seth has just decided to replace his computer. his old computer cost him $1,433 when he bought it exactly…

seth has just decided to replace his computer. his old computer cost him $1,433 when he bought it exactly seven years ago. seth paid for it with his credit card, which has an apr of 11.70%, compounded monthly. he made no other purchases with the card and paid off his balance after two and a half years of making identical monthly payments. the computer consumed about $0.79 of electricity every day. in total, what percentage of the lifetime cost of the computer did the electricity make up? (assume that two out of the seven years were leap years, and round all dollar values to the nearest cent.)\na. 82.157%\nb. 58.500%\nc. 21.718%\nd. 54.898%

seth has just decided to replace his computer. his old computer cost him $1,433 when he bought it exactly seven years ago. seth paid for it with his credit card, which has an apr of 11.70%, compounded monthly. he made no other purchases with the card and paid off his balance after two and a half years of making identical monthly payments. the computer consumed about $0.79 of electricity every day. in total, what percentage of the lifetime cost of the computer did the electricity make up? (assume that two out of the seven years were leap years, and round all dollar values to the nearest cent.)\na. 82.157%\nb. 58.500%\nc. 21.718%\nd. 54.898%

Answer

Explanation:

Step1: Calculate the number of days

The computer was used for 7 years. There are 2 leap - years and 5 non - leap years. A non - leap year has 365 days and a leap year has 366 days. So the total number of days $n=2\times366 + 5\times365=732+1825 = 2557$ days.

Step2: Calculate the total cost of electricity

The computer consumes $0.79$ dollars of electricity per day. So the total cost of electricity $C_{e}=0.79\times2557=$2020.03$.

Step3: Calculate the cost of the computer with credit - card interest

The formula for the monthly payment of a loan is $M = P\frac{r(1 + r)^{n}}{(1 + r)^{n}-1}$, where $P$ is the principal amount, $r$ is the monthly interest rate, and $n$ is the total number of payments. The annual percentage rate (APR) is $11.70%=0.117$, so the monthly interest rate $r=\frac{0.117}{12}=0.00975$. The number of payments $n = 2.5\times12=30$ months and the principal $P = 1433$. First, calculate $(1 + r)^{n}=(1 + 0.00975)^{30}\approx1.32129$. Then $M=1433\times\frac{0.00975\times1.32129}{1.32129 - 1}=1433\times\frac{0.0128826}{0.32129}\approx1433\times0.0401 = 57.4633$. The total amount paid for the computer $C_{c}=M\times n=57.4633\times30=$1723.90$.

Step4: Calculate the percentage of electricity cost

The total lifetime cost of the computer $C = C_{c}+C_{e}=1723.90 + 2020.03=$3743.93$. The percentage of the electricity cost is $\frac{C_{e}}{C}\times100=\frac{2020.03}{3743.93}\times100\approx54.898%$.

Answer:

d. 54.898%