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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]

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New post 11 Aug 2012, 11:54
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Difficulty:

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Question Stats:

56% (03:32) correct 44% (01:33) wrong based on 73 sessions

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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: topic-144735.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 31 Jan 2014, 07:06, edited 2 times in total.
Renamed the topic.
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Re: OG 13 #199 [#permalink]

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New post 11 Aug 2012, 13:00
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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: 0.99999999/1.0001-0.99999991/1.0003 [#permalink]

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New post 31 Jan 2014, 06:19
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Re: 0.99999999/1.0001-0.99999991/1.0003 [#permalink]

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New post 31 Jan 2014, 07:07
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\(\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=\)

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

\(\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=\frac{1-10^{-8}}{1+10^{-4}}-\frac{1-9*10^{-8}}{1+3*10^{-4}}\)

Now apply \(a^2-b^2=(a+b)(a-b)\):

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

Answer: D.

OPEN DISCUSSION OF THIS QUESTION IS HERE: topic-144735.html
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Re: 0.99999999/1.0001-0.99999991/1.0003   [#permalink] 31 Jan 2014, 07:07
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