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# If xyz < 0, is x < 0 ?

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Math Expert
Joined: 02 Sep 2009
Posts: 55732
If xyz < 0, is x < 0 ?  [#permalink]

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27 Jun 2014, 10:21
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Difficulty:

45% (medium)

Question Stats:

64% (01:45) correct 36% (01:40) wrong based on 561 sessions

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If $$xyz < 0$$, is $$x < 0$$?

(1) $$x - y < 0$$

(2) $$x - z < 0$$

Kudos for a correct solution.

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Math Expert
Joined: 02 Sep 2009
Posts: 55732
If xyz < 0, is x < 0 ?  [#permalink]

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27 Jun 2014, 10:22
1
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SOLUTION

If $$xyz < 0$$, is $$x < 0$$?

$$xyz < 0$$ implies that either all three unknowns are negative or one is negative and the remaining two are positive.

(1) $$x - y < 0$$. This means that $$x < y$$. Could $$x$$ be negative? Yes, if $$x$$, $$y$$, and $$z$$ are negative. Could $$x$$ be positive? Yes, if $$x$$ and $$y$$ are positive and $$z$$ is negative. Not sufficient.

(2) $$x - z < 0$$. Basically the same here. We have that $$x < z$$. Could $$x$$ be negative? Yes, if $$x$$, $$y$$, and $$z$$ are negative. Could $$x$$ be positive? Yes, if $$x$$ and $$z$$ are positive and $$y$$ is negative. Not sufficient.

(1)+(2) We have that $$x < y$$ and $$x < z$$. Could $$x$$ be positive? No, because if $$x$$ is positive then from $$x < y$$ and $$x < z$$, both $$y$$ and $$z$$ must be positive but in this case $$xyz$$ will be positive not negative as given in the stem. Therefore $$x$$ must be negative. Sufficient.

Try NEW inequalities PS question.
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Re: If xyz < 0, is x < 0 ?  [#permalink]

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27 Jun 2014, 10:43
2
If $$xyz < 0$$, is $$x < 0$$?

(1) $$x - y < 0$$

(2) $$x - z < 0$$

Attachment:

Untitled.png [ 4.29 KiB | Viewed 5709 times ]

Product of 3 nos can be negative if one of them is negative and other two nos are of the same sign or if all the nos are negative.

Case 1:x=-1,y=2,z=100
Case 2:x=1,z=2,y=-100
Case 3 x=1,y=3,z=-100
Case 4:x=-100,y=-2,z=-1

Consider St 1 says : x<y...So out of the table following cases are possible

Cases 1,3 and 4: For cases 1,4 we see that x<0 but case 3 x>0...So St 1 is insufficient

St 2 says x<z, so we have case 1,2 and 4
If it is case 1 and 4 then x<0 and answer to our question is yes but if it case 2 then answer is no

Combining we see that for both statements Case and 1 and 4 are applicable and for these cases x<0.

Ans is C
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Re: If xyz < 0, is x < 0 ?  [#permalink]

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28 Jun 2014, 01:31
1
1
Bunuel wrote:

If $$xyz < 0$$, is $$x < 0$$?

(1) $$x - y < 0$$

(2) $$x - z < 0$$

Kudos for a correct solution.

Statement one :

x-y<0
x<y

Here x can take value of a positive or a negative number. x can be a negative number and y and z can be positive number or x and y can be positive numbers and z can take a negative value. Statement is insufficient.

Statement two:
x-z<0
x<z

This statement is insufficient to determine sign of x. x,y and z all can be negative or x and z can take positive values and y can be a negative number.

Both statements combined together, x is the smallest number and product of x,y and z is a negative value. so x has to be a negative number. Either x is a negative number and y and z are positive or all the three numbers are negative. In both the cases, x is a negative number. Hence Ans=C
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Joined: 22 Jun 2013
Posts: 35
Re: If xyz < 0, is x < 0 ?  [#permalink]

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28 Jun 2014, 05:53
2

x.y.z < 0

This is possible when :
All 3 (x,y,z) are <0
OR
One of x,y,z < 0

1.
x-y<0
-> x<y
Both x & y can be positive OR x can be negative
Cannot say anything

2.
x-z<0
-> x<z
Again, Both x & z can be positive OR x can be negative
Cannot say anything

1+2
On combining x<y & x<z
This means x is the smallest of all three
therefore it must be negative for x.y.z <0 to hold true.
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Joined: 02 Sep 2009
Posts: 55732
Re: If xyz < 0, is x < 0 ?  [#permalink]

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29 Jun 2014, 12:30
SOLUTION

If $$xyz < 0$$, is $$x < 0$$?

$$xyz < 0$$ implies that either all three unknowns are negative or one is negative and the remaining two are positive.

