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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.

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 \(9999*10^{-4} - 9997*10^{-4} = 2*10^{-4}\)

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.

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

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!

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.
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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...
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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.

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.
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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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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.

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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