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Manager  Joined: 02 Dec 2012
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Question Stats: 52% (02:32) correct 48% (02:22) wrong based on 2668 sessions

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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) $$10^{(-4)}$$

OG 2019 #215 PS00574
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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.
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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)

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

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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)

2 minutes ain't gonna do this math. 3mins may with a high margin of error under exam condition.

Look at the options closely

A. 0.00000001
B. 0.00000003
C. 0.0003
D. 0.0002
E. 0.0001

Only one option is an even number, the rest odd (an esoteric sort of even/odd number)

both fractions in the question are odds
odd minus odd is always even.
So answer must be even.
Only D is even.

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1
4
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*1.0003 = 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.
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26
5
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.
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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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.

Thanks,
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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.
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10
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)
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4
5
I'm going to repost Ron's solution

OR
Just do what a fifth grader would do! Attachment: 00000370.png [ 81.02 KiB | Viewed 14672 times ]

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

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3
3
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
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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...
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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3
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)

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
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Last edited by JeffTargetTestPrep on 03 May 2016, 09:19, edited 2 times in total.
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1
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.
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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) $$10^{(-4)}$$

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

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

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

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

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

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

=> $$2*10^{-4}$$
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VeritasPrepKarishma 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)

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.

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

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

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

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

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

$$.0002 = 2*10^{-4}$$

Here is the video solution to this difficult looking problem: https://www.veritasprep.com/gmat-soluti ... olving_211
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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Bunuel 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)

$$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=\frac{1-10^{-8}}{1+10^{-4}}-\frac{1-9*10^{-8}}{1+3*10^{-4}}$$

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.

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 Math Expert V
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dave13 wrote:
Bunuel 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)

$$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=\frac{1-10^{-8}}{1+10^{-4}}-\frac{1-9*10^{-8}}{1+3*10^{-4}}$$

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.

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

Check below for more:
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Ultimate GMAT Quantitative Megathread

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