Alice and Bruce each bought a refrigerator, and the sum of

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Alice and Bruce each bought a refrigerator, and the sum of their purchases was $900. If twice of what Alice paid was $75 more than what Bruce paid, what did Alice pay for her refrigerator?

A. $275
B. $325
C. $425
D. $575
E. $625

The OA is B.

Please, can anyone explain this PS question? I tried to solve it but I can't get the correct answer. Thanks.

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by Brent@GMATPrepNow » Tue Apr 24, 2018 10:24 am
swerve wrote:Alice and Bruce each bought a refrigerator, and the sum of their purchases was $900. If twice of what Alice paid was $75 more than what Bruce paid, what did Alice pay for her refrigerator?

A. $275
B. $325
C. $425
D. $575
E. $625

The OA is B.

Please, can anyone explain this PS question? I tried to solve it but I can't get the correct answer. Thanks.
This question lends itself nicely to testing the answer choices

We'll start with answer choice C ($425), the middle value.
If Alice paid $425, how much did Bruce pay?
Given: twice of what Alice paid was $75 more than what Bruce paid
Two times $425 = $850, so Bruce paid $775 (since $850 - $75 = $775
COMBINED PAYMENTS = $425 + $775 = $1200
No good - we need the combined payments to be $900
ELIMINATE C
We can also ELIMINATE D and E, because those values will yield an even greater values of the combined payments.

Now let's try B ($325)
Given: twice of what Alice paid was $75 more than what Bruce paid
Two times $325= $650, so Bruce paid $575 (since $650 - $75 = $575
COMBINED PAYMENTS = $325 + $575 = $900
Perfect!!!

Answer: B

Cheers,
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by Brent@GMATPrepNow » Tue Apr 24, 2018 10:24 am
swerve wrote:Alice and Bruce each bought a refrigerator, and the sum of their purchases was $900. If twice of what Alice paid was $75 more than what Bruce paid, what did Alice pay for her refrigerator?

A. $275
B. $325
C. $425
D. $575
E. $625

The OA is B.

Please, can anyone explain this PS question? I tried to solve it but I can't get the correct answer. Thanks.
We can also solve the question algebraically, using 1 variable or 2 variables.
Here's a solution with 1 variable:

Given: Twice of what Alice paid was $75 more than what Bruce paid
Let x = the amount Alice paid
So, 2x - 75 = the amount Bruce paid

Given: The sum of their purchases was $900
So, (the amount Alice paid) + (the amount Bruce paid) = 900
We get: x + (2x - 75) = 900
Simplify: 3x - 75 = 900
Solve: x = 325

Answer: B

Cheers,
Brent
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by GMATGuruNY » Tue Apr 24, 2018 10:31 am
swerve wrote:Alice and Bruce each bought a refrigerator, and the sum of their purchases was $900. If twice of what Alice paid was $75 more than what Bruce paid, what did Alice pay for her refrigerator?

A. $275
B. $325
C. $425
D. $575
E. $625
Let A = Alice's amount and B = Bruce's amount.

Twice of what Alice paid was $75 more than what Bruce paid.
Since twice Alice's amount is equal to $75 more than Bruce's amount, we get:
2A = 75 + B
2A - 75 = B.

The sum of their purchases was $900.
Thus, the two blue values above must sum to $900:
A + (2A-75) = 900
3A = 975
A = 325.

The correct answer is B.
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by Scott@TargetTestPrep » Thu Apr 26, 2018 3:01 pm
swerve wrote:Alice and Bruce each bought a refrigerator, and the sum of their purchases was $900. If twice of what Alice paid was $75 more than what Bruce paid, what did Alice pay for her refrigerator?

A. $275
B. $325
C. $425
D. $575
E. $625
We can let a = the amount Alice paid and b = the amount Bruce paid. Thus we have:

a + b = 900

and

2a = b + 75

Thus, b = 2a - 75. Substituting this into the first equation, we have:

a + 2a - 75 = 900

3a = 975

a = 325

Answer: B

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