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If K is the sum of reciprocals of the consecutive integers from 43 to
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07 Jan 2013, 02:05
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If K is the sum of reciprocals of the consecutive integers from 43 to 48, inclusive, then K is closest in value to which of the following? A. 1/12 B. 1/10 C. 1/8 D. 1/6 E. 1/4 How do we decide between 1/6 and 1/8
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Re: If K is the sum of reciprocals of the consecutive integers from 43 to
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07 Jan 2013, 02:14
fozzzy wrote: If K is the sum of reciprocals of the consecutive integers from 43 to 48, inclusive, then K is closest in value to which of the following?
A. 1/12 B. 1/10 C. 1/8 D. 1/6 E. 1/4
How do we decide between 1/6 and 1/8 Given that \(K=\frac{1}{43}+\frac{1}{44}+\frac{1}{45}+\frac{1}{46}+\frac{1}{47}+\frac{1}{48}\). Notice that 1/43 is the larges term and 1/48 is the smallest term. If all 6 terms were equal to 1/43, then the sum would be 6/43=~1/7, but since actual sum is less than that, then we have that K<1/7. If all 6 terms were equal to 1/48, then the sum would be 6/48=1/8, but since actual sum is more than that, then we have that K>1/8. Therefore, 1/8<K<1/7. So, K must be closer to 1/8 than it is to 1/6. Answer: C. Similar question to practice from OG: misthesumofthereciprocalsoftheconsecutiveintegers143703.htmlHope it helps.
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Re: If K is the sum of reciprocals of the consecutive integers from 43 to
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07 Jan 2013, 02:15
fozzzy wrote: If K is the sum of reciprocals of the consecutive integers from 43 to 48, inclusive, then K is closest in value to which of the following?
A. 1/12 B. 1/10 C. 1/8 D. 1/6 E. 1/4
How do we decide between 1/6 and 1/8 I believe a good approximation would be to take the mean, reciprocal of that and multiply by 6 (No of numbers being added) = \(\frac{6}{45.5}\) which is closest to \(\frac{6}{48}\) (\frac{1}{6} would be \(\frac{6}{36}\) and \(\frac{1}{10}\)would be \(\frac{6}{60}\)) and hence \(\frac{1}{8}\)




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Re: If K is the sum of reciprocals of the consecutive integers from 43 to
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07 Jan 2013, 02:18
fozzzy wrote: If K is the sum of reciprocals of the consecutive integers from 43 to 48, inclusive, then K is closest in value to which of the following?
A. 1/12 B. 1/10 C. 1/8 D. 1/6 E. 1/4
How do we decide between 1/6 and 1/8 The numbers are \(1/43 + 1/44+ 1/45 + 1/46 + 1/47 + 1/48\). The easiest method is to find smart numbers. If you consider each of the numbers as \(1/42\), then there sum will be \(6/42\) or \(1/7\). Remember that since we chose a higher number than those given, hence the actual sum will be smaller than \(1/7\). Now consider each of the numbers \(1/48\). Then in such case, the sum will be \(6/48\) or \(1/8\). Remember that since we chose a smaller number than those given, hence the actual sum will be greater than \(1/8\). Therefore the sum lies between \(1/7\) and \(1/8\). Hence among teh answer choices, the sum is closest to \(1/8\). +1C
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Re: If K is the sum of reciprocals of the consecutive integers from 43 to
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08 Jan 2013, 01:29
fozzzy wrote: If K is the sum of reciprocals of the consecutive integers from 43 to 48, inclusive, then K is closest in value to which of the following?
A. 1/12 B. 1/10 C. 1/8 D. 1/6 E. 1/4
How do we decide between 1/6 and 1/8 Hi, Well all other approaches are correct. Here is one more. A little less calculation intensive. From 1/43 to 1/48, there are 6 #. => We can infer that sum of 6 # of 1/40s > Sum of (1/43+/1/44+......+1/48) > Sum of 6 # of 1/50s =>So, 6/40 > Sum of (1/43+/1/44+......+1/48) > 6/50 => 1/6.66 > Sum of (1/43+/1/44+......+1/48) > 1/8.33 => 1/6.66 > 1/ (6.66< Denominator < 8.33) > 1/8.33 Only option available is C. Answer is 1/8. Shalabh Jain
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Re: If K is the sum of reciprocals of the consecutive integers from 43 to
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26 Apr 2013, 13:30
What is the sum of \(\frac{1}{43}+ ... +\frac{1}{48}\)?
\(\frac{1}{43}(1+\frac{43}{44}+\frac{43}{45}+\frac{43}{46}+\frac{43}{47}+\frac{43}{48})\) we can rewrite as: \(\frac{1}{43}(1+1+1+1+1+1)=\frac{6}{43}\)
6/43 is something more than 7, so is colse to 8 \(\frac{6}{43}=(almost)\frac{1}{8}\) C



