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If x, y, and z are integers, is xz < 0? (1) xy > 0 (2) yz < 0

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If x, y, and z are integers, is xz < 0? (1) xy > 0 (2) yz < 0  [#permalink]

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New post 20 Mar 2018, 06:43
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A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

82% (01:16) correct 18% (00:51) wrong based on 34 sessions

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Re: If x, y, and z are integers, is xz < 0? (1) xy > 0 (2) yz < 0  [#permalink]

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New post 20 Mar 2018, 06:44
Bunuel wrote:
If x, y, and z are integers, is xz < 0?

(1) xy > 0
(2) yz < 0


9. Inequalities



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Re: If x, y, and z are integers, is xz < 0? (1) xy > 0 (2) yz < 0  [#permalink]

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New post 20 Mar 2018, 07:11
Bunuel wrote:
If x, y, and z are integers, is xz < 0?

(1) xy > 0
(2) yz < 0


we are looking for whether xz is less the 0 or not.

statement 1: xy> 0
in this case both x and y can be negative or positive. on the other hand no information pertinent to z is given. insufficient.

statement 2: yz< 0

here we are clear that either y or z is negative but we don't know which of this 2 is negative.

combining 2 statements:
from statement 1 we know x an y both can be positive or negative.

phase 1: lets assume both x and y are positive. in that case in statement 2 z will be negative. in that case xz<0
phase 2: lets assume both x and y are negative. in that case in statement 2 z will be positive. in that case xz< 0

therefore correct answer will be C.
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If x, y, and z are integers, is xz < 0? (1) xy > 0 (2) yz < 0  [#permalink]

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New post 20 Mar 2018, 23:07
Bunuel wrote:
If x, y, and z are integers, is xz < 0?

(1) xy > 0
(2) yz < 0


Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 3 variables (x,y and z) and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.


Conditions 1) & 2)
Since xy > 0 and yz < 0, we have (xz)y^2 < 0 or xz<0.
Both conditions together are sufficient.

Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
Since we don't have information about z, condition 1) is not sufficient.

Condition 2)
Since we don't have information about x, condition 2) is not sufficient.

Therefore, C is the answer.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.
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