If x

Question

If x < y < z, is xyz > 0? (1) xy > 0. (2) xz > 0.

Bunuel · Answer

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 00 , 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.

Spidy001 · Answer

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.

subhashghosh · Answer

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.

thesfactor · Answer

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

Bunuel · Answer

Bumping for review and further discussion.

abid1986 · Answer

what is wrong with my approach

Taking 1 & 2
xy -xz > 0
x(y-z) > 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.

Bunuel · Answer

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: help-with-add-subtract-mult-divid-multiple-inequalities-155290.html Next, even if we had x(y-z) > 0, then it would mean that either both multiples are positive or both multiples are negative: x>0 and y-z>0 ( y>z ). x<0 and y-z<0 ( y

WoundedTiger · Answer

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 and < ) --> a-c>b-d (take the sign of the inequality you subtract from). Example: 3<4 and 5>1 --> 3-5<4-1 . The 2 inequalities in question can be added but not subtracted. For more on Inequalities refer: help-with-add-subtract-mult-divid-multiple-inequalities-155290.html#p1242251 inequality-and-absolute-value-questions-from-my-collection-86939.html#p652806

MarcusFenix · Answer

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?

Bunuel · Answer

Yes, your approach is correct. It's basically the same as in my post here: if-x-y-z-is-xyz-110615.html#p888498

stonecold · Answer

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

almazovniko · Answer

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

Basshead · Answer

If x < y < z, is xyz > 0?

(1) xy > 0.

x and y must have the same signs. However, we don't know anything about z. Insufficient.

(2) xz > 0.

x and z must have the same signs. However, we don't know anything about y. Insufficient.

(1&2) xy > 0 and xz > 0.

x, y, z can all be positive or all negative. Insufficient.

Answer is E.

bumpbot · Answer

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If x < y < z, is xyz > 0?

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