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If xyz < 0, is x < 0 ?

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Math Expert
Joined: 02 Sep 2009
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If xyz < 0, is x < 0 ? [#permalink]  27 Jun 2014, 09:21
Expert's post
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Difficulty:

35% (medium)

Question Stats:

61% (01:56) correct 39% (00:54) wrong based on 70 sessions

If xyz < 0, is x < 0?

(1) x - y < 0

(2) x - z < 0

Kudos for a correct solution.

[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Sep 2009
Posts: 19036
Followers: 3361

Kudos [?]: 24445 [0], given: 2677

If xyz < 0, is x < 0 ? [#permalink]  27 Jun 2014, 09:22
Expert's post
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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.

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Re: If xyz < 0, is x < 0 ? [#permalink]  27 Jun 2014, 09:43
1
KUDOS
If xyz < 0, is x < 0?

(1) x - y < 0

(2) x - z < 0

Attachment:

Untitled.png [ 4.29 KiB | Viewed 639 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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Kudos [?]: 17 [1] , given: 28

Re: If xyz < 0, is x < 0 ? [#permalink]  28 Jun 2014, 00:31
1
KUDOS
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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Kudos [?]: 16 [2] , given: 129

Re: If xyz < 0, is x < 0 ? [#permalink]  28 Jun 2014, 04:53
2
KUDOS

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.
Math Expert
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Re: If xyz < 0, is x < 0 ? [#permalink]  29 Jun 2014, 11:30
Expert's post
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]  13 Jul 2014, 00:32
MAKE A line number to see that the matter is easy
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Re: If xyz < 0, is x < 0 ?   [#permalink] 13 Jul 2014, 00:32
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