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# 0.99999999/1.0001 - 0.99999991/1.0003 =

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26 Dec 2012, 07:36
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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)
[Reveal] Spoiler: OA
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22 Jul 2013, 22:16
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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)

Responding to a pm:
To be honest, I can't think of an alternative method. The fractions are really complicated and need to be simplified before proceeding. For simplification, I think you will need to use a^2 - b^2 = (a - b)(a + b)

All I can suggest is that you can try to solve it without the exponents if that seems easier e.g.

$$\frac{0.99999999}{1.0001}-\frac{.99999991}{1.0003}$$

$$\frac{{1 - .00000001}}{{1 + .0001}}-\frac{{1 - .00000009}}{{1 + .0003}}$$

$$\frac{{1^2 - .0001^2}}{{1 + .0001}}-\frac{{1^2 - 0.0003^2}}{{1 + .0003}}$$

$$\frac{{(1 - .0001)(1 + .0001)}}{{(1 + .0001)}}-\frac{{(1 - .0003)(1 + .0003)}}{{(1 + .0003)}}$$

$$(1 - .0001) - (1 - .0003)$$

$$.0002 = 2*10^{-4}$$
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 34830 Followers: 6482 Kudos [?]: 82629 [44] , given: 10108 Re: 0.99999999/1.0001 - 0.99999991/1.0003 = [#permalink] ### Show Tags 26 Dec 2012, 07:39 44 This post received KUDOS Expert's post 44 This post was BOOKMARKED Walkabout wrote: $$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=$$ (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. _________________ Verbal Forum Moderator Joined: 10 Oct 2012 Posts: 630 Followers: 76 Kudos [?]: 1026 [12] , given: 136 Re: 0.99999999/1.0001 - 0.99999991/1.0003 = [#permalink] ### Show Tags 22 Jul 2013, 23:00 12 This post received KUDOS 3 This post was BOOKMARKED Walkabout wrote: $$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=$$ (A) 10^(-8) (B) 3*10^(-8) (C) 3*10^(-4) (D) 2*10^(-4) (E) 10^(-4) The best solution is already outlined by Bunuel/Karishma. Because this is an Official Problem, I was sure there might be another way to do this. So i did spend some time and realized that 0.99999999 might be a multiple of 1.0001 because of the non-messy options and found this : 9*1.0001 = 9.0009 ; 99*1.0001 = 99.0099 and as because the problem had 9 eight times, we have 9999*1.0001 = 9999.9999. Again, looking for a similar pattern, the last digit of 0.99999991 gave a hint that maybe we have to multiply by something ending in 7, as because we have 1.0003 in the denominator. And indeed 9997*1.0001 = 9999.9991. Thus, the problem boiled down to $$9999*10^{-4} - 9997*10^{-4} = 2*10^{-4}$$ D. Maybe a bit of luck was handy. _________________ Intern Joined: 28 May 2012 Posts: 29 Concentration: Finance, General Management GMAT 1: 700 Q50 V35 GPA: 3.28 WE: Analyst (Investment Banking) Followers: 1 Kudos [?]: 37 [11] , given: 84 Re: 0.99999999/1.0001 - 0.99999991/1.0003 = [#permalink] ### Show Tags 23 Jul 2013, 00:25 11 This post received KUDOS 2 This post was BOOKMARKED Here is my alternative solution for this problem (not for all problems): $$\frac{A}{B} - \frac{C}{D}= \frac{(AD-BC)}{BD}$$. So $$\frac{0.99999999}{1.0001} - \frac{0.99999991}{1.0003}= \frac{(0.99999999*1.0003-0.99999991*1.0001)}{(1.0001*1.0003)}$$. For this case, the ultimate digit of 0.99999999*1.0003-0.99999991*1.0001 is 6 In the denominator, the ultimate digit of 1.0001*1.0003 is 3 Therefore, the ultimate digit of the final result is 2. So it should be 2 * 0.00...01 --> Only D has the last digit of 2. Alternatively, we can calculate each fraction, $$\frac{0.99999999}{1.0001}$$ has last digit of 9, and $$\frac{0.99999991}{1.0003}$$ has last digit of 7, so the final last digit is 2 --> D This is a special problem. For example $$\frac{...6}{4}$$ can have a result of ...4 or ...9. Therefore, in this case we have to calculate as Bunuel did. In general, we can only apply this strategy only if the last digit of divisor is 1, 2 or 3. Intern Joined: 25 Jul 2014 Posts: 20 Concentration: Finance, General Management GPA: 3.54 WE: Asset Management (Venture Capital) Followers: 0 Kudos [?]: 21 [5] , given: 52 Re: 0.99999999/1.0001 - 0.99999991/1.0003 = [#permalink] ### Show Tags 29 Aug 2014, 20:02 5 This post received KUDOS 3 This post was BOOKMARKED This is a crazy question eventhough I got the correct answer... I simply think: 1. In fraction 1, we have 8 decimals devided by 4 decimals, so the result would be a number with 4 decimals 2. Fraction 2, same, so we should have another number with 4 decimals 3. Take these 2 numbers subtract each other, we should have another number with 4 decimals, so answer should be some thing 10^-4 --> eliminate A and B 4. We have an odd number - another odd number, the result should be van even number ---> eliminate C and E IN real test, if I pump into this kind of question, I would just guess and move on Math Expert Joined: 02 Sep 2009 Posts: 34830 Followers: 6482 Kudos [?]: 82629 [2] , given: 10108 Re: 0.99999999/1.0001 - 0.99999991/1.0003 = [#permalink] ### Show Tags 02 Jul 2013, 01:16 2 This post received KUDOS Expert's post 2 This post was BOOKMARKED Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HERE DS questions on Arithmetic: search.php?search_id=tag&tag_id=30 PS questions on Arithmetic: search.php?search_id=tag&tag_id=51 _________________ Manager Joined: 22 Dec 2014 Posts: 50 Followers: 2 Kudos [?]: 28 [1] , given: 182 0.99999999/1.0001 - 0.99999991/1.0003 = [#permalink] ### Show Tags 12 Jul 2015, 07:56 1 This post received KUDOS Walkabout wrote: $$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=$$ (A) 10^(-8) (B) 3*10^(-8) (C) 3*10^(-4) (D) 