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D
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Difficulty:
55%
(hard)
Question Stats:
61% (01:53) correct
39%
(01:58)
wrong
based on 889
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If 6xy = \(x^2\)y + 9y, what is the value of xy?
(1) x = –2
(2) x < 0
(1) x = –2
(2) x < 0
The answer is D, with the following explanation:
The equation in question can be rephrased as follows:
x2y – 6xy + 9y = 0
y(x2 – 6x + 9) = 0
y(x – 3)2 = 0
Therefore, one or both of the following must be true:
y = 0 or
x = 3
It follows that the product xy must equal either 0 or 3y. This question can therefore be rephrased several different ways: "What is y?" or "Does x = 3?"
(1) SUFFICIENT: If x = -2, then:
(–2)2y – 6(–2)y + 9y = 0
4y + 12y + 9y = 0
25y = 0
Therefore y = xy = 0. Tn fact, for any value of x other than 3, y must equal zero.
(2) SUFFICIENT: x is negative. Therefore, x cannot equal 3, and it follows that y = 0. Therefore, xy = 0.
The correct answer is D.
What I did, i simply applied the info from statement 1, and got that -12Y=13Y which can only be true if Y is 0 and therefore, XY is 0.
In the second statement, i did the same and the result came to be the same. Is this faulty reasoning? Also, this is said to be 700-800 level question and my way seems just too easy to solve such a high level problem...
The equation in question can be rephrased as follows:
x2y – 6xy + 9y = 0
y(x2 – 6x + 9) = 0
y(x – 3)2 = 0
Therefore, one or both of the following must be true:
y = 0 or
x = 3
It follows that the product xy must equal either 0 or 3y. This question can therefore be rephrased several different ways: "What is y?" or "Does x = 3?"
(1) SUFFICIENT: If x = -2, then:
(–2)2y – 6(–2)y + 9y = 0
4y + 12y + 9y = 0
25y = 0
Therefore y = xy = 0. Tn fact, for any value of x other than 3, y must equal zero.
(2) SUFFICIENT: x is negative. Therefore, x cannot equal 3, and it follows that y = 0. Therefore, xy = 0.
The correct answer is D.
What I did, i simply applied the info from statement 1, and got that -12Y=13Y which can only be true if Y is 0 and therefore, XY is 0.
In the second statement, i did the same and the result came to be the same. Is this faulty reasoning? Also, this is said to be 700-800 level question and my way seems just too easy to solve such a high level problem...
19 Dec 2010, 17:22
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MisterEko
For statement (1) your reasoning looks OK. As for statement (2) what do you mean by "did the same"?
One can also do this way: \(6xy=x^2y + 9y\) --> \(y(x^2-6x+9)=0\) --> \(y(x-3)^2=0\) --> either \(x=3\) or \(y=0\) (or both).
(1) x = –2 --> so \(x\neq{3}\), then \(y=0\) and \(xy=0\). Sufficient.
(2) x < 0 --> again \(x\neq{3}\), then \(y=0\) and \(xy=0\). Sufficient.
Answer: D.
Check question #1 here: inequality-and-absolute-value-questions-from-my-collection-86939.html for harder version of the same problem.
Hope it helps.
17 Sep 2015, 20:24
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anupamadw
Usually, it is not a good idea to cancel out a variable. You lose a solution if you do.
If you do not cancel out the y, you get that this inequality: 6xy=x^2y + 9y holds when either x = 3 or y = 0.
If you do cancel out the y, you get that this inequality: 6xy=x^2y + 9y holds when x = 3.
You lost the y = 0 value. Hence, it's always better to take out y common and keep it on the side, not cancel it.
