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kondor19780726 [428]
3 years ago
11

With only a​ part-time job and the need for a professional​ wardrobe, Rachel quickly maxed out her credit card the summer after

graduation. With her first​ full-time paycheck in​ August, she vowed to pay ​$260 each month toward paying down her ​$10 comma 574 outstanding balance and not to use the card. The card has an annual interest rate of 16 percent. How long will it take Rachel to pay for her​ wardrobe? Should she shop for a new​ card? Why or why​ not?
Business
1 answer:
choli [55]3 years ago
3 0

Answer:

a. It will take her 5 years to pay for her wardrobe

b. She should shop for a new card once she is done paying for this one.

c. She should shop for a new card after finishing paying for this card since going further into debt with the current card would be a bad idea. This is due to the fact that an annual interest rate of 16% is very high. The best option would therefor to finish her payments on the credit card, then shop for a new card with a lower annual interest rate.

Explanation:

Use the formula below to determine the number of months it would take Rachel to pay off her debt;

C *{1-(1+r)^(-n×t)}/(r/n)=PV

where;

C=annuity

r=annual interest rate

n=number of compounding periods in a year

t=number of years

PV=present value

In our case;

PV=$10,574

C=$260

r=16%=16/100=0.16

n=12

t=unknown

replacing;

260*{1-(1+0.16/12)^(-12×t)}/(0.16/12)=10,574

1-(1+0.16/12)^(-12×t)={10,574×(0.16/12)}/260

1-{1.013^(-12 t)}=0.542

(1-0.542)=1.013^(-12 t)

ln 0.458=-12 t (ln 1.013)

t=-ln 0.458/12×ln 1.013

t=5

It will take her 5 years to pay for her wardrobe

b. She should shop for a new card once she is done paying for this one.

c. She should shop for a new card after finishing paying for this card since going further into debt with the current card would be a bad idea. This is due to the fact that an annual interest rate of 16% is very high. The best option would therefor to finish her payments on the credit card, then shop for a new card with a lower annual interest rate.

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