Question

If the interest is payable quarterly, then find out after what time will Rs.1600 amount to 1852.20 at 20%.
A. 3 months \[\]
B. 6 months\[\]
C. 9 months\[\]
D. 1 year\[\]

Answer 1

Hint: Use the compound interest formula using the given data from the question. The compound interest is the concept where the interest is added into the principal and in the next compound period, the new principal is the sum of the old principal and accumulated interest.

Complete step-by-step solution:
We shall denote the principal as $P$, the compound period as $t$, the rate of interest as $r$, the number of compound periods as $n$, and the amount accumulated after the time period $nt$ as $A$. Then we know from the formula of compound interest,
\[A=P{{\left( 1+\dfrac{r}{n} \right)}^{nt}}\]
We find from the given data in the question that we have to calculate compound period $t$ where the principal$P=1600$, the rate of interest $r=5%=.05$, the number of compound periods $n=1$ quarter and finally the accumulated amount after time period $nt$, $A=1852.20$. Putting all these values in the compound interest formula
\[ 1852.20=P{{\left( 1+\dfrac{r}{n} \right)}^{nt}}=1600{{\left( 1+\dfrac{0.20}{4} \right)}^{4t}}\\
  \Rightarrow 1852.20=1600{{\left( 1+.05 \right)}^{4t}} \\
  \Rightarrow {{\left( 1.05 \right)}^{4t}}=\text{1}.\text{157625=}{{\left( 1.05 \right)}^{3}} \\
  \Rightarrow t=\dfrac{3}{4} \]
So the time period is three-fourth of a year. So the time in months is $\dfrac{3}{4}\times 12=9$ months.
So the correct choice out of the given options is C. 9 months

Note: The question tests your knowledge of compound interest. We need to be careful of while substituting the required values in the formula. Similarly, we can also determine other unknowns in the compound interest formula if data is given sufficiently. Here we can use the alternative method
Taking logarithm both side of the compound interest formula,
\[\log A=\log \left[ P{{\left( 1+\dfrac{r}{n} \right)}^{nt}} \right]\]
Using logarithmic identities we expand above,
\[\log A=\log \left[ P{{\left( 1+\dfrac{r}{n} \right)}^{nt}} \right]=\log P+\log {{\left( 1+\dfrac{r}{n} \right)}^{nt}}=\log P+nt\log \left( 1+\dfrac{r}{n} \right)\]
We know that $\log 1=0$,
\[\log A=\log P+nt\log \dfrac{r}{n}\Rightarrow t=\dfrac{\log A-\log P}{n\log \dfrac{r}{n}}=\dfrac{\log \left( \dfrac{A}{P} \right)}{n\log \dfrac{r}{n}}\]
Putting necessary values,
\[t=\dfrac{\log \left( \dfrac{1852.20}{1600} \right)}{4\log \dfrac{.20}{4}}=0.75\]
So the time period is $0.75\times 12=9$ months.