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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 06 Jan 2018, 05:27
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, 06:53
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 20 Jul 2019, 06:44
VeritasKarishma Bunuel chetan2u ScottTargetTestPrep

I am unable to understand why do we not simply approximate a larger no if we are deducting
a very small no as 1 from it.

E.g. \(999,999^2\) is same as \(10^6\)
since 1000 =\(10^3\) and \(a^m\) *\(a^n\) = \(a^(m+n)\)

So effectively \(999,999^2\) - 1 is same as \(10^6\) (How much effect will 1 have
after deducting from such a huge no: negligible, right?)

Why do we use difference of squares instead of approximation while later is more quicker?
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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New post 21 Jul 2019, 00:20
adkikani wrote:
VeritasKarishma Bunuel chetan2u ScottTargetTestPrep

I am unable to understand why do we not simply approximate a larger no if we are deducting
a very small no as 1 from it.

E.g. \(999,999^2\) is same as \(10^6\)
since 1000 =\(10^3\) and \(a^m\) *\(a^n\) = \(a^(m+n)\)

So effectively \(999,999^2\) - 1 is same as \(10^6\) (How much effect will 1 have
after deducting from such a huge no: negligible, right?)

Why do we use difference of squares instead of approximation while later is more quicker?


Yes, certainly approximation is far more quicker. But here are the options are very very small too. Try approximating and let us know how you plan to arrive at the answer.
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0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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New post 27 Aug 2019, 09:03
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) \(10^{(-4)}\)



OG 2019 #215 PS00574


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

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

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

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

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

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New post 19 Sep 2019, 12:52
Can someone tell me if I did it in the right way:

I rounded off the denominator to 1*10^-4

So now we can safely subtract the two equations:

0.00000008/10^-4

This becomes 2*10^-8+4 which gives D

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

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New post 27 Nov 2019, 14:07
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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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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New post 27 Nov 2019, 21:09
jabhatta@umail.iu.edu wrote:
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.
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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New post 27 Nov 2019, 21:31
VeritasKarishma wrote:
jabhatta@umail.iu.edu wrote:
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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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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New post 27 Nov 2019, 22:02
jabhatta@umail.iu.edu wrote:
VeritasKarishma wrote:
jabhatta@umail.iu.edu wrote:
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


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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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =   [#permalink] 27 Nov 2019, 22:02

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