1. sidney opened a savings account as an annuity for a childs college fund. monthly deposits will be made…

1. sidney opened a savings account as an annuity for a childs college fund. monthly deposits will be made for 18 years, or 216 months, at a 5% average annual rate of return. 2. if sidney deposits $50 each month, the account will have $10,800 in deposits and $6,462.84 interest for a future value in 18 years of $17,262.84. 3. if sidney deposits $100 each month, the account will have $21,600 in deposits and $12,925.69 interest for a future value in 18 years of $34,525.69. 4. the future value in 18 years for $200 monthly deposits is $69,051.37 and $400 monthly deposits is $138,102.75. consider sidneys annuity example above with a 5% average rate of return. 1) which monthly payment will have a future value more than $100,000 in 18 years? $100 $200 $400 2) the monthly payment needed for a future value of $50,000 in 18 years is less than $50 between $100 and $200

1. sidney opened a savings account as an annuity for a childs college fund. monthly deposits will be made for 18 years, or 216 months, at a 5% average annual rate of return. 2. if sidney deposits $50 each month, the account will have $10,800 in deposits and $6,462.84 interest for a future value in 18 years of $17,262.84. 3. if sidney deposits $100 each month, the account will have $21,600 in deposits and $12,925.69 interest for a future value in 18 years of $34,525.69. 4. the future value in 18 years for $200 monthly deposits is $69,051.37 and $400 monthly deposits is $138,102.75. consider sidneys annuity example above with a 5% average rate of return. 1) which monthly payment will have a future value more than $100,000 in 18 years? $100 $200 $400 2) the monthly payment needed for a future value of $50,000 in 18 years is less than $50 between $100 and $200

Answer

  1. Question 1:
    • Explanation:

      • Step1: Recall the future - value of an ordinary annuity formula

        • The formula for the future - value of an ordinary annuity is $F = A\times\frac{(1 + r)^{n}-1}{r}$, where $F$ is the future value, $A$ is the monthly payment, $r$ is the monthly interest rate, and $n$ is the total number of periods. The annual interest rate $i = 5%=0.05$, so the monthly interest rate $r=\frac{0.05}{12}$, and the number of years $t = 18$ years, so the number of periods $n=18\times12 = 216$ months. We want to find $A$ when $F = 100000$.
        • Rearranging the formula for $A$, we get $A=\frac{F\times r}{(1 + r)^{n}-1}$.

      • Step2: Calculate $r$ and substitute values

        • $r=\frac{0.05}{12}\approx0.004167$. Then $(1 + r)^{n}=(1 + 0.004167)^{216}$. Using a calculator, $(1 + 0.004167)^{216}\approx2.45409$.
        • $A=\frac{100000\times0.004167}{2.45409 - 1}=\frac{416.7}{1.45409}\approx286.6$. Since $200\lt286.6\lt400$, the answer is $$200$.
    • Answer: $$200$
  2. Question 2:
    • Explanation:

      • Step1: Use the future - value of an ordinary annuity formula

        • We know $F = 50000$, $r=\frac{0.05}{12}$, and $n = 216$. Using $A=\frac{F\times r}{(1 + r)^{n}-1}$.
        • First, calculate $(1 + r)^{n}=(1+\frac{0.05}{12})^{216}\approx2.45409$.
        • Then $A=\frac{50000\times\frac{0.05}{12}}{2.45409 - 1}=\frac{50000\times0.004167}{1.45409}=\frac{208.35}{1.45409}\approx143.3$. So the monthly payment is less than $$150$.
    • Answer: less than $$150$