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# If x, y and z are positive numbers, is xy+z > x+yz?

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 4904
GPA: 3.82
If x, y and z are positive numbers, is xy+z > x+yz? [#permalink]

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09 Feb 2018, 00:41
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Difficulty:

15% (low)

Question Stats:

85% (01:17) correct 15% (03:17) wrong based on 41 sessions

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[GMAT math practice question]

If $$x, y$$ and $$z$$ are positive numbers, is $$xy+z > x+yz$$?

$$1) x>1$$
$$2) y>1$$
[Reveal] Spoiler: OA

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Intern
Joined: 25 Jul 2017
Posts: 12
Re: If x, y and z are positive numbers, is xy+z > x+yz? [#permalink]

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09 Feb 2018, 01:22
If x,yandz are positive numbers, is xy+z>x+yz?

1)x>1
2)y>1

From the question , we can rearrange:
xy+z>x+yz
xy-x>yz-z
x(y-1)>z(y-1)
(x-z)(y-1)>0

i.e the expression is +ve when
case 1: x-z>0 and y>1
case 2: x-z<0 and y<1

From statement 1 : x> 1 . It is clearly insufficient
From statement 2: y>1 . we dont know anything about x,z. It is insuffucient

Combining, we still do not know about x & z. Hence, E is the answer.

Happy Chocolate Day
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 4904
GPA: 3.82
Re: If x, y and z are positive numbers, is xy+z > x+yz? [#permalink]

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11 Feb 2018, 16:57
=>

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.

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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Re: If x, y and z are positive numbers, is xy+z > x+yz?   [#permalink] 11 Feb 2018, 16:57
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