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M60-25

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M60-25  [#permalink]

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New post 11 Jun 2018, 07:03
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

100% (00:57) correct 0% (00:00) wrong based on 9 sessions

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If \(x, y and z\) are positive numbers, is \(xy+z > x+yz\)?

1) \(x>1\)

2) \(y>1\)

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Math Revolution GMAT Instructor
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Re M60-25  [#permalink]

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New post 11 Jun 2018, 07:03
Official Solution:



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 and y) and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first.



Conditions 1) & 2):

\(xy + z > x + yz\)

⇔ \(xy - x - yz + z > 0\)

⇔ \(x(y-1)-z(y-1) > 0\)

⇔ \((x-z)(y-1) > 0\)

⇔ \(x - z > 0\) since \(y > 1\).



If \(x = 2, y = 2\), and \(z = 1\), then \(xy + z = 5, x + yz = 4\) and \(xy + z > x + yz\).

So, the answer is 'yes'.

If \(x = 2, y = 2\), and \(z = 3\), then \(xy + z = 7\), \(x + yz = 8\) and \(xy + z < x + yz\).

So, the answer is 'no'.

Since the question does not have a unique answer, both conditions together are not sufficient.



Therefore, E 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.

Answer: E
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Re: M60-25  [#permalink]

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New post 10 Aug 2018, 02:59
I think the scenario "If x = 2, y = 2 and z = 3 ..." is incorrect.

Since x – z > 0, x > z.
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Re: M60-25  [#permalink]

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New post 14 Sep 2018, 07:30
Yes, the explanation is not Correct , if x - z >0 then how come x =2 and z=3? Please can someone help ?
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Re: M60-25  [#permalink]

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New post 23 Sep 2018, 12:25
xy+z > x+yz
xy+z-x-yz>0
x(y-1)-z(y-1)>0
(x-z)(y-1)>0?

so for this statement to be true, (x-z) and (y-1) should have the same sign. We need to statements if they provide this info.

from 1: x>1, no info on y and z, so insuff
from 2: y>1, (y-1)>0, we don't have info on (x-z) so insuff


1+2 combined, x>1 and y-1>0, no info on x-z sp insuff. therefore E
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Re M60-25  [#permalink]

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New post 19 Jan 2019, 11:01
I think this is a high-quality question and I don't agree with the explanation. it needs to be c
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Re M60-25   [#permalink] 19 Jan 2019, 11:01
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