Find the Compound Interest on a sum of Rs.1000 at the rate of 10% per annum for 1.5 years when interest is compounded half-yearly

Principal(P)  = Rs 4000

Rate of Interest(r )  = 10%

Time = 2.5 years Here first we take n = 2 years.

A = P( 1 + r /100)n  Where  A = Amount  and n = number of years

A = 4000 (11/10)2

A = 4000 ( 121 / 100 )

A = Rs 4840

The Amount after two years = Rs 4840 Now Principal = Rs 4840

Simple interest for last ½ year is = PRT/100  =  (4840 × 10 × 1 / 2) / 100 = Rs.242

Amount after 2 ½ year =  Rs 4840 +  Rs 242 = Rs 5082

C.I = A - P

C.I = Rs 5082 - Rs 4000 = Rs1082

∴ The compound interest is Rs 1082.

Amount after 11⁄2 years when interest is compounded yearly

$=5000\times {(1+\frac{4}{100})}^1\times (1+\frac{\frac{1}{2}\times 4}{100})=5000\times \frac{104}{100}\times (1+\frac{2}{100})=5000\times \frac{104}{100}\times \frac{102}{100}=50\times 104\times \frac{51}{50}=104\times 51=\text{Rs.}5304$

Compound Interest for 1 1⁄2 years when interest is compounded yearly

= Rs.(5304 - 5000)

Amount after 11⁄2 years when interest is compounded half-yearly

$=\text{P}{(1+\frac{\text{(R/2)}}{100})}^{\text{2T}}=5000{(1+\frac{\left(4/2\right)}{100})}^{2\times \frac{3}{2}}=5000{(1+\frac{2}{100})}^3=5000{\left(\frac{102}{100}\right)}^3=5000\left(\frac{102}{100}\right)\left(\frac{102}{100}\right)\left(\frac{102}{100}\right)=50\times 102\times \frac{51}{50}\times \frac{51}{50}=102\times 51\times \frac{51}{50}=51\times 51\times \frac{51}{25}=\text{Rs.}5306.04$

Compound Interest for 1 1⁄2 years when interest is compounded half-yearly = Rs.(5306.04 - 5000) Difference in the compound interests = (5306.04 - 5000) - (5304 - 5000) = 5306.04 - 5304 = Rs. 2.04