A car loan is offered at $8\% $ annual interest, compounded annually. After the first year, the interest due is $\$ 240$. What is the principal on the loan?
A) $\$ 2400$
B) $\$ 3000$
C) $\$ 4200$
D) $\$ 3500$

Hint: First, assume the principal on the loan.
Evaluate the compound interest by the following formula:
$I = P\left[ {{{\left( {1 + \dfrac{r}{{100}}} \right)}^t} - 1} \right]$
Substitute all the values in the formula and evaluate the principal.

Complete step by step answer:
We are given that a car loan is offered at $8\% $ annual interest and interest is compounded annually. After a year, the interest due is $\$ 240$.
We have to find the principal on the loan.
Let the principal be $P$.
Time and rate are given so we use the formula for calculating the compound interest.
The compound interest by the following formula:
$I = P\left[ {{{\left( {1 + \dfrac{r}{{100}}} \right)}^t} - 1} \right]$
Here, $I$is the compound interest, $P$is the principal, $r$is the rate per annum and $t$ is the time in years.
We have interest due is $\$ 240$, therefore, $I = 240$
The rate of interest annually is $8\% $, therefore, $r = 8$
The time is one year, therefore, $t = 1$
Substitute all the values in the given formula and evaluate the value of $P$.
$
  240 = P\left[ {{{\left( {1 + \dfrac{8}{{100}}} \right)}^1} - 1} \right] \\
  240 = P \times \dfrac{8}{{100}} \\
  P = \$ 3000 \\
 $
Hence, the principal value on the loan for the car is $\$ 3000$.
Therefore, option (B) is correct.

Note: We can use another for this particular question which is shown below:
After one year, the compound interest and the simple interest on the principal value is the same. From second year onwards the value of compound interest is greater than the simple interest.
So, we can use a simple interest formula for this question because time in one year.
The formula is,
$SI = \dfrac{{PRT}}{{100}}$
Substitute $SI = 240,T = 1$and $R = 8$to evaluate the value of $P$.
$
  240 = \dfrac{{P \times 8 \times 1}}{{100}} \\
  P = \$ 3000 \\
 $
Hence, the principal value is $\$ 3000$.