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.

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:

Learn the winning strategy for a high GRE score — what do people who reach a high score do differently? We're going to share insights, tips and strategies from data we've collected from over 50,000 students who used examPAL.

Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.

Re: M is the sum of the reciprocals of the consecutive integers from 201
[#permalink]

Show Tags

04 Sep 2016, 19:55

Hi Manonamission,

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 from 201
[#permalink]

Show Tags

26 Nov 2018, 05:32

mike34170 wrote:

I'd handle this differently:

Numbers range from 201 to 300, so the average is approximatively 250.

Then, the sum of all reciprocals equals to 100 times 1/250, i.e. 100/250 ---> 1/2,5

Answer A

Although you do get the answer, it is just a coincidence as the correct way to solve is given above by bunel. Let me explain why this doesn't work using a simpler example!

Let's say we have the following sum:

1/1 + 1/2 + 1/3 + 1/4 + 1/5.

If we solve this with a calculator, we get 2.283

If we use the approach you suggested for this one, we would have the middle value (1/3) multiplied by 5.

(1/3) * 5 = 5/3 = 1.66

This is not the same as 2.283. So the method doesn't really work when we have fractions.

If we have a sum that doesn't involve fractions, the method does work. For example: