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If x < y < z, is xyz > 0?
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09 Mar 2011, 15:01
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63% (01:33) correct 37% (01:20) wrong based on 176 sessions
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If x < y < z, is xyz > 0? (1) xy > 0. (2) xz > 0.
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Re: 254. If x < y < z, is xyz > 0? (1) xy > 0. (2) xz > 0.
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09 Mar 2011, 15:35
banksy wrote: 254. If x < y < z, is xyz > 0? (1) xy > 0. (2) xz > 0. If x < y < z, is xyz > 0?(1) xy > 0 > x and y have the same sign. Now, if both x and y are positive, then we would have that \(0<x<y<z\), so in this case all three would be positive, which would mean that \(xyz>0\), but if both x and y are negative, then z could be positive as well as negative thus xyz may or may not be positive. Not sufficient. (2) xz > 0 > x and z have the same sign and as \(x < y < z\) then all three have the same sign: if all of them are positive then \(xyz>0\) but if all of them are negative then \(xyz<0\). Not sufficient. (1)+(2) It's still possible all three to be positive as well as negative. Not sufficient. Answer: E.
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Re: 254. If x < y < z, is xyz > 0? (1) xy > 0. (2) xz > 0.
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09 Mar 2011, 16:38
x,y,z need not be integers .
1. xy>0 , doesnt say anything about z. Not sufficient.
2. same thing here , we dont know anything about y . Not sufficient.
Together x>0 , y>0,z>0 xyz >0 x<0, y<0,z<0 xyz <0 , so not sufficient.
Answer is E.



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Re: 254. If x < y < z, is xyz > 0? (1) xy > 0. (2) xz > 0.
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09 Mar 2011, 21:45
From (1) xy > 0 means both have same sign, hence if both are negative: then xyz < 0 if z < 0, and xyz > 0 if z is positive And if both are poitive, then xyz is always > 0. So (1) is not enough. From (2) xz > 0, so here too both x and z have same sign, and hence y will also have same sign But if x,y and z are negative, then xyz < 0 and if x,y and z are positive, then xyz > 0 From(1) and (2), x, y,z can be either all +ve or all ve, so the answer is E.
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Re: 254. If x < y < z, is xyz > 0? (1) xy > 0. (2) xz > 0.
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18 Mar 2011, 05:30
This one is quite simple to be honest...think about it. The question is asking if either x, y or z are negative. If one of them is negative, then xyz = ve. More importantly if z is negative then all three numbers are negative. Statement 1 says xy>0 so they can be either positive or negative. INSUFF Statement 2 says xz>0 ....same as above. 1+2 is the same deal. So E



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Re: If x < y < z, is xyz > 0?
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23 Feb 2014, 05:30
Bumping for review and further discussion.
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Re: 254. If x < y < z, is xyz > 0? (1) xy > 0. (2) xz > 0.
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29 Apr 2014, 03:04
Bunuel wrote: banksy wrote: 254. If x < y < z, is xyz > 0? (1) xy > 0. (2) xz > 0. If x < y < z, is xyz > 0?(1) xy > 0 > x and y have the same sign. Now, if both x and y are positive, then we would have that \(0<x<y<z\), so in this case all three would be positive, which would mean that \(xyz>0\), but if both x and y are negative, then z could be positive as well as negative thus xyz may or may not be positive. Not sufficient. (2) xz > 0 > x and z have the same sign and as \(x < y < z\) then all three have the same sign: if all of them are positive then \(xyz>0\) but if all of them are negative then \(xyz<0\). Not sufficient. (1)+(2) It's still possible all three to be positive as well as negative. Not sufficient. Answer: E. what is wrong with my approach Taking 1 & 2 xy xz > 0 x(yz) > 0 either x > 0 or y > z y > z is discarded because it contradicts the stem therefore x > 0 therefor y & z is also > 0 Hence xyz > 0 ,Sufficient.



