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

 It is currently 18 Jun 2019, 06:48

### 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

# M61-17

Author Message
TAGS:

### Hide Tags

Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 7465
GMAT 1: 760 Q51 V42
GPA: 3.82

### Show Tags

18 Jun 2018, 04:48
00:00

Difficulty:

55% (hard)

Question Stats:

45% (02:24) correct 55% (02:37) wrong based on 11 sessions

### HideShow timer Statistics

If $$a≠b$$, and $$\frac{a^2}{(a-b)}=b$$, which of the following could be $$a$$?

A. -1
B. 0
C. 1
D. 2
E. $$a$$ does not exist

_________________
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" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 7465 GMAT 1: 760 Q51 V42 GPA: 3.82 Re M61-17 [#permalink] ### Show Tags 18 Jun 2018, 04:48 Official Solution: If $$a≠b$$, and $$\frac{a^2}{(a-b)}=b$$, which of the following could be $$a$$? A. -1 B. 0 C. 1 D. 2 E. $$a$$ does not exist $$\frac{a^2}{(a-b)}=b$$ $$a^2=b(a-b)$$ $$a^2=ab – b^2$$ $$a^2 - ab + b^2 = 0$$ Since $$a≠b$$, one of $$a$$ and $$b$$ is not zero. If $$a ≠ 0$$ or $$b ≠ 0$$, $$a^2 - ab + b^2$$ cannot be zero for the following reason: $$a^2 - ab + b^2 = a^2 – 2a(\frac{b}{2}) + b^2 = a^2 – 2a(\frac{b}{2}) + \frac{b^2}{4} + (\frac{3}{4})b^2$$ $$= (a – \frac{b}{2})^2 + (\frac{3}{4})b^2 > 0$$ Thus, there is no pair of real numbers $$(a,b)$$ satisfying the equation $$a^2 - ab + b^2 = 0$$. Therefore, $$a$$ cannot exist, and the answer is E. 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"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"
Manager
Joined: 01 May 2017
Posts: 82
Location: India

### Show Tags

28 Dec 2018, 01:39
a^2/(a-b) = b

a^2 = b(a-b)

a^2 = ab -b^2

a^2 + b^2 -ab = 0

a^2 + b^2 -2ab = -ab

(a-b)^2 = -ab

Case (1) if a != 0 and b!= 0
(a-b)^2 is always positive

case(2) if a = 0
b^2 = 0
implies b = 0
But given a!= b
So from case(1) and case 2 such 'a' doesn't exist
Re: M61-17   [#permalink] 28 Dec 2018, 01:39
Display posts from previous: Sort by

# M61-17

Moderators: chetan2u, Bunuel