2*10^(-4) (E) 10^(-4) $$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=\frac{1-10^{-8}}{1-10^{-4}}-\frac{1-9*10^{-8}}{1-3*10^{-4}}$$ (1) Let $$a=10^{-4}$$ --> (1) equals $$\frac{1-a^{2}}{1-a}-\frac{1-(3a)^{2}}{1-3a}=\frac{(1-a)(1+a)}{1-a}-\frac{(1-3a)(1+3a)}{1-3a}=(1+a)-(1+3a)=1+a-1-3a=-2a=-2*10^{-4}$$ --> Answer: supposed to be D, but I can't explain why there is negative sign in my final answer but the provided answer does not have it!! Please help to find my mistake! Math Forum Moderator Joined: 20 Mar 2014 Posts: 2643 GMAT 1: 750 Q49 V44 GPA: 3.7 WE: Engineering (Aerospace and Defense) Followers: 107 Kudos [?]: 1232 [1] , given: 786 Re: 0.99999999/1.0001 - 0.99999991/1.0003 = [#permalink] ### Show Tags 12 Jul 2015, 08:24 1 This post received KUDOS Expert's post 1 This post was BOOKMARKED Beat720 wrote: Walkabout wrote: $$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=$$ (A) 10^(-8) (B) 3*10^(-8) (C) 3*10^(-4) (D) 2*10^(-4) (E) 10^(-4) $$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}$$=$$\frac{1-10^{-8}}{1-10^{-4}}-\frac{1-9*10^{-8}}{1-3*10^{-4}}$$ (1) Let $$a=10^{-4}$$ --> (1) equals $$\frac{1-a^{2}}{1-a}-\frac{1-(3a)^{2}}{1-3a}=\frac{(1-a)(1+a)}{1-a}-\frac{(1-3a)(1+3a)}{1-3a}=(1+a)-(1+3a)=1+a-1-3a=-2a=-2*10^{-4}$$ --> Answer: supposed to be D, but I can't explain why there is negative sign in my final answer but the provided answer does not have it!! Please help to find my mistake! Correct the denominators above with '-' in red to '+' and you will have the correct answer. $$1.0001 = 1+10^{-4}$$ and not $$1-10^{-4}$$. You have made a similar mistake for the second part. _________________ Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515 Rules for Posting in Quant Forums: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html Writing Mathematical Formulae in your posts: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html#p1096628 GMATCLUB Math Book: http://gmatclub.com/forum/gmat-math-book-in-downloadable-pdf-format-130609.html Everything Related to Inequalities: http://gmatclub.com/forum/inequalities-made-easy-206653.html#p1582891 Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html Director Joined: 12 Sep 2015 Posts: 638 Location: Canada Followers: 66 Kudos [?]: 471 [1] , given: 19 0.99999999/1.0001 - 0.99999991/1.0003 = [#permalink] ### Show Tags 10 Apr 2016, 10:51 1 This post received KUDOS Walkabout wrote: $$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=$$ (A) 10^(-8) (B) 3*10^(-8) (C) 3*10^(-4) (D) 2*10^(-4) (E) 10^(-4) Another approach is to combine the fractions and then use some approximation. First combine the fractions by finding a common denominator. (9999.9999)/(10001) - (9999.9991)/(10003) = (9999.9999)(10003)/(10001)(10003) - (9999.9991)(10001) /(10003)(10001) = [(10003)(9999.9999) - (10001)(9999.9991)] / (10001)(10003) = [(10003)(10^4) - (10001)(10^4)] / (10^4)(10^4) ... (approximately) = [(10003) - (10001)] / (10^4) ... (divided top and bottom by 10^4) = 2/(10^4) = 2*10^(-4) = D Cheers, Brent _________________ Brent Hanneson – Founder of gmatprepnow.com Brent also tutors students for the GMAT Verbal Forum Moderator Joined: 02 Aug 2009 Posts: 4043 Followers: 272 Kudos [?]: 2817 [1] , given: 98 Re: 0.99999999/1.0001 - 0.99999991/1.0003 = [#permalink] ### Show Tags 11 Apr 2016, 07:19 1 This post received KUDOS Expert's post Walkabout wrote: $$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=$$ (A) 10^(-8) (B) 3*10^(-8) (C) 3*10^(-4) (D) 2*10^(-4) (E) 10^(-4) Hi, we should be able to take advantage of choices whereever possible... Ofcourse, I donot think choices here were given to be able to eliminate all except ONE... But then that is what is possible in this Q with the choices given... $$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=$$.. both the terms individually are $$\frac{ODD}{ODD}$$so each term should come out as ODD and $$ODD - ODD =EVEN$$... $$\frac{0.99999999*1.0003-1.0001*0.99999991}{1.0003*1.0001}=$$ should be $$\frac{EVEN}{ODD}$$.. so our answer should be something with the last digit in DECIMALs as some EVEN number.. Only D has a 2 in its ten-thousandths place.. D But ofcourse if we had another choice of same type, we would have had to use a^2-b^2 as done by bunuel But if choice permits, GMAT is all about using the opportunities... _________________ Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372 Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html Intern Joined: 09 Sep 2013 Posts: 19 Followers: 1 Kudos [?]: 1 [0], given: 7 Re: 0.99999999/1.0001 - 0.99999991/1.0003 = [#permalink] ### Show Tags 09 Oct 2013, 17:50 How did we even know to apply a^2 - b^2 = (a - b)(a + b) to this problem? I understand the math, but if I saw this problem on the test I would have never guessed to apply that method. Thanks, C Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6907 Location: Pune, India Followers: 1989 Kudos [?]: 12354 [0], given: 221 Re: 0.99999999/1.0001 - 0.99999991/1.0003 = [#permalink] ### Show Tags 09 Oct 2013, 21:36 runningguy wrote: How did we even know to apply a^2 - b^2 = (a - b)(a + b) to this problem? I understand the math, but if I saw this problem on the test I would have never guessed to apply that method. Thanks, C (a^2 - b^2) is the "mathematical" method i.e. a very clean solution that a Math Prof will give you. With enough experience a^2 - b^2 method will come to you. But since most of us are not Math professors, we could get through using brute force. Two alternative approaches have been given by mau5 and lequanftu26. You may want to give them a thorough read. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: 0.99999999/1.0001 - 0.99999991/1.0003 = [#permalink]

