(x - \frac{3}{10}x\Big) = \frac{7}{10}x$

Hence, for an decrease of 30%, the given number should be multiplied by $\frac{7}{10}$.

7Exercise 9(C)

The price of a fan increases from ₹ 3260 to ₹ 3749. Find the increase per cent in its price.

Show SolutionHide Solution↓

Given:

Original Price = ₹ 3260

New Price = ₹ 3749.

Increase in price = New Price - Original Price

Substituting the values in above, we get:

Increase in price = ₹ 3749 - ₹ 3260 = ₹ 489

Increase % =

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

= $\left( \frac{489}{3260} \times 100 \right)$%

= $\left( \frac{489}{815} \times 25 \right)$%

= $\left( \frac{489}{163} \times 5 \right)$%

= $\left( \frac{3}{1} \times 5 \right)$%

= 15%

Hence, the increase percentage in its price = 15%.

8Exercise 9(C)

The monthly salary of Mr Rakesh is ₹ 32500. After deducting the provident fund, he gets ₹ 29900 per month. What per cent of the salary is deducted as provided fund?

Show SolutionHide Solution↓

Given:

Total Salary = ₹ 32500

Net Salary = ₹ 29900.

Provident fund deduction = Total Salary - Net Salary

Substituting the values in above, we get:

Provident fund deduction = ₹ 32500 - ₹ 29900 = ₹ 2600

Deduction % =

$\left( \frac{\text{Provident fund deduction}}{\text{Total Salary}} \times 100 \right)$%

= $\left( \frac{2600}{32500} \times 100 \right)$%

= $\left( \frac{2600}{325} \times 1 \right)$%

= $\left( \frac{104}{13} \right)$%

= 8%

Percentage of the salary deducted as provided fund = 8%

9Exercise 9(C)

A car was purchased last years for ₹ 415000. Now, its value is ₹ 356900. At what rate is the car depreciating?

Show SolutionHide Solution↓

Given:

Original Value = ₹ 415000

Current Value = ₹ 356900

Depreciation amount = ₹ 415000 - ₹ 356900 = ₹ 58100

Depreciation rate =

$\left( \frac{\text{Depreciation amount}}{\text{Original Value}} \times 100 \right)$%

$\left( \frac{58100}{415000} \times 100 \right)$%

= $\frac{58100}{4150} \times 1$%

= $\frac{5810}{415}$%

= 14%

Depreciation rate of the car is 14%.

10Exercise 9(C)

On decreasing the price of a car by 6%, its value becomes ₹ 249100. What was the original price of the car?

Show SolutionHide Solution↓

Given:

New value = ₹ 249100

Decrease = 6%

Let original price be x.

Since the New value represents the amount left after the decrease, we can say:

94% of the Original Price = New value

94% of x = ₹ 249100

$$\begin{aligned}\frac{94}{100}x = ₹ 249100 \\[1em] \Rightarrow x = ₹ \frac{249100 \times 100}{94} \\[1em] \Rightarrow x = ₹ \frac{2650 \times 100}{1} \\[1em] \Rightarrow x = ₹ 265000\end{aligned}$$

The original price of the car is ₹ 265000

11Exercise 9(C)

On increasing the salary of a man by 12%, his salary is increased by ₹ 2316. What was his original salary?

Show SolutionHide Solution↓

Given:

Increase % = 12%

Increase Amount = ₹ 2316

Let original salary be x.

12% of x = 2316

$$\begin{aligned}\frac{12}{100}x = ₹ 2316 \\[1em] \Rightarrow x = ₹ \frac{2316 \times 100}{12} \\[1em] \Rightarrow x = ₹ \frac{193 \times 100}{1} \\[1em] \Rightarrow x = ₹ 19300\end{aligned}$$

The original salary is ₹ 19300.

12Exercise 9(C)

The salary of Gopal was increased by 10% and then the increased salary was decreased by 10%. Find the net increase or decrease per cent in his original salary.

Show SolutionHide Solution↓

Given:

Increase of salary = 10%

Decrease of salary = 10%

Let original salary be 100.

After 10% increase: 100 + 10 = 110.

After 10% decrease on 110:

110 - (10% of 110)

$$\begin{aligned}= 110 - \Big(\frac{10}{100} \times 110\Big) \\[1em] = 110 - \Big(\frac{1}{10} \times 110\Big) \\[1em] = 110 - \Big(\frac{1}{1} \times 11\Big) \\[1em] = 110 - 11 \\[1em] = 99 \\[1em]\end{aligned}$$

Net change = 100 - 99 = 1

Net decrease = 1%.

