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Three Years Ago, the Population of a Town Was 50000. If the Annual Increase During Three Successive Years Be at the Rate of 4%, 5% and 3% Respectively, Find the Present Population. - Mathematics

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Question

Three years ago, the population of a town was 50000. If the annual increase during three successive years be at the rate of 4%, 5% and 3% respectively, find the present population.

Sum
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Solution

Here, 
P = Initial population = 50, 000
\[ R_1 = 4 % \]
\[ R_2 = 5 % \]
\[ R_3 = 3 % \]
n = Number of years = 3
∴ Population after three years = P \[\left( 1 + \frac{R_1}{100} \right)\left( 1 + \frac{R_2}{100} \right)\left( 1 + \frac{R_3}{100} \right)\]
\[ = 50, 000\left( 1 + \frac{4}{100} \right)\left( 1 + \frac{5}{100} \right)\left( 1 + \frac{3}{100} \right)\]
\[ = 50, 000\left( 1 . 04 \right)\left( 1 . 05 \right)\left( 1 . 03 \right)\]
\[ = 56, 238\]
Hence, the population after three years is 56, 238.

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Chapter 14: Compound Interest - Exercise 14.4 [Page 27]

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RD Sharma Mathematics [English] Class 8
Chapter 14 Compound Interest
Exercise 14.4 | Q 4 | Page 27

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