question 16, 4.c.78 - t part 2 of 3 abe deposits $50 each month for 30 years in an account with an apr of…

question 16, 4.c.78 - t part 2 of 3 abe deposits $50 each month for 30 years in an account with an apr of 3.5%. beatrice deposits $100 each month for 15 years in an account with an apr of 3.5%. a. verify that abe and beatrice deposit the same amount of money during the stated periods of time. how much do they deposit? b. use technology to compute the accumulated balance in each account and explain the results. a. select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. a. abe and beatrice both deposit $18000. b. abe deposits $ and beatrice deposits $ different from abes deposit. b. the accumulated balance in abes account is $ (round to the nearest cent as needed.)
Answer
Explanation:
Step1: Calculate Abe's total deposit
Abe deposits $50 per month for 30 years. There are 12 months in a year, so the number of months $n = 30\times12=360$. The total deposit $A_{1}=50\times360 = 18000$.
Step2: Calculate Beatrice's total deposit
Beatrice deposits $100 per month for 15 years. The number of months $m=15\times 12 = 180$. The total deposit $A_{2}=100\times180=18000$.
Step3: Calculate accumulated - balance for Abe (using the future - value of an ordinary annuity formula $F = P\times\frac{(1 + r)^{n}-1}{r}$, where $P$ is the monthly payment, $r$ is the monthly interest rate, and $n$ is the number of payments. The annual percentage rate (APR) is 3.5%, so the monthly interest rate $r=\frac{0.035}{12}$, and $n = 360$, $P = 50$)
$r=\frac{0.035}{12}\approx0.002917$, $F_{1}=50\times\frac{(1 + 0.002917)^{360}-1}{0.002917}$. $(1 + 0.002917)^{360}\approx2.8338$. $F_{1}=50\times\frac{2.8338 - 1}{0.002917}=50\times\frac{1.8338}{0.002917}\approx31409.77$.
Step4: Calculate accumulated - balance for Beatrice (using the future - value of an ordinary annuity formula with $P = 100$, $r=\frac{0.035}{12}$, and $n = 180$)
$r=\frac{0.035}{12}\approx0.002917$, $n = 180$. $(1 + 0.002917)^{180}\approx1.6909$. $F_{2}=100\times\frac{1.6909 - 1}{0.002917}=100\times\frac{0.6909}{0.002917}\approx23685.29$.
Answer:
a. A. Abe and Beatrice both deposit $18000. b. The accumulated balance in Abe's account is $31409.77.