S.chand Class 8 Maths Solution Chapter 8 Percentage And Its Applications Exercise 8 A

 Exercise 8 A

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Question 1

Write these percentages as fractions and decimals :

(i) $60 \%$

Sol :

$=\dfrac{60}{100}=\dfrac{3}{5}$

=0.6

(ii) $240 \%$

Sol :

$=\dfrac{240}{100}=\dfrac{12}{5}$

=2.4

(iii) $12 \dfrac{1}{2} \%$

Sol :

$=\dfrac{25}{2}\%=\dfrac{25}{2 \times 100}$

$=\dfrac{1}{8}$

=0.125

(iv) $55 \dfrac{2}{3} \%$

Sol :

$=\dfrac{167}{3}\%$

$=\dfrac{167}{3 \times 100}=\dfrac{167}{300}$

=0.557...

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Question 2

Write these numbers as percentages :

(i) $\frac{3}{4}$Sol :

Sol :

$=\dfrac{3}{4} \times 100$

=75%

(ii) $\frac{21}{40}$Sol :

Sol :

$=\dfrac{21}{40} \times 100$

$=\dfrac{105}{2}\%$

$=52\dfrac{1}{2}\%$

(iii) $0.008$

Sol :

$=\dfrac{8}{1000} \times 100$

$=\dfrac{4}{5}\%$

=0.8%

(iv) $0.375$

Sol :

$=\dfrac{375}{1000}\times 100$

$=\dfrac{75}{2}\%$

=37.5%

(v) $1.07$

Sol :

$=\dfrac{107}{100} \times 100$

=107 %

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Question 3

Express the following percentage as ratios:

(i) $200 \%$

Sol :

$=\dfrac{200}{100}=\dfrac{2}{1}$

=2 : 1

(ii) $6 \dfrac{2}{3} \%$

Sol :

$=\dfrac{20}{3}\%=\dfrac{20}{3 \times 100}$

$=\dfrac{1}{15}$

=1 : 15

(iii) $75 \%$

Sol :

$=\dfrac{75}{100}=\dfrac{3}{4}$

=3 : 4

(iv) $33 \dfrac{1}{3} \%$

Sol :

$=\dfrac{100}{3}\%=\dfrac{100}{3 \times 100}$

$=\dfrac{1}{3}$

=1 : 3

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Question 4

Express the following ratios as percentage:

(i) $1: 2$

Sol :

$=\dfrac{1}{2} \times 100$

=50%

(ii) $7: 10$

Sol :

$=\dfrac{7}{10} \times 100$

=70%

(iii) $21: 25$

Sol :

$=\dfrac{21}{25} \times 100$

=84%

(iv) $5: 12$

Sol :

$=\dfrac{5}{12} \times 100$

$=\dfrac{125}{3}\%=41\dfrac{2}{3}\%$

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Question 5

Evaluate :

(i) $19 \%$ of $₹ 4200$

Sol :

$=\dfrac{19}{100} \times 4200$

=798 g

(ii) $75 \%$ of $440 \mathrm{~kg}$

Sol :

$=\dfrac{75}{100} \times 440$ 

=330 g

(iii) $30 \%$ of $20 \mathrm{~g}$

Sol :

$=\dfrac{30}{100} \times 20$ 

=6 g

(iv) $2.5 \%$ of $350 \mathrm{~m}$

Sol :

$=\dfrac{25}{10 \times 100} \times 350$

$=\dfrac{35}{4}$

=8.75 m

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Question 6

Express the first quantity as a percentage of the second :

(i) $20 \mathrm{~g}$ of $4 \mathrm{~kg}$

Sol :

=20 g of 4000g

$=\dfrac{20}{4000} \times 100$

$=\dfrac{1}{2}\%$

=0.5%

(ii) $200 \mathrm{~mL}$ of 5 litres

Sol :

=200 ml of 5000 ml

$=\dfrac{200}{5000} \times 100$

=4%

(iii) $48 \mathrm{~cm}$ of $1 \mathrm{~m}$

Sol :

=48 cm pf 100 cm

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

=48%

(iv) $\frac{2}{3}$ of $\frac{1}{2}$

Sol :

$=\dfrac{\frac{2}{3}}{\frac{1}{3}} \times 100$

$=\dfrac{2}{3} \times \dfrac{3}{1} \times 100$

=200 %

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Question 7

(i) If $14 \%$ of a certain number is 63 , find the number.

