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Re: Is Z less than 0? [#permalink]
20 Mar 2012, 00:21

1

This post received KUDOS

Expert's post

Is z less than 0?

(1) xy>0 and yz<0 --> 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.

Re: Is Z less than 0? [#permalink]
20 Mar 2012, 00:28

Hi,

I hope this helps

We have to prove that Is z < 0? Statement 1. xy>0 and yz<0 => x & y are negative / x & y are positive => z is postive / z is negative Hence Insufficient

2. x>0 Insufficient since nothing said about y or z

Combine X is positive = >X is positive Y is positive Thus Z is negative

Hence sufficient _________________

Giving +1 kudos is a better way of saying 'Thank You'.

Re: Is Z less than 0? [#permalink]
14 Mar 2013, 18:28

Bunuel wrote:

Is z less than 0?

(1) xy>0 and yz<0 --> 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.

Hope it's clear.

for the first statement I concluded that x and y have the same signs and since y times z is negative so z is the opposite sign . I made a chart trying +/- signs for each until I got A ... still cannot quite understand why C ? _________________

Re: Is Z less than 0? [#permalink]
15 Mar 2013, 00:14

Expert's post

TheNona wrote:

Bunuel wrote:

Is z less than 0?

(1) xy>0 and yz<0 --> 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.

Hope it's clear.

for the first statement I concluded that x and y have the same signs and since y times z is negative so z is the opposite sign . I made a chart trying +/- signs for each until I got A ... still cannot quite understand why C ?

Yes, xy > 0, means that x and y are either both positive or both negative. But this does not help us much.

yz < 0 means that y and z have the opposite signs. So, if y is negative then z is positive and if y is positive then z negative. Hence from the first statement z could be positive as well as negative.

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