Question

Meena borrowed Rs. 50,000 at 20% p.a. compounded half-yearly. What amount of money will discharge her debt after $1\dfrac{1}{2}$ years? \[\]
A.Rs. 28275\[\]
B. Rs. 28275\[\]
C. Rs. 66550\[\]
D. Rs. 56275\[\]

Answer 1

Hint: We use the formula of compound interest the accumulated interest $A=P{{\left( 1+\dfrac{r}{100} \right)}^{nt}}$ where original principal sum is $P$, the overall compound period is $t$, the rate of interest per time period as $r$, the number of compound periods or frequency as $n$. We find the interest rate $r$ for half the year by dividing the given interest per annum by 2. We find compound frequency $n$ by dividing given time period in years by $\dfrac{1}{2}$.\[\]

Complete step by step answer:
We find from the given data in the question that Meena borrowed Rs. 50,000 at 20% p.a. compounded half-yearly. We are asked to find the amount that will discharge her dept after $1\dfrac{1}{2}$ years. The amount to discharge her is the same amount she has to pay the loaner with compound interest. \[\]
We have to calculate the accumulated amount $A$ compound period $t$ where we are given the principal $P=50000$ rupees, time period is $t=1$ half-year, the rate of interest per half year is half of rate of interest per annum that is $r=\dfrac{20\%}{2}=10%$ and the number of compound periods is $n=1\dfrac{1}{2}\div \dfrac{1}{2}=3$ half-years. Putting all these values in the compound interest formula we have
\[\begin{align}
  & A=P{{\left( 1+\dfrac{10}{100} \right)}^{nt}} \\
 & \Rightarrow A=50000{{\left( \dfrac{11}{10} \right)}^{3\times 1}} \\
 & \Rightarrow A=50000{{\left( \dfrac{11}{10} \right)}^{3}} \\
 & \Rightarrow A=50000\times \dfrac{1331}{1000} \\
 & \Rightarrow A=66550 \\
\end{align}\]

So, the correct answer is “Option C”.

Note: We must be careful of confusion between simple and compound interest. The compound interest is the concept in depositing and borrowing where the interest is added into the principal and in the next compound period the new principal is the sum of old principal and accumulated interest. We can alternatively solve taking logarithm both side of the compound interest formula $\log A=\log \left[ P{{\left( 1+\dfrac{r}{n} \right)}^{nt}} \right]$. The compound interest $I$ generated after a time period $nt$ is given by $I=A-P$.