"> Word Problem 6 | Job Exam Rare Mathematics (JEMrare)

Word Problem 6

Problem 6: A man invests a certain amount in a project at 4% per anum simple interest. After 5 years he gets Rs 3600. What amount he invested?



Solution:

Suppose investment is ‘x” i-e initial amount P;

P = x Rs

Time Period of investment;

T= 5years

Rate of interest R;

R = 4% annual

We know relation or formula for interest;

I = \frac{P\times R\times T}{100} ----------(1))

Put values;

I =\frac{ x \times 4 \times 5}{100}

\Rightarrow I =\frac{ 20x}{100}

Also we know ;

A = P+I

But A = 3600

Put values of ‘A’, ‘P’ and ‘I”

3600= x+ \frac{20x}{100}

Solving for ‘x’

\Rightarrow 3600= \frac{100x+20x}{100}

\Rightarrow 3600= \frac{120x}{100}

\Rightarrow x = \frac{3600\times 100}{120}

\Rightarrow x = 3000

So, he invested Rs 3000/-

 

 

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