Question

There is $60\% $ increase in an amount in 6 years at simple interest. What will be the compound interest on ${\text{Rs}}12,000$ after 3 years at the same rate?
A) \[{\text{Rs}}2160\]
B) \[{\text{Rs}}3120\]
C) \[{\text{Rs}}3972\]
D) \[{\text{Rs}}6240\]

Answer 1

Hint:
Here, we will use the amount formula and substitute the formula of simple interest in it. Then we will simplify the equation and substitute the given values in the equation to get the rate of interest. Then we will substitute the values of principal, rate of interest and time period in the formula of compound interest to get the required answer.

Formula Used:
We will use the following formulas:
1) $S.I = \dfrac{{P.R.T}}{{100}}$, where $S.I$ is the Simple Interest, $P$ is the Principal, $R$ is the rate of interest per annum and $T$ is the time period.
2) $C.I = P{\left( {1 + \dfrac{R}{{100}}} \right)^n} - P$ , where $C.I$ is the Compound Interest, $P$ is the Principal, $R$ is the rate of interest per annum and $n$ is the time period.

Complete step by step solution:
We know that the total amount, then it is equal to the sum of the principal and the interest taken together.
Amount, $A = P + S.I$
$ \Rightarrow A = P + \dfrac{{P \cdot R \cdot T}}{{100}}$
Substituting $S.I = \dfrac{{P.R.T}}{{100}}$ in the above equation, we get
Taking $P$ common, we get,
$ \Rightarrow A = P\left( {1 + \dfrac{{R \cdot T}}{{100}}} \right)$
Now, we are given that there is $60\% $ increase in an amount in 6 years at simple interest.
Hence, substituting Time, $T = 6$ years

And, if principal was $P = 100\% $, then, $60\% $ increase will make the amount equal to $\left( {100 + 60} \right) = 160\% $ of $P$
Therefore, we get,
$\dfrac{{160}}{{100}}P = P\left( {1 + \dfrac{{R \cdot \left( 6 \right)}}{{100}}} \right)$
$ \Rightarrow \dfrac{{160}}{{100}} = \left( {1 + \dfrac{{6R}}{{100}}} \right)$
We can write the LHS as:
$ \Rightarrow 1 + \dfrac{{60}}{{100}} = \left( {1 + \dfrac{{6R}}{{100}}} \right)$
Comparing the LHS and RHS, clearly,
$R = 10\% $
Hence, the rate of interest per annum, $R = 10\% $
Now, we are required to find the compound interest on ${\text{Rs}}12,000$ after 3 years at the same rate.
Hence, the formula of compound interest is $C.I = P{\left( {1 + \dfrac{R}{{100}}} \right)^N} - P$.
Now, substituting the Principal, $P = {\text{Rs}}12,000$, Time Period, $N = 3$ years and, Rate of interest per annum, $R = 10\% $, we get
$C.I = 12000{\left( {1 + \dfrac{{10}}{{100}}} \right)^3} - 12000$
Simplify the terms inside the bracket, we get
$ \Rightarrow C.I = 12000{\left( {\dfrac{{11}}{{10}}} \right)^3} - 12000$
Taking common and solving further,
$ \Rightarrow C.I = 12000\left[ {{{\left( {\dfrac{{11}}{{10}}} \right)}^3} - 1} \right]$
Applying the exponent on the terms, we get
$ \Rightarrow C.I = 12000\left[ {\dfrac{{1331}}{{1000}} - 1} \right]$
Taking the LCM, we get
$ \Rightarrow C.I = 12000\left[ {\dfrac{{1331 - 1000}}{{1000}}} \right] = 12000\left[ {\dfrac{{331}}{{1000}}} \right]$
Multiplying the terms, we get
$ \Rightarrow C.I = 12 \times 331 = 3972$
Therefore, the required compound interest on ${\text{Rs}}12,000$ after 3 years at the same rate is ${\text{Rs}}3972$

Hence, option C is the correct answer.

Note:
In this question, we have used the formula of Simple Interest as well as Compound Interest. Simple Interest is the interest earned on the Principal or the amount of loan. Its formula is, as we have discussed,
There is another type of interest, which is the Compound Interest. Compound Interest is calculated both on the Principal as well as on the accumulated interest of the previous year. Hence, this is also known as ‘interest on interest’.