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If K is the sum of reciprocals of the consecutive integers from 43 to

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If K is the sum of reciprocals of the consecutive integers from 43 to  [#permalink]

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New post 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  [#permalink]

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New post 07 Jan 2013, 02:14
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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: m-is-the-sum-of-the-reciprocals-of-the-consecutive-integers-143703.html

Hope it helps.
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Re: If K is the sum of reciprocals of the consecutive integers from 43 to  [#permalink]

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New post 07 Jan 2013, 02:15
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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  [#permalink]

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New post 07 Jan 2013, 02:18
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1
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 \(1/8\).
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Re: If K is the sum of reciprocals of the consecutive integers from 43 to  [#permalink]

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New post 08 Jan 2013, 01:29
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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.

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Re: If K is the sum of reciprocals of the consecutive integers from 43 to  [#permalink]

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New post 26 Apr 2013, 13:30
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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
\(\frac{6}{43}=(almost)\frac{1}{8}\)
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Re: If K is the sum of reciprocals of the consecutive integers from 43 to  [#permalink]

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New post 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: m-is-the-sum-of-the-reciprocals-of-the-consecutive-integers-143703.html

Hope 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.
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Re: If K is the sum of reciprocals of the consecutive integers from 43 to  [#permalink]

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New post 27 Apr 2013, 04:14
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1
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: m-is-the-sum-of-the-reciprocals-of-the-consecutive-integers-143703.html

Hope 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  [#permalink]

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New post Updated on: 03 Aug 2014, 22:01
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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
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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  [#permalink]

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New post 28 Aug 2015, 00:27
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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  [#permalink]

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New post 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  [#permalink]

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New post 25 Nov 2017, 15:18
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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

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Re: If K is the sum of reciprocals of the consecutive integers from 43 to  [#permalink]

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