Word Problem 14

Problem 14: The population of a town increases by 5% annually and its present population is 64000. The population after 3 years time will be;

a. 74088      b. 74008    c. 74808     d. 78004

Math for job test -

basic mathematics mcqs with solution


This looks as percent problem - but if seen carefully it is  interest calculation problem

We are given following information;

Increase rate = 5%

Present Population = 64000

Time Duration = 3 Years

Population after 3 years =?

To use the interest formula;

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

We assume here that;

A= Population after 3 years=?

P= Present Population=64000

R= Rate of increase=5

T= 3 (years time)=3

\Rightarrow A=64000[1+\frac{5}{100}] ^{3}

\Rightarrow A=64000[1+\frac{1}{20}] ^{3}

\Rightarrow A=64000[\frac{21}{20}] ^{3}

\Rightarrow A=64000[\frac{21\times21\times21}{20\times20\times20}]

\Rightarrow A=74088

So, the population after three years will be 74088





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