Word Problem 17 (Population of a town)

Problem 17 : During the first year the population of a town increases by 4% and during second year it diminished by 4%. If at the end of the second year its population was 2496 then initially the population was;

a. 2400          b. 2550             c. 25000         d. 2500

basic arithmetic test for jobs, workplace math tests –

basic mathematics mcqs with solution

Solution :

Its a problem of compound interest ! On population we apply compound interest calculator or compound interest formula here  – lets start!

Increase Case



Let us suppose;

y= population at the start of first year=?

R= rate of increase =4%

A=Population at the end of first year=?

T= Time duration in year=1


Since this is case of increase – we use +ve formula

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

\Rightarrow A=y[1+\frac{4}{100}] ^{1}

\Rightarrow A=[\frac{104}{100}]y


Decrease Case


Population diminish in the second year. Here,

P= [\frac{104}{100}]y (Since population at the end of the first year is now initial population for the second year ! )

R= Rate of decrease =4%

T= Time in years= 1

A=2496 (population at the end of second year)


Since this is case of decrease – we use -ve formula;

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

Putting values of variables for second year;

 \Rightarrow A=[\frac{104}{100} ]y[1-\frac{4}{100}] ^{1}

 \Rightarrow A=\frac{104y}{100} [ \frac{96}{100}]

But A=2496

 \Rightarrow 2496=\frac{104y}{100} [ \frac{96}{100}]

 \Rightarrow y =\frac{100}{96} \times \frac{100}{104} \times2496

 \Rightarrow y =2500

So, the population at the beginning of first year was 2500




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