"> Word Problem 8 | Job Exam Rare Mathematics (JEMrare)

Word Problem 8

Problem 8: Two persons namely A and B borrow Rs 4000 at 10 percent interest pa. A borrowed money at the compound interest rate and B at simple interest basis. In both the cases, the interest was calculated on half yearly basis. Who will pay back more than the other, and what amount as extra ?



Solution: 

Two persons are taking loan here and they have to pay back loan with interests – who will pay more interest — is the question!

In this problem one person is applied with Simple Interest and other one is implied with Compound Interest. Moreover, note that interests are being calculated on half yearly basis i-e in one year 2 terms are made for interests calculation.

Person A

Initial amount ;

P= 400

Rate of interest is 10% for one year , so for half year the rate will be 5% i-e

R=5

There are two terms of half year each in one year , so;

T=2

We know  formula of simple interest (SI);

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

I = \frac{400\times 5\times 2}{100}

I = 40

So, the interest which ‘A’ will pay is Rs 40/-

Person B

Same data for ‘B’ as well, i-e

Initial amount ;

P= 400

Rate of interest is 10% for one year , so for half year the rate will be 5% i-e

R=5

There are two terms of half year each in one year , so;

T=2

But he is implied with Compound Interest (CI)

We know Amount  formula for compound interest;

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

Put above values in this;

\Rightarrow A= 400(\frac{100+5}{100 }) ^2

\Rightarrow A= 400(\frac{105}{100 }) ^2

\Rightarrow A= 400(\frac{105\times 105}{100\times 100 })

\Rightarrow A= 4(\frac{105\times 105}{100 })

\Rightarrow A= (\frac{44100}{100 })

\Rightarrow A= 441

Now we know

A=P+I

Put value of ‘A’ and ‘P’ from above, we get;

441=400+I

\Rightarrow I = 41

So, ‘A’ will pay interest of 40 Rs and ‘B’ will pay more than ‘A’ the interest of 41 Rs

Or we can say ‘B’ is paying back Re=1/- more than ‘A’

 

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