0.99999999/1.0001 - 0.99999991/1.0003 = : Problem Solving (PS)

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KarishmaB

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

09 Oct 2013, 21:36

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Expert Reply

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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KarishmaB

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

08 May 2022, 22:00

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woohoo921 wrote:

KarishmaB 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.

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

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

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

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

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

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

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

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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KarishmaB

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

15 Dec 2017, 00:45

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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.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#-soluti ... olving_211

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Kinshook

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

27 Aug 2019, 09:03

1

Kudos

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

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

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

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

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

IMO D

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dave13

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

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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adkikani

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

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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KarishmaB

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

21 Jul 2019, 00:20

Expert Reply

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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jabhatta2

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

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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KarishmaB

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Posts: 14816

Location: Pune, India

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

27 Nov 2019, 21:09

Expert Reply

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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jabhatta2

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Posts: 1376

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

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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KarishmaB

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

27 Nov 2019, 22:02

Expert Reply

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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Vsenth05

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

09 Jul 2021, 09:39

All the given terms are odd, i.e., odd/odd - odd/odd.

odd/odd is always odd. Hence, it becomes odd - odd.

odd - odd is always even. In the given answer options only Option D is even.

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Crytiocanalyst

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

09 Jul 2021, 10:03

It took me 5 min to get to the right choice
0.999999991.0001−.999999911.00030.999999991.0001−.999999911.0003

1−.000000011+.0001−1−.000000091+.00031−.000000011+.0001−1−.000000091+.0003

12−.000121+.0001−12−0.000321+.000312−.000121+.0001−12−0.000321+.0003

(1−.0001)(1+.0001)(1+.0001)−(1−.0003)(1+.0003)(1+.0003)(1−.0001)(1+.0001)(1+.0001)−(1−.0003)(1+.0003)(1+.0003)

(1−.0001)−(1−.0003)
=0.9999-0.9997 = 0.0002
hence IMO D

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

13 Jul 2021, 03:27

how about this estimation approach?

99/1.1 -91/1.3

= 9*10 - 7*10
=90-70
=20

the only option that has 2 is Ans: D

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100mitra

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

08 Sep 2021, 00:19

Correct Option : D = (2 * \(10^{-4}\))

= ( \(10^{-10}\) * (99 - 91) ) / ( \(10^{-4}\) * (1 -3 ) )
= \((10^{-6}\) x (8/2)
= \(2^2\) * \(10^{-6}\)
= (2 * \(10^{-4}\))

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Regor60

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

23 Mar 2022, 13:42

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

Posted from my mobile device

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woohoo921

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

08 May 2022, 16:31

KarishmaB 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.

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

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

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

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

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

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

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

L

Kinshook

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

10 Nov 2022, 07:24

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

IMO D

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

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

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