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S.chand books class 8 maths solution chapter percentage

PERCENTAGE

EXERCISE 7 (A)


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}

= 0.33


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 %

 


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}

 


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

 


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%

 


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

 


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

 


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

 


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}\%

 


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


 

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