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Word Problem 11

Problem 11: The population of a town increases at the rate of 5% yearly. The present population is 9261. Find the population three years ago.

a) 10000                         b) 9000                             c) 8000                              d) 7500

Solution:

simple arithmetic questions and answers

This problem is to be solved using compound interest formula , how? If we assume population as an amount then we can say ” if present amount is 9261, what was the principal amount three years ago if increment has been at the rate of 5% per annum?”

We arrange data as below;

A = 9261

R=5% pa

T=3 years

P=?

using formula of compound interest;

A=P (1+\frac{R}{100})^{T}

Putting values from above data;

\Rightarrow 9261=P (1+\frac{5}{100})^{3}

\Rightarrow 9261=P (\frac{105}{100})^{3}

\Rightarrow 9261=P (\frac{21}{20})^{3}

\Rightarrow 9261=P (\frac{21\times21\times21}{20\times20\times20})

\Rightarrow P=(\frac{20\times20\times20}{21\times21\times21})\times 9261

\Rightarrow P=8000

So, the amount (population)  three years ago was 8000,

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