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(3.999999/2.001) - (3.999996/2.002) = ???

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Bunuel
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(3.999999/2.001) - (3.999996/2.002) = ??? [#permalink] New post  01 Nov 2021, 03:45
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\(\frac{(3.999999)}{(2.001)} - \frac{(3.999996)}{(2.002)} = ?\)


(A) \(\frac{(1)}{(1000)}\)

(B) \(\frac{(1)}{(10000)}\)

(C) \(\frac{(1)}{(100000)}\)

(D) \(\frac{(1)}{(1000000)}\)

(E) \(\frac{(1)}{(10000000)}\)
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chetan2u
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Re: (3.999999/2.001) - (3.999996/2.002) = ??? [#permalink] New post  01 Nov 2021, 04:13
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Bunuel wrote:
\(\frac{(3.999999)}{(2.001)} - \frac{(3.999996)}{(2.002)} = ?\)


(A) \(\frac{(1)}{(1000)}\)

(B) \(\frac{(1)}{(10000)}\)

(C) \(\frac{(1)}{(100000)}\)

(D) \(\frac{(1)}{(1000000)}\)

(E) \(\frac{(1)}{(10000000)}\)



\(\frac{(3.999999)}{(2.001)} - \frac{(3.999996)}{(2.002)} \)

\(\frac{(4-0.000001)}{(2.001)} - \frac{(4-0.000004)}{(2.002)} \)

\(\frac{(2^2-(0.001)^2)}{(2.001)} - \frac{(2^2-(0.002)^2)}{(2.002)} \)

Now we know that \(a^2-b^2=(a-b)(a+b)\).

\(\frac{(2+0.001)(2-0.001)}{2.001}-\frac{(2-0.002)(2+0.002)}{2.002}\)

\(\frac{2.001*1.999}{2.001}-\frac{2.002*1.998}{2.002}\)

\(1.999-1.998=0.001=\frac{1}{1000}\)

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(3.999999/2.001) - (3.999996/2.002) = ??? [#permalink] New post  Updated on: 01 Nov 2021, 05:07
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On taking LCM of the denominators:-

(6 decimals*3 decimals - 6 decimals*3 decimals)/6 decimals

Least significant digits of the two numbers in the numerator are going to be 9*2 = 8 and 6*1 = 6. Since the least significant digit cannot be 0 on subtraction of these two numbers, and each number has 9 decimals, there will be 9 decimals post subtraction.

=> 9 decimals/6 decimals = 3 decimals

So answer is (A) 1/1000

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Originally posted by GMATVirtuoso on 01 Nov 2021, 04:09.
Last edited by GMATVirtuoso on 01 Nov 2021, 05:07, edited 8 times in total.
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errorlogger
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Re: (3.999999/2.001) - (3.999996/2.002) = ??? [#permalink] New post  01 Nov 2021, 05:02
Bunuel wrote:
\(\frac{(3.999999)}{(2.001)} - \frac{(3.999996)}{(2.002)} = ?\)


(A) \(\frac{(1)}{(1000)}\)

(B) \(\frac{(1)}{(10000)}\)

(C) \(\frac{(1)}{(100000)}\)

(D) \(\frac{(1)}{(1000000)}\)

(E) \(\frac{(1)}{(10000000)}\)


If we use approximation, we can see that the result is 4/2 - 4/2 ~ 0
So, the right option should be the answer that is closest to 0.
In this case, 1/1000 - A

chetan2u, is it ok to go with this assumption?
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chetan2u
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Re: (3.999999/2.001) - (3.999996/2.002) = ??? [#permalink] New post  01 Nov 2021, 05:13
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errorlogger wrote:
Bunuel wrote:
\(\frac{(3.999999)}{(2.001)} - \frac{(3.999996)}{(2.002)} = ?\)


(A) \(\frac{(1)}{(1000)}\)

(B) \(\frac{(1)}{(10000)}\)

(C) \(\frac{(1)}{(100000)}\)

(D) \(\frac{(1)}{(1000000)}\)

(E) \(\frac{(1)}{(10000000)}\)


If we use approximation, we can see that the result is 4/2 - 4/2 ~ 0
So, the right option should be the answer that is closest to 0.
In this case, 1/1000 - A

chetan2u, is it ok to go with this assumption?


No, that will not be ok.

Rather A is farthest from 0. Closest to 0 is 1/10000000.

Even approximation through number of decimal places may not be a great idea because 1.5/2 will not give you one decimal place but 2 decimal places as 1.5/2 = 0.75
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Re: (3.999999/2.001) - (3.999996/2.002) = ??? [#permalink] New post  04 Jun 2023, 09:37
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Re: (3.999999/2.001) - (3.999996/2.002) = ??? [#permalink]
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