13Exercise 9(C)

After deducting 4% of a bill, the amount still to be paid is ₹ 1488. How much was the original bill?

Show SolutionHide Solution↓

Given:

Amount still to be paid = ₹ 1488

Percentage of deduction = 4%

The original bill represents the full amount, which is 100%.

The "Amount still to be paid" is what remains after the 4% has been taken away from the 100%.

Remaining Percentage = 100% - 4% = 96%

Let the original bill be x.

Since ₹ 1488 is the amount left after the 4% deduction, it is equal to 96% of the original bill.

96% of x = ₹ 1488

$$\begin{aligned}\phantom{=} \frac{96}{100}x = ₹ 1488 \\[1em] x = ₹ \frac{1488 \times 100}{96} \\[1em] x = ₹ \frac{1488 \times 25}{24} \\[1em] x = ₹ \frac{62 \times 25}{1} \\[1em] x = ₹ 1550\end{aligned}$$

The original bill is ₹ 1550.

14Exercise 9(C)

The weight of a boy was 40 kg. But, it was wrongly measured as 42 kg. Find the error per cent.

Show SolutionHide Solution↓

Given:

Actual weight = 40 kg

Measured weight = 42 kg.

Error = 42 - 40 = 2 kg

$\text {Error}$% = $\left( \frac{\text{Error}}{\text{Actual Value}} \times 100 \right)$%

= $\frac{2}{40} \times 100$%

= $\frac{1}{20} \times 100$%

= $\frac{1}{1} \times 5$%

= 5%

Error percentage = 5%.

11 Mark · Multiple Choice Questions

$\frac{2}{5}$ expressed as a percentage is

Show SolutionHide Solution↓

Given:

Fraction = $\frac{2}{5}$

To convert a fraction to a percentage, multiply by 100.

Percentage = $\left( \frac{2}{5} \times 100 \right)$% = $(2 \times 20)$% = 40%

Hence, option 2 is the correct option.

21 Mark · Multiple Choice Questions

The ratio 1 : 4 expressed as a percentage is

Show SolutionHide Solution↓

Given:

Ratio = 1 : 4

And

1 : 4 = $\frac{1}{4}$

Percentage = $\left( \frac{1}{4} \times 100 \right)$% = 25%

Hence, option 3 is the correct option.

31 Mark · Multiple Choice Questions

$2\frac{1}{7}$% expressed as a fraction is

Show SolutionHide Solution↓

Given:

Percentage = $2\frac{1}{7}$%

$2\frac{1}{7}$% = $\frac{15}{7}$%.

To convert to a fraction, divide by 100:

$\frac{15}{7} \times \frac{1}{100} = \frac{15}{700} = \frac{3}{140}$

Hence, option 3 is the correct option.

41 Mark · Multiple Choice Questions

If x% of 80 = 16, then the value of x is

Show SolutionHide Solution↓

Given:

x% of 80 = 16

$$\begin{aligned}\Rightarrow \frac{x}{100} \times 80 = 16 \\[1em] \Rightarrow x = \frac{16 \times 100}{80} \\[1em] \Rightarrow x = \frac{16 \times 5}{4} \\[1em] \Rightarrow x = 4 \times 5 \\[1em] \Rightarrow x = 20\end{aligned}$$

Hence, option 3 is the correct option.

51 Mark · Multiple Choice Questions

What per cent of a day is 27 minutes?

Show SolutionHide Solution↓

Given:

Part = 27 minutes

Total = 1 day = 24 hours

Convert day to minutes: 24 x 60 = 1440 minutes.

$\text {Percentage} = \left( \frac {\text{Part}}{\text{Total}} \times 100 \right)$%

$\left( \frac{27}{1440} \times 100 \right)$%

= $\frac{2700}{1440}$%

= $\frac{270}{144}$%

= $\frac{15}{8}$% $\quad$ [Dividing 270 and 144 by 18]

= $1\frac{7}{8}$%

Hence, option 1 is the correct option.

61 Mark · Multiple Choice Questions

A number decreased by 15% gives 68. The number is

Show SolutionHide Solution↓

Decreased by 15% becomes 68.

Let the number be x.

The new number is what remains after 15% has been taken away from 100%.

