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Is x^2 + y^2 > 6?

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by mridula » Fri Feb 20, 2009 10:46 am
Is x^2 + y^2 > 6?

(1) (x + y)^2 > 6
(2) xy = 2
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

Can anyone explain to me how to answer this question?

OA is E
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Source: — Data Sufficiency |
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by DanaJ » Fri Feb 20, 2009 11:39 am
1 means that x^2 + y^2 + 2xy > 6. However, since we don't have any info on xy, we can't tell if x^2 + y^2 > 6. So 1 is insufficient.
2 should be examined by using the basic inequality that (x + y)^2 >= 0, giving us that x^2 + y^2 + 2xy >=0. Since xy = 2, then 2xy = 4. Subract 4 from the previous inequality and you get that x^2 + y^2 >= - 4. This however does not help establish if x^2 + y^2 is greater than 6.

Taken together, the two stmts tell us that x^2 + y^2 > 2, but again, this doesn't help us with the initial question. So the OA is ok.
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by mridula » Fri Feb 20, 2009 11:45 am
DanaJ wrote:1 means that x^2 + y^2 + 2xy > 6. However, since we don't have any info on xy, we can't tell if x^2 + y^2 > 6. So 1 is insufficient.
2 should be examined by using the basic inequality that (x + y)^2 >= 0, giving us that x^2 + y^2 + 2xy >=0. Since xy = 2, then 2xy = 4. Subract 4 from the previous inequality and you get that x^2 + y^2 >= - 4. This however does not help establish if x^2 + y^2 is greater than 6.

Taken together, the two stmts tell us that x^2 + y^2 > 2, but again, this doesn't help us with the initial question. So the OA is ok.
Got it! I was just over thinking!!!!! :D.

Lesson Learnt: Sometimes, solutions are easier than what we think they are! :)

Thank You very much.
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