Question

You are considering taking out one of two loans. Loan R has a principal of $17,550, an interest rate of 5. 32% (compounded monthly), and a duration of seven years. Loan S has a principal of $15,925, an interest rate of 6. 07% (compounded monthly), and a duration of nine years. Assuming that you pay back each in monthly intervals, which loan will have a greater lifetime total, and how much greater will it be? Round all dollar values to the nearest cent. A. Loan S’s lifetime total will be $1,274. 12 greater than Loan R’s. B. Loan S’s lifetime total will be $32. 99 greater than Loan R’s. C. Loan R’s lifetime total will be $350. 88 greater than Loan S’s. D. Loan R’s lifetime total will be $58. 96 greater than Loan S’s.

Answer 1

The difference between the monthly payment of R and S is equal to $48.53 by following the compound interest formula. Thus, Loan R's monthly loan amount is greater than Loan S.

What is a Compound interest loan?

Combined interest (or compound interest) is the loan interest or deposit calculated based on both the original interest and accrued interest from earlier periods.

$\rm\,For\,R\\\\P = \ \,17,550\\r\,= 5.32\%\\Time\,= n= 7\,years\\Amount\,paid= [P(1+\dfrac{r}{100\times12})^{n\times12} ]\\=[ 17,550 (1+\dfrac{5.32}{100\times12})^{7\times12} ]\\= [ 17,550 (\dfrac{12.0532}{12})^{84} ]\\\\= [ 17,550 (1.00443^{84} ]\\\\= \ 25,440.48\\\\Total\,monthly\,payment = \rm\,\dfrac{25,440.48}{84}\\\\= \ \, 302.86$

$\rm\,For\,S =\\\\P=\,\ 15,925\\r\,= 6.07\%\\T=n= 9\,years\\\\Amount\,paid\,= [P(1+\dfrac{r}{100\times12})^{n\times12} ]\\\\\= [15,925(1+\dfrac{0.0607}{12})^{9\times12} ]\\\\\\= [15,925(1+\dfrac{0.0607}{12})^{108} ]\\\\=[15,925(1.7247.84)} ]\\\\\= \ 27,467.19\\\\Total\,monthly\,payment =\dfrac{\rm\,\ \,27,469.19}{108}\\\\= \ 254.326$

The difference between the monthly payment of R and S is equal to $48.53.

Hence, Loan R's monthly payment is greater than the loan's monthly payment by $48.53

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