d Word Problems 1 | Job Exam Rare Mathematics (JEMrare) - Solved MCQs
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Word Problems 1

Problem 1: During one year the population of a town increased 5% and during the next it decrease 5%. At the end of the second year the population was 49875. What was the population at the beginning of the first year?



Solution:

Let Population At Beginning of 1st Year=x

We use here formulas of compound interest for increment and decrement.

for increment the formula is;

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

and for decrement the formula is ;

A= P (1-\frac{R}{100})^{T}----(2)

1st Year

Data for first year can be arranged as under;

Principal population P = x

Rate of increase R=5%

Term T = 1 Year

Population at the end of 1st Year A=?

As population increase during first year, put these values in formula(1) above;

A= x(1+\frac{5}{100})^{1}

\Rightarrow A= x+\frac{x}{20}

So Population at the end of 1st year is   x+\frac{x}{20} whch will  now be the principal population for 2nd year:

2nd Year

Principal population P= x+\frac{x}{20}

Rate of decrease R = 5%

Term T = 1 year

Population at the end of 2nd year A =?

As population decreases during 2nd year, put these values in formula (2) above;

A= x+\frac{x}{20}(1-\frac{5}{100})^{1}

A= x+\frac{x}{20}-(x+\frac{x}{20})(\frac{5}{100})}

A= x+\frac{x}{20}-\frac{5x}{100}-\frac{5x}{2000}}

A= \frac{2000x+100x-100x-5x}{2000}

A= \frac{2000x-5x}{2000}

A= \frac{1995x}{2000}

Thus population at the end of 2nd year is  \frac{1995x}{2000}. But population as given at the end of 2nd year is 49875

So,

\frac{1995x}{2000}=49875

\Rightarrow x=\frac{2000}{1995}\times 49875

\Rightarrow x=50000

So population in the beginning was 50000

 

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