Alice and Bruce each bought a refrigerator . . .

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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 .

Experts, may you tell me how to solve this PS question please? I need some help.

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by GMATGuruNY » Thu Nov 16, 2017 8:31 am
VJesus12 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 EconomistGMATTutor » Thu Nov 16, 2017 9:07 am
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 .
Hi VJesus12,
Let's take a look at your question.

This question can be solved using system of linear equations.
Let Alice paid x dollars and Bruce paid y dollars.

The sum of their purchases is $900, therefore,
$$x+y=900$$
$$y=900-x...(i)$$

The question states that:
"Twice of what Alice paid was $75 more than what Bruce paid"
$$2x=75+y...\left(ii\right)$$

Plugin value of y from eq(i) to eq(ii),
$$2x=75+900-x$$
$$2x=975-x$$
$$2x+x=975$$
$$3x=975$$
$$x=\frac{975}{3}$$
$$x=325$$

Since x is the amount that Alice paid for her purchase, therefore, She paid $325.
Hence, option B is correct.

Hope it helps.
I am available if you'd like any follow up.
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by Brent@GMATPrepNow » Thu Nov 16, 2017 9:11 am
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
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

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by Scott@TargetTestPrep » Wed Oct 16, 2019 6:08 pm
VJesus12 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 .

Experts, may you tell me how to solve this PS question please? I need some help.
We can let a = the amount Alice paid for her refrigerator and b = the amount Bruce paid for his refrigerator. Therefore,

a + b = 900

and

2a = b + 75

Isolating b in the second equation, we have b = 2a - 75. Substituting this for b in the first equation, we have:

a + 2a - 75 = 900

3a = 975

a = 325

Answer: B

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