WoundedTiger wrote:
If \(xyz < 0\), is \(x < 0\)?
(1) \(x - y < 0\)
(2) \(x - z < 0\)
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Product of 3 nos can be negative if one of them is negative and other two nos are of the same sign or if all the nos are negative.
Case 1:x=-1,y=2,z=100
Case 2:x=1,z=2,y=-100
Case 3 x=1,y=3,z=-100
Case 4:x=-100,y=-2,z=-1
Consider St 1 says : x<y...So out of the table following cases are possible
Cases 1,3 and 4: For cases 1,4 we see that x<0 but case 3 x>0...So St 1 is insufficient
St 2 says x<z, so we have case 1,2 and 4
If it is case 1 and 4 then x<0 and answer to our question is yes but if it case 2 then answer is no
Combining we see that for both statements Case and 1 and 4 are applicable and for these cases x<0.
Ans is C
Hi,
This method is little longer, but just though to put out so that some users can understand what cases should be considered for such type of questions.
we have given a base condition that xyz<0
we can infer two things form this
either x<0&y<0&z<0 or one of x,y,z is <0.
Now Statement 1:
x-y<0
So we have three cases for this
x<0 & y<0 and |x|>|y|
But since xyz<0 from this we have x<0,y<0,z<0
x<0 & y>0 and |x|>|y|
But since xyz<0 from this we have x<0,y>0,z>0
x>0 & y>0 and |y|>|x|
But since xyz<0 from this we have x<0,y>0,z>0
Now Statement 1:
x-y<0
So we have three cases for this
x<0 & y<0 and |x|>|y| ------(a)
But since xyz<0 from this we have x<0,y<0,z<0
x<0 & y>0 and |x|>|y| ------(b)
But since xyz<0 from this we have x<0,y>0,z>0
x>0 & y>0 and |y|>|x|------(c)
But since xyz<0 from this we have x>0,y>0,z<0
Now Statement 2: x-z<0
So we have three cases for this
x<0 & z<0 and |x|>|z| -----(d)
But since xyz<0 from this we have x<0,y<0,z<0
x<0 & z>0 and |x|>|z|-----(e)
But since xyz<0 from this we have x<0,y>0,z>0
x>0 & z>0 and |z|>|x|-----(f)
But since xyz<0 from this we have x>0,y<0,z<0
Combing 1 and 2 we have
cases a, b, d,e we see that
we get either x<0, y<0, z<0
or
x<0 , y>0, z>0
But in both the cases x<0
Probus
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Probus
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