Problem Solving

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

We want to find 1/201 + 1/202 + 1/203 + . . . + 1/299 + 1/300

NOTE: there are 100 fractions in this sum.

Let's examine the extreme values (1/201 and 1/300)

First consider a case where all of the values are equal to the smallest fraction (1/300)
We get: 1/300 + 1/300 + 1/300 + ... + 1/300 = 100/300 = 1/3
So, the original sum must be greater than 1/3

Now consider a case where all of the values are equal to the biggest fraction (1/201)
In fact, let's go a little bigger and use 1/200
We get: 1/200 + 1/200 + 1/200 + ... + 1/200 = 100/200 = 1/2
So, the original sum must be less than 1/2

Combine both cases to get 1/3 < M < 1/2 = A

Cheers,
Brent

Hi John fran kennedi,

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.

Brent's already pointed out the easiest way to figure out the minimum and maximum values of the sum of the reciprocals. You can actually stop working once you figure out the minimum though:

Since 1/300 < 1/201 and the sum of those 100 terms would be 1/3 AT THE MINIMUM, the only answer that's possible would be A. The extra work that Mitch did just confirms the maximum value of the sum, but it's 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.

GMAT assassins aren't born, they're made,
Rich

To add to what Rich said... make sure that you always scan the answer choices on PS questions before you dive in and start solving. They can often give you clues about how to approach the problem.

I've written an entire article about that here:
https://www.manhattanprep.com/gmat/blog ... -pen-down/