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


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


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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