OG 11 Q D41

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OG 11 Q D41

by skhalilov » Tue May 15, 2007 10:25 am
The question reads: If x+y/z>0, is x<0?

(1) x<y
(2) z<0

The answer in the book is C.

I can see that from (2) we know that x+y has to be negative. But the explanation in the book claims that x has to be lower than -y, thus it both x and y are negative. But what if you have say, x= 1 and y = -2? You would still have a negative nominator.

Any thoughts ??

Thanks!!

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by bww » Tue May 15, 2007 2:16 pm
Ok, so from I) we know that x<y, thus clearly not sufficient. II) we know that z<0, so the denominator is negative, and for the statement to be true, the numerator (x+y) must be negative as well. But, we can't determine if x is pos or neg. II) alone is not sufficient.

Both statements put together, z must be negative, x<y, and x+y<0 to satisfy the statement. I think the answer is yes, X<0 and it takes both statements together to be sufficient. If you let x be a postiive number, y must be a greater positive number and thus does not follow that x+y<0.

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by atarafder » Tue May 15, 2007 7:17 pm
for x+y/z>0 -
1. (x+y) and z are both positive
2. (x+y) and z are both negative

using A - Insufficient
using B - Insufficient

If you combine the two -
x<y - means (x+y) is negative and z is negative - sufficient.