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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

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.

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

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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 [#permalink]

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07 Jan 2013, 02:17

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.

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

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07 Jan 2013, 02:18

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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

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
_________________

Thanks this made it clear I was confused between those 2 options.

Thank you. By the way do not forget to attend following event this weekend to learn how to improve by up to 70 points in 25 days. Pl. click to know more.

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

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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 \(\frac{6}{43}=(almost)\frac{1}{8}\) C _________________

It is beyond a doubt that all our knowledge that begins with experience.

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

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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.

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!

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

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26 Apr 2013, 16:13

My method:

\(\frac{1}{43}\) through \(\frac{1}{48}\) are all very close to \(\frac{1}{50}\) (we are dealing with very small fractions at this point, so the differences are nearly none)

So I added all six together as \(\frac{1}{50}\) each, giving a total of \(\frac{6}{50}\). This reduces to \(\frac{3}{25}\), which is near \(\frac{3}{24}=\frac{1}{8}\)

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.

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.
_________________

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

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02 Jan 2014, 07:06

How about percentage? If we see 1/43 to 1/48 each is greater than 2%. So sum will be slightly greater than 2*6= 12% Now only option C is slightly more than 12%. So answer is C

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

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28 Jan 2015, 10:50

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

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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
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