myhelper: S.chand composite mathematics class 8 solution chapter 7 percentage Exercise 7A

EXERCISE 7 (A)

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QUESTION 1

Write each of these percentages as a fraction and a decimal.

(i) 15%

Sol :

$=\dfrac{15}{100}$

= 0.15

(ii) 84%

Sol :

$=\dfrac{84}{100}$

= 0.84

(iii) 12.5%

Sol :

$=\dfrac{12.5}{100}$

= 0.125

(iv) 120%

Sol :

$=\dfrac{120}{100}$

= 1.2

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

Sol :

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

$=\dfrac{1}{3}$

= 0.33

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QUESTION 2

Write each of the following as percentages.

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

Sol :

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

= 60 %

(ii) $\dfrac{19}{20}$

Sol :

$=\dfrac{19}{20} \times 100$

= 95 %

(iii) $2\dfrac{2}{3}$

Sol :

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

$=266\dfrac{2}{3}$%

(iv) 0.95

Sol :

$=\dfrac{95}{100}$

$=\dfrac{95}{100}\times 100$

= 95 %

(v) 2.575

Sol :

$=\dfrac{2575}{1000}\times 100$

= 257.5 %

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QUESTION 3

Express each of the following ratios as percents.

(i) 1 : 4

Sol :

$=\dfrac{1}{4}\times 100$

= 25%

(ii) 11 : 20

Sol :

$=\dfrac{11}{20}\times 100$

= 55 %

(iii) 21 : 40

Sol :

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

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

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QUESTION 4

Express each of the following percents as ratios.

(i) 45%

Sol :

$=\dfrac{45}{100}$

$=\dfrac{9}{20}$

or 9 : 20

(ii) $16\dfrac{2}{3}\%$

Sol :

$=\dfrac{50}{3}$

$=\dfrac{50}{3}\times 100$

$=\dfrac{1}{6}$

or 1 : 6

(iii) 130%

Sol :

$=\dfrac{130}{100}$

$=\dfrac{13}{10}$

or 13 : 10

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QUESTION 5

Express the first quantity as a percentage of the second quantity.

(i) 10 cm of 1 m

Sol :

1m=100cm

$=\dfrac{10\text{cm}}{100\text{cm}}\times100\%$

= 10%

(ii) 4 kg of 120 kg

Sol :

$=\dfrac{4}{120}\times100\%$

$=\dfrac{400}{120}=\dfrac{10}{3}=3\dfrac{1}{3}\%$

(iii) $\dfrac{1}{10}\text{ of }\dfrac{2}{5}$

Sol :

$=\left(\dfrac{\dfrac{1}{10}}{\dfrac{2}{5}}\right)\times 100\%$

$=\left(\dfrac{1}{10}\div\dfrac{2}{5}\right)\times 100\%$

$=\left(\dfrac{1}{10}\times\dfrac{5}{2}\right)\times 100\%$

$=\dfrac{1}{4}\times100\%$

= 25%

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QUESTION 6

Find the following amounts.

(i) 12% of 600

Sol:

$=\dfrac{12}{100}\times 600$

=12×6

= 72

(ii) $33\dfrac{1}{3}\%$ of 2400 people

Sol :

$=\left(\dfrac{33\dfrac{1}{3}}{100}\right)\times 2400$

$=\left(\dfrac{100}{3}\div\dfrac{100}{1}\right)\times 2400$

$=\left(\dfrac{100}{3}\times\dfrac{1}{100}\right)\times 2400$

$=\left(\dfrac{1}{3}\right)\times 2400$

= 800 peoples

(iii) 48% of 1 litre

Sol :

1l = 100ml

$=\dfrac{48}{100}\times 1000ml$

= 48×10 ml

= 480 ml

(iv) $7\dfrac{1}{7}$ of 3kg 500g

Sol:

1 kg = 1000 g

3 kg = 3000g

3 kg 500 g = 3000 + 500 g

3 kg 500 g = 3500 g

$=\left(\dfrac{7\dfrac{1}{7}}{100}\right)\times 3500g$

$=\left(\dfrac{50}{7}\div\dfrac{100}{1}\right)\times 3500g$

$=\left(\dfrac{50}{7}\times\dfrac{1}{100}\right)\times 3500g$

$=\left(\dfrac{1}{14}\right)\times 3500g$

= 250 g

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QUESTION 7

A metal bar is 2.4 meters long.After heating its length increases by 2%.What is the new length?

Sol :

Original length 2.4 m

A.T.Q

Length increased by 2%

$=\dfrac{2}{100}\times 2.40$

= 0.048 m

New length = Original Length + Length increased by 2%

= 2.4m + 0.048m

= 2.448 m

Hence , new length is equal to 2.448m

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QUESTION 8

A pair of shoes costs ₹ 3000.During sale, its price reduced by 40%.What is the new price after reduction?

Sol :

Original Price = 3000

During sale price reduced by 40% $=\dfrac{40}{100}\times 3000$

= 4×300

= 1200

New Price = Original Price - 40%

= 3000 - 1200

= 1800

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QUESTION 9

Piyush's marks in Mathematics were wrongly read as 65 instead of 85.What is the percentage error in the marks?

Sol :

Error = 85 - 65 = 20

Percentage error $=\dfrac{\text{error}}{85}\times 100$

$=\dfrac{20}{85}\times 100$

$=\dfrac{2000}{85}=23\dfrac{9}{17}\%$

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QUESTION 10

A man gets a 15% hike salary.If his new salary is ₹ 16,100.What was his original salary?

Sol :

Let the original salary be 100

Increase in salary is 15% of 100 $=\dfrac{15}{100}\times 100$ = 15

New salary be 115

When new salary are 115 , original salary are 100

When new salary are 16100 , original salary are $=\dfrac{100}{115}\times 16100$ = 14000

ALTERNATE METHOD

$\text{Amount}=\text{Principal}\left(1+\dfrac{\text{rate of interest}}{100}\right)^n$

$16100=\text{Principal}\left(1+\dfrac{15}{100}\right)^1$

$16100=\text{Principal}\left(\dfrac{100+15}{100}\right)$

$16100=\text{Principal}\left(\dfrac{115}{100}\right)$

$\text{Principal} = 16100\times \dfrac{100}{115}$

$\text{Principal} = \dfrac{1610000}{115}$

Principal = 14000

S.chand composite mathematics class 8 solution chapter 7 percentage Exercise 7A

EXERCISE 7 (A)

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