compound interest ws\ngo to the following website\nhttps://www.thecalculatorsite.com/finance/calculators/comp…

compound interest ws\ngo to the following website\nhttps://www.thecalculatorsite.com/finance/calculators/compoundinterestcalculator.php\nor type the calculator site compound - interest calculator and click on the first option.\nuse the calculator to solve the following problems\n1. if you saved $700 a month for 40 years with an initial deposit of $500, how much be in the account at the following interest rates:\na. 3%\nb. 6%\nc. 9%\nd. 12%\ne. 15%\n2. if you saved $700 a month (the average car payment) for 40 years with an initial deposit of $700, how much would be in the account if it earned 10% per year on average?\n3. if you wanted to become a millionaire, how much money would you have to save every month starting at age 20, assuming a 10% rate of return?\na. by age 30\nb. by age 50\nc. by age 60\n4. if the average inflation is 3% each year for the next 20 years and the current cost of a big mac meal is $8, how much would it cost in 20 years (2045)?

compound interest ws\ngo to the following website\nhttps://www.thecalculatorsite.com/finance/calculators/compoundinterestcalculator.php\nor type the calculator site compound - interest calculator and click on the first option.\nuse the calculator to solve the following problems\n1. if you saved $700 a month for 40 years with an initial deposit of $500, how much be in the account at the following interest rates:\na. 3%\nb. 6%\nc. 9%\nd. 12%\ne. 15%\n2. if you saved $700 a month (the average car payment) for 40 years with an initial deposit of $700, how much would be in the account if it earned 10% per year on average?\n3. if you wanted to become a millionaire, how much money would you have to save every month starting at age 20, assuming a 10% rate of return?\na. by age 30\nb. by age 50\nc. by age 60\n4. if the average inflation is 3% each year for the next 20 years and the current cost of a big mac meal is $8, how much would it cost in 20 years (2045)?

Answer

Explanation:

Step1: Recall compound - interest formula for monthly savings

The future - value of an ordinary annuity (monthly savings) formula is $FVA = P\times\frac{(1 + r)^{n}-1}{r}+PV\times(1 + r)^{n}$, where $P$ is the monthly payment, $r$ is the monthly interest rate, $n$ is the total number of periods, and $PV$ is the present value (initial deposit).

Problem 1a:

$P=$700$, $PV = $500$, $t = 40$ years, so $n=40\times12 = 480$ months, $r=\frac{0.03}{12}=0.0025$

Step2: Calculate future - value of annuity part

$FVA_{annuity}=700\times\frac{(1 + 0.0025)^{480}-1}{0.0025}$ First, calculate $(1 + 0.0025)^{480}=e^{480\times\ln(1.0025)}\approx3.3102$ $FVA_{annuity}=700\times\frac{3.3102 - 1}{0.0025}=700\times\frac{2.3102}{0.0025}=700\times924.08=$646856$

Step3: Calculate future - value of initial deposit

$FVA_{initial}=500\times(1 + 0.0025)^{480}=500\times3.3102=$1655.1$

Step4: Calculate total future value

$FVA = 646856+1655.1=$648511.1$

Problem 1b:

$r=\frac{0.06}{12}=0.005$, $n = 480$ $FVA_{annuity}=700\times\frac{(1 + 0.005)^{480}-1}{0.005}$ $(1 + 0.005)^{480}=e^{480\times\ln(1.005)}\approx10.9574$ $FVA_{annuity}=700\times\frac{10.9574 - 1}{0.005}=700\times\frac{9.9574}{0.005}=700\times1991.48=$1394036$ $FVA_{initial}=500\times(1 + 0.005)^{480}=500\times10.9574=$5478.7$ $FVA=1394036 + 5478.7=$1399514.7$

Problem 1c:

$r=\frac{0.09}{12}=0.0075$, $n = 480$ $(1 + 0.0075)^{480}=e^{480\times\ln(1.0075)}\approx38.3376$ $FVA_{annuity}=700\times\frac{38.3376 - 1}{0.0075}=700\times\frac{37.3376}{0.0075}=700\times4978.35=$3484845$ $FVA_{initial}=500\times(1 + 0.0075)^{480}=500\times38.3376=$19168.8$ $FVA=3484845+19168.8=$3504013.8$