85% of x = 68

$$\begin{aligned}\frac{85}{100}x = 68 \\[1em] \Rightarrow x = \frac{68 \times 100}{85} \\[1em] \Rightarrow x = \frac{4 \times 100}{5} \quad \text{[Dividing 68 and 85 by 17]} \\[1em] \Rightarrow x = \frac{4 \times 20}{1} \quad \text{[Dividing 100 and 5 by 5]} \\[1em] \Rightarrow x = 80\end{aligned}$$

Hence, option 2 is the correct option.

71 Mark · Multiple Choice Questions

A number increased by 25% gives 45. The number is

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Given:

Increased by 25% becomes 45.

Let the number be x.

The new number is the total after adding 25% to 100%.

125% of x = 45

$$\begin{aligned}\Rightarrow \frac{125}{100}x = 45 \\[1em] \Rightarrow x = \frac{45 \times 100}{125} \\[1em] \Rightarrow x = \frac{45 \times 4}{5} \quad \text{[Dividing 100 and 125 by 25]} \\[1em] \Rightarrow x = \frac{9 \times 4}{1} \quad \text{[Dividing 45 and 5 by 5]} \\[1em] \Rightarrow x = 36\end{aligned}$$

Hence, option 4 is the correct option.

81 Mark · Multiple Choice Questions

After an increase of 15%, the salary of a person becomes ₹ 51750. His original salary was

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Given:

Salary becomes ₹ 51750 after 15% increase.

Let original salary be x.

115% of x = ₹ 51750

$$\begin{aligned}\Rightarrow \frac{115}{100}x = ₹ 51750 \\[1em] \Rightarrow x = ₹ \frac{51750 \times 100}{115} \\[1em] \Rightarrow x = ₹ \frac{51750 \times 20}{23} \quad \text{[Dividing 100 and 115 by 5]} \\[1em] \Rightarrow x = ₹ \frac{2250 \times 20}{1} \quad \text{[Dividing 51750 and 23 by 23]} \\[1em] \Rightarrow x = ₹ 45000\end{aligned}$$

Hence, option 1 is the correct option.

91 Mark · Multiple Choice Questions

In an examination, 95% of the total examinees passed. If the number of failures was 36, how many examinees were there?

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Given:

Passed = 95%

Failures = 36

Failure percentage = 100% - 95% = 5%.

Let total examinees be x.

5% of x = 36

$$\begin{aligned}\Rightarrow\frac{5}{100}x = 36 \\[1em] \Rightarrow x = \frac{36 \times 100}{5} \\[1em] \Rightarrow x = \frac{36 \times 20}{1} \\[1em] \Rightarrow x = 720\end{aligned}$$

Hence, option 3 is the correct option.

101 Mark · Multiple Choice Questions

The value of a machine depreciates 15% annually. If its present value is ₹ 51000, what was its value one year earlier?

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Given:

Present value = ₹ 51000

Depreciation = 15%

Let the value one year ago be x.

Depreciation means the value is taken away from 100%: 100% - 15% = 85%

85% of x = 51000

$$\begin{aligned}\Rightarrow \frac{85}{100}x = ₹ 51000 \\[1em] \Rightarrow x = \frac{51000 \times 100}{85} \\[1em] \Rightarrow x = 600 \times 100 \\[1em] \Rightarrow x = ₹ 60000\end{aligned}$$

Hence, option 4 is the correct option.

1Mental Maths

Fill in the blanks :

(i) Add 15% of 60 to 40. ...............

(ii) A student scored 32 marks out of 40 in Maths. Express this as a per cent. ...............

(iii) If 40% of the length of a cloth is 120 cm, what is the whole length of the cloth? ...............

(iv) What per cent of 2.8 kg is 980 g? ...............

(v) Express 0.74 as a per cent. ...............

Show SolutionHide Solution↓

(i) Add 15% of 60 to 40. 49

(ii) A student scored 32 marks out of 40 in Maths. Express this as a per cent.80%

(iii) If 40% of the length of a cloth is 120 cm, what is the whole length of the cloth? 300 cm

(iv) What per cent of 2.8 kg is 980 g? 35%

(v) Express 0.74 as a per cent. 74%

Explanation

(i) Given:

Original number = 40

Percentage to add = 15% of 60

Calculate 15% of 60:

$$\begin{aligned}= \frac{15}{100} \times 60 \\[1em] = \frac{15}{5} \times 3 \\[1em] = \frac{3}{1} \times 3 \\[1em] = 9\end{aligned}$$

Add this to 40:

40 + 9 = 49

(ii) Given:

Scored = 32

Total = 40

$\text {Percentage}= \left( \frac{\text{Scored}}{\text{Total}} \times 100 \right)$%

$$\begin{aligned}= \frac{32}{40} \times 100 \\[1em] = \frac{4}{5} \times 100 \\[1em] = \frac{4}{1} \times 20 \\[1em]\end{aligned}$$

= 80%

(iii) Given:

40% of length = 120 cm.

