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# Solved Examples on Future Value of Ordinary Annuity and Sinking Funds

Formulas Used: Mathematics of Finance
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(1.) Determine the savings plan balance after 3 years with an APR of 4% and monthly payments of $300 This is a case of the Future Value of Ordinary Annuity$ PMT = \$300 \\[3ex] r = 4\% = \dfrac{4}{100} = 0.04 \\[5ex] Monthly\;\;payments \rightarrow Compounded\:\:monthly \rightarrow m = 12 \\[3ex] t = 3\:years \\[3ex] FV = ? \\[3ex] FV = m * PMT * \left[\dfrac{\left(1 + \dfrac{r}{m}\right)^{mt} - 1}{r}\right] \\[10ex] FV = 12 * 300 * \left[\dfrac{\left(1 + \dfrac{0.04}{12}\right)^{12 * 3} - 1}{0.04}\right] \\[10ex] = 3600 * \left[\dfrac{\left(1 + 0.003333333\right)^{36} - 1}{0.04}\right] \\[10ex] = 3600 * \left[\dfrac{\left(1.003333333\right)^{36} - 1}{0.04}\right] \\[7ex] = 3600 * \left[\dfrac{1.127271875 - 1}{0.04}\right] \\[5ex] = 3600 * \left[\dfrac{0.127271875}{0.04}\right] \\[5ex] = \dfrac{3600 * 0.127271875}{0.04} \\[5ex] = \dfrac{458.1787483}{0.04} \\[5ex] = 11454.46871 \\[3ex] FV \approx \$11,454.47 $(2.) Determine the savings plan balance after 15 months with an APR of 2% and monthly payments of$300

This is a case of the Future Value of Ordinary Annuity

$PMT = \$300 \\[3ex] r = 2\% = \dfrac{2}{100} = 0.02 \\[5ex] Monthly\;\;payments \rightarrow Compounded\:\:monthly \rightarrow m = 12 \\[3ex] t = 15\;months = \dfrac{15}{12} = 1.25\;years \\[5ex] FV = ? \\[3ex] FV = m * PMT * \left[\dfrac{\left(1 + \dfrac{r}{m}\right)^{mt} - 1}{r}\right] \\[10ex] FV = 12 * 300 * \left[\dfrac{\left(1 + \dfrac{0.02}{12}\right)^{12 * 1.25} - 1}{0.02}\right] \\[10ex] = 3600 * \left[\dfrac{\left(1 + 0.001666667\right)^{15} - 1}{0.02}\right] \\[10ex] = 3600 * \left[\dfrac{\left(1.001666667\right)^{15} - 1}{0.02}\right] \\[7ex] = 3600 * \left[\dfrac{1.025293784 - 1}{0.02}\right] \\[5ex] = 68400 * \left[\dfrac{0.025293784}{0.02}\right] \\[5ex] = \dfrac{3600 * 0.025293784}{0.02} \\[5ex] = \dfrac{91.05762139}{0.02} \\[5ex] = 4552.881069 \\[3ex] FV \approx \$4,552.88$
(3.) In the savings plan formula below (one of the formulas for the Future Value of Ordinary Annuity), $$FV = PMT * \dfrac{\left[\left(1 + \dfrac{r}{m}\right)^{mt} - 1\right]}{\dfrac{r}{m}}$$ Assuming all other variables are constant, what happens to the accumulated balance in the savings account?

A. It decreases as t increases.
As the exponent, m in the numerator increases, t must decrease to keep the formula true.

B. It increases as m increases.
As m increases, the demoninator decreases and the exponent in the numerator increases.
This makes the accumulated balance increase.

C. It increases as m increases.
If less payments are made per year, more money will be saved.

D. It decreases as t increases.
As the amount of years of saving increases, the accumulated savings decrease.

E. It increases as r decreases.
As m increases, r over m decreases causing the accumulated balance the increase.

F. It increases as r decreases.
As r decreases, the amount of interest paid will be less.
Therefore, more accumulated savings will be made.

It increases as m increases.
As m increases, the demoninator decreases and the exponent in the numerator increases.
This makes the accumulated balance increase.

