Question

a man took a loan of ₹60000 from a bank. if the rate of interest is 10٪ pa, find the difference in amounts he would be paying after 1½ years if the interest is compounded 1. annually 2. half yearly​

Answer 1

Answer:

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Answer 2

Given:-

• Principal = Rs60000
• Rate = 10% p.a.
• Times = 1½ years.

To find:-

• The difference between the amounts the man will have to pay if the interest is compounded annually and half-yearly.

Solution:-

1. Interest compounded annually,

We know,

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

Hence,

A = $\sf{60000\bigg(1+\dfrac{10}{100}\bigg)^1\bigg(1+\dfrac{10}{200}\bigg)^{2\times\dfrac{1}{2}}}$

A = $\sf{60000\bigg(1+\dfrac{1}{10}\bigg)^1\bigg(1+\dfrac{1}{20}\bigg)^1}$

=> A = $\sf{60000\bigg(\dfrac{10+1}{10}\bigg)^1\bigg(\dfrac{20+1}{20}\bigg)^1}$

=> A = $\sf{60000\bigg(\dfrac{11}{10}\bigg)\bigg(\dfrac{21}{20}\bigg)}$

=> A = $\sf{69300}$

Therefore, the amount when the interest is compounded annually is Rs.69300

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2. Interest compounded half-yearly,

Here let us convert the time into improper fraction,

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

We know,

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

Hence,

A = $\sf{60000\bigg(1+\dfrac{10}{200}\bigg)^{2\times \dfrac{3}{2}}}$

=> A = $\sf{60000\bigg(1+\dfrac{1}{20}\bigg)^3}$

=> A = $\sf{60000\bigg(\dfrac{20+1}{20}\bigg)^3}$

=> A = $\sf{60000\bigg(\dfrac{21}{20}\bigg)\bigg(\dfrac{21}{20}\bigg)\bigg(\dfrac{21}{20}\bigg)}$

=> A = 7.5 × 21 × 21 × 21

=> A = 69457.5

Therefore the amount when the interest is compounded half-yearly is Rs.69457.5

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Now,

Difference in both the amounts:-

= Rs.(69457.5 - 69300)

= Rs.157.5

Therefore, the difference between the amount when the interest is compounded annually and half-yearly is Rs.157.5

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Explore More!!

The formula used to find the amount when the interest is compounded annually:-

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

The formula used to find amount when the interest is compounded half-yearly:-

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

The formula used to find amount when the interest is compounded quarterly:-

• $\sf{A = P\bigg(1+\dfrac{r}{400}\bigg)^{4n}}$

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