Word Problem 16 (Population Increase and decrease)

Simple & Compound Interest 0

Problem 16: The population of a town increases at the rate of 5% every year. The population is 8000. In how many years it becomes 9261?

a. 2 years b. 3 years c. 4 years d.3-1/ 2 years

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Solution :

This problem involving compound interest.

Here we are given;

R=Increase rate=5%

P=Present Population= 8000

A= Increased Population=9261

T= time duration for increase=?

We will use the formula of compound interest calculation;

$A=P[1+\frac{R}{100}] ^{T}$

Putting values of variables in this formula;

$\Rightarrow 9261=8000[1+\frac{5}{100}] ^{T}$

$\Rightarrow 9261=8000[\frac{105}{100}] ^{T}$

$\Rightarrow 9261=8000[1.05] ^{T}$

$\Rightarrow \frac{9261}{8000}=[1.05] ^{T}$

$\Rightarrow 1.57625=[1.05] ^{T}$

Taking logarithm on both sides;

$\Rightarrow log(1.57625)=log [1.05] ^{T}$

$\Rightarrow log(1.57625)=T log [1.05]$

$\Rightarrow 0.0635=T \times 0.0211892$

$\Rightarrow T =\frac{0.0635628}{0.0211892}=3$

So, population became 9261 in time of three years.

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