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655-705 (Hard)|   Algebra|      
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Bunuel
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(1.00001)(0.99999) - (1.00002)(0.99998) = 04 May 2021, 05:45
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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55% (hard)

Question Stats:

63% (02:03) correct 37% (01:59) wrong based on 289 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. \(3(10^{-5})\)
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C
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(1.00001)(0.99999) - (1.00002)(0.99998) = 08 May 2021, 02:00
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Bunuel
\((1.00001)(0.99999) - (1.00002)(0.99998) =\)

A. 0

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

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

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

E. \(3(10^{-5})\)
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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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(1.00001)(0.99999) - (1.00002)(0.99998) = Updated on: 22 Feb 2025, 04:12
Originally posted by GMATinsight on 04 May 2021, 06:09.
Last edited by Bunuel on 22 Feb 2025, 04:12, edited 2 times in total.
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Bunuel
\((1.00001)(0.99999) - (1.00002)(0.99998) =\)

A. 0

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

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

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

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

\((1.00001)(0.99999) - (1.00002)(0.99998)\) = \((1+.00001)(1-.00001)\) - \((1+.00002)(1-.00002)\) = \((1^2-.00001^2)\) - \((1^2-.00002^2)\)

\( = -1^2*(10^{-5})^2\) + \(2^2*(10^{-5})^2\) = \(3*10^{-10}\)

Answer: Option C
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(1.00001)(0.99999) - (1.00002)(0.99998) = 04 May 2021, 06:15
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Bunuel
\((1.00001)(0.99999) - (1.00002)(0.99998) =\)

A. 0

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

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

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

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


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

I. We can see that the terms are equidistant from 1, so let us use that for \((1-a)(1+a)=1^2-a^2\)

\((1.00001)(0.99999) - (1.00002)(0.99998) =\)
\((1+0.00001)(1-0.00001) - (1+0.00002)(1-0.00002) =\)
\(1^2-(0.00001)^2-(1^2-0.00002^2)=\)
\(0.00002^2-0.00001^2=(0.00002+0.00001)(0.00002-0.00001)= \)
\( 0.00003*0.00001=3*10^{-5}*1*10^{-5}=3*10^{-10}\)


II. Make use of options

Both turn out to be product of two numbers with 5 digits after decimal. So we are looking at an answer with 5+5 or 10 digits after decimal or 10^{-5}.

Only B and C left.

Now the digit at the rightmost decimal place is 1*9-2*8=9-6=3

Option C gives the correct digit as 3.

C
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(1.00001)(0.99999) - (1.00002)(0.99998) = 04 May 2021, 06:30
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\((1.00001)(0.99999) = (1+ 10^{-5})(1-10^{-5})\)
and
\((1.00002)(0.99998) = (1+ 2*10^{-5})(1-2*10^{-5})\)

we know \((a-b)(a+b) =(a^2 - b^2)\)
so the above equations can be simplified as
\((1+ 10^{-5})(1-10^{-5}) = [1^2 - (10^{-5})^2]\)
&
\((1+ 2*10^{-5})(1-2*10^{-5}) = [1^2 - (2*(10^{-5})^2] \)

\([1^2 - (10^{-5})^2] - [1^2 - (2*(10^{-5})^2] = [10^{-10} -(4*{10^{-10})]\)

\(= 3*10^{-10}\) Option C
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(1.00001)(0.99999) - (1.00002)(0.99998) = 16 May 2021, 02:04
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I realized this type of questions somehow has appeared in my last two GMAT attempts and I got frozen by it.. any way to get similar practice?
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(1.00001)(0.99999) - (1.00002)(0.99998) = 27 Jul 2021, 03:46
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Bunuel
\((1.00001)(0.99999) - (1.00002)(0.99998) =\)

A. 0

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

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

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

E. \(3(10^{-5})\)
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VeritasKarishma can you pls expand the below in detailed step by step way

so i got this and got stuck how to proceed further \((1-10^{-5})^2 - (1-2*10^{-5})^2\)

BrentGMATPrepNow hi Brent maybe you could answer :grin:
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(1.00001)(0.99999) - (1.00002)(0.99998) = 04 Jul 2022, 08:47
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Use aˆ{2} - bˆ{2} in the question stem,

=>(1+10ˆ{-5}) * (1-10ˆ{-5}) - (1+2*10ˆ{-5}) * (1-2* 10ˆ{-5})
=> (1-10ˆ{-10}) - (1-2*10ˆ{-10})
=> -10ˆ{-10} - 4* 10ˆ{-10}
=> 3 * 10ˆ{-10}, Option C
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(1.00001)(0.99999) - (1.00002)(0.99998) = 13 Jun 2025, 07:55
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