Question

Suppose for the principal $P$, rate $R\%$ and time $T$, the simple interest is $S$ and compound interest is $C$.

Consider the possibilities. (i) $C>S$ (ii) $C=S$ (iii) $C<S$ Then:

• A. Only (i) is correct
• B. either (i) or (ii) is correct
• C. either (ii) or (iii) is correct
• D. Only (iii) is correct

Answer 1

The correct option is B

either (i) or (ii) is correct

Explanation for correct option:

Identify the relationship between compound interest and simple interest for certain amount with having rate and duration same

Case 1. $C>S$
This is possible if interest is compounded quarterly or half-yearly.

for example: If interest is compounded half early

Let, Principal ($P$) $=₹100$

Rate ($R$) $=10\%$

Time ($T$) $=1years$

Simple interest ($S$) $=\frac{P\times R\times T}{100}$

$=\frac{100\times 10\times 1}{100}$

$=₹10$

Since, Amount $=P{\left(1+\frac{{\displaystyle \frac{R}{2}}}{100}\right)}^{2T}$

$=100{\left(1+\frac{5}{100}\right)}^2$

$$\begin{gathered} =100{\left(\frac{105}{100}\right)}^2 \\ =₹110.25 \end{gathered}$$

Compound interest $C=\mathrm{Amount}–\mathrm{Principal}$

$C=110.25-100$

$C=₹10.25$

So, $C>S$

Case 2. $C=S$
This is possible if interest is compounded yearly.

For example:

Simple interest ($S$) $=\frac{P\times R\times T}{100}$

$=\frac{100\times 10\times 1}{100}$

$=₹10$

Since, Amount $=P{\left(1+\frac{R}{100}\right)}^T$

$=100{\left(1+\frac{10}{100}\right)}^1$

$=100\left(\frac{11}{10}\right)$

$=100\times \frac{11}{10}=₹110$

Compound interest $C=\mathrm{Amount}–\mathrm{Principal}$

$C=110-100$

$C=₹10$

So, $C=S$
Case 3. $C<S$
It can not be true because simple interest cannot be greater than compound interest.
Hence, Option B is correct option.