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a○b is defined as 1/(a+b) – 1/a.

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a○b is defined as 1/(a+b) – 1/a.  [#permalink]

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New post 22 Aug 2019, 01:04
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Question Stats:

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[GMAT math practice question]

\(a○b\) is defined as \(\frac{1}{(a+b)} – \frac{1}{a}.\)

If \(x = (1+a)○(1-a)\) and \(y = (1-a)○(1+a)\), then what is the value of \(x*y\)?

\(A. \frac{1}{4}\)

\(B. \frac{1}{3}\)

\(C. \frac{1}{2}\)

\(D. 1\)

\(E. \frac{3}{4}\)

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Re: a○b is defined as 1/(a+b) – 1/a.  [#permalink]

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New post 22 Aug 2019, 01:14
MathRevolution wrote:
[GMAT math practice question]

\(a○b\) is defined as \(\frac{1}{(a+b)} – \frac{1}{a}.\)

If \(x = (1+a)○(1-a)\) and \(y = (1-a)○(1+a)\), then what is the value of \(x*y\)?

\(A. \frac{1}{4}\)

\(B. \frac{1}{3}\)

\(C. \frac{1}{2}\)

\(D. 1\)

\(E. \frac{3}{4}\)


x = 1/(1+a+1-a) - 1/(1+a) = 1/2-1/(1+a)
y = 1/(1-a+1+a) - 1/(1-a) = 1/2 - 1/(1-a)

x*y = 1/4 - (1/(1+a) + 1/(1-a))1/2 + 1/(1+a)(1-a)
= 1/4 -2/(1-a^2)*1/2 =1/(1-a^2)
= 1/4
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Re: a○b is defined as 1/(a+b) – 1/a.  [#permalink]

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New post 22 Aug 2019, 02:42
MathRevolution wrote:
[GMAT math practice question]

\(a○b\) is defined as \(\frac{1}{(a+b)} – \frac{1}{a}.\)

If \(x = (1+a)○(1-a)\) and \(y = (1-a)○(1+a)\), then what is the value of \(x*y\)?

\(A. \frac{1}{4}\)

\(B. \frac{1}{3}\)

\(C. \frac{1}{2}\)

\(D. 1\)

\(E. \frac{3}{4}\)


Let \(a=0\).

\(x = (1+a)○(1-a) = (1+0)○(1-0) = 1○1 = \frac{1}{1+1} – \frac{1}{1} = \frac{1}{2} - 1 = -\frac{1}{2}\)

\(y = (1-a)○(1+a) = (1-0)○(1+0) = 1○1 = \frac{1}{1+1} – \frac{1}{1} = \frac{1}{2} - 1 = -\frac{1}{2}\)

Thus:
\(xy = -\frac{1}{2} * -\frac{1}{2} = \frac{1}{4}\)


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Re: a○b is defined as 1/(a+b) – 1/a.  [#permalink]

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New post 22 Aug 2019, 03:03
GMATGuruNY wrote:
MathRevolution wrote:
[GMAT math practice question]

\(a○b\) is defined as \(\frac{1}{(a+b)} – \frac{1}{a}.\)

If \(x = (1+a)○(1-a)\) and \(y = (1-a)○(1+a)\), then what is the value of \(x*y\)?

\(A. \frac{1}{4}\)

\(B. \frac{1}{3}\)

\(C. \frac{1}{2}\)

\(D. 1\)

\(E. \frac{3}{4}\)


Let \(a=0\).

\(x = (1+a)○(1-a) = (1+0)○(1-0) = 1○1 = \frac{1}{1+1} – \frac{1}{1} = \frac{1}{2} - 1 = -\frac{1}{2}\)

\(y = (1-a)○(1+a) = (1-0)○(1+0) = 1○1 = \frac{1}{1+1} – \frac{1}{1} = \frac{1}{2} - 1 = -\frac{1}{2}\)

Thus:
\(xy = -\frac{1}{2} * -\frac{1}{2} = \frac{1}{4}\)



Hey GMATGuruNY

What is the logic behind taking a=0?
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Re: a○b is defined as 1/(a+b) – 1/a.  [#permalink]

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New post 22 Aug 2019, 03:57
NandishSS wrote:
Hey GMATGuruNY

What is the logic behind taking a=0?


Letting a=0 makes the math easier.
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Re: a○b is defined as 1/(a+b) – 1/a.  [#permalink]

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New post 25 Aug 2019, 18:52
=>

\(x = (1+a)○(1-a) = \frac{1}{(1+a+1-a)} – \frac{1}{(1+a)} = \frac{1}{2} – \frac{1}{(1+a)} = \frac{(a-1)}{(2(a+1))}.\)

\(y = (1-a)○(1+a) = \frac{1}{(1-a+1+a)} – \frac{1}{(1-a)} = \frac{1}{2} – \frac{1}{(1-a)} = \frac{(-a-1)}{(2(1-a))}.\)

So, \(xy = [\frac{(a-1)}{(2(a+1))}]*[\frac{(-a-1)}{(2(1-a))}] = \frac{[-(a-1)(a+1)]}{[-4(a+1)(a+1))]} = \frac{1}{4}.\)

Therefore, A is the answer.
Answer: A
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Re: a○b is defined as 1/(a+b) – 1/a.   [#permalink] 25 Aug 2019, 18:52
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