Let the whole length be x.

$$\begin{aligned}\frac{40}{100} \times x = 120 \\[1em] \Rightarrow x = \frac{120 \times 100}{40} \\[1em] \Rightarrow x = 3 \times 100 \\[1em] \Rightarrow x = 300 \text{ cm}\end{aligned}$$

(iv) Given:

Part = 980 g

Total = 2.8 kg

Convert total to grams:

1 kg = 1000 grams

2.8 kg = 2.8 x 1000 g = 2800g

$\text {Percentage} = \left( \frac {\text{Part}}{\text{Total}} \times 100 \right)$%

= $\left( \frac{980}{2800} \times 100 \right)$%

= $\frac{980}{28}$%

= 35%

(v) Given:

Decimal = 0.74

To convert a decimal to a percentage, multiply by 100.

0.74 x 100 = 74%

2Mental Maths

Write true (T) or false (F) :

(i) $\frac{1}{4}$ can be expressed as 0.25%.

(ii) 1 : 5 expressed as a per cent is 20%.

(iii) 100% of 1 is 100 and 1% of 100 is 100.

(iv) 16% of ₹ 25 is ₹ 10.

(v) 18 marks out of 30 marks is more than 50 marks out of 80 marks.

Show SolutionHide Solution↓

(i) False
Reason — To express a fraction as a percentage, we multiply by 100.

$\frac{1}{4} \times 100 = 25$%.

0.25 is the decimal form, but as a percentage, it is 25% not 0.25%.

(ii) True
Reason — The ratio 1 : 5 is written as the fraction $\frac{1}{5}$.

Percentage = $\Big( \frac{1}{5} \times 100 \Big)$% = 20%.

(iii) False
Reason —

100% of 1 = $\frac{100}{100} \times 1 = 1$

1% of 100 = $\frac{1}{100} \times 100 = 1$

Neither calculation results in 100. So the statement is false.

(iv) False
Reason — 16% of ₹ 25

$$\begin{aligned}= ₹ \Big(\frac{16}{100} \times 25\Big) \\[1em] = ₹ \frac{16}{4} \\[1em] = ₹ 4\end{aligned}$$

The value ₹ 10 is incorrect. So the statement is false.

(v) False
Reason — We must compare the percentages:

Percentage = $\Big(\frac {\text{Scored Marks}}{\text{Total Marks}} \times 100\Big)$%

Percentage of first score

= $\Big(\frac{18}{30} \times 100\Big)$%

= $\Big(\frac{18}{3} \times 10\Big)$%

= $\Big(\frac{6}{1} \times 10\Big)$%

= 60%

Percentage of second score

= $\Big(\frac{50}{80} \times 100\Big)$%

= $\Big(\frac{50}{4} \times 5\Big)$%

= $\Big(\frac{25}{2} \times 5\Big)$%

= $\frac{125}{2}$%

= 62.5%

Since 60% < 62.5%, the statement that the first is "more" is false.

1Case Study Based Questions

Radhika is in the Chemistry Laboratory and is working on a practical. She mixed three salts A, B and C taking 150 gm of A and 300 gm each of B and C to prepare a mixture X.

(1) The percentage of salt A in mixture X is :

  1. 15%
  2. 20%
  3. 25%
  4. 30%

(2) By what percent is salt C more than salt A in mixture X ?

  1. 50%
  2. 100%
  3. 150%
  4. 200%

(3) Radhika added 250 gm of salt B to mixture X to prepare a new mixture Y. By what per cent did salt B increase ?

  1. $63\frac{1}{3}$%

  2. 75%

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

  4. 121%

(4) The percentage of salt A in mixture Y is :

Show SolutionHide Solution↓

Given:

Salt A = 150 gm

Salt B = 300 gm

Salt C = 300 gm

Total weight of Mixture X = 150 + 300 + 300 = 750 gm

(1)
$\text {Percentage} = \Big( \frac{\text{Salt A}}{\text{Total weight of Mixture X}} \times 100 \Big)$%

= $\Big( \frac{150}{750} \times 100 \Big)$%

= $\Big( \frac{1}{5} \times 100 \Big)$%

= 20%

Hence, option 2 is the correct option.