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Re: If K is the sum of reciprocals of the consecutive integers from 43 to
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26 Apr 2013, 14:06
Bunuel wrote: fozzzy wrote: If K is the sum of reciprocals of the consecutive integers from 43 to 48, inclusive, then K is closest in value to which of the following?
A. 1/12 B. 1/10 C. 1/8 D. 1/6 E. 1/4
How do we decide between 1/6 and 1/8 Given that \(K=\frac{1}{43}+\frac{1}{44}+\frac{1}{45}+\frac{1}{46}+\frac{1}{47}+\frac{1}{48}\). Notice that 1/43 is the larges term and 1/48 is the smallest term. If all 6 terms were equal to 1/43, then the sum would be 6/43=~1/7, but since actual sum is less than that, then we have that K<1/7. If all 6 terms were equal to 1/48, then the sum would be 6/48=1/8, but since actual sum is more than that, then we have that K>1/8. Therefore, 1/8<K<1/7. So, K must be closer to 1/8 than it is to 1/6. Answer: C. Similar question to practice from OG: misthesumofthereciprocalsoftheconsecutiveintegers143703.htmlHope it helps. Bunuel, I understand your method. However, how can we know that the distance between k and 1/8 is shorter than the distance between k and 1/6. For example, if k were almost 1/7, we would have to calculate the distance between 1/8 and 1/7 and also the distance between 1/7 and 1/6. I make this comment because the GMAT Prep explains that point, but it does that in a complex way. Thanks!



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Re: If K is the sum of reciprocals of the consecutive integers from 43 to
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27 Apr 2013, 04:14
danzig wrote: Bunuel wrote: fozzzy wrote: If K is the sum of reciprocals of the consecutive integers from 43 to 48, inclusive, then K is closest in value to which of the following?
A. 1/12 B. 1/10 C. 1/8 D. 1/6 E. 1/4
How do we decide between 1/6 and 1/8 Given that \(K=\frac{1}{43}+\frac{1}{44}+\frac{1}{45}+\frac{1}{46}+\frac{1}{47}+\frac{1}{48}\). Notice that 1/43 is the larges term and 1/48 is the smallest term. If all 6 terms were equal to 1/43, then the sum would be 6/43=~1/7, but since actual sum is less than that, then we have that K<1/7. If all 6 terms were equal to 1/48, then the sum would be 6/48=1/8, but since actual sum is more than that, then we have that K>1/8. Therefore, 1/8<K<1/7. So, K must be closer to 1/8 than it is to 1/6. Answer: C. Similar question to practice from OG: misthesumofthereciprocalsoftheconsecutiveintegers143703.htmlHope it helps. Bunuel, I understand your method. However, how can we know that the distance between k and 1/8 is shorter than the distance between k and 1/6. For example, if k were almost 1/7, we would have to calculate the distance between 1/8 and 1/7 and also the distance between 1/7 and 1/6. I make this comment because the GMAT Prep explains that point, but it does that in a complex way. Thanks! Even if K=1/7, still the distance between 1/8 and 1/7 is less than the distance between 1/7 and 1/6.
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Re: If K is the sum of reciprocals of the consecutive integers from 43 to
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Updated on: 03 Aug 2014, 22:01
I did in this way \(\frac{1}{43} + \frac{1}{44} + \frac{1}{45} + \frac{1}{46} + \frac{1}{47} + \frac{1}{48}\)
\(= (\frac{1}{43} + \frac{1}{48}) + (\frac{1}{44} + \frac{1}{47}) + (\frac{1}{45} + \frac{1}{46})\) .... Grouping the denominator's whose addition is same (91)
\(= \frac{1}{24} + \frac{1}{24} + \frac{1}{24}\) (Approx)
\(= \frac{3}{24}\) (Approx)
\(= \frac{1}{8}\) Answer = C
Originally posted by PareshGmat on 29 Apr 2013, 00:55.
Last edited by PareshGmat on 03 Aug 2014, 22:01, edited 1 time in total.