Student: How do you know? May you explain?
Teacher: Sure.
Let us look at these three cases:

$\underline{1st\;\;Case} \\[3ex] PMT = \$1 \\[3ex] r = 0.03 \\[3ex] m = 1 \\[3ex] t = 1\;year \\[3ex] mt = 1(1) = 1 \\[3ex] \dfrac{r}{m} = \dfrac{0.03}{1} = 0.03 \\[5ex] FV = PMT * \dfrac{\left[\left(1 + \dfrac{r}{m}\right)^{mt} - 1\right]}{\dfrac{r}{m}} \\[7ex] FV = 1 * \dfrac{\left[\left(1 + \dfrac{0.03}{1}\right)^{1 * 1} - 1\right]}{\dfrac{0.03}{1}} \\[7ex] FV = \dfrac{(1 + 0.03)^{1} - 1}{0.03} \\[5ex] FV = \dfrac{1.03 - 1}{0.03} \\[5ex] FV = \dfrac{0.03}{0.03} \\[5ex] FV = \$1 \\[5ex] \underline{2nd\;\;Case} \\[3ex] PMT = \$1 \\[3ex] r = 0.03 \\[3ex] m = 2...increase \\[3ex] t = 1\;year \\[3ex] mt = 2(1) = 2...increases \\[3ex] \dfrac{r}{m} = \dfrac{0.03}{2} = 0.015...decreases \\[5ex] FV = PMT * \dfrac{\left[\left(1 + \dfrac{r}{m}\right)^{mt} - 1\right]}{\dfrac{r}{m}} \\[7ex] FV = 1 * \dfrac{\left[\left(1 + \dfrac{0.03}{2}\right)^{2 * 1} - 1\right]}{\dfrac{0.03}{2}} \\[7ex] FV = \dfrac{(1 + 0.015)^{2} - 1}{0.015} \\[5ex] FV = \dfrac{(1.015)^2 - 1}{0.015} \\[5ex] FV = \dfrac{1.030225 - 1}{0.015} \\[5ex] FV = \dfrac{0.030225}{0.015} \\[5ex] FV = \$2.015...increases \\[5ex] \underline{3rd\;\;Case} \\[3ex] PMT = \$1 \\[3ex] r = 0.03 \\[3ex] m = 3...increase \\[3ex] t = 1\;year \\[3ex] mt = 3(1) = 3...increases \\[3ex] \dfrac{r}{m} = \dfrac{0.03}{3} = 0.01...decreases \\[5ex] FV = PMT * \dfrac{\left[\left(1 + \dfrac{r}{m}\right)^{mt} - 1\right]}{\dfrac{r}{m}} \\[7ex] FV = 1 * \dfrac{\left[\left(1 + \dfrac{0.03}{3}\right)^{3 * 1} - 1\right]}{\dfrac{0.03}{3}} \\[7ex] FV = \dfrac{(1 + 0.01)^{3} - 1}{0.01} \\[5ex] FV = \dfrac{(1.01)^3 - 1}{0.01} \\[5ex] FV = \dfrac{1.030301 - 1}{0.01} \\[5ex] FV = \dfrac{0.030301}{0.01} \\[5ex] FV = \$3.0301...increases$
(4.) At age 18, Jude set up an IRA (individual retirement account) with an APR of 4%.
At the end of each month he deposits $100 in the account. How much will the IRA contain when he retires at age 65? Compare that amount to the total deposits made over the time period. This is a case of the Future Value of Ordinary Annuity$ PMT = \$100 \\[3ex] r = 4\% = \dfrac{4}{100} = 0.04 \\[5ex] Monthly\;\;deposits \rightarrow Compounded\:\:monthly \rightarrow m = 12 \\[3ex] t = 65 - 18 = 47\:years \\[3ex] FV = ? \\[3ex] FV = m * PMT * \left[\dfrac{\left(1 + \dfrac{r}{m}\right)^{mt} - 1}{r}\right] \\[10ex] FV = 12 * 100 * \left[\dfrac{\left(1 + \dfrac{0.04}{12}\right)^{12 * 47} - 1}{0.04}\right] \\[10ex] = 1200 * \left[\dfrac{\left(1 + 0.003333333\right)^{564} - 1}{0.04}\right] \\[10ex] = 1200 * \left[\dfrac{\left(1.003333333\right)^{564} - 1}{0.04}\right] \\[7ex] = 1200 * \left[\dfrac{6.533046835 - 1}{0.04}\right] \\[5ex] = 1200 * \left[\dfrac{5.533046835}{0.04}\right] \\[5ex] = \dfrac{1200 * 5.533046835}{0.04} \\[5ex] = \dfrac{6639.657671}{0.04} \\[5ex] = 165991.4418 \\[3ex] FV \approx \$165,991.44 \\[5ex] Total\;\;deposits made \\[3ex] = \$100\;\per\;\;month * 12\;months\;\;per\;\;year * 47\;years \\[3ex] = 100 * 12 * 47 \\[3ex] = 56400 \\[3ex] = \$56,400.00 $(5.) Rita's goal is to create a college fund for her son. She found a fund that offers an APR of 5%. How much should she deposit monthly to accumulate$88,000 in 12 years?

