Question

If a compound interest on a certain sum for 2 years is Rs.21. What could be the simple interest?
A. Rs.20
B. Rs.16
C. Rs.18
D. Rs.20.50

Answer 1

Hint: Compound interest and simple interest is given as:
 \[C.I. = P{\left( {1 + \dfrac{r}{{100}}} \right)^n} - P \]

 \[S.I. = \dfrac{{P \times R \times T}}{{100}} \]

Where, P is the principal amount, r is interest rate in % and n is time period and similarly, R and T in the second expression are also representing the interest rate and time period in years. Use the above identities to get the required answer.
Don't try to get the exact value of rate of interest or principal amount, just try to observe the relation calculated from the compound interest formula and get S.I. directly with that relation by writing 21 as \[10 \times 2 + 1 \] .

Complete step-by-step answer:
As we know the formula of compound interest is given as:
 \[C.I. = P{\left( {1 + \dfrac{r}{{100}}} \right)^n} - P \] . . . . . . . . . . . .(i)

Where, P is the principal amount, r is the interest rate in percentage for time n years.
And similarly, simple interest is given by the relation:
 \[S.I. = \dfrac{{P \times R \times T}}{{100}} \] . . . . . . . . . . . .(ii)
Where, P is the principal amount at which interest rate R% is acting for a period of time T years.
So, now coming to the question, we are given compound interest on a certain amount for 2 years as Rs.2 and hence, we need to find simple interest on the same amount for the same interest and time as well.
So, let 'P' be the sum for which C.I. is given in the problem. Hence, using equation (i), we have:
 \[\begin{array}{l}n = 2\,years\\C.I. = 21\,Rs.\\r = ?\end{array} \]
So, we get:
 \[C.I. = P{\left( {1 + \dfrac{r}{{100}}} \right)^n} - P \]
 \[21 = P{\left( {1 + \dfrac{r}{{100}}} \right)^2} - P \] . . . . . . . . . . . .(iii)
Now, as we know the algebraic identity of \[{\left( {a + b} \right)^2} \] is given as:
 \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab \]

So, expanding \[{\left( {1 + \dfrac{r}{{100}}} \right)^2} \] we get:
\[\begin{array}{l}21 = P\left( {1 + \left( {\dfrac{r}{{100}}} \right) + \dfrac{{2r}}{{100}}} \right) - P\\21 = P + \dfrac{{P{r^2}}}{{10000}} + \dfrac{{2rP}}{{100}} - P\end{array}\]
Or \[21 = \dfrac{{2\Pr }}{{100}} + \dfrac{{P{r^2}}}{{10000}}\]
Or \[\dfrac{{{\mathop{\rm P}\nolimits} \left( {2r} \right)}}{{100}} + \dfrac{{P{r^2}}}{{10000}} = 21\] . . . . . . . . . . . .(iv)
Now, if we break 21 as \[10 + 10 + 1 \] i.e. \[\left( {2 \times 10 + 1} \right) \] , then it gives two equal parts 10 and another 1. And hence, we can write equation (iv) as:

\[\dfrac{{2\Pr }}{{100}} + \dfrac{{P{r^2}}}{{10000}} = 10 \times 2 + 1\]

So, on comparison, we get:
\[\dfrac{{2\Pr }}{{100}} = 20\,\,and\,\,\dfrac{{P{r^2}}}{{10000}} = 1\] . . . . . . . . . . . .(v)
 \[S.I. = \dfrac{{P \times 2 \times r}}{{100}} = \dfrac{{2\Pr }}{{100}} \]

Hence, we can put the value of \[\dfrac{{2\Pr }}{{100}} \] as 20 from equation (v). So, we get:
 \[S.I. = 20\,\,Rs. \]

Hence, option (A) is the correct answer.

Note: One may try to get exact value of r or P from the equation \[\dfrac{{{\mathop{\rm P}\nolimits} \left( {2r} \right)}}{{100}} + \dfrac{{P{r^2}}}{{10000}} = 21\] which can be a complex approach. We can find the value of S.I. for the given criteria by writing 21 as \[20 \times 2 + 1 \] , which is the key point of the question.
One may go wrong if she/he uses formula to calculate C.I. using formula
 \[C.I. = P{\left( {1 + \dfrac{r}{{100}}} \right)^n} \] , which is wrong. So, take care of the formula for C.I. as well, which is given as \[C.I. = P{\left( {1 + \dfrac{r}{{100}}} \right)^n} - P \] .