Question

Question :
1. In a factory, the production of bicycles rose to 13,31000 from 10,00,000 in 3 years. Find the rate of growth per annum.

2. The population of a town is increasing at the rate of 10 % per annum. What will be the population of town after three years if the present population is 15000?

3. The difference between the compound Interest and Simple Interest on a certain sum for 2 years at 6 % per annum is Rs. 90. Find the Sum.
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Answer 1

Solution - 1

$\sf \: Present \: population \: = \: 13,31,000$

$\sf \: Previous \: Population \: = \: 10,00,000$

$\sf \: Number \: of \: years \: (n) \: = \: 3$

$\sf \: Present \: production \: = \: Previous \: production \: \bigg(1 \: + \: \dfrac{R}{100} \bigg)^{n}$

$\sf 13,31,000 = 10,00,000 \: = \: \: \bigg(1 \: + \: \dfrac{R}{100} \bigg)^{3}$

$\sf \dfrac{13,31,000}{10,00,000} \: = \: \: \bigg(1 \: + \: \dfrac{R}{100} \bigg)^{3}$

$\sf \dfrac{13,31}{10,00} \: = \: \: \bigg(1 \: + \: \dfrac{R}{100} \bigg)^{3}$

$\sf \bigg(\dfrac{11}{10} \bigg)^{3} \: = \: \: \bigg(1 \: + \: \dfrac{R}{100} \bigg)^{3}$

$\sf \bigg(\dfrac{11}{10} \bigg) \: = \: \: \bigg(1 \: + \: \dfrac{R}{100} \bigg)$

$\sf \bigg(\dfrac{11}{10} \: - 1 \bigg) \: = \: \: \bigg( \dfrac{R}{100} \bigg)$

$\sf \bigg(\dfrac{1}{10} \bigg) \: = \: \: \bigg( \dfrac{R}{100} \bigg)$

$\sf \bigg( \dfrac{100}{10} \bigg) \: = \: R$

$\bf \: R \: = \: 10$

$\sf \: Hence, \: the \: required \: rate \: of \: growth \: = \: 10 \: \% \: annum.$

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Solution - 2

$\sf \: \bf Given :$ ↴

$\sf \: Present \: population \: (P) \: = \: 15000.$

$\sf \: Rate \: of \: Growth \: or \: Increase \: = \: 10 \: \% \: per \: annum.$

$\sf \: Time \: (n) \: = \: 3 \: years.$

$\sf \: Population \: after \: 3 \: years \: = \: P \bigg( \: 1 \: + \: \dfrac{R}{100}\bigg)^{n}$

$\: \sf \: = \: 15000 \: \bigg( 1 \: + \: \dfrac{10}{100} \bigg)^{n}$

$\: \sf \: = \: 15000 \: \bigg( \dfrac{11}{100} \bigg)^{3}$

$\: \sf \: = \: 15000 \: \times \: \bigg( \dfrac{11}{100} \: \times \: \dfrac{11}{100} \: \times \: \dfrac{11}{100} \bigg)$

$\sf \: \: \: \longrightarrow \: 19,965$

$\sf \: Hence \: population \: after \: 3 \: years \: = \: Rs. \: 19,965.$

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Solution - 3

$\sf \: \bf \: Given :$ ↴

$\sf \: Principal \: = \: Rs. 90$

$\sf \: Time \: (n) \: = \: 2 \: years$

$\sf \: Rate \: = \: 6 \: \%$

$\sf \: Difference \: = P \: \bigg(\dfrac{R}{100} \bigg)^{2}$

$\sf \: 90 \: = \: \dfrac{P \: \times \: 6 \times \: 6}{100 \: \times 100}$

$\sf \: P\: = \: \dfrac{90 \: \times \: 100\times \: 100}{ 100\times \: 100}$

$\bf \: = \: Rs. \: 25000$

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