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

EXERCISE 7 (A)


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



QUESTION 1

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

(i) 15%

Sol :

=15100

= 0.15

(ii) 84%

Sol :

=84100

= 0.84

(iii) 12.5%

Sol :

=12.5100

= 0.125

(iv) 120%

Sol :

=120100

= 1.2

(v) 3313%

Sol :

=1003×1100

=13

= 0.33


Q2 | Ex-7A | Class-8 | Schand Composite Mathematics | Percentage | myhelper

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QUESTION 2

Write each of the following as percentages.

(i) 35

Sol :

35×100

= 60 %

(ii) 1920

Sol :

=1920×100

= 95 %

(iii) 223

Sol :

=223×100

=26623%

(iv) 0.95

Sol :

=95100

=95100×100

= 95 %

(v) 2.575

Sol :

=25751000×100

= 257.5 %

 


Q3 | Ex-7A | Class-8 | Schand Composite Mathematics | Percentage | myhelper

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QUESTION 3

Express each of the following ratios as percents.

(i) 1 : 4

Sol :

=14×100

= 25%


(ii) 11 : 20

Sol :

=1120×100

= 55 %


(iii) 21 : 40

Sol :

=2140×100

=5212 %

 


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QUESTION 4

Express each of the following percents as ratios.

(i) 45%

Sol :

=45100

=920

or 9 : 20


(ii) 1623%

Sol :

=503

=503×100

=16

or 1 : 6

 


(iii) 130%

Sol :

=130100

=1310

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

=10cm100cm×100%

= 10%


(ii) 4 kg of 120 kg

Sol :

=4120×100%

=400120=103=313%

 


(iii) 110 of 25

Sol :

=(11025)×100%

=(110÷25)×100%

=(110×52)×100%

=14×100%

= 25%

 


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QUESTION 6

Find the following amounts.

(i) 12% of 600

Sol:

=12100×600

=12×6

= 72

 


(ii) 3313% of 2400 people

Sol :

=(3313100)×2400

=(1003÷1001)×2400

=(1003×1100)×2400

=(13)×2400

= 800 peoples

 


(iii) 48% of 1 litre

Sol :

1l = 100ml

=48100×1000ml

= 48×10 ml

= 480 ml

 


(iv) 717 of 3kg 500g

Sol:

1 kg = 1000 g

3 kg = 3000g

3 kg 500 g = 3000 + 500 g

3 kg 500 g = 3500 g

=(717100)×3500g

=(507÷1001)×3500g

=(507×1100)×3500g

=(114)×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%

=2100×2.40

= 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% =40100×3000

= 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 =error85×100

=2085×100

=200085=23917%

 


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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 =15100×100 = 15

New salary be 115

When new salary are 115 , original salary are 100

When new salary are 16100 , original salary are =100115×16100 = 14000

 

ALTERNATE METHOD

Amount=Principal(1+rate of interest100)n

16100=Principal(1+15100)1

16100=Principal(100+15100)

16100=Principal(115100)

Principal=16100×100115

Principal=1610000115

Principal = 14000


 

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