Alex deposited \(x\) dollars into a new account that earned \(8\) percent annual interest, compounded annually. One year later Alex deposited an additional \(x\) dollars into the account. If there were no other transactions and if the account contained \(w\) dollars at the end of two years, which of the following expresses \(x\) in terms of \(w?\)
A. \(\dfrac{w}{1+1.08}\)
B. \(\dfrac{w}{1.08+1.16}\)
C. \(\dfrac{w}{1.16+1.24}\)
D. \(\dfrac{w}{1.08+1.08^2}\)
E. \(\dfrac{w}{1.08^2+1.08^2}\)
Answer: D
Source: Official Guide
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Alex deposited \(x\) dollars into a new account that earned \(8\) percent annual interest, compounded annually. One year
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In the first year, the initial deposit of $x earns 8% interest, so after one year, the account holds 1.08x dollars.
Then an additional $x is deposited in the account, so the account now contains 1.08x + x = x(1.08 + 1) dollars.
This amount now earns 8% interest over the second year, so at the end of two years, the account holds (1.08)(x)(1.08 + 1) dollars. Since this amount is w, we can solve the following for x:
w = (1.08)(x)(1.08 + 1)
x = w/[(1.08)(1.08 + 1)]
x = w/(1.08^2 + 1.08)
and the answer is D.
Then an additional $x is deposited in the account, so the account now contains 1.08x + x = x(1.08 + 1) dollars.
This amount now earns 8% interest over the second year, so at the end of two years, the account holds (1.08)(x)(1.08 + 1) dollars. Since this amount is w, we can solve the following for x:
w = (1.08)(x)(1.08 + 1)
x = w/[(1.08)(1.08 + 1)]
x = w/(1.08^2 + 1.08)
and the answer is D.
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