Question

Arif took a loan of 80,000 from a bank. If the rate of interest is 10% per annum find the difference in amounts he would be paying after 1 1/2 years if the interest is
(i) compounded annually."
(i) compounded half yearly.​

Answer 1

Given:-

• Principal = 80000
• Rate = 10%
• Time = $\sf{1\dfrac{1}{2}\:years}$

To Find:-

• The difference between the amounts he would pay amount after 1 and half year if the interest is:-
• (i) compounded annually
• (ii) compounded half-yearly

Solution:-

(i) compounded annually

We know,

$\sf{A = P\bigg(1+\dfrac{r}{100}\bigg)^n}$

Hence,

$\sf{A = 80000\bigg(1+\dfrac{10}{100}\bigg)^1\bigg(1+\dfrac{10}{200}\bigg)^{2\times\dfrac{1}{2}}}$

= $\sf{A = 80000\bigg(1+\dfrac{1}{10}\bigg)^1\bigg(1+\dfrac{1}{20}\bigg)^1}$

= $\sf{A = 80000\bigg(\dfrac{10+1}{10}\bigg)^1\bigg(\dfrac{20+1}{20}\bigg)^1}$

= $\sf{A = 80000\bigg(\dfrac{11}{10}\bigg)\bigg(\dfrac{21}{20}\bigg)}$

= $\sf{A_1 = 92400}$

Therefore the amount after 1 anda half year if the interest is compounded annually will be Rs.92400.

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(ii) Compounded half-yearly

Let us first convert the time into it's improper form,

$\sf{1\dfrac{1}{2} = \dfrac{3}{2}}$

We know,

$\sf{A = P\bigg(1+\dfrac{r}{200}\bigg)^{2n}}$

Hence,

$\sf{A = 80000\bigg(1+\dfrac{10}{200}\bigg)^{2\times \dfrac{3}{2}}}$

= $\sf{A = 80000\bigg(1+\dfrac{1}{20}\bigg)^3}$

=> $\sf{A = 80000\bigg(\dfrac{20+1}{20}\bigg)^3}$

=> $\sf{A = 80000\bigg(\dfrac{21}{20}\bigg)^3}$

=> $\sf{A = 80000\bigg(\dfrac{21}{20}\bigg)\bigg(\dfrac{21}{20}\bigg)\bigg(\dfrac{21}{20}\bigg)}$

=> $\sf{A_2 = 92610}$

Therefore the amount after 1 and half year of the amount is compounded half-yearly will be Rs.92610.

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Now Let us find the difference.

Difference:-

Since A₂ > A₁

Therefore,

A₂ - A₁

= 92610 - 92400

= 210

Therefore the difference between the amounts if the compound interest is compounded annually and half-yearly will be of Rs.210.

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