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Re: If K is the sum of reciprocals of the consecutive integers from 43 to
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28 Aug 2015, 00:27
One way would be to find the middle terms. Since total terms is 5. Middle term will be 3rd term. i.e. 1/45. Which should be the approximate (but less) than original mean. 1/45 * 5 = 1/9. So you know that the sum will be very lose to 1/9 but just a little more. 1/8 is the closest and also the correct answer.
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Re: If K is the sum of reciprocals of the consecutive integers from 43 to
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29 Sep 2015, 13:24
swanidhi wrote: One way would be to find the middle terms. Since total terms is 5. Middle term will be 3rd term. i.e. 1/45. Which should be the approximate (but less) than original mean. 1/45 * 5 = 1/9. So you know that the sum will be very lose to 1/9 but just a little more. 1/8 is the closest and also the correct answer. There are actually 6 terms. Anyway, your approach may work in this case.



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Re: If K is the sum of reciprocals of the consecutive integers from 43 to
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25 Nov 2017, 15:18
fozzzy wrote: If K is the sum of reciprocals of the consecutive integers from 43 to 48, inclusive, then K is closest in value to which of the following?
A. 1/12 B. 1/10 C. 1/8 D. 1/6 E. 1/4
We want the approximate sum of 1/43 + 1/44 + 1/45 + . . . + 1/48 Let's make the following observations about the upper and lower values: Upper values: If all 6 fractions were 1/43, the sum would be (6)(1/43) = 6/43 ~ 6/42 = 1/7 Lower values: If all 6 fractions were 1/48, the sum would be (6)(1/48) = 6/48 = 1/8 From this we can conclude that 1/8 < K < 1/7 If K is between 1/8 and 1/7, then K must be closer to 1/8 than it is to 1/6 Answer = C Cheers, Brent
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Re: If K is the sum of reciprocals of the consecutive integers from 43 to
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12 Jul 2019, 03:29
all 1/42...would have led to 6/42=1/7
all 1/48 would have led to 6/48=1/8
Clearly it will be nearer to 1/8. than anything else. Hence C



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Re: If K is the sum of reciprocals of the consecutive integers from 43 to
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26 Sep 2019, 00:25
total numbers = 4843+1=6 Average (48+43)/2=45.5 Sum of recriproal = 6*10/455 which is approximately 6/45 which is 1/7.5 –> closest is 1/8
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Re: If K is the sum of reciprocals of the consecutive integers from 43 to
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17 Oct 2019, 19:31
fozzzy wrote: If K is the sum of reciprocals of the consecutive integers from 43 to 48, inclusive, then K is closest in value to which of the following? A. 1/12 B. 1/10 C. 1/8 D. 1/6 E. 1/4 How do we decide between 1/6 and 1/8 We see that K = 1/43 + 1/44 + 1/45 + 1/46 + 1/47 + 1/48. We see that K is a sum of 6 numbers. Since each number is less than 1/42 and greater than or equal to 1/48, a lower estimate for K is 6 x 1/48 = 6/48 = ⅛, and an upper estimate for K is 6 x 1/42 = 6/42 = 1/7. In other words, K is between 1/7 and 1/8. Therefore, K is closest to 1/8. Answer: C
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Re: If K is the sum of reciprocals of the consecutive integers from 43 to
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