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]
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- bubbliiiiiiii
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Rule: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
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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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.
If you know our own innovative logics to find the answer, you don't need to actually solve the problem.
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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.
If you know our own innovative logics to find the answer, you don't need to actually solve the problem.
www.mathrevolution.com
l The one-and-only World's First Variable Approach for DS and IVY Approach for PS that allow anyone to easily solve GMAT math questions.
l The easy-to-use solutions. Math skills are totally irrelevant. Forget conventional ways of solving math questions.
l The most effective time management for GMAT math to date allowing you to solve 37 questions with 10 minutes to spare
l Hitting a score of 45 is very easy and points and 49-51 is also doable.
l Unlimited Access to over 120 free video lessons at https://www.mathrevolution.com/gmat/lesson
l Our advertising video at https://www.youtube.com/watch?v=R_Fki3_2vO8
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As shown in my post above, the equation in the question prompt (4xy = x²y + 4y) requires that x=2 or that y=0.sandipgumtya wrote:Hi Mitch,
didn't understand how u r getting statement 2 as sufficient.
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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Followed here and elsewhere by over 1900 test-takers.
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As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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