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Explanation needed

Expert replies
by silivest60 » Fri Nov 30, 2007 10:37 am
I think (E) is the answer for this one, but I'm not sure....
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Source: — Data Sufficiency |

by StarDust845 » Mon Dec 03, 2007 4:22 pm
The answer is C.

Here is an explanation, let me know if you don't follow my argument anywhere. It is a lengthy to explain to someone, but is easy in its own right.

1) a and b can't be 0.
2) a and b MUST both be +ve or -ve (otherwise we can't get values for
(-a, b) and (-b, a) so that they lie in the same quadrant).
3) We can infer from 2 that that the points (-a, b) and (-b, a) MUST lie either in Quadrant II or IV, going anticlock wise in a graph.

4) Given xy > 0, i.e. x and y can't be 0 and both of them have to be positive or negative.

If both are +ve then, (-x, y) lies in Quadrant II, if both are -ve then (-x, y) lies in Quadrant IV. But they need not necessarily lie in the same quadrant as those two points (-a, b) and (-b, a).

So the answer to the GMAT questions can't be A or D. Now we have to examine the second statement.

5) ax > 0 i.e. x can't be 0 and both a and x are positve or -ve. We can't solve the problem from this only, because y can then be +ve or -ve or 0 so that (-x, y) can always go to a diferent quad or no quad at all - Remember if y is 0, the point (-x,y) doesn't belong to any quad.

Now let's combine both, this means xy > 0 and ax > 0. From 4) above we concluded that (-x,y) can go to quadrant II or IV but we don't know which one. Now add another restriction i.e. ax > 0

xy > 0 and ax > 0 implies

6) x > 0 and y > 0 --> in this case (-x , y) belongs to Quadrant II and because 'a' MUST be > 0 (ax > 0) and 'b' MUST be > 0 from (2) above
(-a, b) and (-b, a) belongs to Quad II.

7) another possibility is x < 0 and y <0> following similar reasoning as in 6) they belong to the same quadrant IV.

Hence the answer to the question is (C), they like in the same quadrant.

Calista.
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THANK YOU

by silivest60 » Mon Dec 03, 2007 7:12 pm
BRAVO Calista!
Great job! Your solution is logical and fluent....I have no problems understanding it...Let me just correct a minor thing....So, statement (1) gives you B, C or E...instead of A or D.
I guess what makes this question difficult is the number of variables that you have to keep track of....
thanks again,
Sorin
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my mistake....

by silivest60 » Mon Dec 03, 2007 7:17 pm
Sorry Calista
my mistake....I miread the first time. You did say "can't be A or D". So it's OK!
Thanks
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