SOLUTION

Is z less than 0?

(1) xy > 0 and yz < 0 --> the first inequality is useless for us since it has no z. From yz < 0 we can say that y and z have opposite sings, but we don't know the sign of y to deduce something about z. Not sufficient.

(2) x > 0. Not sufficient.

(1)+(2) From xy > 0 we can say that x and y have the same sign, and since from (2) x > 0 then y > 0 too. Now, from yz < 0 we have that y and z have opposite sings and as y > 0 then z < 0. Sufficient.

Answer: C.
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Is z < 0?

(1) xy > 0 which means
either x > 0 and y > 0
or x < 0 and y < 0

So either y > 0 or y < 0

If y > 0 and yz < 0 then z < 0
but if y < 0 and yz < 0 then z > 0

Hence Statement (1) is Not Sufficient

(2) x > 0
This statement by itself gives no information about z, so this is Not Sufficient

(1) & (2)
xy > 0 and yz < 0 and x > 0

if we take xy > 0 and x > 0 then y > 0
We also have yz < 0, as y > 0, so z < 0

So (1) & (2) together is Sufficient. Hence answer is C
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Is z less than 0?

Statement 1: xy > 0 and yz < 0.

xy > 0 means x and y have the same sign. i.e. if x>0 then y>0 and if x<0 then y<0

also, yz<0 means y and z have the opp. sign. Now since y can be either greater than 0 or less than zero, hence z can also attain two different value greater than zero when y is less than zero and less than zero when y is greater than zero.

Hence not sufficient

Statement 2 x > 0

this doesn't tell us anything about z hence not sufficient

combining statement 1 and 2 we know that both x and y are greater than zero. hence z will be less than zero. Thus sufficient.

hence C

I have done this in a different way.

Statement 1: xy > 0 and yz < 0
If you multiply both the terms, result would be (positive)(negative) = negative
(xy)(yz) < 0
xy^2z < 0
y^2 will always be positive due to even power
a) if x<0 then z>0 to make the above equation <0
b) if x>0 then z<0 to make the above equation <0
Two choices so not sufficient

Statement 2: x > 0. Not sufficient. No information about y and z

Statement 1 + 2 : Clearly shows that x>0 so z <0 - Sufficient.

Bunnel: Please let me know if I have missed anything.

St1) x and z will have opposite signs. But don’t know which is negative
St2) No info about z. clearly INSUF
1+2-->per st2 x is positive and since st1 tells us that x and z have opposite signs z has to be negative. SUFF
Ans C
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1) xy/yz<0
x/z<0
z<0 or z>0
2) x>0
1+2) x/z<0 , x>0, so z<0

Let's analyse each statement:
S1. We can form 2 cases here
a) x and y are both positive; one of y or z is positive and one negative- if y is positive, z will have to be negative
b) x and y are both negative; one of y or z is positive and one negative - if y is negative, z will have to be positive

Therefore, since 2 different cases of y are possible, this statement is not sufficient. Strike off A and D.

S2. this doesn't really say anything about z. Not sufficient straight-away. Strike off B.

combining the 2, we can clearly see that y will have to be +ve since x is (and for xy to be >0, they both have to be either positive or both negative). There is only 1 case in our workings above that specify what will happen if y is positive; z will be negative (since the second part of S1 has said yz<0).

The answer to the question is z<0 is a firm Yes through C, hence that is our final answer.

Hope that Is helpful
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