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705-805 Level|   Algebra|   Arithmetic|                                 
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0.99999999/1.0001 - 0.99999991/1.0003 = 25 Feb 2018, 17:25
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Magoosh video solution:





https://gmat.magoosh.com/forum/2827-0-9 ... 991-1-0003
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0.99999999/1.0001 - 0.99999991/1.0003 = 22 Aug 2019, 04:04
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Walkabout
\(\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=\)


(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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\(\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}= \frac{(1 - .00000001)}{(1+.0001)} - \frac{(1 - .00000009)}{(1+.0003)} = (1-.0001) - (1-.0003) = .0002 = 2*10^-4\)

IMO D
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0.99999999/1.0001 - 0.99999991/1.0003 = 27 Nov 2019, 14:07
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chetan2u VeritasKarishma Gladiator59 Bunuel generis

Hi Experts -- i see the answer is D in this question but i do have a question regarding "Estimating" this

Give estimation is a strategy for the GMAT -- i estimated :

1.0001 == estimated down to 1.0000
1.0003 == estimated down to 1.0000

Hence i was left with

\(\frac{0.99999999}{1}\) - \(\frac{0.99999991}{1}\) =

0.99999999 −0.99999991 =0.00000008 or 8 * \(10^{-8}\)

Question : why isn't the estimation method getting me close /somewhat close to option D ...option D seems 1000 times larger than my estimation (option B and option C sees closer)

Any idea where the logic is wrong in this case for such a HUGE difference ?
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0.99999999/1.0001 - 0.99999991/1.0003 = 27 Nov 2019, 21:09
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jabhatta@umail.iu.edu
chetan2u VeritasKarishma Gladiator59 Bunuel generis

Hi Experts -- i see the answer is D in this question but i do have a question regarding "Estimating" this

Give estimation is a strategy for the GMAT -- i estimated :

1.0001 == estimated down to 1.0000
1.0003 == estimated down to 1.0000

Hence i was left with

\(\frac{0.99999999}{1}\) - \(\frac{0.99999991}{1}\) =

0.99999999 −0.99999991 =0.00000008 or 8 * \(10^{-8}\)

Question : why isn't the estimation method getting me close /somewhat close to option D ...option D seems 1000 times larger than my estimation (option B and option C sees closer)

Any idea where the logic is wrong in this case for such a HUGE difference ?
Show more

Note that all the numbers are very close to 1. So you cannot estimate some of them to 1 and not others. All options are very very close to each other too. So you cannot estimate here.
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0.99999999/1.0001 - 0.99999991/1.0003 = 27 Nov 2019, 21:31
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jabhatta@umail.iu.edu
chetan2u VeritasKarishma Gladiator59 Bunuel generis

Hi Experts -- i see the answer is D in this question but i do have a question regarding "Estimating" this

Give estimation is a strategy for the GMAT -- i estimated :

1.0001 == estimated down to 1.0000
1.0003 == estimated down to 1.0000

Hence i was left with

\(\frac{0.99999999}{1}\) - \(\frac{0.99999991}{1}\) =

0.99999999 −0.99999991 =0.00000008 or 8 * \(10^{-8}\)

Question : why isn't the estimation method getting me close /somewhat close to option D ...option D seems 1000 times larger than my estimation (option B and option C sees closer)

Any idea where the logic is wrong in this case for such a HUGE difference ?

Note that all the numbers are very close to 1. So you cannot estimate some of them to 1 and not others. All options are very very close to each other too. So you cannot estimate here.
Show more

Hi karishma

Thank you so much for replying

Understand every answer is close to 1

But you would agree though the estimation method is OFF by times 1000 compared to option D

Any idea why the estimation is so off by any chance

I think every one would make the estimation that 1.0001 can be written up as 1.0000 and the same with 1.0003 as 1.0000

Posted from my mobile device
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0.99999999/1.0001 - 0.99999991/1.0003 = 27 Nov 2019, 22:02
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jabhatta@umail.iu.edu
VeritasKarishma
jabhatta@umail.iu.edu
chetan2u VeritasKarishma Gladiator59 Bunuel generis

