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# M24 Q 7 explanation

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M24 Q 7 explanation [#permalink]  12 Mar 2009, 13:49
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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

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

If $$a$$ is small, $$\frac{4}{2 + a}$$ is close to $$2 - a$$ . This is because $$(2 - a)(2 + a) = 4 - a^2$$ , which is close to 4 if $$a$$ is small. So, $$\frac{4}{2.001} = \frac{4}{2 + 0.001}$$ is approximately $$2 - 0.001 = 1.999$$ .

I am posting the OE because I did not understand the OE. I arrived at the solution though. I need to understand the above logic of

4/2+a is close to 2-a
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Re: M24 Q 7 explanation [#permalink]  12 Mar 2009, 18:10
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icandy wrote:
Which of the following is closest to $$\frac{4}{2.001}$$ ?

(C) 2008 GMAT Club - m24#7

* 1.997
* 1.998
* 1.999
* 2.000
* 2.001

If $$a$$ is small, $$\frac{4}{2 + a}$$ is close to $$2 - a$$ . This is because $$(2 - a)(2 + a) = 4 - a^2$$ , which is close to 4 if $$a$$ is small. So, $$\frac{4}{2.001} = \frac{4}{2 + 0.001}$$ is approximately $$2 - 0.001 = 1.999$$ .

I am posting the OE because I did not understand the OE. I arrived at the solution though. I need to understand the above logic of

4/2+a is close to 2-a

4/(2+a) = {4*(2-a)}/ {(2+a)(2-a)} = 4(2-a)/{4-a^2} ~ 4(2-a)/4 = 2-a

since a is too small.. a^2 .. is negligible.. ~zero.
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Re: M24 Q 7 explanation [#permalink]  12 Mar 2009, 18:46
x2suresh wrote:
icandy wrote:
Which of the following is closest to $$\frac{4}{2.001}$$ ?

(C) 2008 GMAT Club - m24#7

* 1.997
* 1.998
* 1.999
* 2.000
* 2.001

If $$a$$ is small, $$\frac{4}{2 + a}$$ is close to $$2 - a$$ . This is because $$(2 - a)(2 + a) = 4 - a^2$$ , which is close to 4 if $$a$$ is small. So, $$\frac{4}{2.001} = \frac{4}{2 + 0.001}$$ is approximately $$2 - 0.001 = 1.999$$ .

I am posting the OE because I did not understand the OE. I arrived at the solution though. I need to understand the above logic of

4/2+a is close to 2-a

4/(2+a) = {4*(2-a)}/ {(2+a)(2-a)} = 4(2-a)/{4-a^2} ~ 4(2-a)/4 = 2-a

since a is too small.. a^2 .. is neglible.. ~zero.

K Thanks. Essentially, the solution is equating 4 and (4-a^2).
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Re: M24 Q 7 explanation [#permalink]  04 Jan 2010, 06:42
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I guessed this and got it wrong

we can easily eliminate choice D and E.

but I feel the official explanation is good but not great because it may lead to traps as the difference in answer choices are too narrow and small.

I tried to solve it arithematically and got the answer in less than 2 mins but I was not in any pressure to solve this in less than 2 mins
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Re: M24 Q 7 explanation [#permalink]  11 Jun 2011, 14:43
$$(4/(2+0.001))*((2-0.001)/(2-0.001))$$

=$$(4/(2^2-0.001^2)))*(1.999)$$

= 1.999 as first part's denominator is very close to 4 as 0.001^2 is very small.

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Re: M24 Q 7 explanation [#permalink]  12 Jan 2012, 05:12
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Alternative method, D and E are not possible since denominator >2.
So, pick the value in the center, 1.998 and multiply by 2.001 = 3.997998. Since number is less than 4.
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Re: M24 Q 7 explanation [#permalink]  13 Jan 2012, 16:03
icandy wrote:
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

[Reveal] Spoiler: OA
C

Source:

If $$a$$ is small, $$\frac{4}{2 + a}$$ is close to $$2 - a$$ . This is because $$(2 - a)(2 + a) = 4 - a^2$$ , which is close to 4 if $$a$$ is small. So, $$\frac{4}{2.001} = \frac{4}{2 + 0.001}$$ is approximately $$2 - 0.001 = 1.999$$ .

I am posting the OE because I did not understand the OE. I arrived at the solution though. I need to understand the above logic of

4/2+a is close to 2-a

My approach was (I had to think a while - I did not solve this in 2 mins)

4/2.001 = (2.001+1.999)/2.001 which then = 1 + (1.999/2.001)

Now I definitely know that the 2nd part is less than 1 and is equal to 1999/2001.

Courtesy of Manhattan GMAT FDP => If the fraction is originally smaller than 1, the fraction increases in value as it approaches 1. ie., 10/11 < 11/12 < 1011/1012
Taking it in reverse 1011/1012 > 11/12 > 10/11

In our case 1999/2001 > 999/1001 (subtracting 1000)
Instead subtract 1001 to get 998/1000. Now 1999/2001 > 998/1000 (0.998)

The 2nd part now is less than 1 and greater then 0.998 which gives our answer C.
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Re: M24 Q 7 explanation [#permalink]  30 Aug 2012, 16:25
vkredi wrote:
icandy wrote:
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

[Reveal] Spoiler: OA
C

Source:

If $$a$$ is small, $$\frac{4}{2 + a}$$ is close to $$2 - a$$ . This is because $$(2 - a)(2 + a) = 4 - a^2$$ , which is close to 4 if $$a$$ is small. So, $$\frac{4}{2.001} = \frac{4}{2 + 0.001}$$ is approximately $$2 - 0.001 = 1.999$$ .

