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

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Difficulty:   55% (hard)

Question Stats: 45% (02:24) correct 55% (02:37) wrong based on 11 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 V
Joined: 16 Aug 2015
Posts: 7465
GMAT 1: 760 Q51 V42 GPA: 3.82

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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.

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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
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M61-17

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