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Which of the following is closest to \(\frac{4}{2.001}\)?

A. 1.997
B. 1.998
C. 1.999
D. 2.000
E. 2.001


The answer options are extremely close to each other, necessitating a very precise calculation. We can't simply round off 2.001 to 2 and assert that \(\frac{4}{2.001}\) is closest to 2 from the options given, due to this small difference.

Let's express 2.001 as 2 + 0.001 and multiply both the numerator and the denominator by the conjugate, 2 - 0.001:

    \(\frac{4}{2.001}=\)

    \(=\frac{4}{2+0.001}=\)

    \(=\frac{4(2-0.001)}{(2+0.001)(2-0.001)}=\)

    \(=\frac{4(2-0.001)}{4-0.001^2}\).

Now, \(0.001^2=0.000001\) is a very small number. Compared to 4, this value is practically negligible (do note that while we couldn't ignore 0.001 earlier, we can overlook 0.000001 now because it is 1000 times smaller than 0.001). Hence, we can approximate the equation as:

    \(\frac{4(2-0.001)}{4-0.001^2} \approx \)

    \(\approx \frac{4(2-0.001)}{4}=\)

    \(= 2-0.001 =\)

    \(=1.999\).

Therefore, \(\frac{4}{2.001}\) is closest to 1.999 among the provided options.

Answer: C
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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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Bunuel
fireinbelly
Which of the following is closest to \(\frac{4}{2.001}\)?

A. 1.997
B. 1.998
C. 1.999
D. 2.000
E. 2.001

I have seen some questions where if the value is .001 or very small value then we usually ignore that value...however the answer seems to be C? why cant we say that 4/2.001 is very close to 4/2 as .001 is negligible here hence the answer would be D in that case. Can someone please help me understanding this concept? Thanks.


M24-07


\(\frac{4}{2.001}=\frac{4}{2+0.001}=\frac{4(2-0.001)}{(2+0.001)(2-0.001)}=\frac{4(2-0.001)}{4-0.001^2}\).

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.

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.
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fireinbelly
Bunuel
fireinbelly
Which of the following is closest to \(\frac{4}{2.001}\)?

A. 1.997
B. 1.998
C. 1.999
D. 2.000
E. 2.001

I have seen some questions where if the value is .001 or very small value then we usually ignore that value...however the answer seems to be C? why cant we say that 4/2.001 is very close to 4/2 as .001 is negligible here hence the answer would be D in that case. Can someone please help me understanding this concept? Thanks.


M24-07


\(\frac{4}{2.001}=\frac{4}{2+0.001}=\frac{4(2-0.001)}{(2+0.001)(2-0.001)}=\frac{4(2-0.001)}{4-0.001^2}\).

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.

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!
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Bunuel
fireinbelly
Which of the following is closest to \(\frac{4}{2.001}\)?

A. 1.997
B. 1.998
C. 1.999
D. 2.000
E. 2.001

I have seen some questions where if the value is .001 or very small value then we usually ignore that value...however the answer seems to be C? why cant we say that 4/2.001 is very close to 4/2 as .001 is negligible here hence the answer would be D in that case. Can someone please help me understanding this concept? Thanks.


M24-07


\(\frac{4}{2.001}=\frac{4}{2+0.001}=\frac{4(2-0.001)}{(2+0.001)(2-0.001)}=\frac{4(2-0.001)}{4-0.001^2}\).

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

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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VeritasPrepKarishma
fireinbelly
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.

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.

Thanks Karishma....it really helps!
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Re: Which of the following is closest to 4/2.001?

I used Numbers concept: Not sure if its the right approach though.

Multiply both Numerator and denominator by 1000

4000/2001= Since the denominator is not of the form 2^m * 5^n The decimal will not terminate.

Looking at the options Only C is Non terminating
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