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

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New post 24 Nov 2012, 11:39
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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]

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New post 24 Nov 2012, 12:28
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(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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New post 24 Nov 2012, 12:33
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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]

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New post 24 Nov 2012, 12:41
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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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New post 30 May 2013, 02:51
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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, 03:12, edited 3 times in total.

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New post 30 May 2013, 02:59

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New post 17 Nov 2016, 15:31
In short, this is a very messy way of dealing with factoring.

Think (a+1)(a-1), etc.

1. We need to rewrite decimals, so they are in fractional form.

{[100,001/100,000][999,999/100,000]} - {[100,002/100,000][99998/100,000]}

2. Trick here is to recognize that we can write each of the numerator as a process applied to 100,000 (i.e. x, -, +, etc)

{[(100,000+1)(100,000-1)]/(10,000,000,000)} - {(100,000+2)(100,000-2)/(10,000,000,000)}

3. Simplify by using knowledge that (a+1)(a-1) = a^2 - 1 & (a+2)(a-2) = a^2 -4

{(10,000,000,000-1) -(10,000,000,000-4)}/(10^10) --> 3x10^(-10)

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

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New post 10 Jan 2017, 04:35
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?


1.00001*0.99999 .... ends in 9, and 1.00002*0.99998 ends in 6 , thus answer has to be in the form 3*10^-x , option c or E

but since multiplication of each term around the subtraction operation will yield 10^-x < 10^-5 thus x > 5 .... thus option c

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

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New post 03 Nov 2017, 03:47
Does anyone know of similar question types for additional practise?

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New post 03 Nov 2017, 04:25
Edofarmer wrote:
Does anyone know of similar question types for additional practise?


Similar questions:
https://gmatclub.com/forum/1-130872.html
https://gmatclub.com/forum/999-999-2-1- ... 05499.html
https://gmatclub.com/forum/topic-143744.html
https://gmatclub.com/forum/3-109271.html
https://gmatclub.com/forum/topic-144735.html
https://gmatclub.com/forum/the-value-of ... 30682.html
https://gmatclub.com/forum/1-0-0001-0-0 ... 59398.html

For more check the questions under the tag of Arithmetic
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Re: (1.00001)(0.99999) - (1.00002)(0.99998) = [#permalink]

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New post 06 Nov 2017, 17:30
anon1 wrote:
\((1.00001)(0.99999) - (1.00002)(0.99998) =\)

A. 0

B. \(10^{-10}\)

C. \(3(10^{-10})\)

D. \(10^{-5}\)

E. \(3(10^{-5})\)


Let’s rewrite the expressions using scientific notation:

(1 + 10^-5)(1 - 10^-5) - (1 + 2 x 10^-5)(1 - 2 x 10^-5)

1^2 - (10^-5)^2 - [1^2 - 2^2 x (10^-5)^2]

1 - 10^-10 - 1 + 4 x 10^-10

3 x 10^-10

Answer: C
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Re: (1.00001)(0.99999) - (1.00002)(0.99998) =   [#permalink] 06 Nov 2017, 17:30
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