Question

There is $60\%$ increase in an amount in $6years$at simple interest. What will be the compound interest of$Rs12000$after $3years$ at the same rate?

Answer 1

Step 1: Given Information

$60\%$ increase in the amount in $6Years$ at Simple Interest

find CI of $Rs12000$after $3years$ at the same rate

We know that,

$SI=\frac{(PxRxT)}{100}$

Where,

$$\begin{gathered} SI=SimpleInterest \\ P=Principle \\ R=RateofInterest \\ T=Timeperiod \end{gathered}$$

Now,

$A=amount$

$A=SI+P$

Step 2: Find the amount

Let $RsP$be the Principal

$A=P+60\%ofP$ ( As per given)

$A=P+\left(\frac{60}{100}\right)P$

$A=\frac{100P+60P}{100}$ ( On taking LCM )

$A=\frac{160P}{100}$ (Simplify $\frac{160P}{100}$by dividing by a common factor $20$in numerator and denominator separately)

$A=\frac{8P}{5}$--------------- (i)

Step 3: Find the rate of interest $R$

$A=SI+P$

$\frac{8P}{5}=$$SI+P$ ( Put the value of $A$from (i) )

$\frac{8P}{5}=$$\frac{(PxRxT)}{100}+P$( Use Formula Of $SI$)

$\frac{8P}{5}=$$\frac{(PxRx6)}{100}+P$( Put the value of time $6Years$)

$\frac{8P}{5}=$$P\left\{\frac{1\times R\times 6}{100}+1\right\}$ ( Taking $P$common from both sides)

$\frac{8P}{5P}=\frac{R\times 6}{100}+1$

$\frac{8}{5}-1=\frac{R\times 6}{100}$

$\frac{8-5}{5}=\frac{R\times 6}{100}$

$\frac{3}{5}\times \frac{100}{6}=R$

$10\%=R$

Or, $R=10\%$----------- (ii)

Step 4: Find Compound Interest

For Compound Interest

$A=P{\left(1+\frac{R}{100}\right)}^n$where $n=timeperiod$

$A=12000{\left(1+\frac{10}{100}\right)}^3$ {putting the value of R from equation (ii)}

$A=12000{\left(1+\frac{1}{10}\right)}^3$

$A=12000{\left(\frac{11}{10}\right)}^3$

$A=12000{\left(\frac{11\times 11\times 11}{10\times 10\times 10}\right)}^{}$

$A=12\times 1331$

$A=₹15,972$

$CompoundInterest\left(CI\right)=Amount\left(A\right)–Principal\left(P\right)$

$CI=15972–12000$

$CI=₹3972$

Hence, $CI=₹3972$ for $3years$at the same rate.