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01 Mar 2014, 23:48
I just rounded it up, did some questionable math and got lucky, it would seem.

1/1.0001 - 1/1.0003 = ?
1/1.0001 = 1/1.0003
1.0003(1) = 1.0001(1)
1.0003-1.0001=?
0.0002

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Re: 0.99999999/1.0001 - 0.99999991/1.0003 = [#permalink]

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03 Mar 2014, 00:14
actionj wrote:
I just rounded it up, did some questionable math and got lucky, it would seem.

1/1.0001 - 1/1.0003 = ?
1/1.0001 = 1/1.0003
1.0003(1) = 1.0001(1)
1.0003-1.0001=?
0.0002

Can you elaborate your method?? I did some approximation & landed up with a wrong answer
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03 Mar 2014, 00:17
Looking at such problems, how to decide (upon looking at the options) if some values are to be approximated or problem has to be solved calculas?
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03 Mar 2014, 00:59
The decision to use the method I used was based on a lack of knowledge to apply any other method. I don't know how to elaborate my method that much more. I rounded both the 0.9999999 up to 1, then cross multiplied, then subtracted to get to the answer. As per my post above, was luck more than anything that I got the correct answer.
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 = [#permalink]

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07 Jun 2014, 02:41
Bunuel/ Karishma,

Could you please share similar type of sums for practice?
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05 Nov 2014, 23:09
MulanQ wrote:
This is a crazy question eventhough I got the correct answer...

I simply think:
1. In fraction 1, we have 8 decimals devided by 4 decimals, so the result would be a number with 4 decimals
2. Fraction 2, same, so we should have another number with 4 decimals
3. Take these 2 numbers subtract each other, we should have another number with 4 decimals, so answer should be some thing 10^-4 --> eliminate A and B
4. We have an odd number - another odd number, the result should be van even number ---> eliminate C and E

IN real test, if I pump into this kind of question, I would just guess and move on

This approach sounds waaaay better to me. Nice thinking.
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 = [#permalink]

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14 May 2015, 07:01
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Just do what a fifth grader would do!

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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =   [#permalink] 14 May 2015, 07:01

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