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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: 6815
GMAT 1: 760 Q51 V42
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:

89% (01:44) correct 11% (01:44) wrong based on 64 sessions

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[GMAT math practice question]

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

$$1) x>1$$
$$2) y>1$$

_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Intern Joined: 25 Jul 2017 Posts: 12 Re: If x, y and z are positive numbers, is xy+z > x+yz? [#permalink] ### Show Tags 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: 6815 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If x, y and z are positive numbers, is xy+z > x+yz? [#permalink] ### Show Tags 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. 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
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Re: If x, y and z are positive numbers, is xy+z > x+yz? &nbs [#permalink] 11 Feb 2018, 16:57
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