Question

Anish took a loan of Rs.80,000 from a bank at 10% rate of interest. Find the difference in amount that he would be paying after 1.5 years if : i) the interest is compounded annually. ii) the interest is compounded half yearly.​

Answer 1

Given:-

• Principal = 80000
• Rate of interest = 10%
• Times = 1.5 years

To Find:-

• The amount after the interest is compounded annually.
• The amount after the interest is compounded half yearly.

Solution:-

(i) The amount when the interest is compounded annually.

We know,

The formula of amount:-

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

Hence,

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

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

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

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

= $\sf{A = 400 \times 11\times 21}$

= $\sf{A = 92400}$

Therefore, Amount after 1.5 years when the interest is compounded annually will be Rs.92400.

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(ii) The amount when the interest is compounded half-yearly.

We know,

The formula of amount when the interest is compounded annually:-

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

Now,

Since we are given the times as 1.5 years we can also write it like:-

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

Hence,

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

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

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

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

$\sf{A = 80000\times \dfrac{21}{20}\times \dfrac{21}{20}\times \dfrac{21}{20}}$

$\sf{A = 10\times 21\times 21\times 21}$

$\sf{A = 92610}$

Therefore, Amount after 1.5 years when the interest is compounded half-yearly will be Rs.92610.

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