(1) $$x - y < 0$$. This means that $$x < y$$. Could $$x$$ be negative? Yes, if $$x$$, $$y$$, and $$z$$ are negative. Could $$x$$ be positive? Yes, if $$x$$ and $$y$$ are positive and $$z$$ is negative. Not sufficient.

(2) $$x - z < 0$$. Basically the same here. We have that $$x < z$$. Could $$x$$ be negative? Yes, if $$x$$, $$y$$, and $$z$$ are negative. Could $$x$$ be positive? Yes, if $$x$$ and $$z$$ are positive and $$y$$ is negative. Not sufficient.

(1)+(2) We have that $$x < y$$ and $$x < z$$. Could $$x$$ be positive? No, because if $$x$$ is positive then from $$x < y$$ and $$x < z$$, both $$y$$ and $$z$$ must be positive but in this case $$xyz$$ will be positive not negative as given in the stem. Therefore $$x$$ must be negative. Sufficient.

Kudos points given to correct solutions above.

Try NEW inequalities PS question.
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Re: If xyz < 0, is x < 0 ?  [#permalink]

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13 Jul 2014, 01:32
MAKE A line number to see that the matter is easy
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Re: If xyz < 0, is x < 0 ?  [#permalink]

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17 Nov 2015, 10:21
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If xyz<0 , is x<0 ?

(1) x−y<0

(2) x−z<0

There are 3 variables (x,y,z) and one equation (xyz<0) in the original condition, 2 equations in the given conditions, so there is high chance (C) will be our answer.
Looking at the conditions together,
x<y and x<z.
If x>0, then yz>0, so this does not satisfy xyz<0 (out of scope). Hence x has to be x<0 . This answers the question 'yes' and is therefore sufficient, making the answer (C).

For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: If xyz < 0, is x < 0 ?  [#permalink]

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15 Sep 2018, 12:18
WoundedTiger wrote:
If $$xyz < 0$$, is $$x < 0$$?

(1) $$x - y < 0$$

(2) $$x - z < 0$$

Attachment:
Untitled.png

Product of 3 nos can be negative if one of them is negative and other two nos are of the same sign or if all the nos are negative.

Case 1:x=-1,y=2,z=100
Case 2:x=1,z=2,y=-100
Case 3 x=1,y=3,z=-100
Case 4:x=-100,y=-2,z=-1

Consider St 1 says : x<y...So out of the table following cases are possible

Cases 1,3 and 4: For cases 1,4 we see that x<0 but case 3 x>0...So St 1 is insufficient

St 2 says x<z, so we have case 1,2 and 4
If it is case 1 and 4 then x<0 and answer to our question is yes but if it case 2 then answer is no

Combining we see that for both statements Case and 1 and 4 are applicable and for these cases x<0.

Ans is C

Hi,

This method is little longer, but just though to put out so that some users can understand what cases should be considered for such type of questions.

we have given a base condition that xyz<0
we can infer two things form this
either x<0&y<0&z<0 or one of x,y,z is <0.

Now Statement 1:
x-y<0
So we have three cases for this
x<0 & y<0 and |x|>|y|
But since xyz<0 from this we have x<0,y<0,z<0
x<0 & y>0 and |x|>|y|
But since xyz<0 from this we have x<0,y>0,z>0
x>0 & y>0 and |y|>|x|
But since xyz<0 from this we have x<0,y>0,z>0

Now Statement 1:
x-y<0
So we have three cases for this
x<0 & y<0 and |x|>|y| ------(a)
But since xyz<0 from this we have x<0,y<0,z<0
x<0 & y>0 and |x|>|y| ------(b)
But since xyz<0 from this we have x<0,y>0,z>0
x>0 & y>0 and |y|>|x|------(c)
But since xyz<0 from this we have x>0,y>0,z<0

Now Statement 2: x-z<0
So we have three cases for this
x<0 & z<0 and |x|>|z| -----(d)
But since xyz<0 from this we have x<0,y<0,z<0
x<0 & z>0 and |x|>|z|-----(e)
But since xyz<0 from this we have x<0,y>0,z>0
x>0 & z>0 and |z|>|x|-----(f)
But since xyz<0 from this we have x>0,y<0,z<0

Combing 1 and 2 we have
cases a, b, d,e we see that

we get either x<0, y<0, z<0
or
x<0 , y>0, z>0

But in both the cases x<0

Probus
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Re: If xyz < 0, is x < 0 ?   [#permalink] 15 Sep 2018, 12:18
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