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3 years ago

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Suppose an individual makes an initial investment of $3,000 in an account that earns 6.6%, compounded monthly, and makes additio

nal contributions of $100 at the end of each month for a period of 12 years. After these 12 years, this individual wants to make withdrawals at the end of each month for the next 5 years (so that the account balance will be reduced to $0). (Round your answers to the nearest cent.)
(a) How much is in the account after the last deposit is made?
(b) How much was deposited?
(c) What is the amount of each withdrawal?
(d) What is the total amount withdrawn?
I get A and C. If you could explain B and D I'd appreciate it.

1 answer:

makkiz [27]3 years ago

7 0

Answer:

b) $17,400

d) $33,517.20

Step-by-step explanation:

a) $28,482.19 . . . . future value of all deposits

__

b) The initial deposit was $3000, and there were 144 deposits of $100 each, for a total of ...

$3000 +144×100 = $17,400 . . . . total deposited

__

c) $558.62

__

d) 60 monthly withdrawals were made in the amount $558.62, for a total of ...

60×$558.62 = $33,517.20 . . . . total withdrawn

_____

<em>Additional information about (a) and (c)</em>

(a) The future value of the initial deposit is the deposit multiplied by the interest multiplier over the period.

A = P(1 +r/n)^(nt) = 3000(1 +.066/12)^(12·12) = 3000·1.0055^144 ≈ 6609.065

The future value of $100 deposits each month is the sum of the series of 144 terms with common ratio 1.0055 and initial value 100.

A = 100(1.055^144 -1)/0.0055 ≈ 21,873.123

So, the total future value is ...

$6609.065 +21873.123 ≈ $28482.188 ≈ $28,482.19

__

(c) The withdrawal amount can be found using the same formula used for loan payments:

A = P(r/n)/(1 -(1 +r/n)^(-nt)) = $28482.19(.0055)/(1 -1.0055^-60) ≈ $558.62

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