Hi Experts -- i see the answer is D in this question but i do have a question regarding "Estimating" this

Give estimation is a strategy for the GMAT -- i estimated :

1.0001 == estimated down to 1.0000
1.0003 == estimated down to 1.0000

Hence i was left with

\(\frac{0.99999999}{1}\) - \(\frac{0.99999991}{1}\) =

0.99999999 −0.99999991 =0.00000008 or 8 * \(10^{-8}\)

Question : why isn't the estimation method getting me close /somewhat close to option D ...option D seems 1000 times larger than my estimation (option B and option C sees closer)

Any idea where the logic is wrong in this case for such a HUGE difference ?

Note that all the numbers are very close to 1. So you cannot estimate some of them to 1 and not others. All options are very very close to each other too. So you cannot estimate here.

Hi karishma

Thank you so much for replying

Understand every answer is close to 1

But you would agree though the estimation method is OFF by times 1000 compared to option D

Any idea why the estimation is so off by any chance

I think every one would make the estimation that 1.0001 can be written up as 1.0000 and the same with 1.0003 as 1.0000

Posted from my mobile device
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Note that .99999999 is also almost 1. In fact it is closer to 1 than 1.0001. So if using approximation, the first fraction becomes 1/1 = 1. Same thing for second fraction. So if using approximation, you will get 1 -1 = 0. Note that every option is very close to 0.

The point is this:
You cannot say that 99/102 = 99/100 = .99
when your options have .99, .98. .97, .96, .95

Actually, the answer is .97 here. The margin of error is so small between the numbers and options that approximation will not give you the correct answer.
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0.99999999/1.0001 - 0.99999991/1.0003 = 14 Mar 2021, 16:10
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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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0.99999999/1.0001 - 0.99999991/1.0003 = 23 Mar 2022, 13:42
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I can't imagine an easier approach than this. Skip all the decimals and estimations:

The numerators are multiples of 10^-8 and the denominators 10^-4.

Set X=10^-4

So left side is (1-X^2/(1+X) which factors to (1+X)*(1-X)/(1+X) which equals

1-X

Right side is (1-9X^2)/(1+3X), which factors to (1+3X)*(1-3X)/(1+3X), which equals

1-3X

Adding the two results

1-X+3X-1 = 2X

Since X= 10^-4, answer is

2*10^-4

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0.99999999/1.0001 - 0.99999991/1.0003 = 08 May 2022, 16:31
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KarishmaB
Walkabout
\(\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=\)

(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.

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

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

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

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

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

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

KarishmaB
This is very helpful, thank you! To clarify my understanding of scientific notation...why is it that we do not count the zero when moving the decimal place right and left for expanding scientific notation? Is my understanding correct that you always start at the first non-zero number?

For example, I know that:
10^-4=0.0001 (but I get stuck because I would think you start at the zero in moving over the decimal e.g., to get .001)
and
10^4=10,000 (but I get stuck because I would think you start at the zero again to get 100,000)

Thank you again :)
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0.99999999/1.0001 - 0.99999991/1.0003 = 08 May 2022, 22:00
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KarishmaB
Walkabout
\(\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=\)

(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.

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

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

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

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

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

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

KarishmaB
This is very helpful, thank you! To clarify my understanding of scientific notation...why is it that we do not count the zero when moving the decimal place right and left for expanding scientific notation? Is my understanding correct that you always start at the first non-zero number?

For example, I know that:
10^-4=0.0001 (but I get stuck because I would think you start at the zero in moving over the decimal e.g., to get .001)
and
10^4=10,000 (but I get stuck because I would think you start at the zero again to get 100,000)

Thank you again :)
Show more

Don't think of it as scientific notation. Think of it as exponents.

\(10^{-4} = \frac{1}{10^4} = \frac{1}{10*10*10*10} = \frac{1}{10,000}\)

10^4 is simply 10 multiplied with itself total four times i.e. 10*10*10*10 = 10,000
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0.99999999/1.0001 - 0.99999991/1.0003 = 21 Aug 2024, 06:46
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ofk
I just rounded it up,
1/1.0001-1/1.0003
=(1.003-1.001)/(1.003*1.001)
=0.002/1
=2*(10^-4)
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Keep in mind that rounding small numbers when small numbers appear as the answer choices is a risky strategy.

