John has been saving $x annually since 3 years ago, and the

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John has been saving $x annually since 3 years ago, and the interest rate for his savings increased r% annually. After saving the money for 3 years as such, he withdrew all the money in the account and spent it all in one year. If he spent a constant amount per each month, how much money did he spend each month in terms of x and r?

(A) x/2 (1+r/100)

(B) x/4 (1+r/100)

(C) x/6 (1+r/100)

(D) x/4 (1+r/50)

(E) x/4 (1+r/25)

OA=D

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by [email protected] » Sun May 14, 2017 11:13 am
Hi ziyuenlau,

What is the source of this question? I ask because it's rather vaguely-worded for an 'interest rate' question. It's unclear when money was put into the bank account (was it there from the very 1st day or was it put in at the end of each year?). It's also not explained how the interest is calculated (simple or compound) nor what the 'starting interest rate' was (before the increases). GMAT question writers carefully word their questions to avoid ambiguity in the language - and this prompt is lacking the specifics to be considered a realistic example of what you'd see on Test Day.

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by elias.latour.apex » Mon May 15, 2017 1:50 pm
The question starts: John has been saving $x annually since 3 years ago...

This is incorrect English grammar. It should read: John has been saving $x annually for 3 years...

It seems that this question was constructed by someone who does not speak English perfectly.
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hazelnut01 wrote:
Sat May 13, 2017 5:51 pm
John has been saving $x annually since 3 years ago, and the interest rate for his savings increased r% annually. After saving the money for 3 years as such, he withdrew all the money in the account and spent it all in one year. If he spent a constant amount per each month, how much money did he spend each month in terms of x and r?

(A) x/2 (1+r/100)

(B) x/4 (1+r/100)

(C) x/6 (1+r/100)

(D) x/4 (1+r/50)

(E) x/4 (1+r/25)

OA=D
The problem is worded poorly. Let’s assume the annual interest rate is a constant r%.

The $x he saved 3 years ago is now x(1 + 3r/100).

The $x he saved 2 years ago is now x(1 + 2r/100).

The $x he saved 1 year ago is now x(1 + r/100).

Therefore, the total he can withdraw now is x(3 + 6r/100) = 3x(1 + 2r/100) = 3x(1 + r/50). Since he withdraws this amount in 12 equal payments, he receives 3x(1 + r/50)/12 = x(1 + r/50)/4 in each payment.

Answer: D

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