Question

Danny Metzger's parents invested $1600 when he was born. This money is to be used for Danny's college education and is to be withdrawn in four equal annual payments beginning when Danny is age 19. Find the amount that will be available each year, if money is worth 7%, compounded annually. (Round your answer to the nearest cent.)

Answer 1

Answer:

$\begin{array}{ccl}Year&&End \ of \ year \ balance\\1&&\ 4,483.18\\2&& \ 3,088.68\\3&&\ 1,596.57\\4&&0\end{array}$

Step-by-step explanation:

The initial amount invested, P = $1,600

Danny's age at which the amount, A, is to be used for his college, t = 19 years

The number of equal annual payments, m, to be withdrawn from the amount = 4

The compound interest on the account, r = 7% = 0.07

Let, A, represent the amount at the year the annual withdrawals starts to be made, we have;

$A = P \times \left(1+\dfrac{r}{n} \right)^{n \times t}$

n = The number of times the interest is applied annually = 1

Therefore;

$A = 1,600 \times \left(1+\dfrac{0.07}{1} \right)^{1 \times 19} \approx 5,786.444$

The amount, m, is withdrawn at start of Danny's first year in college to give the amount in the account = A - m

The amount in the account at the end of the first year with compound interest, r = (A - m)×(1 + r)¹ = (A - m)×(1 + r)

At the stat of the second year, the second withdrawal is made to give the starting amount = (A - m)×(1 + r) - m

The amount in the account at the end of the second year = ((A - m)×(1 + r) - m)×(1 + r)

At the start of the third year, the amount in the account =  ((A - m)×(1 + r) - m)×(1 + r) - m

At the end of the third year, we have the amount in the account  = (((A - m)×(1 + r) - m)×(1 + r) - m) × (1 + r)

At the start of the forth year, the last yearly installment is withdrawn from the account and we have 0 balance in the account.

Therefore, on the fourth year, we have the amount in the account = (((A - m)×(1 + r) - m)×(1 + r) - m) × (1 + r) - m = 0

(A - m)×(1 + r)³ - m×(1 + r)² - m × (1 + r) - m = 0

A×(1 + r)³ - m×(1 + r)³ - m×(1 + r)² - m × (1 + r) - m = 0

A×(1 + r)³ = m×(1 + r)³ + m×(1 + r)² + m×(1 + r) + m = m × ((1 + r)³ + (1 + r)² + (1 + r) + 1)

A×(1 + r)³ = m × ((1 + r)³ + (1 + r)² + (1 + r) + 1)

m = A×(1 + r)³/((1 + r)³ + (1 + r)² + (1 + r) + 1)

∴ m = 5,786.444×(1 + 0.07)³/((1 + 0.07)³ + (1 + 0.07)² + (1 + 0.07) + 1) ≈ 1,596.56

The amount withdrawn annually, m ≈ $1,596.56

The amount in the account at the end of the each year is given as follows;

First year = (A - m)×(1 + r) = (5,786.444 - 1,596.56)×(1 + 0.07) ≈ 4,483.17588

Second year = ((A - m)×(1 + r) - m)×(1 + r) = ((5,786.444 - 1,596.56)×(1 + 0.07) - 1,596.56)×(1 + 0.07) = 3,088.68

Third year = (((A - m)×(1 + r) - m)×(1 + r) - m) × (1 + r) = (((5,786.444 - 1,596.56)×(1 + 0.07) - 1,596.56)×(1 + 0.07) - 1,596.56) × (1 + 0.07) = 1,596.57

At the end of the first year, we have $4,483.17588

At the end of the second year, we have $3,088.68

At the end of the third year, we have $1,596.57

At the end of the fourth year, we have 0