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Re: 254. If x < y < z, is xyz > 0? (1) xy > 0. (2) xz > 0.
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29 Apr 2014, 04:01
abid1986 wrote: Bunuel wrote: banksy wrote: 254. If x < y < z, is xyz > 0? (1) xy > 0. (2) xz > 0. If x < y < z, is xyz > 0?(1) xy > 0 > x and y have the same sign. Now, if both x and y are positive, then we would have that \(0<x<y<z\), so in this case all three would be positive, which would mean that \(xyz>0\), but if both x and y are negative, then z could be positive as well as negative thus xyz may or may not be positive. Not sufficient. (2) xz > 0 > x and z have the same sign and as \(x < y < z\) then all three have the same sign: if all of them are positive then \(xyz>0\) but if all of them are negative then \(xyz<0\). Not sufficient. (1)+(2) It's still possible all three to be positive as well as negative. Not sufficient. Answer: E. what is wrong with my approach Taking 1 & 2 xy xz > 0 x(yz) > 0either x > 0 or y > zy > z is discarded because it contradicts the stem therefore x > 0 therefor y & z is also > 0 Hence xyz > 0 ,Sufficient. First of all you cannot subtract xz > 0 from xy > 0, because the signs of the inequalities are in the same direction, you can only add them. Adding/subtracting/multiplying/dividing inequalities: helpwithaddsubtractmultdividmultipleinequalities155290.htmlNext, even if we had x(yz) > 0, then it would mean that either both multiples are positive or both multiples are negative: \(x>0\) and \(yz>0\) (\(y>z\)). \(x<0\) and \(yz<0\) (\(y<z\)). Hope it helps.
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Re: 254. If x < y < z, is xyz > 0? (1) xy > 0. (2) xz > 0.
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29 Apr 2014, 04:13
abid1986 wrote: Bunuel wrote: banksy wrote: 254. If x < y < z, is xyz > 0? (1) xy > 0. (2) xz > 0. If x < y < z, is xyz > 0?(1) xy > 0 > x and y have the same sign. Now, if both x and y are positive, then we would have that \(0<x<y<z\), so in this case all three would be positive, which would mean that \(xyz>0\), but if both x and y are negative, then z could be positive as well as negative thus xyz may or may not be positive. Not sufficient. (2) xz > 0 > x and z have the same sign and as \(x < y < z\) then all three have the same sign: if all of them are positive then \(xyz>0\) but if all of them are negative then \(xyz<0\). Not sufficient. (1)+(2) It's still possible all three to be positive as well as negative. Not sufficient. Answer: E. what is wrong with my approach Taking 1 & 2 xy xz > 0 x(yz) > 0 either x > 0 or y > z
y > z is discarded because it contradicts the stem therefore x > 0 therefor y & z is also > 0 Hence xyz > 0 ,Sufficient. Note From Bunuel: ADDING/SUBTRACTING INEQUALITIES: You can only add inequalities when their signs are in the same direction:If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) > \(a+c>b+d\). Example: \(3<4\) and \(2<5\) > \(3+2<4+5\). You can only apply subtraction when their signs are in the opposite directions:If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\). The 2 inequalities in question can be added but not subtracted. For more on Inequalities refer: helpwithaddsubtractmultdividmultipleinequalities155290.html#p1242251inequalityandabsolutevaluequestionsfrommycollection86939.html#p652806
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Re: If x < y < z, is xyz > 0?
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30 Apr 2014, 16:33
The way I approached this question is:
(1) xy > 0  This tells us that xy are both positive or both negative but doesn't tell us anything about z; hence insufficient.
(2) xz > 0  Same as (1) doesn't tell us anything about y.
Combined x,y,z might all be positive in which case xyz will be > 0 or all be negative is which case xyz < 0; hence insufficient.
Is this a correct approach or am I missing something?



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Re: If x < y < z, is xyz > 0?
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01 May 2014, 09:54
MarcusFenix wrote: The way I approached this question is:
(1) xy > 0  This tells us that xy are both positive or both negative but doesn't tell us anything about z; hence insufficient.
(2) xz > 0  Same as (1) doesn't tell us anything about y.
Combined x,y,z might all be positive in which case xyz will be > 0 or all be negative is which case xyz < 0; hence insufficient.
Is this a correct approach or am I missing something? Yes, your approach is correct. It's basically the same as in my post here: ifxyzisxyz110615.html#p888498
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Re: If x < y < z, is xyz > 0?
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13 Mar 2016, 08:09
This Question is similar to the case of is PQR even here we can make test cases for P,Q,R which shows us => PQR>0 or PQR <0 hence E
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Re: If x < y < z, is xyz > 0?
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03 Jan 2019, 20:18
What a silly mistake. Got caught up figuring out if it was = 0 in any of the cases. Together they dont. So picked C. But +/ were still possibilities.... thats what happens when you take a break from studying :D




Re: If x < y < z, is xyz > 0?
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03 Jan 2019, 20:18