Unless possessing some particular insight, one could have rounded the denominators also, leading to an incorrect answer.

Posted from my mobile device
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0.99999999/1.0001 - 0.99999991/1.0003 = 05 Mar 2025, 21:57
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Is there a link for more similar question? If yes, can you share it please?
Bunuel
Walkabout
\(\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=\)

(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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0.99999999/1.0001 - 0.99999991/1.0003 = 05 Mar 2025, 23:50
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werty_123
Is there a link for more similar question? If yes, can you share it please?
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Try checking Arithmetic tag.
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0.99999999/1.0001 - 0.99999991/1.0003 = 18 May 2025, 02:48
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Quote:
\(\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=\)


(A) \(10^{(-8)}\)

(B) \(3*10^{(-8)}\)

(C) \(3*10^{(-4)}\)

(D) \(2*10^{(-4)}\)

(E) \(10^{(-4)}\)
Show more

The optimal method i.e. difference of squares didn't strike me, and I used some estimation method. Had the options been more trickier or closer, I might have fallen into some trap. Here it goes:

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

= \(\frac{(1 - 0.00000001)}{1.0001}-\frac{(1 - 0.00000009)}{1.003}\)

= \(\frac{1}{1.0001} - \frac{0.00000001}{1.0001}-\frac{1}{1.003}+\frac{0.00000009}{1.003}\)

= \(\frac{1}{1.0001} - \)(small number ≈ 0) \( - \frac{1}{1.0003} + \) (small number ≈ 0)

= \(\frac{1}{1.0001} - \frac{1}{1.0003} - \) (small number ≈ 0) \( + \) (small number ≈ 0)

= \(\frac{(1.0003 - 1.0001)}{1.0001 * 1.0003}- 0 \)

Now, 1.0001 * 1.0003 ≈ 1.

= 0.0002

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

= option D.
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0.99999999/1.0001 - 0.99999991/1.0003 = 02 Jul 2025, 09:48
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The logical way I thought about this is that factoring all the powers of 10 out, from both the denominator and the numerator, would leave us with a 10^-4. The numbers inside the brackets would then be 99999999/10001 - 99999991/10003. Since for the first term, the unit digit is likely to be 9, the unit digit of the second number is likely to be 7, so 9-7=2, and we have the 10^-4 outside
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0.99999999/1.0001 - 0.99999991/1.0003 = 06 Jul 2025, 02:05
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a number is even or odd only if it is an integer, how did you conclude that the fractions deduce to an odd/even number?
please help!
Ekland
Walkabout
\(\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=\)

(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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0.99999999/1.0001 - 0.99999991/1.0003 = 06 Jul 2025, 02:39
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ashKing12
a number is even or odd only if it is an integer, how did you conclude that the fractions deduce to an odd/even number?
please help!
Ekland
Walkabout
\(\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=\)

(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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That solution is not correct. Ignore it.
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0.99999999/1.0001 - 0.99999991/1.0003 = 04 Aug 2025, 01:02
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Kinshook
Walkabout
\(\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=\)


(A) \(10^{(-8)}\)

(B) \(3*10^{(-8)}\)

(C) \(3*10^{(-4)}\)

(D) \(2*10^{(-4)}\)

(E) \(10^{(-4)}\)


OG 2019 #215 PS00574

\(\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}= \frac{(1 - .00000001)}{(1+.0001)} - \frac{(1 - .00000009)}{(1+.0003)} = (1-.0001) - (1-.0003) = .0002 = 2*10^-4\)

IMO D
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I guess first order approximation was utilised here. Was the \( 8 * 10^{-8} \) (in your final answer) ignored due to being too small?

My working:
\(\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}= 1-(10^{-8}+10^{-4}) - (1-(9*10^{-8}+3*10^{-4}) = 2*10^{-4}+8*10^{-8} ≈ 2*10^{-4} \)
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0.99999999/1.0001 - 0.99999991/1.0003 = 21 Oct 2025, 07:35
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