If w + x <0, is w - y >0?
(1) x+y < 0
(2) y < x < w
please explanations ...
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Hi Carlo,Carlo75 wrote:If w + x <0>0?
(1) x+y < 0
(2) y < x < y
please explanations ...
Could you type the question again? Statement 2, of course, doesn't make much sense as it's written! I've just learned myself that using < and > can call up html tags, so you should click the 'Disable HTML in this post' button before submitting to make things read correctly.
As per question,
w+x<0> w+x-y<y> w-y<-(x+y).
Stem1: x+y<0> w-y <x>0. But we can not say w-y is <or> 0. Hence, this is not sufficient.
Stem2: y<x<w> Add -y to all, 0<x-y<w> w-y >0. Hence, this is sufficient.
So, answer is B.
w+x<0> w+x-y<y> w-y<-(x+y).
Stem1: x+y<0> w-y <x>0. But we can not say w-y is <or> 0. Hence, this is not sufficient.
Stem2: y<x<w> Add -y to all, 0<x-y<w> w-y >0. Hence, this is sufficient.
So, answer is B.
There was a formatting problem in my earlier reply. pl refer this.
As per question,
w+x<0. this can be writen as (w+x-y)<y then, (w-y)<-(x+y).
Stem1: x+y<0. So (w-y)<x> 0. But we can't say (w-y) is <or> 0. Hence, this is not sufficient.
Stem2: y<x<w. Add -y to all, 0<(x-y)<w>0. Hence, this is sufficient.
So, answer is B.
As per question,
w+x<0. this can be writen as (w+x-y)<y then, (w-y)<-(x+y).
Stem1: x+y<0. So (w-y)<x> 0. But we can't say (w-y) is <or> 0. Hence, this is not sufficient.
Stem2: y<x<w. Add -y to all, 0<(x-y)<w>0. Hence, this is sufficient.
So, answer is B.
