Question

Montoya Construction needs to borrow $375,000 to build a road to install utilities in a small subdivision. It borrows the funds at 8% for 90 days. When interest rates were high in 1980, the interest rate would have been 20%. Find the difference in interest between the two rates.

Answer 1

Step 1

Given;

$\begin{gathered} \text{Principal(p)= \ 375000} \\ \text{First rate = 8\%=}\frac{8}{100}=0.08 \\ \text{Second rate = 20\%= }\frac{20}{100}=0.2 \\ \text{Time}=\frac{90}{365}=\frac{18}{73} \end{gathered}$

Required; To find the difference in interest between the two periods.

Step 2

State the formula for simple interest

$A=P(1+rt)$

Step 3

Find the interest when the rate is 8%

$\begin{gathered} A=375000(1+(0.08\times\frac{18}{73}) \\ A=375000(1+\frac{36}{1825}) \\ A=\text{\ }382397.26 \end{gathered}$

Therefore the interest is given as;

$A-P=382397.26-375000=\text{\ }7397.26$

Step 4

Find the interest in 1980 with a 20% rate

$\begin{gathered} A=375000(1+(0.2\times\frac{18}{73}) \\ A=\text{\ }393493.15 \end{gathered}$

The interest is given as;

$A-p=393493.15-375000=\text{\ }18493.15\text{ }$

Step 5

Find the difference in interest between the two rates.

$\text{\ }18493.15-\text{\ }7397.26=\text{\ }11095.89$

Hence, the difference in interest between the two rates = $11095.89