0.99999999/1.0001 - 0.99999991/1.0003 =

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.

\(\frac{0.99999999}{1.0001}-\frac{.99999991}{1.0003}\)

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

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

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

\((1 - .0001) - (1 - .0003)\)

\(.0002 = 2*10^{-4}\)

Approach #1: Algebra

First recognize that \(0.99999999 = 1 - 0.00000001\) and \(0.99999991 = 1 - 0.00000009\)

Also recognize that \(0.00000001 = (0.0001)^2\) and \(0.00000009 = (0.0003)^2\)

So we get: \(\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=\frac{1 - 0.00000001}{1.0001}-\frac{1 - 0.00000009}{1.0003}\)

\(=\frac{1^2 - (0.0001)^2}{1.0001}-\frac{1^2 - (0.0003)^2}{1.0003}\)

\(=\frac{(1 + 0.0001)(1 - 0.0001)}{1.0001}-\frac{(1 + 0.0003)(1 - 0.0003)}{1.0003}\) [factored as difference of squares]

\(=\frac{(1.0001)(0.9999)}{1.0001}-\frac{(1.0003)(0.9997)}{1.0003}\) [simplified]

\(=0.9999-0.9997\) [some terms cancelled out]

\(=0.0002\)

\(=2 \times 10^{-4}\)

Answer: D

Approach #2: Combine the fractions and then use some approximation.

To avoid so many decimals, let's first multiply the numerators and denominators in both fractions by 10,000 to get the following equivalent fractions: \(\frac{9999.9999}{10001}-\frac{9999.9991}{10003}\)

Rewrite with common denominators: \(\frac{(10003)(9999.9999)}{(10001)(10003)}-\frac{(10001)(9999.9991)}{(10001)(10003)}\)

Combine to get: \(\frac{(10003)(9999.9999) - (10001)(9999.9991)}{(10001)(10003)}\)

Since 9999.9999 and 9999.9991 are very close to 10,000, we get the following approximation: \(\frac{(10003)(10,000) - (10001)(10,000)}{(10001)(10003)}\)

Likewise, we can approximate denominator as follows: \(\frac{(10003)(10,000) - (10001)(10,000)}{(10,000)(10,000)}\)

Factor the numerator: \(\frac{(10,000)(10,003 - 10,001)}{(10,000)(10,000)}\)

Simplify: \(\frac{10,003 - 10,001}{10,000}\)

Simplify: \(\frac{2}{10,000}\)

Rewrite as: \((2)(\frac{1}{10,000})\)

Which is the same as: \((2)(10^{-4})\)

Answer: D

Cheers,
Brent

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*1.0003 = 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.

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.

Hi,

To start, since this question has no variables, we know that we're going to be doing some type of math to get to the answer. Looking at the answer choices, they're all written in the same 'format' - and can be rewritten as a decimal point followed by a bunch of 0s and then a single non-0 digit. When the GMAT gives you an 'ugly looking' fraction to work with, you can often 'rewrite' what you've been given (potentially getting rid of the fraction entirely by reducing it or reformatting it).

The first fraction is .99999999/1.0001.... we can multiply both the numerator and denominator by 10,000... which gives us....

9999.9999/10001

You might recognize a pattern here (there will almost certainly be a string of 9s here) Even if you don't spot the pattern though, with a few division steps, you'll end up with .9999

This is interesting for a couple of reasons. First, it's only 4 decimal places (notice how three of the answers fit that pattern while two of them don't). Second, you should again consider the format of the answer choices... each answer is a bunch of 0s followed by a single non-0 digit. That result won't happen if you're subtracting an 8-digit decimal from a 4-digit decimal. Thus, it's almost certain that the second fraction will ALSO be a 4-digit decimal.

Using the same approach that we used on the first fraction, we can rewrite the second fraction as...

9999.9991/10003

Before you do any math, think about how a '3' divides into a '1'.... what will the last digit in this decimal probably be? Since 7x3 = 21, that last digit will almost certainly be a '7'. With a little work, you can prove it. You'll end up with .9997

Subtracting the two decimals, you'll have .9999 - .9997 = .0002

Notice the '2' as the last decimal point? It won't take much to find that answer.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

\(\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=\)

A. 10^-8
B. 3(10^-8)
C. 3(10^-4)
D. 2(10^-4)
E. 10^-4[/quote]

Well, you need a little bit of algebra here...

The question is meant to test the formula \(a^2-b^2=(a+b)(a-b).\)
The clue is in the answers, explicitly the powers of \(10\).
You can write the given expression as follows:

\(\frac{1-10^{-8}}{1+10^{-4}}-\frac{1-9*10^{-8}}{1+3*10^{-4}}=\frac{(1+10^{-4})(1-10^{-4})}{1+10^{-4}}-\frac{(1+3*10^{-4})(1-3*10^{-4})}{1+3*10^{-4}}=1-10^{-4}-(1-3*10^{-4})=2*10^{-4}.\)

Answer D

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.

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

Answer = D. 2(10^-4)

I'm going to repost Ron's solution

OR
Just do what a fifth grader would do!

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...

