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Expert replies
by bubbliiiiiiii » Tue Aug 25, 2015 4:35 am
If x and y are integers and 4xy = x2y + 4y, what is the value of xy?

(1) y − x = 2
(2) x3 < 0

OA B

My question:

[spoiler]Why can't we cancel out y in question stem and work with a quadratic equation in x?[/spoiler]
Regards,

Pranay
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Source: — Data Sufficiency |

by GMATGuruNY » Tue Aug 25, 2015 5:16 am
bubbliiiiiiii wrote:If x and y are integers and 4xy = x²y + 4y, what is the value of xy?

(1) y − x = 2
(2) x³ < 0
Rule:
If an equation includes an unknown that could be equal to 0, DO NOT DIVIDE BY THE UNKNOWN.

Here, it is possible that y=0, so we cannot divide 4xy = x²y + 4y by y.
Instead, we should COLLECT LIKE TERMS, as follows:
0 = x²y - 4xy + 4y
0 = y(x² - 4y + 4)
0 = y(x-2)².
The right side will be equal to 0 if y=0 or if x=2.
Question stem, rephrased:
If it is required that x=2 or that y=0, what is the value of xy?

Statement 1: y − x = 2
If x=2, then y=4.
In this case, xy = 2*4 = 8.
If y=0, then x=-2.
In this case, xy = (-2)(0) = 0.
Since xy can be different values, INSUFFICIENT.

Statement 2: x³ < 0
Since x=2 is not possible, the condition in red requires that y=0.
Thus, xy = (x)(0) = 0.
SUFFICIENT.

The correct answer is B.
Last edited by GMATGuruNY on Wed Aug 26, 2015 9:35 am, edited 1 time in total.
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by Max@Math Revolution » Wed Aug 26, 2015 6:43 am
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.

If x and y are integers and 4xy = x2y + 4y, what is the value of xy?

(1) y - x = 2
(2) x3 < 0

==> using the original condition we can get (x^2)y-4xy+4y=0, y(x^2-4x+4)=0, y(x-2)^2=0 therefore y=0 or x=2. Thus we have two variables (x,y) and one equation (4xy = x2y + 4y) meaning we would need one more equation, therefore D is likely the answer. .

When we actually solve the problem,

in case of 1), x=2, y=4 or y=0, x=-2 makes it not unique, thus is not sufficient
in case of 2, x^3<0==> x<0 makes it impossible for x to have value 2, thus y = 0 and xy=0. This is sufficient
Therefore the answer is B.



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by sandipgumtya » Wed Aug 26, 2015 7:56 am
Hi Mitch,
didn't understand how u r getting statement 2 as sufficient.
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by GMATGuruNY » Wed Aug 26, 2015 9:56 am
sandipgumtya wrote:Hi Mitch,
didn't understand how u r getting statement 2 as sufficient.
As shown in my post above, the equation in the question prompt (4xy = x²y + 4y) requires that x=2 or that y=0.

Statement 2: x³ < 0
Here, it is not possible that x=2.
Since the equation in the question prompt requires that x=2 or that y=0, and Statement 2 indicates that x≠2, it must be true that y=0.
Since y=0, xy=0.
Since the value of xy can be determined, Statement 2 is SUFFICIENT.
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