maria is going to take out a loan with a principal of $19,700. she has narrowed down her options to two…

Explanation:

Step1: Calculate monthly payment formula

The formula for the monthly payment of a loan is (M = P\times\frac{r(1 + r)^{n}}{(1 + r)^{n}-1}), where (P) is the principal, (r) is the monthly interest rate, and (n) is the total number of payments. For Bank M: (P=$19700), annual interest rate (i = 7.1%=0.071), so monthly interest rate (r=\frac{0.071}{12}), and (n = 5\times12=60) [M_{M}=19700\times\frac{\frac{0.071}{12}(1+\frac{0.071}{12})^{60}}{(1+\frac{0.071}{12})^{60}-1}] [r_{M}\approx\frac{0.071}{12}\approx0.00592], ((1 + r_{M})^{60}\approx(1.00592)^{60}\approx1.439) [M_{M}=19700\times\frac{0.00592\times1.439}{1.439 - 1}=19700\times\frac{0.00852}{0.439}\approx19700\times0.0194\approx$382] The total lifetime cost (C_{M}=M_{M}\times60=$382\times60=$22920)

For Bank N: (P = $19700), annual interest rate (i=7.8% = 0.078), so monthly interest rate (r=\frac{0.078}{12}=0.0065), and (n=4\times12 = 48) [M_{N}=19700\times\frac{0.0065(1 + 0.0065)^{48}}{(1+0.0065)^{48}-1}] ((1 + 0.0065)^{48}\approx(1.0065)^{48}\approx1.377) [M_{N}=19700\times\frac{0.0065\times1.377}{1.377-1}=19700\times\frac{0.00895}{0.377}\approx19700\times0.0237\approx$467] The total lifetime cost (C_{N}=M_{N}\times48=$467\times48=$22416)

Step2: Compare monthly payments and lifetime costs

We have (M_{M}\approx$382\lt M_{N}\approx$467) (lower monthly payment for Bank M) and (C_{M}=$22920\gt C_{N}=$22416) (lower lifetime cost for Bank N)

Answer:

B. Maria should choose Bank M's loan if she cares more about lower monthly payments, and she should choose Bank N's loan if she cares more about the lowest lifetime cost.