S.chand books class 8 maths solution chapter 7 Percentage exercise 7B

EXERCISE 7 (B)

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Q1 | Ex-7B | Class-8 | Schand Composite Mathematics | Percentage | myhelper

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Question 1

What percent of a day is the time duration of 3 hours?

Sol :

Total hours in a day = 24 hours

Given hours = 3 hours in percentage

$=\dfrac{3}{24}\times 100\%$

$=\dfrac{300}{24}=12\dfrac{1}{2}\%$

Hence , $12\dfrac{1}{2}\%$ of a day represent 3 hours

 

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Q2 | Ex-7B | Cl-8 | Schand Composite Mathematics | Percentage | myhelper

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Question 2

About 67% of a human body's total weight is water.If Aparna weights 70 kg, how much of her weight is water?

Sol :

According to question:

67% of 70 kg

$=\dfrac{67}{100}\times 70$

$=\dfrac{4690}{100}$

= 46.90 kg

Hence , 46.90 kg is water from 70 kg of total body weight

 

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Q3 | Ex-7B | Cl-8 | Schand Composite Mathematics | Percentage | myhelper

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Question 3

A person's salary increases from ₹ 7,200 to ₹ 8,100.What is the percentage increase in his salary?

Sol :

According to question:

Increase in salary = 8100 - 7200 = 900

Percentage increase $=\dfrac{900}{7200}\times 100$

$=\dfrac{90000}{7200}=\dfrac{900}{72}=\dfrac{100}{8}$ $=\dfrac{50}{4}=\dfrac{25}{2}=12\dfrac{1}{2}$

$12\dfrac{1}{2}$ is the percentage increase in his salary

 

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Question 4

A fruit seller had some apples.He sells 40% apples and still has 420 apples.How many apples did he originally have?

Sol :

Let the total number of apples is x

As , already mentioned 40% already sold which means 60% of total apples is equal to 420

⇒$\dfrac{60}{100}\times x = 420$

⇒$x=\dfrac{420\times 100}{60}$

⇒x=700

Hence , the total number of apples is 700

 

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Q5 | Ex-7B | Cl-8 | Schand Composite Mathematics | Percentage | myhelper

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Question 5

A water pipe is cut into pieces.The longer piece is 70% of the length of the pipe.By how much percent is the longer piece longer than the shorter piece?

Sol :

Let the water pipe's original total length be 100 m

Then Length of longer piece of pipe is 70% of original $=\dfrac{70}{100}\times 100m$ = 70 m

And Length of shorter piece of pipe is 30% of original $=\dfrac{30}{100}\times 100m$ = 30 m

As you can see here Longer piece is long by 70 - 30 = 40 m than shorter piece

In percentage the longer piece longer than the shorter piece

$=\dfrac{40m}{30}\times 100\%$

$=\dfrac{400}{3}\%$

 

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Question 6

If a number is increased by 20% and the resulting number is again increased by 20%, what percent is the total increase?

Sol :

Let the original number be 100 and Increased by 20%

Which is = 100 + 20% of 100

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

= 120

And again 20% increased

= 120 + 20% of 120

$= 120 + \dfrac{20}{100}\times 120$

= 120 + 24

= 144

Total increase = 144 - 100 = 44

Percentage increase $=\dfrac{44}{100}\times 100\%$

= 44 %

 

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Q7 | Ex-7B | Cl-8 | Schand Composite Mathematics | Percentage | myhelper

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Question 7

If A's income is 50% less than that of B, then B's income is what percent more than that of A?

Sol :

Let the income of B is 100

Then , A.T.Q A's income is

A = B - 50% of B

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

= 100 - 50

A = 50

 

Difference in B's income and A's income = 100 - 50 = 50

Percentage increase in B's income as compare to A's income $=\dfrac{\text{difference}}{\text{A's income}}\times 100\%$

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

= 100 %

 

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Q8 | Ex-7B | Cl-8 | Schand Composite Mathematics | Percentage | myhelper

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Question 8

If the price of petrol be raised by 20%, then what is the percentage by which a car owner must reduce its consumption so as not to increase his expenditure on petrol?