I am posting the OE because I did not understand the OE. I arrived at the solution though. I need to understand the above logic of

4/2+a is close to 2-a

My approach was (I had to think a while - I did not solve this in 2 mins)

4/2.001 = (2.001+1.999)/2.001 which then = 1 + (1.999/2.001)

Now I definitely know that the 2nd part is less than 1 and is equal to 1999/2001.

Courtesy of Manhattan GMAT FDP => If the fraction is originally smaller than 1, the fraction increases in value as it approaches 1. ie., 10/11 < 11/12 < 1011/1012
Taking it in reverse 1011/1012 > 11/12 > 10/11

In our case 1999/2001 > 999/1001 (subtracting 1000)
Instead subtract 1001 to get 998/1000. Now 1999/2001 > 998/1000 (0.998)

The 2nd part now is less than 1 and greater then 0.998 which gives our answer C.

I like your thinking and I think it's all correct, except how would you know whether 1+1999/2001 is closer to 2.000 or to 1.999?
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Re: M24 Q 7 explanation [#permalink]  30 Aug 2012, 19:02
Another simple method guys:

4/(2.001) = 4/(2+0.001)

Multiply top and bottom by (2-0.001)
The formula here is (a-b)(a+b) = a^2-b^2. We have all learnt this from OG.

^ denotes: raised to the power of

You get
4(2-0.001)/(2^2 - (10^-3)^2)

The denominator can be rounded off to 4 because 10^-6 is negligeble; ateast compared 10^-3 in the numerator.

So the equation shortens to :
4 (2-0.001)/(4) = 2-0.001 = 1.999.

This method is beautiful because it doesn't need any of the hardcore maths and uses only the techniques described in the OG.
I hope this answers the question whether the answer is closer to 1.999 or 2.000. It is obviously closer to 1.999 because the 10^-6 is negligeble compared to 10^-3.
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Re: M24 Q 7 explanation [#permalink]  14 Jan 2013, 05:36
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1
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icandy wrote:
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

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

If $$a$$ is small, $$\frac{4}{2 + a}$$ is close to $$2 - a$$ . This is because $$(2 - a)(2 + a) = 4 - a^2$$ , which is close to 4 if $$a$$ is small. So, $$\frac{4}{2.001} = \frac{4}{2 + 0.001}$$ is approximately $$2 - 0.001 = 1.999$$ .

I am posting the OE because I did not understand the OE. I arrived at the solution though. I need to understand the above logic of

4/2+a is close to 2-a

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

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

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: $$\frac{4(2-0.001)}{4-0.001^2}\approx{\frac{4(2-0.001)}{4}}=2-0.001=1.999$$.

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Re: M24 Q 7 explanation [#permalink]  14 Jan 2013, 06:08
2.001 = 2 ( 1 + (10^-3)/2)

1/(x+a) = (x+a)^-1 = x-a

4/2.001 = 4/ 2 ( 1 + (10^-3)/2) = 2 ( 1 - (10^-3)/2) = 2 - 0.001 = 1.999
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Re: M24 Q 7 explanation [#permalink]  14 Jan 2013, 08:54
icandy wrote:
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

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

If $$a$$ is small, $$\frac{4}{2 + a}$$ is close to $$2 - a$$ . This is because $$(2 - a)(2 + a) = 4 - a^2$$ , which is close to 4 if $$a$$ is small. So, $$\frac{4}{2.001} = \frac{4}{2 + 0.001}$$ is approximately $$2 - 0.001 = 1.999$$ .

I am posting the OE because I did not understand the OE. I arrived at the solution though. I need to understand the above logic of

4/2+a is close to 2-a

Just go by the gut. 4/2 would be 2 so D and E are gone. And since the denominator is slightly bigger, the result would be minutely less than 2. Thus, 1.999.
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Re: M24 Q 7 explanation [#permalink]  07 Jan 2014, 05:10
I got it wrong too

icandy ur good

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Re: M24 Q 7 explanation [#permalink]  07 Jan 2014, 06:12
4000/2001 approx 1.999

C it should be

Any alternative to long division?
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Re: M24 Q 7 explanation [#permalink]  07 Jan 2014, 06:16
Expert's post
idinuv wrote:
4000/2001 approx 1.999

C it should be

Any alternative to long division?

Check here: m24-q-7-explanation-76513.html#p1168628

Hope it helps.
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Re: M24 Q 7 explanation [#permalink]  20 Apr 2014, 02:22
1.999*2.001 has the structure of (2-a)(2+a) = 4-a2 (a=0.001)
In A and B, the products of 2.001 with 1.998 and 1.997 are smaller than that in C-> A and B are out

Eliminate E for the same reason when compared with D. D = 2*2.001 = 4+0.002. 0.002 > (0.001)^2 -> choose C
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Re: M24 Q 7 explanation [#permalink]  23 Feb 2015, 08:29
Very tough one.
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Re: M24 Q 7 explanation   [#permalink] 23 Feb 2015, 08:29
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# M24 Q 7 explanation

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