If the compound interest on a sum at 14$\frac{2}{3}$% per annum for 3 years is ₹1,680, find the simple interest for the same period and rate.

The correct option is ₹1,456.

Step 1: Understand the given data

Time period ($t$) = 3 years
Compound Interest ($CI$) = ₹1,680
Rate of interest ($r$) = 14$\frac{2}{3}$% per annum

Step 2: Convert the rate of interest into a fraction

$r=14\frac{2}{3}\%=\frac{44}{3}\%=\frac{44}{3\×100}=\frac{11}{75}$

Step 3: Relate Amount, Principal, and Compound Interest

The formula for the total amount ($A$) under compound interest is:

$A=P{(1+\frac{r}{100})}^t$

Substituting $\frac{r}{100}=\frac{11}{75}$ and $t=3$:

$A=P{(1+\frac{11}{75})}^3=P{\left(\frac{86}{75}\right)}^3$

Compound Interest ($CI$) is given by $CI=A-P$:

$CI=P[{\left(\frac{86}{75}\right)}^3-1]$

Calculating the cubes:

${86}^3=636056$

${75}^3=421875$

So,

$CI=P[\frac{636056}{421875}-1]=P\left(\frac{636056-421875}{421875}\right)=P\left(\frac{214181}{421875}\right)$

We are given that $CI=1680$:

$1680=P\left(\frac{214181}{421875}\right)$

$P=\frac{1680\×421875}{214181}$

Step 4: Calculate the Simple Interest (SI)

The formula for Simple Interest is:

$SI=\frac{P\×r\×t}{100}$

Since $\frac{r}{100}=\frac{11}{75}$ and $t=3$:

$SI=P\&times;\frac{11}{75}\&times;3=P\&times;\frac{\mathrm{33<!--\; <mn-->75}}{}$

Substituting the expression for $P$:

$SI=\left(\frac{1680\&times;421875}{214181}\right)\&times;\frac{33}{75}$

Since $421875/75=5625$:

$SI=\frac{1680\&times;5625\&times;33}{214181}=\frac{311850000}{214181}\&approx;1456$

Thus, the simple interest for the same period and rate is ₹1,456.