Chapter 4, Problem 2TP

Solution

To find: The difference between daily compounded interest and monthly compounded interest.

Option (A) is correct.

Given:

The principal amount $\left(P\right)=\$1000$ , annual interest rate $\left(r\right)=3\%$ , and time in(years) $t=2$.

Concept used:

If $P$ be the principal amount, r is the rate of interest (annually), and compounded $n$ times per year then the amount $A$ after $t$ years is given by the equation:

  $A=P{\left(1+\frac{r}{n}\right)}^{n\cdot t}$

Calculation:

Substitute the known values in the following data to compute interest.

  $\begin{array}{c}{A}_{daily}=P{\left(1+\frac{r}{n_{daily}}\right)}^{n_{daily}\cdot t}\\ =1000{\left(1+\frac{0.03}{365}\right)}^{\left(365\right)\cdot 2}\\ =1000{\left(1+0.000082\right)}^{730}\\ =1000{\left(1.000082\right)}^{730}\end{array}$

Therefore,

  $\begin{array}{c}\approx 1000\left(1.06183\right)\\ \approx \$1061.83\end{array}$

Similarly, solve the compound interest in monthly for two years.

  $\begin{array}{c}{A}_{monthly}=P{\left(1+\frac{r}{n_{monthly}}\right)}^{n_{monthly}\cdot t}\\ =1000{\left(1+\frac{0.03}{12}\right)}^{12\cdot 2}\\ =1000{\left(1+0.0025\right)}^{24}\\ =1000{\left(1.0025\right)}^{24}\end{array}$

Therefore,

  $\begin{array}{l}\approx 1000\left(1.06176\right)\\ \approx \$1061.76\end{array}$

Thus, the difference between the amount of compound daily from compound monthly.

  $\begin{array}{c}{A}_{difference}=A_{daily}-A_{monthly}\\ =1061.83-1061.76\\ =\$0.07\end{array}$

Conclusion:

Hence, the difference between the amount of compound interest daily from compound interest monthly is $\$0.07$.