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Examples 6 to 10

June 13, 2020
Category: Examples 6 to 10

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(6) What percent is :

(i) 120 of 80 (ii) 125 ml. of 7.5 litres (iii) 10 gm of 1 kg ?

Soln. (i) Let, x % of 80 = 120

$\therefore \frac{x}{100} \times$ 80 = 120

$\therefore$ x = $\frac{120 \times 100}{80}$ = 150 $\therefore$ x % = 150 %

(ii) Let, x % of 7500 = 125 ( $\because$ 7.5 litre = 7500 ml. )

$\therefore \frac{x}{100} \times$ 7500 = 125

x = $\frac{125 \times 100}{7500}$ = $\frac{5}{3}$ = $1\frac{2}{3}$ $\therefore$ x % = $1\frac{2}{3}$ %

(iii) Let, x % of 1000 = 10 ( $\because$ 1 kg. = 1000 gm. )

$\therefore \frac{x}{100} \times$ 1000 = 10

$\therefore$ x = $\frac{10 \times 100}{1000}$ = 1000 $\therefore$ x % = 1 %

(7) If A’s monthly income is 50 % more than that of B, how much percent is B, how much percent is B’s income less than that of A?

Soln. Required percentage = $\left [\frac{50}{\left (100+50 \right )}\times100 \right ]$ % = 33 $\frac{1}{3}$ %

(8) If A’s height is 20 % less than that of B, how much percent is B’s height more than that of A?

Soln. Required percent = $\left [\frac{20}{\left (100-20 \right )}\times100 \right ]$ % = 25 %

(9) If the price of rice is increased by 10 %, find by how much percent a householder must reduce her consumption of rice so as not to increase the expenditure.

Soln. Reduction in consumption = $\left [\frac{10}{\left (100+10 \right )}\times100 \right ]$ % = 9 $\frac{1}{11}$ %.

(10) The population of a town is 14400. If it increases annually at 20 %, what will be its population 2 years hence ? What was it 2 years ago ?

Soln.

Population after 2 years

= 14400 $\times \frac{120}{100} \times \frac{120}{100}$ = 20736

Population 2 years ago

= 14400 $\times \frac{100}{120} \times \frac{100 }{120}$ = 10000

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Examples 6 to 10

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