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M24 Q 7 explanation [#permalink]
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12 Mar 2009, 14:49
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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 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 \(2  0.001 = 1.999\) . The correct answer is C. 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 2a



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Re: M24 Q 7 explanation [#permalink]
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12 Mar 2009, 19:10
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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 \(2  0.001 = 1.999\) . The correct answer is C.
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 2a 4/(2+a) = {4*(2a)}/ {(2+a)(2a)} = 4(2a)/{4a^2} ~ 4(2a)/4 = 2a since a is too small.. a^2 .. is negligible.. ~zero.
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Re: M24 Q 7 explanation [#permalink]
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12 Mar 2009, 19:46
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 \(2  0.001 = 1.999\) . The correct answer is C.
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 2a 4/(2+a) = {4*(2a)}/ {(2+a)(2a)} = 4(2a)/{4a^2} ~ 4(2a)/4 = 2a since a is too small.. a^2 .. is neglible.. ~zero. K Thanks. Essentially, the solution is equating 4 and (4a^2).



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Re: M24 Q 7 explanation [#permalink]
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04 Jan 2010, 07:42
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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]
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11 Jun 2011, 15:43
\((4/(2+0.001))*((20.001)/(20.001))\)
=\((4/(2^20.001^2)))*(1.999)\)
= 1.999 as first part's denominator is very close to 4 as 0.001^2 is very small.
Answer is C.



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Re: M24 Q 7 explanation [#permalink]
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12 Jan 2012, 06:12
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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. Answer is C.



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Re: M24 Q 7 explanation [#permalink]
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13 Jan 2012, 17:03
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 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 \(2  0.001 = 1.999\) . The correct answer is C. 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 2a 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]
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30 Aug 2012, 17:25
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 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 \(2  0.001 = 1.999\) . The correct answer is C. 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 2a 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]
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30 Aug 2012, 20:02
Another simple method guys:
4/(2.001) = 4/(2+0.001)
Multiply top and bottom by (20.001) The formula here is (ab)(a+b) = a^2b^2. We have all learnt this from OG.
^ denotes: raised to the power of
You get 4(20.001)/(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 (20.001)/(4) = 20.001 = 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]
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14 Jan 2013, 06:36
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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 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 \(2  0.001 = 1.999\) . The correct answer is C. 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 2a 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 \(\frac{4}{2.001}=\frac{4}{2+0.001}=\frac{4(20.001)}{(2+0.001)(20.001)}=\frac{4(20.001)}{40.001^2}\). Now, since \(0.001^2\) is very small number then \(40.001^2\) is very close to 4 itself, so \(0.001^2\) is basically negligible in this case and we can write: \(\frac{4(20.001)}{40.001^2}\approx{\frac{4(20.001)}{4}}=20.001=1.999\). Answer: C.
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Re: M24 Q 7 explanation [#permalink]
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14 Jan 2013, 07:08
2.001 = 2 ( 1 + (10^3)/2)
1/(x+a) = (x+a)^1 = xa
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]
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14 Jan 2013, 09:54
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 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 \(2  0.001 = 1.999\) . The correct answer is C. 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 2a 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]
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07 Jan 2014, 06:10
I got it wrong too icandy ur good
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Re: M24 Q 7 explanation [#permalink]
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07 Jan 2014, 07:12
4000/2001 approx 1.999
C it should be
Any alternative to long division?



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Re: M24 Q 7 explanation [#permalink]
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07 Jan 2014, 07:16



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Re: M24 Q 7 explanation [#permalink]
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20 Apr 2014, 03:22
1.999*2.001 has the structure of (2a)(2+a) = 4a2 (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]
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23 Feb 2015, 09:29
Very tough one.
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