(2)

Difference = Salt C - Salt A

Substituting the values in above, we get:

Difference = 300 - 150 = 150 gm

$\text {Percentage} = \Big( \frac{\text{Difference}}{\text{Value of A}} \times 100 \Big)$%

= $\Big( \frac{150}{150} \times 100 \Big)$%

= 100%

Hence, option 2 is the correct option.

(3)

Original B = 300 gm

Amount added = 250 gm

$\text {Increase}$% = $\Big( \frac{\text{Added amount}}{\text{Original B}} \times 100 \Big)$%

= $\Big( \frac{250}{300} \times 100 \Big)$%

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

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

Hence, option 3 is the correct option.

(4)

New total weight (Mixture Y) = 750 + 250 = 1000 gm

Weight of Salt A remains 150 gm

$\text {Percentage} = \Big( \frac{\text{Salt A}}{\text{Total Weight (Mixture Y)}} \times 100 \Big)$%

= $\Big( \frac{150}{1000} \times 100 \Big)$%

= $\Big( \frac{150}{10} \times 1 \Big)$%

= 15%

Hence, option 1 is the correct option.

2Case Study Based Questions

Today is Eid. Ameena is very happy. Everyone in her family gave her a few coins in her bag. She now has 25 one rupee coins, 20 two rupee coins and 15 five rupee coins in her bag.

(1) The number of five rupee coins is what per cent of the total number of coins in the bag ?

  1. 15%
  2. 20%
  3. 25%
  4. 30%

(2) The value of two rupee coins is what per cent of the total value of coins in the bag ?

  1. $4\frac{3}{4}$%

  2. $33\frac{1}{3}$%

  3. 32%

  4. $28\frac{4}{7}$%

(3) The number of five rupee coins is what per cent less than that of two rupee coins ?

  1. $17\frac{1}{2}$%

  2. 20%

  3. 25%

  4. $33\frac{1}{3}$%

(4) The value of five rupee coins is what per cent more than that of two rupee coins?

Show SolutionHide Solution↓

Given:

₹ 1 coins: 25 (Value = 1 x ₹ 25 = ₹ 25)

₹ 2 coins: 20 (Value = 2 x ₹ 20 = ₹ 40)

₹ 5 coins: 15 (Value = 5 x ₹ 15 = ₹ 75)

Total number of coins = 25 + 20 + 15 = 60 coins

Total value of coins = 25 + 40 + 75 = ₹ 140

(1)
$\text {Percentage} = \Big( \frac{\text{Number of 5 rupee coins}}{\text{Total number of coins}} \times 100 \Big)$%

= $\Big( \frac{15}{60} \times 100 \Big)$%

= $\Big( \frac{1}{4} \times 100 \Big)$%

= 25%

Hence, option 3 is the correct option.

(2)
$\text {Percentage} = \Big( \frac{\text{Value of 2 rupee coins}}{\text{Total value of coins}} \times 100 \Big)$%

= $\Big( \frac{40}{140} \times 100 \Big)$%

= $\Big( \frac{2}{7} \times 100 \Big)$%

= $\Big( \frac{200}{7} \Big)$%

= $28\frac{4}{7}$%

Hence, option 4 is the correct option.

(3)

Difference in number = Number of ₹ 2 coins - Number of ₹ 5 coins

Substituting the values in above, we get:

Difference = 20 - 15 = 5

Percentage =

$\Big( \frac{\text{Difference in number}}{\text{Number of ₹ 2 coins}} \times 100 \Big)$%

$\Big( \frac{5}{20} \times 100 \Big)$%

= $\Big( \frac{1}{4} \times 100 \Big)$%

= 25%

Hence, option 3 is the correct option.

(4)

Difference in value = Value of ₹ 5 coins - Value of ₹ 2 coins

Substituting the values in above, we get:

Difference in value = ₹ 75 - ₹ 40 = ₹ 35

Percentage =

$\Big( \frac{\text{Difference in value}}{\text{Value of ₹ 2 coins}} \times 100 \Big)$%

= $\Big( \frac{35}{40} \times 100 \Big)$%

= $\Big( \frac{7}{8} \times 100 \Big)$%

= $\Big( \frac{7}{2} \times 25 \Big)$%

= $\Big( \frac{175}{2} \Big)$%

= 87.5%

= $87\frac{1}{2}$%

Hence, option 4 is the correct option.

1Assertions and Reasons

Assertion: 1% of 100 is 1 and 100% of 1 is also 1.

