BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Integers Properties

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Moderator
Posts: 772
Joined: Wed Aug 30, 2017 6:29 pm
Followed by:6 members

Integers Properties

by BTGmoderatorRO » Sun Dec 24, 2017 9:27 am
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

OA is A

What does inclusive stands for there and what is the best approach to solving this, an Expert pls.
Top
Source: — Problem Solving |
User avatar
Master | Next Rank: 500 Posts
Posts: 228
Joined: Thu Apr 20, 2017 1:02 am
Location: Global
Thanked: 32 times
Followed by:3 members
GMAT Score:770

by elias.latour.apex » Sun Dec 24, 2017 11:36 am
Inclusive means that it includes the first term (1/201) and the last term (1/300).

The simplest solution to this is to estimate the result. Since there are 100 terms and the middle term is 1/250, the answer is likely to be near 100/250, which reduces to 2/5. That is answer choice (A).
Elias Latour
Verbal Specialist @ ApexGMAT
blog.apexgmat.com
+1 (646) 736-7622
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Sun Dec 24, 2017 12:27 pm
Roland2rule 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
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
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Post new topic Post Reply
• Page 1 of 1