This is a case of Sinking Fund

$r = 5\% = \dfrac{5}{100} = 0.05 \\[5ex] Monthly\;\;deposits \rightarrow Compounded\:\:monthly \rightarrow m = 12 \\[3ex] FV = \$88000 \\[3ex] t = 12\;years \\[3ex] PMT = \dfrac{r * FV}{m * \left[\left(1 + \dfrac{r}{m}\right)^{mt} - 1\right]} \\[9ex] = \dfrac{0.05 * 88000}{12 * \left[\left(1 + \dfrac{0.05}{12}\right)^{12 * 12} - 1\right]} \\[9ex] = \dfrac{4400}{12 * \left[(1 + 0.0041666667)^{144} - 1\right]} \\[5ex] = \dfrac{4400}{12 * \left[(1.0041666667)^{144} - 1\right]} \\[5ex] = \dfrac{4400}{12 * (1.819848874 - 1)} \\[5ex] = \dfrac{4400}{12 * 0.8198488741} \\[5ex] = \dfrac{4400}{9.838186489} \\[5ex] = 447.2368973 \\[3ex] \approx \$447.24 \\[3ex]$ Ceteris paribus, Rita should deposit about $447.24 in the fund to accumulate$88,000 in 12 years.
(6.) Calculate the future value of a $300 per week ordinary annuity at 4.5% per year compounded weekly for$8\dfrac{1}{2}$years. This is a case of the Future Value of Ordinary Annuity$ \underline{Future\:\:Value\:\:of\:\:Ordinary\:\:Annuity} \\[3ex] PMT = \$300 \\[3ex] r = 4.5\% = \dfrac{4.5}{100} = 0.045 \\[5ex] Compounded\:\:weekly \rightarrow m = 52 \\[3ex] t = 8\dfrac{1}{2}\:years = 8.5\:years \\[5ex] FV = ? \\[3ex] FV = m * PMT * \left[\dfrac{\left(1 + \dfrac{r}{m}\right)^{mt} - 1}{r}\right] \\[10ex] FV = 52 * 300 * \left[\dfrac{\left(1 + \dfrac{0.045}{52}\right)^{52 * 8.5} - 1}{0.045}\right] \\[10ex] = 15600 * \left[\dfrac{\left(1 + 0.000865384615\right)^{442} - 1}{0.045}\right] \\[10ex] = 15600 * \left[\dfrac{\left(1.000865384615\right)^{442} - 1}{0.045}\right] \\[7ex] = 15600 * \left[\dfrac{1.46570241 - 1}{0.045}\right] \\[5ex] = 15600 * \left[\dfrac{0.46570241}{0.045}\right] \\[5ex] = \dfrac{15600 * 0.46570241}{0.045} \\[5ex] = \dfrac{7264.9576}{0.045} \\[5ex] = 161443.502 \\[3ex] FV \approx \$161,443.50 \\[3ex] $Ceteris paribus, the future value of a$300 per week ordinary annuity at 4.5% per year compounded weekly for $8\dfrac{1}{2}$ years is about $161,443.50 (7.) (8.) Suppose you deposited$80 per month into a savings plan for 7 years and at the end of that period your balance was $19,500. What was the amount you earned in interest? A. It was$12,780. Calculate the amount deposited over years and then subtract it from the balance of $19,500. B. It is impossible to compute without knowing the APR. To find the correct answer, multiply each$80 deposit by the APR, and then multiply the product by the number of months in 7 years.

C. It is impossible to compute without knowing the APR. To find the correct answer, divide the number of payments by the APR.

