S.chand publication New Learning Composite mathematics solution of class 7 Chapter 8 Percentage and Its Applications Exercise 8B

 Exercise 8B

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Q1 | Ex-8B |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 1

Find.

(a) 25% of 32

Sol :

$\frac{25}{100}\times 32$

=8

(b) $16\frac{1}{3}%$ of 390

Sol :

$=\frac{49}{300}\times 390$

=63.7

(c) 50% of 29

Sol :

$=\frac{50}{100}\times 29$

=14.5

(d) $33\frac{1}{3}%$ of 150

Sol :

$=\frac{100}{300}\times 150$

=50

(e) 20% of 775

Sol :

$=\frac{20}{100}\times 775$

=155

(f) 15% of 3000

Sol :

$=\frac{15}{100}\times 3000$

=450

(g) 28% of 1225

Sol :

$=\frac{28}{100}\times 1225$

=343

(h) 18% of 2650

Sol :

$=\frac{18}{100}\times 2650$

=477

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Q2 | Ex-8B |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 2

(a) 10% of Rs. 775

Sol :

$=\frac{10}{100}\times 775$

=77.5

Ans : 77.5 Rs

(b) 15% of 800 kg

Sol :

$=\frac{15}{100}\times 800$

=120

Ans : 120 kg

(c) 45% of 180 L

Sol :

$=\frac{45}{100}\times 180$

=81 L

(d) 37% of Rs. 680

Sol :

$=\frac{37}{100}\times 680$

=251.6 Rs

(e) 120% of Rs. 65

Sol :

$=\frac{120}{100}\times 65$

=78 Rs

(f) $16\frac{2}{3}$% of 480 m

Sol :

$=\frac{500}{300}\times 480$

=80m

(g) 36% of 1.5 km = ____ m

Sol :

$=\frac{36}{100}\times 1500$

=540m

(h) $27\frac{1}{2}$% of 8 m = ____ cm

Sol :

$=\frac{55}{200}\times 800$

=200 cm

(i) $66\frac{2}{3}$% of 2 days = ____ h.

Sol :

$=\frac{200}{300}\times 48$ [2 days=48h]

=32 h

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Q3 | Ex-8B |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 3

A certain can of ice tea contains 4% of the recommended daily allowance of sodium. The recommended daily allowance is 2500 miligrams. How many miligrams of sodium are in the can of iced tea?

Sol :

∴Quantity of sodium are in the iced tea

$=\left(\frac{4}{100}\times 2500\right)$mg

=100

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Q4 | Ex-8B |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 4

27% of a fertiliser is superphosphate. How much superphosphate (in kg) is in 2 tonnes of the mixture?

Sol :

2 tonnes=2000 kg

∴Quantity of superphosphate$=\left(\frac{27}{100}\times 2000\right)$kg

=540

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Q5 | Ex-8B |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 5

In a referendum, 68% of the 85,800 voters voted YES and the rest voted No. How many voted NO?

Sol :

Percentage of voter voted =100-68=32

∴Number of voters voted $=\frac{32}{100}\times 85800$

=27456