Question

Arif took a loan of Rs 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after ${\displaystyle 1\frac{1}{2}}$ years if the interest is

(1) Compounded annually

(2) Compounded half yearly

Answer 1

1) P = Rs 80,000

R = 10% per annum

n =  ${\displaystyle 1\frac{1}{2}}$ years

The amount for 1 year and 6 months can be calculated by first calculating the amount for 1 year using the compound interest formula, and then calculating the simple interest for 6 months on the amount obtained at the end of 1 year.

Firstly, the amount for 1 year has to be calculated.

A = Rs ${\displaystyle \left[80000{(1+\frac{10}{100})}^1\right]}$

= Rs ${\displaystyle \left[80000(1+\frac{1}{10})\right]=Rs(80000\times \frac{11}{10})=Rs88000}$

By taking Rs 88,000 as principal, the SI for the next ${\displaystyle \frac{1}{2}}$ year will be calculated.

S.I = ${\displaystyle \frac{\text{P x R x T}}{100}}$ = Rs ${\displaystyle \left(\frac{88000\times 10\times \frac{1}{2}}{100}\right)}$  = Rs 4400

Interest for the first year = Rs 88000 − Rs 80000 = Rs 8,000

And interest for the next ${\displaystyle \frac{1}{2}}$ year = Rs 4,400

Total C.I. = Rs 8000 + Rs 4,400 = Rs 1,2400

A = P + C.I. = Rs (80000 + 12400) = Rs 92,400

2) The interest is compounded half yearly.

Rate = 10% per annum = 5% per half year

There will be three half years in ${\displaystyle 1\frac{1}{2}}$ years.

A = Rs${\displaystyle \left[80000{(1+\frac{5}{100})}^3\right]=Rs\left[80000{(1+\frac{1}{20})}^3\right]}$

= Rs${\displaystyle (80000\times \frac{21}{20}\times \frac{21}{20}\times \frac{21}{20})}$ = Rs 92610

Difference between the amounts = Rs 92,610 − Rs 92,400 = Rs 210