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

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(1.00001)(0.99999) - (1.00002)(0.99998) = [#permalink] New post 24 Nov 2012, 10:39
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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)

[Reveal] Spoiler:
The solution suggests to turn all the expressions into scientific notation and then simplify but I was wondering if anyone here had any mind boggling tricks to do this in a more elegant manner?
[Reveal] Spoiler: OA
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Re: (1.00001)(0.99999) - (1.00002)(0.99998) = [#permalink] New post 24 Nov 2012, 11:28
(1.00001)(0.99999) - (1.00002)(0.99998) = (100001/100000)*(99999/100000) - (100002/100000)*(99998/100000) =
Right here from the get go we see that the denominator of the expression above is 100000*100000 = 10^(-10) - Denominator, and we live it right there and focus on the numerator.
100001*99999 = (100000+1)*99999=100000*99999+99999
100002*99998 = (100000+2)*99998 = 100000*99998+2*99998
= 100000*99999+99999-100000*99998-2*99998 = 100000+99999-2*99998 = 100000+99999-2*(99999-1)=100000+99999-2*99999+2=100000-99999+2=3 - Numerator
Or, 3*(10^(-10) = 3/10^(10)
The numbers might look intimidating and cumbersome, but a little manipulation goes a long way. Or, once you got the denominator you can cross out some answer choices and guess in case you running out of time.
Please, feel free to correct me, if I went awry
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Re: (1.00001)(0.99999) - (1.00002)(0.99998) = [#permalink] New post 24 Nov 2012, 11:33
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anon1 wrote:
(1.00001)(0.99999) - (1.00002)(0.99998) =

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

[Reveal] Spoiler:
The solution suggests to turn all the expressions into scientific notation and then simplify but I was wondering if anyone here had any mind boggling tricks to do this in a more elegant manner?


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}.

Answer: C.
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Re: (1.00001)(0.99999) - (1.00002)(0.99998) = [#permalink] New post 24 Nov 2012, 11:41
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anon1 wrote:
(1.00001)(0.99999) - (1.00002)(0.99998) =

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

[Reveal] Spoiler:
The solution suggests to turn all the expressions into scientific notation and then simplify but I was wondering if anyone here had any mind boggling tricks to do this in a more elegant manner?


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

= (1 + 10^{-5})(1 - 10^{-5}) - (1 + 2*10^{-5})(1 - 2*10^{-5})

= 1 - 10^{-10} - 1 + 4*10^{-10}

=3*10^{-10}

Answer is hence C.

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(1.00001)(0.99999) - (1.00002)(0.99998) = [#permalink] New post 30 May 2013, 01:51
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I wrote it in a simple way:

(1 + \frac{1}{10^5}) (1 - \frac{1}{10^5}) - (1 + \frac{2}{10^5}) (1 - \frac{2}{10^5})

= 1 - \frac{1}{10^{10}} - (1 - \frac{4}{10^{10}})

= \frac{4}{10^{10}} - \frac{1}{10^{10}}

= \frac{3}{10^{10}}

Answer = C
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Last edited by PareshGmat on 27 Aug 2014, 02:12, edited 3 times in total.
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Re: (1.00001)(0.99999) - (1.00002)(0.99998) = [#permalink] New post 30 May 2013, 01:59
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Re: (1.00001)(0.99999) - (1.00002)(0.99998) = [#permalink] New post 26 Aug 2014, 21:47
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Re: (1.00001)(0.99999) - (1.00002)(0.99998) =   [#permalink] 26 Aug 2014, 21:47
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