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Re: If y < x < z and xyz < 0, is xy > 0? [#permalink]
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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Expert Reply
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]
1
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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]
1
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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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Re: 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


(1) sufic
yz>0: y,z - - or + +
yxz: - (-) - xy>0
yxz: + (+) + invalid

(2) insufic
xz>0: x,z - - or + +
yxz: (-) - - xy>0
yxz: (-) + + xy<0

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