Question

determine the monthly payment of a loan for $3,000 at 7.5% interest compounded monthly for 36 months
a. $93.32
b. $95.40
c. $211.33
d. $783.60

Explanation:

Step1: Identify the formula for loan payment

The formula for the monthly payment \( M \) of a loan with principal \( P \), monthly interest rate \( r \), and number of months \( n \) is:
\[ M = \frac{P \cdot r \cdot (1 + r)^n}{(1 + r)^n - 1} \]
Here, \( P = 3000 \), the annual interest rate is \( 7.5\% = 0.075 \), so the monthly interest rate \( r=\frac{0.075}{12}=0.00625 \), and the number of months \( n = 36 \).

Step2: Calculate \( (1 + r)^n \)

\[ (1 + 0.00625)^{36}=(1.00625)^{36}\approx1.251797 \]

Step3: Calculate the numerator

\[ P\cdot r\cdot(1 + r)^n=3000\times0.00625\times1.251797 = 18.75\times1.251797\approx23.47 \]

Step4: Calculate the denominator

\[ (1 + r)^n-1 = 1.251797 - 1=0.251797 \]

Step5: Calculate the monthly payment

\[ M=\frac{23.47}{0.251797}\approx93.21 \]
Wait, there might be a miscalculation. Let's recalculate the numerator correctly:
\[ P\cdot r\cdot(1 + r)^n=3000\times0.00625\times(1.00625)^{36} \]
First, \( 3000\times0.00625 = 18.75 \)
Then, \( 18.75\times(1.00625)^{36} \)
\( (1.00625)^{36}\approx e^{0.00625\times36}=e^{0.225}\approx1.2523 \) (using the approximation \( e^x\approx1 + x+\frac{x^2}{2}+\frac{x^3}{6} \) for \( x = 0.225 \))
So \( 18.75\times1.2523\approx18.75\times1.2523 = 23.48 \)
Denominator: \( 1.2523-1 = 0.2523 \)
Then \( M=\frac{23.48}{0.2523}\approx93.06 \), which is close to option a. $93.32$ (maybe due to more precise calculation of \( (1.00625)^{36} \))

Let's use a more accurate calculation for \( (1.00625)^{36} \):
Using the formula for compound interest, \( (1 + \frac{0.075}{12})^{36} \)
We can calculate step - by - step:
\( (1.00625)^1 = 1.00625 \)
\( (1.00625)^2=1.00625\times1.00625 = 1.0125390625 \)
\( (1.00625)^4=(1.0125390625)^2\approx1.0252807617 \)
\( (1.00625)^8=(1.0252807617)^2\approx1.0511303711 \)
\( (1.00625)^{16}=(1.0511303711)^2\approx1.1048690479 \)
\( (1.00625)^{32}=(1.1048690479)^2\approx1.2206944244 \)
\( (1.00625)^{36}=(1.00625)^{32}\times(1.00625)^4\approx1.2206944244\times1.0252807617\approx1.251797 \)
Now, numerator: \( 3000\times0.00625\times1.251797 = 18.75\times1.251797 = 18.75\times1.251797 = 23.4703 \)
Denominator: \( 1.251797 - 1 = 0.251797 \)
\( M=\frac{23.4703}{0.251797}\approx93.21 \), which is closest to option a. $93.32$ (the slight difference is due to more precise decimal calculations in the actual financial formula or calculator - based computation)

Answer:

a. $93.32$