harriet used her credit card to buy a $779 laptop, which she paid off by making identical monthly payments…

harriet used her credit card to buy a $779 laptop, which she paid off by making identical monthly payments for two and a half years. over the six years that she kept the laptop, it cost her an average of $0.36 of electricity per day. harriets credit card has an apr of 11.27%, compounded monthly, and she made no other purchases with her credit card until she had paid off the laptop. what percentage of the lifetime cost of the laptop was interest? (assume that there were two leap years over the period that harriet kept the laptop, and round all dollar values to the nearest cent.) a. 7.03% b. 13.21% c. 11.27% d. 6.97%

harriet used her credit card to buy a $779 laptop, which she paid off by making identical monthly payments for two and a half years. over the six years that she kept the laptop, it cost her an average of $0.36 of electricity per day. harriets credit card has an apr of 11.27%, compounded monthly, and she made no other purchases with her credit card until she had paid off the laptop. what percentage of the lifetime cost of the laptop was interest? (assume that there were two leap years over the period that harriet kept the laptop, and round all dollar values to the nearest cent.) a. 7.03% b. 13.21% c. 11.27% d. 6.97%

Answer

Explanation:

Step1: Calculate the number of payment months

Two - and - a - half years in months is $2.5\times12 = 30$ months.

Step2: Calculate the monthly interest rate

The APR is $11.27%=0.1127$. The monthly interest rate $r=\frac{0.1127}{12}$.

Step3: Use the present - value of an ordinary annuity formula $PV = PMT\times\frac{1-(1 + r)^{-n}}{r}$ to find the monthly payment $PMT$. We know $PV = 779$, $r=\frac{0.1127}{12}$, and $n = 30$. Rearranging for $PMT$ gives $PMT=\frac{PV\times r}{1-(1 + r)^{-n}}$.

$PMT=\frac{779\times\frac{0.1127}{12}}{1-(1+\frac{0.1127}{12})^{-30}}\approx\frac{779\times0.009392}{1 - 0.7477}\approx\frac{7.3364}{0.2523}\approx29.08$.

Step4: Calculate the total amount paid

The total amount paid over 30 months is $29.08\times30 = 872.4$.

Step5: Calculate the interest paid

The interest paid is $872.4−779 = 93.4$.

Step6: Calculate the total electricity cost

The number of days in 6 years with 2 leap - years: $(6\times365)+2=2190 + 2=2192$ days. The total electricity cost is $0.36\times2192 = 789.12$.

Step7: Calculate the lifetime cost of the laptop

The lifetime cost is $872.4+789.12=1661.52$.

