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

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If y < x < z and xyz < 0, is xy > 0? [#permalink]

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

1) yz > 0
2) xz > 0
[Reveal] Spoiler: OA

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Re: If y < x < z and xyz < 0, is xy > 0? [#permalink]

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

As xyz < 0, we know that either x,y, & z are all negative or only y is negative.

Question asks whether xy > 0, which basically means whether only y is negative?

1) yz > 0

Case 1, both y and z are negative.

In this case x also becomes negative. therefore xy is positive.

Case 2. both y and z are positive.

In this case the original condition x*y*z < 0 does not hold true, hence this case is not possible.

As only 1 case is possible, this statement is sufficient.


2) xz > 0

Case 1

Both x and z are negative and y is also negative, so x*y becomes positive

Case 2

Only x and z are positive and y is negative, so x*y becomes negative

As there are two possibilities, this statement is not sufficient.

Answer is A
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Re: If y < x < z and xyz < 0, is xy > 0? [#permalink]

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New post 17 Apr 2017, 18:37
1) yz > 0

Case 1, both y and z are negative.

In this case x also becomes negative. therefore xy is positive.

Case 2. both y and z are positive.

In this case the original condition x*y*z < 0 does not hold true, hence this case is not possible.


can someone explain on this?

Thanks
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Re: If y < x < z and xyz < 0, is xy > 0? [#permalink]

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himanshu22 wrote:
1) yz > 0

Case 1, both y and z are negative.

In this case x also becomes negative. therefore xy is positive.

Case 2. both y and z are positive.

In this case the original condition x*y*z < 0 does not hold true, hence this case is not possible.


can someone explain on this?

Thanks



Hi...
CASE II does not exit..
Because for xyz to be NEGATIVE, y has to be -ive in all cases.

Solution:-

Given that xyz<0..
Two cases..
a) all three x,y and z are NEGATIVE
b) only one is NEGATIVE and other two are POSITIVE.. AND if only one is NEGATIVE, it has to be the smallest value, which is y here

So y is NEGATIVE for sure and NONE is 0

Let's see the statements...
1)yz>0...
Since y is NEGATIVE, z will also be NEGATIVE..
As negative*negative = positive
And z alone cannot be NEGATIVE, even x will be NEGATIVE then only xyz<0..
So xz>0
Suff

2) xz>0...
Both x and z can be -ive, then xy will be>0
or both can be +ive, then xy will be<0..
Different answers possible
Insufficient

A
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Re: If y < x < z and xyz < 0, is xy > 0? [#permalink]

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A in my opinion.

If xyz<0 then 2 cases
1. Y<0
2. All <0

From statement 1. yz>0 means z is negative and hence case 2 is satisfied. So statement 2 is sufficient.

From statement 2.xz>0 gives no information of sign about x. Hence insufficient


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Re: If y < x < z and xyz < 0, is xy > 0? [#permalink]

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

1) yz > 0
2) xz > 0

lets evaluate
y < x < z and xyz < 0 this info is given and hold true

product of 3 numbers will be negative when all three are negative or either of them is negative
also
y < x < z
when all three negative above eq can hold true
but for single variable to be negative
+ve <+ve<-ve (never possible)
+ve <-ve<+ve(never possible)
only possible case is when y is negative

first statement
yz>0
both positive or both negative
if both are positive then x<0 for xyz<0 to hold true but x cannot be negative as per above
So all three are negative hence xy>0 Sufficient

St 2
xz > 0
both positive or both negative
so both can be positive and y can be negative above discussion
xy<0
all three can be negative and xy>0
Not sufficient

answer is A
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Re: If y < x < z and xyz < 0, is xy > 0? [#permalink]

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

Let test cases to achieve the info in prompt:

We need either 3 -ve numbers or 1 -ve number.

Case 1: -3 < -2 < -1 .....xyz < 0

Case 2: -3 < 2 < 3......xyz < 0

No other cases

1) yz > 0

Case 1 is viable, while case 2 is not viable.

X must be -ve........xy>0......Answer is always Yes

Sufficient

2) xz > 0

Both cases are viable.

X is either -ve or +ve

Insufficient

Answer: A
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If y < x < z and xyz < 0, is xy > 0? [#permalink]

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GMATPrepNow wrote:
If y < x < z and xyz < 0, is xy > 0?

1) yz > 0
2) xz > 0


Target question: Is xy > 0?

Given: y < x < z and xyz < 0
If xyz < 0, then there are only TWO POSSIBLE CASES:
case a: all 3 numbers are negative (x, y and z are all NEGATIVE)
case b: 1 number is negative and the other 2 numbers are positive. Since y < x < z, then it must be the case that y is NEGATIVE, and x and z are POSITIVE

Statement 1: yz > 0
Let's compare this information with our given information.
The statement 1 information satisfies the conditions in case a. Reason: In case a, y and z are both negative, so yz > 0. So, case a IS possible
Conversely, this same information does not satisfy the conditions in case b. Reason: In case b, y is negative and z is positive, so yz < 0. So, case b is NOT possible
Since only case a is possible, we can be certain that x, y and z are all NEGATIVE, which means xy > 0
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: xz > 0
We'll compare this information with our given information.
The statement 2 information satisfies the conditions in case a. Reason: In case a, y and z are both negative, so xz > 0. So, case a IS possible
This information ALSO satisfies the conditions in case b. Reason: In case b, x and z are both positive, so xz > 0. So, case b IS possible
So, BOTH cases are possible.
In case a, x, y and z are all NEGATIVE, which means xy > 0
In case b, y is negative and x and z are positive, which means xy < 0
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT


Answer:
[Reveal] Spoiler:
A


Cheers,
Brent
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If y < x < z and xyz < 0, is xy > 0?   [#permalink] 18 Apr 2017, 06:37
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