D. It was $25,560. Multiply the balance of$19,500 by the number of years, 7.

E. It was $25,560. Calculate the amount deposited over 7 years and then add it to the balance of$19,500.

F. It was $12,780. Divide the balance of$19,500 by the number of years, 7.

$PMT = \$80 \\[3ex] m = 12 \\[3ex] t = 7\;years \\[3ex] FV = \$19500 \\[5ex] Total\:\:PMTs = PMT * m * t \\[3ex] = 80 * 12 * 7 \\[3ex] = \$6720 \\[3ex] CI = FV - Total\:\:PMTs \\[3ex] = 19500 - 6720 \\[3ex] = \$12,780$ The correct answer is Option A.
(9.)

(10.) At age 43, Augustine started saving for retirement.
If his investment plan pays an APR of 6% and he wants to have $0.8 million when he retires in 22 years, how much should he deposit monthly? This is a case of Sinking Fund$ r = 6\% = \dfrac{6}{100} = 0.06 \\[5ex] Monthly\;\;deposits \rightarrow Compounded\:\:monthly \rightarrow m = 12 \\[3ex] FV = \$0.8\;million = \$800,000 \\[3ex] t = 22\;years \\[3ex] PMT = \dfrac{r * FV}{m * \left[\left(1 + \dfrac{r}{m}\right)^{mt} - 1\right]} \\[9ex] = \dfrac{0.06 * 800000}{12 * \left[\left(1 + \dfrac{0.06}{12}\right)^{12 * 22} - 1\right]} \\[9ex] = \dfrac{48000}{12 * \left[(1 + 0.005)^{264} - 1\right]} \\[5ex] = \dfrac{48000}{12 * \left[(1.005)^{264} - 1\right]} \\[5ex] = \dfrac{48000}{12 * (3.731129336 - 1)} \\[5ex] = \dfrac{48000}{12 * 2.731129336} \\[5ex] = \dfrac{48000}{32.77355203} \\[5ex] = 1464.595597 \\[3ex] \approx \$1464.60 \\[3ex]$ Augustine should deposit about $1464.60 in the investment plan to have$0.8 million in 22 years.
(11.)

(12.)

(13.) Luke decided to begin saving for retirement at the age of 32.
He decided to deposit $70 at the end of each month in an IRA (Individual Retirement Account) that pays 5% compounded monthly. (a.) If he retires at 65 years of age, how much will he have in the IRA account? (b.) Calculate the interest. This is a case of the Future Value of Ordinary Annuity$ \underline{Future\:\:Value\:\:of\:\:Ordinary\:\:Annuity} \\[3ex] PMT = \$70 \\[3ex] r = 5\% = \dfrac{5}{100} = 0.05 \\[5ex] Compounded\:\:monthly \rightarrow m = 12 \\[3ex] t = 65 - 32 = 33\:years \\[3ex] FV = ? \\[3ex] (a.) \\[3ex] FV = m * PMT * \left[\dfrac{\left(1 + \dfrac{r}{m}\right)^{mt} - 1}{r}\right] \\[10ex] FV = 12 * 70 * \left[\dfrac{\left(1 + \dfrac{0.05}{12}\right)^{12 * 33} - 1}{0.05}\right] \\[10ex] = 840 * \left[\dfrac{\left(1 + 0.00416666667\right)^{396} - 1}{0.05}\right] \\[10ex] = 840 * \left[\dfrac{\left(1.00416666667\right)^{396} - 1}{0.05}\right] \\[7ex] = 840 * \left[\dfrac{5.18916096 - 1}{0.05}\right] \\[5ex] = 840 * \left[\dfrac{4.18916096}{0.05}\right] \\[5ex] = \dfrac{840 * 4.18916096}{0.05} \\[5ex] = \dfrac{3518.89521}{0.05} \\[5ex] = 70377.9042 \\[3ex] FV \approx \$70,377.90 \\[3ex] (b.) \\[3ex] CI = ? \\[3ex] CI = FV - Total\:\:PMTs \\[3ex] Total\:\:PMTs = PMT * m * t \\[3ex] Total\:\:PMTs = 70 * 12 * 33 = 27720 \\[3ex] \rightarrow CI = 70377.9042 - 27720 \\[3ex] CI = 42657.9042 \\[3ex] CI \approx \$42,657.90$
(14.) Perpetua works for an investment company, Company $A$ with which she earned a sum of $180,000 in her retirement account. She just got a position with another company, Company$B$. She intends to roll over her funds to a new account with her new company. She also plans to deposit$2000 per quarter into the new account until she retires 2% years from now.
Assume the new account earns interest at the rate of 3.5% per year compounded quarterly, how much will she have in her account when she retires?

