Chapter 3, Problem 5P

To determine

The interest and total amount due at the end of the loan for both simple and compound interest.

Answer to Problem 5P

The interest and total amount due at the end of the loan for both simple and compound interest is shown in the table below.

	$SI$	$A_{\text{s}}$	$CI$	$A_{\text{c}}$
(a)	$\$100$	$\$1100$	$\$1102.50$	$\$102.5$
(b)	$\$450$	$\$1950$	$\$2007.34$	$\$507.34$
(c)	$\$10000$	$\$20000$	$\$25937.42$	$\$15937.42$
(d)	$\$562500$	$\$81250$	$\$203426.54$	$\$178426.54$
(e)	$\$191000$	$\$238750$	$\$1830620.40$	$\$1782870.40$

Explanation of Solution

Given:

	Loan	Years	Rate
(a)	$\$1000$	$2$	$5\%$
(b)	$\$1500$	$5$	$6\%$
(c)	$\$10,000$	$10$	$10\%$
(d)	$\$25,000$	$15$	$15\%$
(e)	$\$47,750$	$20$	$20\%$

Concept used:

Write the expression to calculate simple interest.

$SI=\frac{P\times R\times T}{100}$ ...... (I)

Here, simple interest is $SI$, principal amount is

$P$, interest rate is

$R$ and time period is $T$.

Write the expression to calculate the amount due to simple interest.

$A_{\text{s}}=P+SI$ ...... (II)

Here, amount due to simple interest is $A_{\text{s}}$.

Write the expression to calculate compound interest.

$CI=P{\left(1+\frac{R}{100}\right)}^T$ ...... (III)

Here, compound interest is $CI$.

Write the expression to calculate the amount due to compound interest.

$A_c=CI-P$ ...... (IV)

Here, the amount due to compound interest is $A_{\text{c}}$. $R$.

Calculation:

Calculate the simple interest.

Substitute $\$1000$ for $P$, $5$ for and $2$ for $T$ in Equation (I).

$\begin{array}{c}SI=\frac{\$1000\times 5\times 2}{100}\\ =\$100\end{array}$

Calculate the amount due to simple interest.

Substitute $\$100$ for $SI$ and $\$1000$ for $P$ in Equation (II).

$\begin{array}{c}{A}_s=\$1000+\$100\\ =\$1100\end{array}$

Calculate the compound interest.

Substitute $\$1000$ for $P$, $5\%$ for $R$ and $2$ for $T$ in Equation (III).

$\begin{array}{c}CI=\$1000{\left(1+\frac{5}{100}\right)}^2\\ =\$1102.50\end{array}$

Calculate the amount due to compound interest.

Substitute $\$1102.5$ for $CI$ and $\$1000$ for $P$ in Equation (IV).

$\begin{array}{c}{A}_{\text{c}}=\$1102.50-\$1000\\ =\$102.5\end{array}$

The values for the loan values are calculated and shown below in the table.

	Loan	Years	Rate	SI	$A_{\text{s}}$	CI	$A_{\text{c}}$
(a)	$\$1000$	$2$	$5\%$	$\$100$	$\$1100$	$\$1102.50$	$\$102.5$
(b)	$\$1500$	$5$	$6\%$	$\$450$	$\$1950$	$\$2007.34$	$\$507.34$
(c)	$\$10,000$	$10$	$10\%$	$\$10000$	$\$20000$	$\$25937.42$	$\$15937.42$
(d)	$\$25,000$	$15$	$15\%$	$\$562500$	$\$81250$	$\$203426.54$	$\$178426.54$
(e)	$\$47,750$	$20$	$20\%$	$\$191000$	$\$238750$	$\$1830620.40$	$\$1782870.40$

Here, SI is calculated from Equation (I), $A_{\text{s}}$ is calculated from Equation (II), $CI$ is calculated from Equation (III) and $A_{\text{c}}$ is calculated from Equation (IV).

Conclusion:

Thus, the interest and total amount is shown in the following table.

	$SI$	$A_{\text{s}}$	$CI$	$A_{\text{c}}$
(a)	$\$100$	$\$1100$	$\$1102.50$	$\$102.5$
(b)	$\$450$	$\$1950$	$\$2007.34$	$\$507.34$
(c)	$\$10000$	$\$20000$	$\$25937.42$	$\$15937.42$
(d)	$\$562500$	$\$81250$	$\$203426.54$	$\$178426.54$
(e)	$\$191000$	$\$238750$	$\$1830620.40$	$\$1782870.40$