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M61-17

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Math Revolution GMAT Instructor
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Joined: 16 Aug 2015
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M61-17  [#permalink]

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New post 18 Jun 2018, 03:48
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

38% (02:50) correct 63% (02:35) wrong based on 8 sessions

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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

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Math Revolution GMAT Instructor
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Re M61-17  [#permalink]

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New post 18 Jun 2018, 03:48
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
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Re: M61-17  [#permalink]

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New post 28 Dec 2018, 00:39
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
GMAT Club Bot
Re: M61-17 &nbs [#permalink] 28 Dec 2018, 00:39
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