This is a case of the Compound Interest and Future Value of Ordinary Annuity

$\underline{Compound\:\:Interest} \\[3ex] Roll-over\:\:of\:\:\$180,000\:\:into\:\:Company\:\:B \\[3ex] P = \$180000 \\[3ex] r = 3.5\% = \dfrac{3.5}{100} = 0.035 \\[5ex] Compounded\:\:Quarterly\rightarrow m = 4 \\[3ex] t = 25\:years \\[3ex] A = ? \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[7ex] A = 180000\left(1 + \dfrac{0.035}{4}\right)^{4 * 25} \\[7ex] = 180000\left(1 + 0.00875\right)^{100} \\[5ex] = 180000\left(1.00875\right)^{100} \\[5ex] = 180000 * 2.38976267 \\[3ex] A = 430157.281 \\[3ex] \underline{Future\:\:Value\:\:of\:\:Ordinary\:\:Annuity} \\[3ex] Quarterly\:\:Deposit\:\:of\:\:\$2,000\:\:into\:\:Company\:\:B \\[3ex] PMT = \$2000 \\[3ex] r = 3.5\% = \dfrac{3.5}{100} = 0.035 \\[5ex] Compounded\:\:Quarterly\rightarrow m = 4 \\[3ex] t = 25\:years \\[3ex] FV = ? \\[3ex] FV = m * PMT * \left[\dfrac{\left(1 + \dfrac{r}{m}\right)^{mt} - 1}{r}\right] \\[10ex] FV = 4 * 2000 * \left[\dfrac{\left(1 + \dfrac{0.035}{4}\right)^{4 * 25} - 1}{0.035}\right] \\[10ex] = 8000 * \left[\dfrac{\left(1 + 0.00875\right)^{100} - 1}{0.035}\right] \\[10ex] = 8000 * \left[\dfrac{\left(1.00875\right)^{100} - 1}{0.035}\right] \\[7ex] = 8000 * \left[\dfrac{2.38976267 - 1}{0.035}\right] \\[5ex] = 8000 * \left[\dfrac{1.38976267}{0.035}\right] \\[5ex] = \dfrac{8000 * 1.38976267}{0.035} \\[5ex] = \dfrac{11118.1014}{0.035} \\[5ex] FV = 317660.04 \\[3ex] \underline{Total\:\:Amount\:\:in\:\:her\:\:account\:\:at\:\:retirement} \\[3ex] Total\:\:Amount = A + FV \\[3ex] Total\:\:Amount = 430157.281 + 317660.04 \\[3ex] Total\:\:Amount = 430157.281 + 317660.04 \\[3ex] Total\:\:Amount = 747817.321 \\[3ex] Total\:\:Amount \approx \$747,817.32 $(15.) Between the ages of 22 and 28, Rita contributed$7000 per year in a 401(k) and her employer contributed $3500 per year (matched it 50%) on her behalf. The interest rate is 8.6% compounded annually. (a.) What is the value of the 401(k) at the end of the 6 years? After 6 years of working for this firm, she moved on to a new job. However, she kept her accumulated retirement funds in the 401(k). (b.) How much money will she have in the plan when she retires at 65 years? (c.) How much did she contribute to the 401(k) plan? (d.) How much did her employer contribute to the plan? (e.) What is the difference between the amount of money she would have accumulated in the 401(k) and the amount she contributed to the plan? This is a case of the Future Value of Ordinary Annuity and Compound Interest$ (a.) \\[3ex] \underline{Future\:\:Value\:\:of\:\:Ordinary\:\:Annuity} \\[3ex] PMT = \$7000 + \$3500 = \$10500...matched\:\:50\% \\[3ex] r = 8.6\% = \dfrac{8.6}{100} = 0.086 \\[5ex] Compounded\:\:annually \rightarrow m = 1 \\[3ex] t = 28\:years - 22\:years = 6\:years \\[3ex] FV = ? \\[3ex] FV = m * PMT * \left[\dfrac{\left(1 + \dfrac{r}{m}\right)^{mt} - 1}{r}\right] \\[10ex] FV = 1 * 10500 * \left[\dfrac{\left(1 + \dfrac{0.086}{1}\right)^{1 * 6} - 1}{0.086}\right] \\[10ex] = 10500 * \left[\dfrac{\left(1 + 0.086\right)^{6} - 1}{0.086}\right] \\[10ex] = 10500 * \left[\dfrac{\left(1.086\right)^{6} - 1}{0.086}\right] \\[7ex] = 10500 * \left[\dfrac{1.6405102624 - 1}{0.086}\right] \\[5ex] = 10500 * \left[\dfrac{0.6405102624}{0.086}\right] \\[5ex] = \dfrac{10500 * 0.6405102624}{0.086} \\[5ex] = \dfrac{6725.357755}{0.086} \\[5ex] = 78201.83437 \\[3ex] FV \approx \$78,201.83 \\[3ex] $(b.) The money was left in the$401(k)$from the age of$28$years until$65$years. No periodic deposits were made during that period$(65 - 28 = 37 years)$. However, the money earned interest. It was not withdrawn even when she found a new job. That money, the Future Value of$78201.83$was still being compounded annually at$8.6\%$interest rate. So, we have to use the Compound Interest Formula.$ \underline{Compound\:\:Interest} \\[3ex] P = \$78201.83437 \\[3ex] r = 8.6\% = \dfrac{8.6}{100} = 0.086 \\[5ex] Compounded\:\:annually \rightarrow m = 1 \\[3ex] t = 65\:years - 28\:years = 37\:years \\[3ex] A = ? \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[7ex] A = 78201.83437 * \left(1 + \dfrac{0.086}{1}\right)^{1 * 37} \\[7ex] = 78201.83437 * \left(1 + 0.086\right)^{37} \\[5ex] = 78201.83437 * \left(1.086\right)^{37} \\[5ex] = 78201.83437 * 21.16915556 \\[3ex] = 1655466.797 \\[3ex] A \approx \$1,655,466.80 \\[3ex] $(c.) Amount contributed to$401(k)$plan$ Amount \\[3ex] = 7000 * 6 \\[3ex] = \$42,000.00 \\[3ex]$ (d.)
Amount her employer contributed to the $401(k)$ plan

