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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 03 Sep 2018, 22:43
3
Hi,

To start, since this question has no variables, we know that we're going to be doing some type of math to get to the answer. Looking at the answer choices, they're all written in the same 'format' - and can be rewritten as a decimal point followed by a bunch of 0s and then a single non-0 digit. When the GMAT gives you an 'ugly looking' fraction to work with, you can often 'rewrite' what you've been given (potentially getting rid of the fraction entirely by reducing it or reformatting it).

The first fraction is .99999999/1.0001.... we can multiply both the numerator and denominator by 10,000... which gives us....

9999.9999/10001

You might recognize a pattern here (there will almost certainly be a string of 9s here) Even if you don't spot the pattern though, with a few division steps, you'll end up with .9999

This is interesting for a couple of reasons. First, it's only 4 decimal places (notice how three of the answers fit that pattern while two of them don't). Second, you should again consider the format of the answer choices... each answer is a bunch of 0s followed by a single non-0 digit. That result won't happen if you're subtracting an 8-digit decimal from a 4-digit decimal. Thus, it's almost certain that the second fraction will ALSO be a 4-digit decimal.

Using the same approach that we used on the first fraction, we can rewrite the second fraction as...

9999.9991/10003

Before you do any math, think about how a '3' divides into a '1'.... what will the last digit in this decimal probably be? Since 7x3 = 21, that last digit will almost certainly be a '7'. With a little work, you can prove it. You'll end up with .9997

Subtracting the two decimals, you'll have .9999 - .9997 = .0002

Notice the '2' as the last decimal point? It won't take much to find that answer.

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

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New post 22 Aug 2019, 04:04
1
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


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

IMO D
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =   [#permalink] 22 Aug 2019, 04:04

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