Reason: x% = $\frac{x}{100}$ and x% of y = $\frac{xy}{100}$.

Show SolutionHide Solution↓

Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Explanation

Assertion:

1% of 100 = $\frac{1}{100} \times 100 = 1$

100% of 1 = $\frac{100}{100} \times 1 = 1$

Both are true. So, Assertion is true.

Reason:

x% = $\frac{x}{100}$ and x% of y = $\frac{xy}{100}$.

This is the correct formula and directly explains the assertion.

Hence, option 1 is the correct option.

2Assertions and Reasons

Assertion: A man travelled 60 km by car and 240 km by train. He travelled 20% of the journey by car and 80% of the journey by train.

Reason: Per cent change = $\frac{Actual\ change}{Original\ value} \times 100$.

Show SolutionHide Solution↓

Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).

Explanation

Assertion:

Total distance = 60 km + 240 km = 300 km.

Car% = $\Big( \frac{\text{Distance travelled by car}}{\text{Total distance}} \times 100 \Big)$%

= $\Big(\frac{60}{300} \times 100\Big)$%

= $\Big(\frac{60}{3} \times 1\Big)$%

= 20%

Train% = $\Big( \frac{\text{Distance travelled by train}}{\text{Total distance}} \times 100 \Big)$%

= $\Big(\frac{240}{300} \times 100\Big)$%

= $\Big(\frac{240}{3} \times 1\Big)$%

= 80%

Both percentages match. So, Assertion is True.

This formula in Reason is correct for calculating percentage change (increase or decrease). Reason is True.

To prove the Assertion, we didn't use the formula for "Per cent change." We used the formula for "Part as a percentage of a whole." The Reason is a true statement, but it does not explain how we got the 20% and 80%.

Hence, option 2 is the correct option.

3Assertions and Reasons

Assertion: If we multiply a number by $\frac{2}{5}$, it will increase by 20%.

Reason: Increase or decrease value is always calculated on the original value.

Show SolutionHide Solution↓

Assertion (A) is false but Reason (R) is true.

Explanation

Assertion:

Multiplying a number by $\frac{2}{5}$ is the same as multiplying by 0.4 (which is 40%).

If you multiply a number by 0.4, you are making it smaller (40% of its original size). To increase a number, the multiplier must be greater than 1.

An increase of 20% would mean multiplying by 1.20 or $\frac{6}{5}$.

Therefore, the statement is false. Assertion is False.

Reason:

Increase or decrease value is always calculated on the original value.

This is a fundamental rule of percentage applications. Reason is True.

Hence, option 4 is the correct option.

1Competency Focused Questions

A sum of money was divided among three labourers — Dinesh, Prakash and Kamal in such a way that Dinesh gets 3 parts, Prakash gets 4 parts and Kamal gets 3 parts. The percentage share of Dinesh, Prakash, and Kamal respectively are:

Show SolutionHide Solution↓

Given:

Dinesh's share = 3 parts

Prakash's share = 4 parts

Kamal's share = 3 parts

Total parts = 3 + 4 + 3 = 10 parts

Dinesh's percentage =

$\left( \frac{\text{Dinesh's parts}}{\text{Total parts}} \times 100 \right)$%

= $\left( \frac{3}{10} \times 100 \right)$%

= $(3 \times 10)$%

= 30%

Prakash's percentage =

$\left( \frac{\text{Prakash's parts}}{\text{Total parts}} \times 100 \right)$%

= $\left( \frac{4}{10} \times 100 \right)$%

= $(4 \times 10)$%

= 40%

Kamal's percentage =

$\left( \frac{\text{Kamal's parts}}{\text{Total parts}} \times 100 \right)$%

= $\left( \frac{3}{10} \times 100 \right)$%

= $(3 \times 10)$%

= 30%

Hence, option 3 is the correct option.

2Competency Focused Questions

A school team won 6 games this year against 4 games won last year. What is the per cent increase?

Show SolutionHide Solution↓

Given:

Games won last year = 4

Games won this year = 6

Increase in games won = This year - Last year

Substituting the values above, we get:

Increase in games won = 6 - 4 = 2

Increase % =

$\left( \frac{\text{Increase in games won}}{\text{Games won last year}} \times 100 \right)$%

= $\left( \frac{2}{4} \times 100 \right)$%

= $\left( \frac{1}{2} \times 100 \right)$%

= 50%

Hence, option 4 is the correct option.

