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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.

Re: 254. If x < y < z, is xyz > 0? (1) xy > 0. (2) xz > 0. [#permalink]

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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

Re: 254. If x < y < z, is xyz > 0? (1) xy > 0. (2) xz > 0. [#permalink]

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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(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

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(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.

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.

Re: 254. If x < y < z, is xyz > 0? (1) xy > 0. (2) xz > 0. [#permalink]

Show Tags

29 Apr 2014, 04:13

1

This post received KUDOS

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(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.

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 \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

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