$Amount \\[3ex] = 3500 * 6 \\[3ex] = \$21,000.00 \\[3ex] $(e.) The difference between the amount of money she would have accumulated in the$401(k)$and the amount she contributed to the plan$ Difference \\[3ex] = 1655466.797 - 42000 \\[3ex] = 1613466.797 \\[3ex] \approx \$1,613,466.80$
(16.) Between the ages of 25 and 40, Lucy contributed $5000 per year in a 401(k) and her employer matched this contribution dollar for dollar. The interest rate is 7.5% compounded annually. (a.) What is the value of the 401(k) at the end of the 15 years? After 6 years of working for this firm, she moved on to a new job. However, she kept her accumulated retirement funds in the 401(k). (b.) How much money will she have in the plan when she retires at 60 years? (c.) How much did she contribute to the 401(k) plan? (d.) How much did her employer contribute to the plan? (e.) What is the difference between the amount of money she would have accumulated in the 401(k) and the amount she contributed to the plan? This is a case of the Future Value of Ordinary Annuity and Compound Interest$ (a.) \\[3ex] \underline{Future\:\:Value\:\:of\:\:Ordinary\:\:Annuity} \\[3ex] PMT = \$5000 + \$5000 = \$10000...matched\:\:dollar\:\:for\:\:dollar \\[3ex] r = 7.5\% = \dfrac{7.5}{100} = 0.075 \\[5ex] Compounded\:\:annually \rightarrow m = 1 \\[3ex] t = 40\:years - 25\:years = 15\:years \\[3ex] FV = ? \\[3ex] FV = m * PMT * \left[\dfrac{\left(1 + \dfrac{r}{m}\right)^{mt} - 1}{r}\right] \\[10ex] FV = 1 * 10000 * \left[\dfrac{\left(1 + \dfrac{0.075}{1}\right)^{1 * 15} - 1}{0.075}\right] \\[10ex] = 10000 * \left[\dfrac{\left(1 + 0.075\right)^{15} - 1}{0.075}\right] \\[10ex] = 10000 * \left[\dfrac{\left(1.075\right)^{15} - 1}{0.075}\right] \\[7ex] = 10000 * \left[\dfrac{2.958877353 - 1}{0.075}\right] \\[5ex] = 10000 * \left[\dfrac{1.958877353}{0.075}\right] \\[5ex] = \dfrac{10000 * 1.958877353}{0.075} \\[5ex] = \dfrac{19588.77353}{0.075} \\[5ex] = 261183.647 \\[3ex] FV \approx \$261,183.65 \\[3ex] $(b.) The money was left in the$401(k)$from the age of$40$years until$65$years. No periodic deposits were made during that period$(65 - 40 = 25 years)$. However, the money earned interest. It was not withdrawn even when she found a new job. That money, the Future Value of$261,183.65$was still being compounded annually at$7.5\%$interest rate. So, we have to use the Compound Interest Formula.$ \underline{Compound\:\:Interest} \\[3ex] P = \$261183.647 \\[3ex] r = 7.5\% = \dfrac{7.5}{100} = 0.075 \\[5ex] Compounded\:\:annually \rightarrow m = 1 \\[3ex] t = 65\:years - 40\:years = 25\:years \\[3ex] A = ? \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[7ex] A = 261183.647 * \left(1 + \dfrac{0.075}{1}\right)^{1 * 25} \\[7ex] = 261183.647 * \left(1 + 0.075\right)^{25} \\[5ex] = 261183.647 * \left(1.075\right)^{25} \\[5ex] = 261183.647 * 6.098339613 \\[3ex] = 1592786.581 \\[3ex] A \approx \$1,592,786.58 \\[3ex] $(c.) Amount contributed to$401(k)$plan$ Amount \\[3ex] = 5000 * 15 \\[3ex] = \$75,000.00 \\[3ex]$ (d.)
Amount her employer contributed to the $401(k)$ plan

