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

Re: If K is the sum of reciprocals of the consecutive integers [#permalink]
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 \(\frac{1}{8}\) _________________

Did you find this post helpful?... Please let me know through the Kudos button.

Re: If K is the sum of reciprocals of the consecutive integers [#permalink]
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]
07 Jan 2013, 02:18

1

This post received KUDOS

Expert's post

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.

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Re: If K is the sum of reciprocals of the consecutive integers [#permalink]
26 Apr 2013, 13:30

2

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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]
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]
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 \(\frac{3}{24}=\frac{1}{8}\)

Re: If K is the sum of reciprocals of the consecutive integers [#permalink]
27 Apr 2013, 04:14

Expert's post

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.

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]
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]
28 Jan 2015, 10:50

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