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0.99999999/1.0001 - 0.99999991/1.0003 =

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

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New post 10 Apr 2016, 09:51
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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)


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,
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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)



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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New post 03 May 2016, 08:00
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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)


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, 08:19, edited 2 times in total.
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 = [#permalink]

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New post 05 Oct 2016, 08:19
It can be solved in 10 sec using digital sum method http://www.sjsu.edu/faculty/watkins/Digitsum.htm
digital sum(0.99999999)/digital sum(1.0001)−digital sum(0.99999991)/digital sum(1.0003)=(0/2)-(1/4)=-7=2

A. 0.00000001 digital sum is 1
B. 0.00000003 digital sum is 3
C. 0.0003 digital sum is 3
D. 0.0002 digital sum is 2 only this option matches
E. 0.0001 digital sum is 1
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 = [#permalink]

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New post 10 Jun 2017, 20:27
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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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Re: 0.99999999/1.0001 - 0.99999991/1.0003 = [#permalink]

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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.

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

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

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

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

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

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


hi mam

your answer to the question is awesome as usual. Anyway, what about rounding up...? Can the principles of rounding up be applicable here to problems as such on the exam day ...?

thanks in advance, mam....
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0.99999999/1.0001 - 0.99999991/1.0003 = [#permalink]

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New post 30 Sep 2017, 10:07
10^4*10^-8 [(10^8-1)/(10^4+1) - (10^8-9)/(10^4+3)]

=10^-4 [10^4-1-10^4-3]
=2*10^-4
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 = [#permalink]

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New post 11 Oct 2017, 07:11
The calculations can be very simple.
1. Round 0.99999999 to 1. As it will give you the approximate answer.
2. Round 0.99999991 to 1 also.
Now take the LCM, the fraction will be 0.0002/1.0000003. Divide, the answer will be approximately 2 x 10e-4. If you want to check the answer from a calculator, then the answer will be 1.9999999999999x10e-4, which is approximately to 2.
This method works for me. Hope it works for you all.
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New post 19 Nov 2017, 03:49
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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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Re: 0.99999999/1.0001 - 0.99999991/1.0003 = [#permalink]

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New post 14 Dec 2017, 23:45
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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0.99999999/1.0001 - 0.99999991/1.0003 = [#permalink]

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New post 06 Jan 2018, 04:27
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 :-)
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 = [#permalink]

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New post 06 Jan 2018, 05:53
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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



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

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New post 07 Feb 2018, 00:42
its too confusing so could anyone described me as school kid please !
I,very appreciate him !
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 = [#permalink]

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New post 07 Feb 2018, 00:56
mkumar26 wrote:
its too confusing so could anyone described me as school kid please !
I,very appreciate him !


There are many different approaches on previous tow pages. I selected 6 which are most elaborate:
https://gmatclub.com/forum/topic-144735.html#p1161158
https://gmatclub.com/forum/topic-144735.html#p1249169
https://gmatclub.com/forum/topic-144735.html#p1249221
https://gmatclub.com/forum/topic-144735 ... l#p1670771
https://gmatclub.com/forum/topic-144735 ... l#p1671128
https://gmatclub.com/forum/topic-144735 ... l#p1680083

Please go through them carefully and if still something remains unclear ask specific question.
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 = [#permalink]

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New post 22 Feb 2018, 04:38
Nice point.it also dint occured to me that ODD -EVEN rules are applicable for integers not for fractions.





ENGRTOMBA2018 wrote:
Nez 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)


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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A conceptual mistake. Odd/even are ONLY for integers. You can not define a fraction to be odd or even, you were lucky in this question.

You do not need anything more than what Bunuel has mentioned in his post http://gmatclub.com/forum/topic-144735.html#p1161158 (frankly you should know this method of attack for fractions to save yourself a lot of time!).

This is a straightforward question that can be solved within 2 minutes if you adopt proper strategies.
Re: 0.99999999/1.0001 - 0.99999991/1.0003 =   [#permalink] 22 Feb 2018, 04:38

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