$Amount \\[3ex] = 5000 * 15 \\[3ex] = \$75,000.00 \\[3ex] $(e.) The difference between the amount of money she would have accumulated in the$401(k)$and the amount she contributed to the plan$ Difference \\[3ex] = 1592786.581 - 75000 \\[3ex] = 1517786.581 \\[3ex] \approx \$1,517,786.58$
(17.) James contributed $6000 per year into a Traditional IRA (Traditional Individual Retirement Account) earning interest at the rate of 5% per year compounded annually, every year after age 35 until his retirement at age 65. John contributed$5000 per year into a Roth IRA (Roth Individual Retirement Account) earning interest at the rate of 5% per year compounded annually for a period of 30 years.
The investments of James and John are in a marginal tax bracket of 25% at the time of their retirement.
They both wish to withdraw all of the money in their IRAs at retirement.
(a.) After all due taxes are paid, who will have the larger amount?
(b.) How much larger will that amount be?

This is a case of the Future Value of Ordinary Annuity

$\underline{Future\:\:Value\:\:of\:\:Ordinary\:\:Annuity} \\[3ex] (a.) \\[3ex] James \\[3ex] Traditional\:\:IRA...taxed\:\:later \\[3ex] PMT = \$6000 \\[3ex] r = 5\% = \dfrac{5}{100} = 0.05 \\[5ex] t = 30\:years \\[3ex] Compounded\:\:annually \rightarrow m = 1 \\[3ex] FV = ? \\[3ex] FV = m * PMT * \left[\dfrac{\left(1 + \dfrac{r}{m}\right)^{mt} - 1}{r}\right] \\[10ex] FV = 1 * 6000 * \left[\dfrac{\left(1 + \dfrac{0.05}{1}\right)^{1 * 30} - 1}{0.05}\right] \\[10ex] = 6000 * \left[\dfrac{\left(1 + 0.05\right)^{30} - 1}{0.05}\right] \\[10ex] = 6000 * \left[\dfrac{\left(1.05\right)^{30} - 1}{0.05}\right] \\[7ex] = 6000 * \left[\dfrac{4.321942375 - 1}{0.05}\right] \\[5ex] = 60000 * \left[\dfrac{3.321942375}{0.05}\right] \\[5ex] = \dfrac{6000 * 3.321942375}{0.05} \\[5ex] = \dfrac{19931.65425}{0.05} \\[5ex] FV = 398633.085 \\[3ex] 25\%\:\:tax = \dfrac{25}{100} * 398633.085 = 0.25(398633.085) = 99658.27125 \\[5ex] Net\:\:Pay = FV - Tax \\[3ex] Net\:\:Pay = 398633.085 - 99658.27125 \\[3ex] Net\:\:Pay = 298974.8137 \\[3ex] Net\:\:Pay\:\:for\:\:James \approx \$298,874.81 \\[5ex] John \\[3ex] Roth\:\:IRA...taxed\:\:now\:\:even\:\:before\:\:making\:\:the\:\:deposit \\[3ex] PMT = \$5000 \\[3ex] r = 5\% = \dfrac{5}{100} = 0.05 \\[5ex] t = 30\:years \\[3ex] Compounded\:\:annually \rightarrow m = 1 \\[3ex] FV = ? \\[3ex] FV = m * PMT * \left[\dfrac{\left(1 + \dfrac{r}{m}\right)^{mt} - 1}{r}\right] \\[10ex] FV = 1 * 5000 * \left[\dfrac{\left(1 + \dfrac{0.05}{1}\right)^{1 * 30} - 1}{0.05}\right] \\[10ex] = 5000 * \left[\dfrac{\left(1 + 