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svetoff [14.1K]
3 years ago
8

A woman worked for 30 years before retiring. At the end of the first year of employment she deposited 5000 into an account for h

er retirement. At the end of each subsequent year of employment, she deposited 3% more than the prior year. The woman made a total of 30 deposits. She will withdraw 50,000 at the beginning of the first year of retirement and will make annual withdrawals at the beginning of each subsequent year for a total of 30 withdrawals. Each of these subsequent withdrawals will be 3% more than the prior year. The final withdrawal depletes the account. The account earns a constant annual effective interest rate. Calculate the account balance after the final deposit and before the first withdrawal.
Business
1 answer:
Mrrafil [7]3 years ago
8 0

Answer:

$797,837

Explanation:

the first withdrawal is $50,000

the second is $51,500

and so on...

the formula that used to solve the interest rate earned by the annuity is:

$50,000 x {[(1 + i)³⁰ - (1 + 3%)³⁰] / [(1 + i)³⁰ x (i - 3%)]} x (1 + i) = $5,000 x {[(1 + i)³⁰ - (1 + 3%)³⁰] / (i - 3%)}

we start to simplify the equation by cancelling  {[(1 + i)³⁰ - (1 + 3%)³⁰] / (i - 3%)}

[$50,000 x (1 + i)] / (1 + i)³⁰ = $5,000

now we cancel $5,000 on each side:

[10 x (1 + i)] / (1 + i)³⁰ = 1

now lets take away (1 + i):

10 / (1 + i)²⁹ = 1

things get a little bit more simple now:

10 = (1 + i)²⁹

²⁹√10 = ²⁹√(1 + i)²⁹

1.082636734 = 1 + i

i = 1.082636734 - 1 = 0.082636734 = 8.2636734%

now we replace i in any equation:

= $50,000 x {[(1 + 0.082636734)³⁰ - 1.03³⁰] / [(1 + 0.082636734)³⁰ x (0.082636734 - 0.03)]} x (1 + 0.082636734)

= $50,000 x  {[10.82636738 - 2.427262471] / [10.82636738 x 0.052636734]} x (1 + 0.082636734)

= $50,000 x  {8.399104909 / 0.56986462} x (1.082636734)

= $50,000 x 14.73877236 x 1.082636734

= $797,837

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