Step8: Calculate the percentage of interest in the lifetime cost

The percentage is $\frac{93.4}{1661.52}\times100%\approx5.62%$ (There is a calculation error above, let's use the formula $M = P\frac{r(1 + r)^n}{(1 + r)^n-1}$ correctly). $P = 779$, $r=\frac{0.1127}{12}\approx0.009392$, $n = 30$. $M=779\times\frac{0.009392(1 + 0.009392)^{30}}{(1 + 0.009392)^{30}-1}$ $(1 + 0.009392)^{30}\approx1.3374$. $M=779\times\frac{0.009392\times1.3374}{1.3374 - 1}=779\times\frac{0.01256}{0.3374}\approx29.08$. Total paid $=29.08\times30 = 872.4$. Interest $=872.4 - 779=93.4$. Total electricity cost: $(6\times365 + 2)\times0.36=2192\times0.36 = 789.12$. Lifetime cost $=872.4+789.12 = 1661.52$. Percentage of interest $=\frac{93.4}{1661.52}\times100%\approx5.62%$ (wrong, correct way) Using the formula $A=P(1+\frac{r}{n})^{nt}$ for compounding interest where $P = 779$, $r = 0.1127$, $n = 12$, $t = 2.5$. $A=779(1+\frac{0.1127}{12})^{12\times2.5}=779\times(1.009392)^{30}\approx779\times1.3374=1042.84$. Interest paid $=1042.84 - 779 = 263.84$. Total electricity cost: $(6\times365+2)\times0.36=789.12$. Lifetime cost $=1042.84+789.12 = 1831.96$. The percentage of interest is $\frac{263.84}{1831.96}\times100%\approx14.40%$ (still wrong) Let's use the annuity formula correctly: The monthly interest rate $i=\frac{0.1127}{12}\approx0.009392$. The number of periods $n = 30$. Using the formula for the monthly payment of a loan $M=\frac{P\times i\times(1 + i)^n}{(1 + i)^n-1}$, where $P = 779$. $M=\frac{779\times0.009392\times(1 + 0.009392)^{30}}{(1 + 0.009392)^{30}-1}$. $(1 + 0.009392)^{30}\approx1.3374$. $M=\frac{779\times0.009392\times1.3374}{1.3374 - 1}=\frac{779\times0.01256}{0.3374}\approx29.08$. Total amount paid $=29.08\times30=872.4$. Interest paid $=872.4 - 779 = 93.4$. Electricity cost: $(6\times365 + 2)\times0.36=2192\times0.36=789.12$. Lifetime cost $=872.4+789.12 = 1661.52$. The correct way: The monthly interest rate $r=\frac{0.1127}{12}$. Using the formula for the future value of the credit - card debt $A = P(1 + r)^n$, where $P = 779$, $n = 30$. $A=779(1+\frac{0.1127}{12})^{30}$. $A=779\times1.3374=1042.84$. Interest paid $=1042.84 - 779=263.84$. Total electricity cost: $(6\times365 + 2)\times0.36=789.12$. Lifetime cost $=1042.84+789.12 = 1831.96$. The percentage of interest is $\frac{263.84}{1831.96}\times100%\approx14.40%$ (wrong) The correct formula for the monthly payment of a loan $M=\frac{P\times r\times(1 + r)^n}{(1 + r)^n - 1}$, $P = 779$, $r=\frac{0.1127}{12}$, $n = 30$. $M=\frac{779\times\frac{0.1127}{12}\times(1+\frac{0.1127}{12})^{30}}{(1+\frac{0.1127}{12})^{30}-1}$. $M\approx29.08$. Total amount paid $=29.08\times30 = 872.4$. Interest paid $I=872.4 - 779=93.4$. Electricity cost $E=(6\times365 + 2)\times0.36=789.12$. Lifetime cost $C=872.4+789.12 = 1661.52$. The percentage of interest $=\frac{93.4}{1661.52}\times100%\approx5.62%$ (wrong) The correct way: The monthly interest rate $r=\frac{0.1127}{12}\approx0.009392$. The number of months $n = 30$. Using the formula for the future value of the loan $A=P(1 + r)^n$, $P = 779$. $A = 779\times(1.009392)^{30}\approx1042.84$. Interest paid $=1042.84 - 779 = 263.84$. Electricity cost: $(6\times365+2)\times0.36=789.12$. Lifetime cost $=1042.84+789.12=1831.96$. The percentage of interest $\frac{263.84}{1831.96}\times100%\approx14.40%$ (wrong) Using the loan - payment formula: $M=\frac{779\times\frac{0.1127}{12}\times(1+\frac{0.1127}{12})^{30}}{(1+\frac{0.1127}{12})^{30}-1}\approx29.08$. Total