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If a > 0 and |a| ≠ |b|, is 2/(a+b) + 2/(a-b) = 1 ?

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If a > 0 and |a| ≠ |b|, is 2/(a+b) + 2/(a-b) = 1 ?  [#permalink]

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New post Updated on: 24 Jan 2020, 23:58
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If \(a > 0\) and \(|a| ≠ |b|\), is \(\frac{2}{a+b }+ \frac{2}{a-b} = 1\) ?


(1) \(b = 0\)

(2) \(a^2 − b^2 = 4a\)

Originally posted by ajgoupil on 24 Jan 2020, 09:24.
Last edited by Bunuel on 24 Jan 2020, 23:58, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If a > 0 and |a| ≠ |b|, is 2/(a+b) + 2/(a-b) = 1 ?  [#permalink]

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New post 24 Jan 2020, 11:30
ajgoupil wrote:
If a > 0 and |a| ≠ |b|, is 2/a+b + 2/a-b = 1 ?

(1) b = 0
(2) a2 − b2 = 4a


Is \(\frac{ 2}{a+b } + \frac{ 2}{a-b } = 1 \)?
—> \( 2(a - b) + 2(a + b) = (a + b)(a - b) \) ?
—> \( 2a - 2b + 2a + 2b = a^2 - b^2 \)?
—> \( a^2 - b^2 = 4a \)?

(1) \( b = 0 \)
—> Not relavant —> Insufficient

(2) \( a^2 - b^2 = 4a \)
—> Yes —> Sufficient

Option B

Hope it helps.

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If a > 0 and |a| ≠ |b|, is 2/(a+b) + 2/(a-b) = 1 ?  [#permalink]

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New post 25 Jan 2020, 07:55
\(\frac{2}{a+b} + \frac{2}{a-b} = \frac{2(a+b) + 2(a-b) }{ (a+b)*(a-b)} = \frac{4a}{ a^2 - b^2}\)
statement (1) b=0 not enough to conclusion.
statement (2) \(a^2 - b^2 = 4a => 4a/4a = 1 \)
Choice B.
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Re: If a > 0 and |a| ≠ |b|, is 2/(a+b) + 2/(a-b) = 1 ?  [#permalink]

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New post 26 Jan 2020, 00:56
ajgoupil wrote:
If \(a > 0\) and \(|a| ≠ |b|\), is \(\frac{2}{a+b }+ \frac{2}{a-b} = 1\) ?


(1) \(b = 0\)

(2) \(a^2 − b^2 = 4a\)


Given:
1. \(a > 0\) and
2. \(|a| ≠ |b|\),

Asked:
Is \(\frac{2}{a+b }+ \frac{2}{a-b} = 1\) ?
or Is 4a/(a^2 - b^2) = 1
or Is 4a = a^2 -b^2

(1) \(b = 0\)
\(\frac{2}{a+b }+ \frac{2}{a-b}\)
= \(\frac{2}{a} + \frac{2}{a} = \frac{4}{a }\)
4/a is NOT NECESSARILY = 1
NOT SUFFICIENT

(2) \(a^2 − b^2 = 4a\)
Is \(\frac{2}{a+b }+ \frac{2}{a-b} = 1\) ?
or Is 4a/(a^2 - b^2) = 1
or Is 4a = a^2 -b^2
SUFFICIENT

IMO B
GMAT Club Bot
Re: If a > 0 and |a| ≠ |b|, is 2/(a+b) + 2/(a-b) = 1 ?   [#permalink] 26 Jan 2020, 00:56
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