Sol :

Let the original price of petrol is 100

New price = 100 + 20% of 100

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

= 120

Difference between prices = 120 - 100 = 20

Percentage decrease of use of petrol $=\dfrac{\text{Difference}}{\text{New price}}\times 100\%$

$=\dfrac{20}{120}\times 100\%$

$=\dfrac{1}{6}\times 100\%=\dfrac{100}{6}$ $=\dfrac{50}{3}=16\dfrac{2}{3}$

Hence , the should reduce its consumption by $16\dfrac{2}{3}$

 

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Q9 | Ex-7B | Cl-8 | Schand Composite Mathematics | Percentage | myhelper

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Question 9

In an examination, 1100 boys and 900 girls appeared. 50% of the boys and 40% of the girls passed the examination.What is the percentage of candidates who failed?

Sol :

Failed in boys = 50% of 1100 $=\dfrac{50}{100}\times 1100$ = 50×11 = 550

Failed in girls = 60% of 900 $=\dfrac{60}{100}\times 900$ = 60×9 = 540

Total percentage of failed students $=\dfrac{\text{total failed students}}{\text{total students}}\times 100$

$=\dfrac{550+540}{1100+900}\times 100\%$

$=\dfrac{1090}{2000}\times 100\%$

= 54.5%

 

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Q10 | Ex-7B | Cl-8 | Schand Composite Mathematics | Percentage | myhelper

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Question 10

A candidate who scores 30% fails by 5 marks, while another candidate who scores 40% marks get 10 more than minimum pass marks.What are the minimum pass marks required to pass?

Sol :

Let x be the total marks . According to question

30% of x + 5  = minimum passing marks ..(i)

Also , minimum passing marks also equal to 40% of x - 10 ..(ii)

From (i) and (ii) , we get

⇒30% of x + 5  = 40% of x - 10

⇒+5+10=40% of x - 30% of x

⇒$15 = \dfrac{40}{100}\times x - \dfrac{30}{100}\times x$

⇒$15 = \dfrac{40x-30x}{100}$

⇒1500=10x

⇒$x = \dfrac{1500}{10}=150$

Total marks = x = 150 And

Minimum passing marks is equal to 30% of x + 5  [from(i)]

$=\dfrac{30}{100}\times 150 + 5$

= 3×15 + 5

= 45 + 5

= 50

Hence , minimum passing marks is equal to 50

 

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Q11 | Ex-7B | Class-8 | Schand Composite Mathematics | Percentage | Chapter 7 | myhelper

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Question 11

Out of his income, Mr.Raj spends 20% on house rents and 70% of the rest on household expenditure.If he saves ₹ 3,600 per month, then what is his total income per month?

Sol :

Let the total income be x

A.T.Q

Spending on house = 20% of x  $=\dfrac{20}{100}\times x=\dfrac{x}{5}$

Remaining = Total income - spending on house

$= x - \dfrac{x}{5}=\dfrac{5x-x}{5}=\dfrac{4x}{5}$ ..(i)

 

Spending on households = 70% of remaining [from (i)]

$=\dfrac{70}{100}\times \dfrac{4x}{5}=\dfrac{7 \times 4x}{10\times 5}$ $=\dfrac{28x}{50}=\dfrac{14x}{25}$

Remaining = remaining [from(i)] - spending on households

$=\dfrac{4x}{5}-\dfrac{14x}{25}=\dfrac{5(4x)-14x}{25}$ $=\dfrac{20x-14x}{25}=\dfrac{6x}{25}$

And this remaining must be equal to his savings

⇒$\dfrac{6x}{25}=3600$

⇒$x=\dfrac{3600\times 25}{6}=\dfrac{90000}{6}$

⇒x=15000

Hence , total income is equal to 15000

 

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