Question

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

Answer 1

$Principle=Rs80,000$

$Rate=10\%$

$T=Semiannual=\frac{1}{2}year$

$Compound\mathrm{int}erest=P\left[1+\frac{R}{100}\right]$

$=80,000{\left[1+\frac{10}{100}\right]}^{\frac{1}{2}}......\left(1\right)$

$=80,000{\left[\frac{11}{10}\right]}^{\frac{1}{2}}$

$Payingafter\frac{3}{2}years$

$T=\frac{3}{2}years$

$C.I.=80,000{\left[1+\frac{10}{100}\right]}^{\frac{3}{2}}........\left(2\right)$

$Differenceinammountsbe$

$80,000{\left[1+\frac{10}{100}\right]}^{\frac{3}{2}}-80,000{\left[1+\frac{10}{100}\right]}^{\frac{1}{2}}$

$80,000{\left[\frac{11}{10}\right]}^{\frac{3}{2}}-80,000{\left[\frac{11}{10}\right]}^{\frac{1}{2}}$

$80,000\left[{\left(\frac{11}{10}\right)}^{\frac{3}{2}}-{\left(\frac{11}{10}\right)}^{\frac{1}{2}}\right]$

$80,000{\left(\frac{11}{10}\right)}^{\frac{1}{2}}\left[\frac{11}{10}-1\right]$

$80,000{\left(\frac{11}{10}\right)}^{\frac{1}{2}}\left(\frac{1}{10}\right)$

$80,000\times \sqrt{\frac{11}{10}}$

$8000\times 1.048$

$8384$