Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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]

Show Tags

07 Jan 2013, 03:15

4

This post received KUDOS

1

This post was BOOKMARKED

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]

Show Tags

07 Jan 2013, 03:17

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]

Show Tags

07 Jan 2013, 03:18

2

This post received KUDOS

1

This post was BOOKMARKED

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]

Show Tags

26 Apr 2013, 14:30

3

This post received KUDOS

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]

Show Tags

26 Apr 2013, 15:06

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]

Show Tags

26 Apr 2013, 17:13

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]

Show Tags

02 Jan 2014, 08:06

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]

Show Tags

28 Aug 2015, 01:27

2

This post received KUDOS

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.
_________________

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

Show Tags

29 Sep 2015, 14:24

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.
_________________

Consider giving me Kudos if I helped, but don´t take them away if I didn´t!