Sol :

Let the required number be x

14% of x=63

$\dfrac{14}{100} \times x=63$

$x=\dfrac{63 \times 100}{14}$

x=450

(ii) If $12 \frac{1}{2} \%$ of a certain amount is $₹ 62.50$, find the amount.

Sol :

Let the amount be x

$12 \frac{1}{2} \%$ of x=62.50

$\dfrac{25}{2}\% \times x$=62.50

$\dfrac{25}{2\times 100} \times x$=62.50

$x=\dfrac{62.50}{100} \times \dfrac{2}{25} \times 100$

x=500

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Question 8

There were 4200 people at a concert and 1400 of these were female. What percentage were male ?

Sol :

Total number of people=4200

Number of female=1400

Number of male=4200-1400=2800

Percentage of male $\dfrac{2800}{4200} \times 100$

$=\dfrac{200}{3} \%=66\dfrac{2}{3}\%$

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Question 9

The cost of a theatre ticket is increased by $40 \%$ for a special movie. What is the new price if the normal price was ₹ 160.

Sol :

Normal price was ₹ 160

Increase =40% of 160$

$=\dfrac{40}{100} \times 160$

=4×16=64

New price=160+64=224

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Question 10

A motorcycle originally cost $₹ 40,000$. Its cost has decreased by $22 \%$. What is its cost now ?

Sol :

Original cost of motorcycle=40000

Decrease=22% of 40000

$=\dfrac{22}{100} \times 40000$

=22×400=8800

New cost=40000-8800=31200

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Question 11

The price of a garment has been reduced by $15 \%$ in a sale to $₹ 306$. Find its original price.

Sol :

Let the original cost be x

According to question,

x-15% of x=306

$x-\dfrac{15}{100}\times x=306$

$x-\dfrac{3x}{20}=306$

$\dfrac{20x-3x}{20}=306$

$\dfrac{17x}{20}=306$

$x=\dfrac{306 \times 20}{17}$

x=360

Multiple Choice Questions (MCQs)

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Question 12

What percent of $\sqrt{0.0169}$ is $0.0117$ ?

(a) 0.9

(b) 0.1

(c) 0.09

(d) 9

Sol :

x% of $\sqrt{0.0169}=0.0117$

$\dfrac{x}{100} \times \sqrt{\dfrac{169}{10000}}=\dfrac{117}{10000}$

$\dfrac{x}{100} \times \dfrac{13}{100}=\dfrac{117}{10000}$

$x \times 13=\dfrac{117}{10000} \times 100 \times 100$

$x=\dfrac{117}{13}$

x=9

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Question 13

What per cent is $3 \%$ of $5 \%$ ?

(a) $15 \%$

(b) $30 \%$

(c) $50 \%$

(d) $60 \%$

Sol :

Let the value be 100

x% of 5%=3%

$\dfrac{x}{100} \times \dfrac{5}{100} \times 100=\dfrac{3}{100} \times 100$

$\dfrac{x}{100} \times 5=3$

$x=\dfrac{3}{5} \times 100$

x=60

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Question 14

A piece of elastic was stretched by $24 \%$ to a length of $31 \mathrm{~cm}$. Find its unstretched original length.

(a) $24 \mathrm{~cm}$

(b) $26 \mathrm{~cm}$

(c) $25 \mathrm{~cm}$

(d) $20 \mathrm{~cm}$

Sol :

Let the original length be x 

According to question,

x+24% of x=31

$x+\dfrac{24}{100} \times x=31$

$\dfrac{100x+24x}{100}=31$

$\dfrac{124x}{100}=31$

$x=\dfrac{31}{124}\times 100$

x=25

High Order Thinking Skills (HOTS)

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Question 15

The sum of two numbers is $\dfrac{28}{25}$ of the first number. The second number is what percent of the first?

Sol :

Let , First number be x

Then , Second number=y

$x+y=\dfrac{28}{25}$of x

$y=\dfrac{28x}{25}-x$

$y=\dfrac{28x-25x}{25}$

$y=\dfrac{3x}{25}$

$\dfrac{y}{x}=\dfrac{3}{25}$

$\dfrac{y}{x}=\dfrac{3}{25} \times 100$

$\dfrac{y}{x}=12%$

 y=12% of x