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:

M is the sum of the reciprocals of the consecutive integers from 201 to 300, inclusive. Which of the following is true?

(A) 1/3 < M < 1/2 (B) 1/5 < M < 1/3 (C) 1/7 < M < 1/5 (D) 1/9 < M < 1/7 (E) 1/12 < M < 1/9

Solution:

Let's first analyze the question. We are trying to find a potential range for M, and M is equal to the sum of the reciprocals from 201 to 300, inclusive. Thus, M is:

1/201 + 1/202 + 1/203 + …+ 1/300

There is no way the GMAT would ever expect us to do this math, and that is exactly why the answer choices in are in the form of an inequality. Thus, we do not need to know the EXACT value of M. The easiest way to determine the RANGE of M is to use easy numbers that can quickly be manipulated.

Note that 1/200 is greater than each of the addends and that 1/300 is less than or equal each of the addends. Therefore, instead of trying to add together 1/201 + 1/202 + 1/203 + …+ 1/300, we are instead going to add 1/200 one hundred times and 1/300 one hundred times. These two sums will give us a high estimate of M and a low estimate of M. Again, we are adding 1/200, one hundred times, and 1/300, one hundred times, because there are 100 numbers from 1/201 to 1/300.

Instead of actually adding each one of these values one hundred times, we will simply multiply each value by 100. We have:

1/300 x 100 = 1/3

1/200 x 100 = 1/2

We see that M is between 1/3 and 1/2.

Answer A.
_________________

Scott Woodbury-Stewart Founder and CEO

GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions

Re: M is the sum of the reciprocals ... [#permalink]

Show Tags

04 Sep 2016, 09:16

Manonamission wrote:

M is the sum of the reciprocals of the consecutive integers from 201 to 300, inclusive. Which of the following is true? (A) 1/3<M<1/2 (B) 1/5<M<1/3 (0 1/7<M<1/5 (D) 1/9<M<1/7 (E) 1/12<M<1/9

Minimum value of M > 1/300 +1/300 + .. 100 times = 100/300 =1/3

Max Value of M < 1/200 +1/200 + .. 100 times = 100/200 = 1/2.

Re: M is the sum of the reciprocals ... [#permalink]

Show Tags

04 Sep 2016, 10:30

M is the sum of the reciprocals of the consecutive integers from 201 to 300, inclusive. Which of the following is true? (A) 1/3<M<1/2 (B) 1/5<M<1/3 (0 1/7<M<1/5 (D) 1/9<M<1/7 (E) 1/12<M<1/9

1/200+1/300=5/600=1/120 (1/120)/2=1/240 100*1/240=5/12=M 1/3<5/12/<1/2 A

In this prompt, the answer choices are "ranges"; this usually means that there's a way to avoid doing lots of math and instead use patterns and logic to save you time. You can actually stop working once you figure out the minimum value...

Since 1/300 < 1/201 and the sum of those 100 terms would be (100)(1/300) = 1/3 AT THE MINIMUM, the only answer that's possible would be A. The extra work that you could do to find the maximum value just confirms what the sum would be at that 'level', but but that work is unnecessary.

As you continue to study, be mindful of how the answer choices are written - they can sometimes provide a huge hint into the fastest way to answer the question.

M is the sum of the reciprocals of the consecutive integers [#permalink]

Show Tags

24 Oct 2016, 11:11

I wanted to take this approach but I'm not sure how correct the formula is...

\(Sn = (100/2) * [(2)(1/300) + (100-1)(1/n)]\)

The sum isn't coming out to a value between 1/3 and 1/2, but I normally use this formula when it comes to summation for sequences. I think my issue is figuring out what the relationship (or increment) should be for the n, n + 1, etc.