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

_________________

Probus

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