GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 23 Oct 2019, 12:37

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# M60-25

Author Message
TAGS:

### Hide Tags

Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 8033
GMAT 1: 760 Q51 V42
GPA: 3.82

### Show Tags

11 Jun 2018, 07:03
00:00

Difficulty:

25% (medium)

Question Stats:

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

### HideShow timer Statistics

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 $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8033 GMAT 1: 760 Q51 V42 GPA: 3.82 Re M60-25 [#permalink] ### Show Tags 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 _________________ 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$79 for 1 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"
Intern
Joined: 11 Nov 2013
Posts: 26

### Show Tags

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.
Intern
Joined: 01 Apr 2018
Posts: 11
Location: India
WE: Consulting (Consulting)

### Show Tags

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 ?
Senior Manager
Joined: 27 Aug 2014
Posts: 345
Location: Netherlands
Concentration: Finance, Strategy
Schools: LBS '22, ISB '21
GPA: 3.9
WE: Analyst (Energy and Utilities)

### Show Tags

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
Intern
Joined: 18 Nov 2014
Posts: 12

### Show Tags

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
Re M60-25   [#permalink] 19 Jan 2019, 11:01
Display posts from previous: Sort by

# M60-25

Moderators: chetan2u, Bunuel