3Competency Focused Questions

If 55% of 1000 : 60% of 2000 = k : 2k + 2, then the value of k is:

Show SolutionHide Solution↓

Given:

55% of 1000 : 60% of 2000 = k : 2k + 2

First, calculate 55% of 1000:

55% of 1000 = $\frac{55}{100} \times 1000$

$$\begin{aligned}= 55 \times 10 \\[1em] = 550\end{aligned}$$

Next, calculate 60% of 2000:

60% of 2000 = $\frac{60}{100} \times 2000$

$$\begin{aligned}= 60 \times 20 \\[1em] = 1200\end{aligned}$$

So, the equation becomes:

550 : 1200 = k : (2k + 2)

Writing the ratios as fractions:

$\frac{550}{1200} = \frac{k}{2k + 2}$

By cross multiplying, we get:

$$\begin{aligned}550 \times (2k + 2) = 1200 \times k \\[1em] \Rightarrow 1100k + 1100 = 1200k \\[1em] \Rightarrow 1200k - 1100k = 1100 \\[1em] \Rightarrow 100k = 1100 \\[1em] \Rightarrow k = \frac{1100}{100} \\[1em] \Rightarrow k = 11\end{aligned}$$

Hence, option 4 is the correct option.

4Competency Focused Questions

The marks in a test decreased from 40 to 30. The percentage decrease is:

Show SolutionHide Solution↓

Given:

Original marks = 40

New marks = 30

Decrease in marks = Original marks - New marks

Substituting the values above, we get:

Decrease in marks = 40 - 30 = 10

Decrease % =

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

= $\left( \frac{10}{40} \times 100 \right)$%

= $\left( \frac{1}{4} \times 100 \right)$%

= 25%

Hence, option 3 is the correct option.

5Competency Focused Questions

Which of the following is greater than 16.3%?

Show SolutionHide Solution↓

To find which option is greater than 16.3%, we convert each option into a percentage and compare.

(1) 163 out of 1000

= $\left( \frac{163}{1000} \times 100 \right)$%

= $\frac{163}{10}$%

= 16.3%

This is equal to 16.3%, not greater.

(2) 113 out of 250

= $\left( \frac{113}{250} \times 100 \right)$%

= $\left( \frac{113}{25} \times 10 \right)$%

= $\frac{1130}{25}$%

= 45.2%

This is greater than 16.3%.

(3) 8 out of 50

= $\left( \frac{8}{50} \times 100 \right)$%

= $(8 \times 2)$%

= 16%

This is less than 16.3%.

(4) 39 out of 750

= $\left( \frac{39}{750} \times 100 \right)$%

= $\left( \frac{39}{75} \times 10 \right)$%

= $\frac{390}{75}$%

= 5.2%

This is less than 16.3%.

Hence, option 2 is the correct option.

6Competency Focused Questions

What percentage of the given figure is shaded?

What percentage of the given figure is shaded? Percentage, Foundation Mathematics R.S. Aggarwal ICSE Class 7.
Show SolutionHide Solution↓

From the figure, we observe:

Total number of squares = 36

Number of fully shaded squares = 12

Number of half shaded squares = 12

Half-shaded squares combine to make 6 full squares

Total number of shaded squares = 12 + 6 = 18

Percentage of shaded part =

$\left( \frac{\text{Number of shaded squares}}{\text{Total number of squares}} \times 100 \right)$%

= $\left( \frac{18}{36} \times 100 \right)$%

= $\left(\frac{1}{2} \times 100\right)$%

= 50%

Hence, option 1 is the correct option.

7Competency Focused Questions

₹50 are to be divided between Mahima and Meenu. Mahima gets 60%. What does Meenu get?

Show SolutionHide Solution↓

Given:

Total amount = ₹ 50

Mahima's share = 60%

Meenu's share % = 100% - 60% = 40%

Meenu's amount =

40% of ₹ 50

$$\begin{aligned}= ₹ \left( \frac{40}{100} \times 50 \right) \\[1em] = ₹ \left( \frac{40}{2} \times 1 \right) \\[1em] = ₹ 20\end{aligned}$$

Hence, option 2 is the correct option.

8Competency Focused Questions

By how much is 80% of 40 greater than $\frac{4}{5}$ of 25?

