Alex deposited \(x\) dollars into a new account that earned \(8\) percent annual interest, compounded annually. One year

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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
Source: — Problem Solving |

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Gmat_mission wrote:
Fri Jan 08, 2021 4:49 am
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
Solution:

We start by determining the new value of the x dollars Alex deposited into his account that earned 8 percent annual interest. At the end of the first year the amount of money in the account was 1.08x, and then he added another x dollars to the account, so the account then had a total value of 1.08x + x dollars. The 1.08x + x dollars earned another 8 percent interest for the year. Thus, the total value of his account at the end of the second year is:

1.08(1.08x + x) = (1.08^2)x + 1.08x

Since the new total value is equal to w, we can set up the following equation:

w = (1.08^2)x + 1.08x

Now we must get x in terms of w.

w = x(1.08^2 + 1.08)

w / (1.08^2 + 1.08) = x

Answer: D

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