hi ,
I'll suggest you a slightly more accurate way to do this..
\(\frac{4}{2.001}=\frac{4}{2+0.001}\))..
multiply both numerator and denominator by 2-.001
\(\frac{4(2-0.001)}{(2+0.001)(2-0.001)}\)
=\(\frac{4(1.999)}{(4-0.000001)}\)....
0.000001 being very small can be neglected, so the equation becomes..
\(\frac{4(1.999)}{4}=1.999\) ans C
hope it helped
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Hope this helps

Which of the following is closest to 4/2.001?

(A) 1.997
(B) 1.998
(C) 1.999
(D) 2.000
(E) 2.001

We are asked 4/2.001 is closest to what , hence we cannot assume or ignore the 0.001 as extremely small .

We would need to consider it . Back to the question

4 is divided by a number 2 the answer would be 2 exactly . In this case 4 is divided by a number which is slightly bigger than 2 i.e 2.001

Hence by value = ( numerator / denominator ) if the denominator increases the value would decrease .

In this case since denominator has increased by 0.001 the value would decrease slightly to less than 2 and hence the answer would be 1.999 .

Hope my explanation helps !!

hi, you cannot assume that here since all vaues are very very close...
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\(\frac{4}{2.001}=\frac{4}{2+0.001}=\frac{4(2-0.001)}{(2+0.001)(2-0.001)}=\frac{4(2-0.001)}{4-0.001^2}\).

Now, since \(0.001^2\) is very small number then \(4-0.001^2\) is very close to 4 itself, so \(0.001^2\) is basically negligible in this case and we can write: \(\frac{4(2-0.001)}{4-0.001^2} \approx \frac{4(2-0.001)}{4}=2-0.001=1.999\).

Answer: C.
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Hi Bunuel,

Thanks for the solution. I understood the way to approach it and why the answer is C.

I have a small query - while solving as you mentioned that 4 - 0.002^2 is equal to 4 only because 0.002^2 is negligible here....However in the main question where 2 - 0.002 is written 0.002 is not negligible.
Is it not negligible only because it is not sufficiently smaller no.? If it were 0.002^2 in the main question instead of 0.002 then could that have been negligible?

Please share your valuable inputs.

hi fireinbelly,
if i may try to explain you..
look at the answer values they are varying to .001(for eg 2.000 or 1.999 or 1.998)... so yes .001 is important here....
but .001^2=.000001, which is way to small as compared to .001 .... so can be left..
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ok...thank you chetan2u!

Dear Bunuel, how the denominator (2+0.001) become equal to (2+0.001)(2-0.001)

Dear Bunuel, how the denominator (2+0.001) become equal to (2+0.001)(2-0.001)[/quote]

hi,
both denominator and numerator are multiplied by (2-.001) to basically simploiify the equation
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Responding to a pm:

I think you got your answer. Whether something is negligible depends on with what we are comparing it. When working with 1.999 and 2.001, then .001 is not negligible. But .000001 is. Similarly, if the options were far apart such as 2, 3 etc then we know that we can ignore .001.
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Thanks Karishma....it really helps!

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