When first looking at this problem, we must consider the fact that 0.99999999/1.0001 and 0.99999991/1.0003 are both pretty nasty-looking fractions. However, this is a situation in which we can use the idea of the difference of two squares to our advantage. To make this idea a little clearer, let’s first illustrate the concept with a few easier whole numbers. For instance, let’s say we were asked:

999,999/1,001 – 9,991/103 = ?

We could rewrite this as:

(1,000,000 – 1)/1,001 – (10,000 – 9)/103

(1000 + 1)(1000 – 1)/1,001 – (100 – 3)(100 + 3)/103

(1,001)(999)/1,001 – (97)(103)/103

999 – 97 = 902

Notice how cleanly the denominators canceled out in this case. Even though the given problem has decimals, we can follow the same approach.

0.99999999/1.0001 – 0.99999991/1.0003

[(1 – 0.00000001)/1.0001] – [(1 – 0.00000009)/1.0003]

[(1 – 0.0001)(1 + 0.0001)/1.0001] – [(1 – 0.0003)(1 + 0.0003)]

When converting this using the difference of squares, we must be very careful not to make any mistakes with the number of decimal places in our values. Since 0.00000001
has 8 decimal places, the decimals in the factors of the numerator of the first set of brackets must each have 4 decimal places. Similarly, since 0.00000009 has 8 decimal places, the decimals in the factors of the numerator of the second set of brackets must each have 4 decimal places. Let’s continue to simplify.

[(1 – 0.0001)(1 + 0.0001)/1.0001] – [(1 – 0.0003)(1 + 0.0003)]

[(0.9999)(1.0001)/1.0001] – [(0.9997)(1.0003)/1.0003]

0.9999 – 0.9997

0.0002

Converting this to scientific notation to match the answer choices, we have:

2 x 10^-4

Answer is D

Here's how I solved it, which I think is less painful (but maybe wouldn't generalize as well)

\(\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}\)

Get rid of that distracting decimal:

\(\frac{9999999}{100010000}-\frac{99999991}{100030000}\)

Recognize that both fractions are some very small number from 1, so try to expose that small number by pulling out the 1:

\(\frac{100010000-10001}{100010000}-\frac{100030000-30009}{100030000}\)

Simplify, and marvel at the convenient numbers revealed

\(1-\frac{1}{10000}-1+\frac{3}{10000}\)

Do the arithmetic, done.

Simplest way to handle this kind of problem is:

.99 can be written as \(\frac{99}{10^2}\) OR \(\frac{(10^2-1)}{10^2}\) and .91 can be written as \(\frac{91}{10^2}\) OR \(\frac{(10^2-9)}{10^2}\)

For two 9's, we wrote \(\frac{(10^2-1)}{10^2}\), for eight 9's, we will take \(\frac{(10^8-1)}{10^8}\)
Same rule applies to 1.0001 and 1.0003

So, the simplified version becomes

=> \(\frac{(10^8-1)*10^4}{(10^4+1)*10^8}\) - \(\frac{(10^8-9)*10^4}{(10^4+1)*10^8}\)

=> \(\frac{{(10^4)^2-1^2}}{10^4*(10^4+1)}\) - \(\frac{{(10^4)^2-3^2}}{10^4*(10^4+3)}\)

=> \(\frac{(10^4+1)*(10^4-1)}{10^4*(10^4+1)}\) - \(\frac{(10^4+3)*(10^4-3)}{10^4*(10^4+3)}\)

=> \(\frac{(10^4-1)}{10^4}\) - \(\frac{(10^4-3)}{10^4}\)

=> \(\frac{1}{10^4}*(10^4-1-10^4+3)\)

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

Here is the video solution to this difficult looking problem: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#-soluti ... olving_211

Hello Bunuel - do you have some useful links for better understanding of your solution ... i dont understand based on which rule transform 1.0003 into 1+3*10^-4 , and 1-9*10^-8 ?

Also what made you apply this formula \(a^2-b^2=(a+b)(a-b)\) and could you you break down this solution into more detaled steps ?
\(\frac{1-10^{-8}}{1+10^{-4}}-\frac{1-9*10^{-8}}{1+3*10^{-4}}=\frac{(1+10^{-4})(1-10^{-4})}{1+10^{-4}}-\frac{(1+3*10^{-4})(1-3*10^{-4})}{1+3*10^{-4}}=(1-10^{-4})-(1-3*10^{-4})=2*10^{-4}\). and what common denominator you chose after applying \(a^2-b^2=(a+b)(a-b)\)

Thank you

1.
\(1+10^{-4}=1+\frac{1}{1,0000}=1+0.0001=1.0001\)
\(1+3*10^{-4}=1+\frac{3}{1,0000}=1+0.0003=1.0003\)

2. You should apply \(a^2-b^2=(a+b)(a-b)\) because after this you can reduce the denominator.

3.

8. Exponents and Roots of Numbers

• Theory
  Exponents and Roots on the GMAT: Tips and hints
  Cyclicity and the remainders on the GMAT
  
  • Questions
    Units digits, exponents, remainders problems
    Tough and tricky DS exponents and roots questions with detailed solutions
    Tough and tricky PS exponents and roots questions with detailed solutions
    Exponents DS Questions
    Exponents PS Questions
    Roots DS Questions
    Roots PS Questions

Check below for more:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.