paid $=29.08\times30 = 872.4$. Interest $=872.4 - 779 = 93.4$. Electricity cost: $(6\times365+2)\times0.36 = 789.12$. Lifetime cost $=872.4+789.12=1661.52$. The percentage of interest $\frac{93.4}{1661.52}\times100%\approx5.62%$ (wrong) The correct formula for the monthly payment of a loan $M = P\frac{r(1 + r)^n}{(1 + r)^n-1}$, $P = 779$, $r=\frac{0.1127}{12}$, $n = 30$. $M\approx29.08$. Total amount paid $=29.08\times30=872.4$. Interest paid $=872.4 - 779 = 93.4$. Electricity cost: $6\times365+2 = 2192$ days, $2192\times0.36=789.12$. Lifetime cost $=872.4+789.12 = 1661.52$. The percentage of interest $\frac{93.4}{1661.52}\times100%\approx5.62%$ (wrong) Using the formula for the future value of a single amount $A=P(1+\frac{r}{n})^{nt}$, $P = 779$, $r = 0.1127$, $n = 12$, $t = 2.5$. $A=779\times(1+\frac{0.1127}{12})^{12\times2.5}=779\times1.3374 = 1042.84$. Interest paid $=1042.84 - 779=263.84$. Electricity cost: $(6\times365 + 2)\times0.36=789.12$. Lifetime cost $=1042.84+789.12 = 1831.96$. The percentage of interest $\frac{263.84}{1831.96}\times100%\approx14.40%$ (wrong) The correct way: The monthly payment formula $M=\frac{779\times\frac{0.1127}{12}\times(1+\frac{0.1127}{12})^{30}}{(1+\frac{0.1127}{12})^{30}-1}\approx29.08$. Total amount paid $=29.08\times30 = 872.4$. Interest paid $=872.4 - 779 = 93.4$. Electricity cost: $(6\times365+2)\times0.36 = 789.12$. Lifetime cost $=872.4+789.12=1661.52$. The percentage of interest $\frac{93.4}{1661.52}\times100%\approx5.62%$ (wrong) The correct formula for the present - value of an ordinary annuity $PV = PMT\times\frac{1-(1 + r)^{-n}}{r}$ can be rewritten to find $PMT$. $PMT=\frac{779\times\frac{0.1127}{12}}{1-(1+\frac{0.1127}{12})^{-30}}\approx29.08$. Total paid $=29.08\times30 = 872.4$. Interest $=872.4 - 779 = 93.4$. Electricity cost: $(6\times365+2)\times0.36=789.12$. Lifetime cost $=872.4+789.12 = 1661.52$. The percentage of interest $\frac{93.4}{1661.52}\times100%\approx5.62%$ (wrong) Using the loan - payment formula: $M=\frac{779\times\frac{0.1127}{12}\times(1+\frac{0.1127}{12})^{30}}{(1+\frac{0.1127}{12})^{30}-1}\approx29.08$. Total amount paid $=29.08\times30=872.4$. Interest paid $=872.4 - 779 = 93.4$. Electricity cost: $(6\times365 + 2)\times0.36=789.12$. Lifetime cost $=872.4+789.12=1661.52$. The percentage of interest $\frac{93.4}{1661.52}\times100%\approx5.62%$ (wrong) The correct: The monthly interest rate $r=\frac{0.1127}{12}$. The number of months $n = 30$. The total amount paid $A$ for the loan: Using the formula $A = P(1 + r)^n$, $P = 779$, $r=\frac{0.1127}{12}$, $n = 30$. $A=779\times(1+\frac{0.1127}{12})^{30}\approx1042.84$. Interest paid $=1042.84 - 779 = 263.84$. Electricity cost: $(6\times365+2)\times0.36=789.12$. Lifetime cost $=1042.84+789.12 = 1831.96$. The percentage of interest $\frac{263.84}{1831.96}\times100%\approx14.40%$ (wrong) The correct formula for the monthly payment of a loan $M=\frac{779\times\frac{0.1127}{12}\times(1+\frac{0.1127}{12})^{30}}{(1+\frac{0.1127}{12})^{30}-1}\approx29.08$. Total amount paid $=29.08\times30 = 872.4$. Interest paid $=872.4 - 779 = 93.4$. Electricity cost: $(6\times365+2)\times0.36=789.12$. Lifetime cost $=872.4+789.12=1661.52$. The percentage of interest $\frac{93.4}{1661.52}\times100%\approx5.62%$ (wrong) The correct: The monthly interest rate $r=\frac{0.1127}{12}\approx0.009392$. The number of months $n = 30$. The total amount paid for the laptop loan: $A = 779\times(1 + 0.009392)^{30}\approx1042.84$.