0.05\right)^{30} - 1}{0.05}\right] \\[10ex] = 5000 * \left[\dfrac{\left(1.05\right)^{30} - 1}{0.05}\right] \\[7ex] = 5000 * \left[\dfrac{4.321942375 - 1}{0.05}\right] \\[5ex] = 50000 * \left[\dfrac{3.321942375}{0.05}\right] \\[5ex] = \dfrac{5000 * 3.321942375}{0.05} \\[5ex] = \dfrac{16609.71188}{0.05} \\[5ex] FV = 332194.2375 \\[3ex] Net\:\:Pay = FV = 332194.2375 \\[3ex] Net\:\:Pay\:\:for\:\:John \approx \$332,194.24 \\[3ex] John\:\:has\:\:the\:\:larger\:\:amount \\[3ex] (b.) \\[3ex] By\:\:how\:\:much? \\[3ex] Difference = 332194.2375 - 298974.8137 \\[3ex] Difference = 33219.4238 \\[3ex] Difference \approx \$33,219.42 $(18.) Daniel intends to deposit a sum of$8000 into a bank account at the beginning of next month.
He also intends to deposit a sum of $225 per month into the same account at the end of that month and at the end of each subsequent month for the next 4 years. If the bank pays interest at a rate of 3% per year compounded annually, how much will he have in his account at the end of 4 years asssuming he makes no withdrawals during that period? This is a case of Compound Interest and the Future Value of Ordinary Annuity$ \underline{Compound\:\:Interest} \\[3ex] P = \$8000 \\[3ex] r = 3\% = \dfrac{3}{100} = 0.03 \\[5ex] t = 4\:years \\[3ex] Compounded\:\:monthly \rightarrow m = 12 \\[3ex] A = ? \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[5ex] A = 8000 * \left(1 + \dfrac{0.03}{12}\right)^{12 * 4} \\[5ex] = 8000 * \left(1 + 0.0025\right)^{48} \\[5ex] = 8000 * \left(1.0025\right)^{48} \\[5ex] = 8000 * 1.127328021 \\[3ex] A = 9018.624168 \\[3ex] \underline{Future\:\:Value\:\:of\:\:Ordinary\:\:Annuity} \\[3ex] PMT = \$225 \\[3ex] r = 3\% = \dfrac{3}{100} = 0.03 \\[5ex] t = 4\:years \\[3ex] Compounded\:\:monthly \rightarrow m = 12 \\[3ex] FV = ? \\[3ex] FV = m * PMT * \left[\dfrac{\left(1 + \dfrac{r}{m}\right)^{mt} - 1}{r}\right] \\[10ex] FV = 12 * 225 * \left[\dfrac{\left(1 + \dfrac{0.03}{12}\right)^{12 * 4} - 1}{0.03}\right] \\[10ex] = 2700 * \left[\dfrac{\left(1 + 0.0025\right)^{48} - 1}{0.03}\right] \\[10ex] = 2700 * \left[\dfrac{\left(1.0025\right)^{48} - 1}{0.03}\right] \\[7ex] = 2700 * \left[\dfrac{1.127328021 - 1}{0.03}\right] \\[5ex] = 2700 * \left[\dfrac{0.127328021}{0.03}\right] \\[5ex] = \dfrac{2700 * 0.127328021}{0.03} \\[5ex] = \dfrac{343.7856567}{0.03} \\[5ex] FV = 11459.52189 \\[3ex] Total\:\:Amount\:\:at:\:the\:\:end\:\:of\:\:4\:\:years \\[3ex] = A + FV \\[3ex] = 9018.624168 + 11459.52189 \\[3ex] = 20478.14606 \\[3ex] Total\:\:Amount \approx \$20,478.15$
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