Problem 1d:

$r=\frac{0.12}{12}=0.01$, $n = 480$ $(1 + 0.01)^{480}=e^{480\times\ln(1.01)}\approx164.464$ $FVA_{annuity}=700\times\frac{164.464 - 1}{0.01}=700\times\frac{163.464}{0.01}=700\times16346.4=$11442480$ $FVA_{initial}=500\times(1 + 0.01)^{480}=500\times164.464=$82232$ $FVA=11442480+82232=$11524712$

Problem 1e:

$r=\frac{0.15}{12}=0.0125$, $n = 480$ $(1 + 0.0125)^{480}=e^{480\times\ln(1.0125)}\approx718.603$ $FVA_{annuity}=700\times\frac{718.603 - 1}{0.0125}=700\times\frac{717.603}{0.0125}=700\times57408.24=$40185768$ $FVA_{initial}=500\times(1 + 0.0125)^{480}=500\times718.603=$359301.5$ $FVA=40185768+359301.5=$40545069.5$

Problem 2:

$P = 700$, $PV=700$, $r=\frac{0.1}{12}\approx0.00833$, $n = 40\times12 = 480$ $FVA_{annuity}=700\times\frac{(1 + 0.00833)^{480}-1}{0.00833}$ $(1 + 0.00833)^{480}=e^{480\times\ln(1.00833)}\approx12.799$ $FVA_{annuity}=700\times\frac{12.799 - 1}{0.00833}=700\times\frac{11.799}{0.00833}=700\times1416.45=$991515$ $FVA_{initial}=700\times(1 + 0.00833)^{480}=700\times12.799=$8959.3$ $FVA=991515+8959.3=$1000474.3$

Problem 3a:

We want $FVA=$1000000$, $t = 10$ years, $n = 10\times12=120$ months, $r=\frac{0.1}{12}\approx0.00833$, $PV = 0$ From $FVA = P\times\frac{(1 + r)^{n}-1}{r}$, we can solve for $P$: $P=\frac{FVA\times r}{(1 + r)^{n}-1}=\frac{1000000\times0.00833}{(1 + 0.00833)^{120}-1}$ $(1 + 0.00833)^{120}=e^{120\times\ln(1.00833)}\approx2.707$ $P=\frac{1000000\times0.00833}{2.707 - 1}=\frac{8330}{1.707}\approx$4880$

Problem 3b:

$t = 30$ years, $n = 30\times12 = 360$ months, $r=\frac{0.1}{12}\approx0.00833$, $PV = 0$ $P=\frac{1000000\times0.00833}{(1 + 0.00833)^{360}-1}$ $(1 + 0.00833)^{360}=e^{360\times\ln(1.00833)}\approx19.837$ $P=\frac{1000000\times0.00833}{19.837 - 1}=\frac{8330}{18.837}\approx$442$

Problem 3c:

$t = 40$ years, $n = 40\times12 = 480$ months, $r=\frac{0.1}{12}\approx0.00833$, $PV = 0$ $P=\frac{1000000\times0.00833}{(1 + 0.00833)^{480}-1}$ $(1 + 0.00833)^{480}=e^{480\times\ln(1.00833)}\approx12.799$ $P=\frac{1000000\times0.00833}{12.799 - 1}=\frac{8330}{11.799}\approx$706$

Problem 4:

Use the compound - growth formula for inflation $FV=PV\times(1 + i)^{n}$, where $PV = 8$, $i = 0.03$, $n = 20$ $FV=8\times(1 + 0.03)^{20}$ $(1 + 0.03)^{20}=e^{20\times\ln(1.03)}\approx1.8061$ $FV=8\times1.8061=$14.45$

Answer:

1a. $$648511.1$ 1b. $$1399514.7$ 1c. $$3504013.8$ 1d. $$11524712$ 1e. $$40545069.5$ 2. $$1000474.3$ 3a. $$4880$ 3b. $$442$ 3c. $$706$ 4. $$14.45$