A series of 10 end-of-year deposits is made that begins with $5,000 at the end of year 1 and decreases at the rate of $300 per year with 12% interest.
a) What amount could be withdrawn at t = 10? $
Round entry to the nearest dollar. Tolerance is ±4.
b) What uniform annual series of deposits (n = 10) would result in the same accumulated balance at the end of year 10? $
Round entry to the nearest dollar. Tolerance is ±4.

Respuesta :

Answer:

a) The amount that could be withdrawn at t = 10 is $68,872.

b) The uniform annual series of deposits is $3,925.

Explanation:

a) What amount could be withdrawn at t = 10? $ Round entry to the nearest dollar. Tolerance is ±4.

Note: See the attached excel for the calculation of the future value in year 10.

From the attached excel file, we have:

Future value in year 10 = $68,872

Therefore, the amount that could be withdrawn at t = 10 is $68,872.

b) What uniform annual series of deposits (n = 10) would result in the same accumulated balance at the end of year 10? $ Round entry to the nearest dollar. Tolerance is ±4.

To calculate the uniform annual series of deposits, we use the formula for calculating the future value of ordinary annuity for as follows:

FV = M * (((1 + r)^n - 1) / r) ................................. (1)

Where,

FV = Future value of the amount deposited in after 10 years = $68,872

M = Uniform annual series of deposits = ?

r = Interest rate = 12%, or 0.12

n = number of year = 10 years

Substituting the values into equation (1) and solve for M , we have:

$68,872 = M * (((1 + 0.12)^10 - 1) / 0.12)

$68,872 = M * 17.5487350695351

M = $68,872 / 17.5487350695351

M = $3,925

Therefore, the uniform annual series of deposits is $3,925.

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