Show SolutionHide Solution↓

First, calculate 80% of 40:

80% of 40 = $\frac{80}{100} \times 40$

$$\begin{aligned}= \frac{80}{10} \times 4 \\[1em] = 8 \times 4 \\[1em] = 32\end{aligned}$$

Next, calculate $\frac{4}{5}$ of 25:

$$\begin{aligned}\frac{4}{5} \text{ of } 25 = \frac{4}{5} \times 25 \\[1em] = 4 \times 5 \\[1em] = 20\end{aligned}$$

Difference = 80% of 40 - $\frac{4}{5}$ of 25

Substituting the values above, we get:

Difference = 32 - 20 = 12

Hence, option 2 is the correct option.

9Competency Focused Questions

Surabhi's both parents are working. Her father's monthly salary is ₹ 40,000 and her mother's monthly salary is ₹ 24,000. Her father spends 50% of his salary and her mother spends 40% of her salary.

(i) Both the parents together spend 40% + 50% = 90% of their total monthly income.

(ii) Both the parents together save less than 50% of their total income.

Show SolutionHide Solution↓

Given:

Father's salary = ₹ 40,000

Mother's salary = ₹ 24,000

Father spends = 50% of his salary

Mother spends = 40% of her salary

Total monthly income = ₹ 40,000 + ₹ 24,000 = ₹ 64,000

Father's spending =

50% of ₹ 40,000

$$\begin{aligned}= ₹ \left( \frac{50}{100} \times 40000 \right) \\[1em] = ₹ \left( \frac{1}{2} \times 40000 \right) \\[1em] = ₹ 20,000\end{aligned}$$

Mother's spending =

40% of ₹ 24,000

$$\begin{aligned}= ₹ \left( \frac{40}{100} \times 24000 \right) \\[1em] = ₹ \left( \frac{2}{5} \times 24000 \right) \\[1em] = ₹ (2 \times 4800) \\[1em] = ₹ 9,600\end{aligned}$$

Total spending = ₹ 20,000 + ₹ 9,600 = ₹ 29,600

Total spending % =

$\left( \frac{\text{Total spending}}{\text{Total income}} \times 100 \right)$%

= $\left( \frac{29600}{64000} \times 100 \right)$%

= $\left( \frac{29600}{640} \right)$%

= $\frac{2960}{64}$%

= 46.25%

Checking statement (i):

Both parents together spend 46.25% of their total monthly income, not 90%. Percentages of different amounts cannot be simply added together because they are calculated on different bases (40000 and 24000).

So, statement (i) is wrong.

Total savings = Total income - Total spending

Total savings = ₹ 64,000 - ₹ 29,600 = ₹ 34,400

Percentage of total income saved =

$\left( \frac{\text{Total savings}}{\text{Total income}} \times 100 \right)$%

= $\left( \frac{34400}{64000} \times 100 \right)$%

= $\left( \frac{34400}{640} \right)$%

= $\frac{3440}{64}$%

= 53.75%

Checking statement (ii):

Both parents together save 53.75%, which is more than 50% of their total income.

So, statement (ii) is wrong.

Hence, option 4 is the correct option.

10Competency Focused Questions

Kamajeet earned ₹ 1,00,000 per month. He used to spend 20% of his salary on food each month. Due to pandemic, his monthly salary was reduced by 20%. But his expenditure on food remained the same.

(i) He spends 25% of his new salary on food after the pandemic.

(ii) His reduced salary is ₹ 80,000.

Show SolutionHide Solution↓

Given:

Original salary = ₹ 1,00,000

Food expenditure = 20% of salary

Salary reduction = 20%

Food expenditure =

20% of ₹ 1,00,000

$$\begin{aligned}= ₹ \left( \frac{20}{100} \times 100000 \right) \\[1em] = ₹ (20 \times 1000) \\[1em] = ₹ 20,000\end{aligned}$$

Salary reduction amount =

20% of ₹ 1,00,000

$$\begin{aligned}= ₹ \left( \frac{20}{100} \times 100000 \right) \\[1em] = ₹ (20 \times 1000) \\[1em] = ₹ 20,000\end{aligned}$$

New salary = Original salary - Salary reduction

New salary = ₹ 1,00,000 - ₹ 20,000 = ₹ 80,000

Checking statement (ii):

His reduced salary is ₹ 80,000.

So, statement (ii) is correct.

Food expenditure remains the same at ₹ 20,000.

Percentage of new salary spent on food =

$\left( \frac{\text{Food expenditure}}{\text{New salary}} \times 100 \right)$%

= $\left( \frac{20000}{80000} \times 100 \right)$%

= $\left( \frac{1}{4} \times 100 \right)$%

= 25%

Checking statement (i):

He spends 25% of his new salary on food after the pandemic.

So, statement (i) is correct.

Hence, option 3 is the correct option.

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