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# GMATClub Tests M28-03

Author Message
Current Student
Joined: 26 Dec 2011
Posts: 6
Schools: Yale '17 (A)
Followers: 0

Kudos [?]: 3 [0], given: 3

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19 Dec 2013, 17:02
If $$x \neq y$$ and $$y = \frac{x^2-y^2}{x-y}$$, then what is the value of y?

1) $$x+y = 3$$

2) $$x(y-3) = 0$$

[Reveal] Spoiler:
A

My question is why not D? below is my logic

Through manipulation we get:
$$y = (x+y)$$
Thus, x = 0 and y = (x+y)

Statement 1 gives the very value of x+y. Sufficient.
Statement 2 prompts two scenarios:
$$x(y-3) = 0$$

$$x= 0$$
OR
$$(y-3) = 0$$

Scenario A
X = 0

OR

Scenario B
Y = 3

From Scenario A we find that if X = 0, y can take any value

From Scenarios B we find that if y = 3, then x can take any value; BUT since we know from the question stem that X has to be equal to zero then if Y = 3, X = 0.
So now looking back at Scenario A, if X = 0, then y *cannot* take any value because it must be equal to 3. Hence D.
Math Expert
Joined: 02 Sep 2009
Posts: 36508
Followers: 7063

Kudos [?]: 92855 [0], given: 10528

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19 Dec 2013, 23:28
NateTheGreat11 wrote:
If $$x \neq y$$ and $$y = \frac{x^2-y^2}{x-y}$$, then what is the value of y?

1) $$x+y = 3$$

2) $$x(y-3) = 0$$

[Reveal] Spoiler:
A

My question is why not D? below is my logic

Through manipulation we get:
$$y = (x+y)$$
Thus, x = 0 and y = (x+y)

Statement 1 gives the very value of x+y. Sufficient.
Statement 2 prompts two scenarios:
$$x(y-3) = 0$$

$$x= 0$$
OR
$$(y-3) = 0$$

Scenario A
X = 0

OR

Scenario B
Y = 3

From Scenario A we find that if X = 0, y can take any value

From Scenarios B we find that if y = 3, then x can take any value; BUT since we know from the question stem that X has to be equal to zero then if Y = 3, X = 0.
So now looking back at Scenario A, if X = 0, then y *cannot* take any value because it must be equal to 3. Hence D.

That's not correct. Why cannot y be 1 for (2)? Or 0, or 100?
_________________
Re: GMATClub Tests M28-03   [#permalink] 19 Dec 2013, 23:28
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# GMATClub Tests M28-03

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