Question

The compound interest on Rs. 30,000 at 7% per annum Rs. 4347. The period (in years) is
(a)2
(b)$2\dfrac{1}{2}$
(c) 3
(d)4

Answer 1

Hint: We have to use the formula used to calculate compound interest on the principle

amount which is given by $CI=P\left[ {{\left( 1+\dfrac{r}{100n} \right)}^{nt}}-1 \right]$ where

CI= Compound interest

P= Principle amount

r= rate of interest (in percentage)

n= number of times interest is compounded in a year

t= time (in years)

Complete step by step answer:
In the question we are given all of the information except ‘t’ which we have to calculate.

Hence, we can use the following formula to solve this question.

$CI=P\left[ {{\left( 1+\dfrac{r}{100n} \right)}^{nt}}-1 \right]$

From the question we have,

P=30,000

r=7%

n=1 (since it is written rate is 7% per annum means interest is compounded once in a year)

CI=4,347

On substituting these values in the question we get,

$CI=P\left[ {{\left( 1+\dfrac{r}{100n} \right)}^{nt}}-1 \right]\Rightarrow 4347=30000\left[

{{\left( 1+\dfrac{7}{100\times 1} \right)}^{1\times t}}-1 \right]$

On simplifying we get,

$4347=30000\left[ {{\left( 1+\dfrac{7}{100\times 1} \right)}^{1\times t}} \right]-30000$

Adding 30000 both sides we get,

$\begin{align}

  & 34347=30000{{\left( 1+\dfrac{7}{100} \right)}^{t}} \\

 & \Rightarrow {{\left( 1+\dfrac{7}{100} \right)}^{t}}=\dfrac{34347}{30000} \\

\end{align}$

On further simplification we get,

$\begin{align}

  & {{(1+0.07)}^{t}}=1.1449 \\

 & \Rightarrow {{(1.07)}^{t}}=1.1449 \\

\end{align}$

Taking log both sides we get,

$\log {{(1.07)}^{t}}=\log 1.1449$

Since, t is the power of 1.07 we can take it outside of logarithm

$\begin{align}

  & t\log 1.07=\log 1.1449 \\

 & \Rightarrow t=\dfrac{\log 1.1449}{\log 1.07} \\

\end{align}$

Now we need to use a calculator to calculate this value. Therefore, we have $t=2.00$

Since, it was time in years therefore the period is 2 years.

Hence, option (a) is the correct answer.

Note: The formula of CI originally comes from $CI=A-P$ and $A=P{{\left( 1+\dfrac{r}{100n} \right)}^{nt}}$ . When we put this value of Amount in $CI=A-P$ we get the formula used above. We can see that the amount 30,000 is not written as principal amount and only after thoroughly understanding the question we can write what is the amount, principal amount and compound interest in the question.