(1.00001)(0.99999) - (1.00002)(0.99998) = : GMAT Problem Solving (PS)
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# (1.00001)(0.99999) - (1.00002)(0.99998) =

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Director
Joined: 14 Sep 2005
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18 Jan 2006, 04:22
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Difficulty:

75% (hard)

Question Stats:

51% (02:28) correct 49% (01:36) wrong based on 197 sessions

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(1.00001)(0.99999) - (1.00002)(0.99998) =

A. 0
B. 10^-10
C. 3(10^-10)
D. 10^-5
E. 3(10^-5)

OPEN DISCUSSION OF THIS QUESTION IS HERE: 1-142988.html
[Reveal] Spoiler: OA

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Auge um Auge, Zahn um Zahn !

Last edited by Bunuel on 03 Jun 2014, 10:10, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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18 Jan 2006, 06:21
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(1.00001*0.99999)-(1.00002*0.99998)

= (1+0.00001)(1-0.00001) - (1+0.00002)(1-0.00002)

= 1 - (0.00001)^2 - 1 + (0.00002)^2

= -(0.0000000001) + (0.0000000004)

= 0.0000000003
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Joined: 20 Nov 2005
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Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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18 Jan 2006, 15:33
(1.00001*0.99999)-(1.00002*0.99998)

= (1 + 0.00001)*(1-0.00001)-(1+ 0.00002) * (1-0.00002)
= (x^2 - y^2) - (x^2 - z^2) = z^2 - y^2
where
x = 1, y = 10^(-5), z = 2* 10^(-5)

= 4*10^(-10) - 10^(-10) = 3 * 10^(-10)
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Re: What is the value of (1.00001*0.99999)-(1.00002*0.99998) [#permalink]

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03 Jun 2014, 09:58
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Hi Bunuel, pls could you update following answer choices from official source?

1. 0
2. 10^-10
3. 3*10^-10
4. 10^-5
5. 3*10^-5

Thanks!
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Re: (1.00001)(0.99999) - (1.00002)(0.99998) = [#permalink]

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03 Jun 2014, 10:14
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(1.00001)(0.99999) - (1.00002)(0.99998) =

A. 0
B. 10^-10
C. 3(10^-10)
D. 10^-5
E. 3(10^-5)

Apply $$a^2-b^2=(a+b)(a-b)$$.

$$1.00001*0.99999-1.00002*0.99998=(1+0.00001)(1-0.00001)-(1+0.00002)(1-0.00002)=$$
$$=1^2-0.00001^2-1^2+0.00002^2=0.00002^2-0.00001^2$$.

Next, $$0.00002^2-0.00001^2=(0.00002+0.00001)(0.00002-0.00001)=0.00003*0.00001=3*10^{-5}*10^{-5}=3*10^{-10}$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: 1-142988.html
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Re: (1.00001)(0.99999) - (1.00002)(0.99998) =   [#permalink] 03 Jun 2014, 10:14
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