ganesh999 wrote:
Bunuel wrote:
gmatbull wrote:
If w + x < 0, is w - y . 0?
(1) x + y < 0
(2) y < x < w
I don't know why i missed this question; with regards to (1)
Please explain your steps.
If w + x < 0 , is w - y > 0 ?Question: is \(w>y\)?
(1) x + y < 0 --> for this statement best way would be to pick numbers: on DS questions when plugging numbers,
goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.
If \(x=0\), \(w=-1\) and \(y=-2\) then the answer would be YES but if \(x=0\), \(w=-2\) and \(y=-1\) then the answer would be. Not sufficient.
(2) \(y<x<w\) --> ignore \(x\) --> \(y<w\), directly tells us the answer. Sufficient.
Answer: B.
Hello Buneul,
I'm struck with the following approach.
For the condition w > y to hold,
I undertook the following approach.
w + x < 0 , w - y > 0
I subtracted both of them
I got x + y < 0.
So i deduced that for w - y > 0 to hold, we need the condition x + y < 0, condition that is present in the answer choice.
and hence I wrote D as the answer.
Am I missing something?
Hello
See, the thing is that you are assuming w-y to be > 0.
The catch here is that IF w+x < 0 and also IF w-y > 0, THEN x+y < 0.
BUT it does NOT mean that IF w+x < 0 and IF x+y < 0, then w-y will be > 0.
Eg, consider w=-4, x=-5, y=-3. Here w+x is negative and x+y is negative, but w-y is not greater than 0, rather w is less than y here.