-
silivest60
- Junior | Next Rank: 30 Posts
- Posts: 29
- Joined: Sat Oct 27, 2007 9:49 pm
- Location: Vancouver, BC, Canada
- Thanked: 1 times
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.
