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0.99999999/1.0001-0.99999991/1.0003

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0.99999999/1.0001-0.99999991/1.0003 [#permalink] New post 11 Aug 2012, 12:54
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40% (02:06) correct 60% (01:06) wrong based on 0 sessions
This is from OG 13, #199. The explanation is tedious and time consuming, I hope I don't see anything like this, because I'd be tempted to guess and move on.

\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=

A. 10^-8
B. 3(10^-8)
C. 3(10^-4)
D. 2(10^-4)
E. 10^-4
[Reveal] Spoiler: OA

Last edited by Bunuel on 11 Aug 2012, 14:49, edited 1 time in total.
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Re: OG 13 #199 [#permalink] New post 11 Aug 2012, 14:00
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RoyHalladay wrote:
This is from OG 13, #199. The explanation is tedious and time consuming, I hope I don't see anything like this, because I'd be tempted to guess and move on.

\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=

A. 10^-8
B. 3(10^-8)
C. 3(10^-4)
D. 2(10^-4)
E. 10^-4


Well, you need a little bit of algebra here...

The question is meant to test the formula a^2-b^2=(a+b)(a-b).
The clue is in the answers, explicitly the powers of 10.
You can write the given expression as follows:

\frac{1-10^{-8}}{1+10^{-4}}-\frac{1-9*10^{-8}}{1+3*10^{-4}}=\frac{(1+10^{-4})(1-10^{-4})}{1+10^{-4}}-\frac{(1+3*10^{-4})(1-3*10^{-4})}{1+3*10^{-4}}=1-10^{-4}-(1-3*10^{-4})=2*10^{-4}.

Answer D
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Re: OG 13 #199   [#permalink] 11 Aug 2012, 14:00
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