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 350,000 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:

Re: average of four distinct positive integers [#permalink]
04 Sep 2009, 22:01

1

This post received KUDOS

Expert's post

sharkk wrote:

We have two equations: I2 + I3 = 100 and I1 + I4 = 140 [strike]50 + 50 = 100 and 40 + 100 = 140 => 1 number is less than 50 50 + 50 = 100 and 50 + 90 = 140 => 0 numbers less than 50 40 + 60 = 100 and 40 + 100 = 140 => 2 numbers are less than 50[/strike] Eliminated because we need distinct 4 numbers 40 + 60 = 100 and 50 + 90 = 140 => 1 number is less than 50 40 + 60 = 100 and 30 + 110 = 140 => 2 numbers are less than 50

Where am I thinking wrong?

The sum of the second and third biggest numbers (as the median is average of the second and third largest numbers average) must be 100, average=50 second large<50<third large. First one<second<50. In your post you have sum of smallest and third biggest sum 100 and smallest and biggest sum 140 which is wrong. _________________

Re: average of four distinct positive integers [#permalink]
04 Sep 2009, 19:55

Expert's post

If the average of four distinct positive integers is 60, how many integers of these four are smaller than 50?

Note: we have four distinct positive integers and x1+x2+x3+x4=240.

(1) One of the integers is 200 --> x1+x2+x3+200=240 --> x1+x2+x3=40, hence three integers are less than 50. Sufficient.

(2) The median of the four integers is 50 --> Median of this set would be the average of middle numbers: x2+x3=100 --> x2<50 (as integers are distinct). x1<x2, hence we have two integers less than 50. Sufficient.

Answer: D.

BUT there is a problem with this question: from (1) we got that there are 3 integers less than 50 and from (2) we got that there are 2 integers less than 50. In DS statements never contradict so either of the statement should be changed. _________________

Re: average of four distinct positive integers [#permalink]
04 Sep 2009, 21:42

We have two equations: I2 + I3 = 100 and I1 + I4 = 140 [strike]50 + 50 = 100 and 40 + 100 = 140 => 1 number is less than 50 50 + 50 = 100 and 50 + 90 = 140 => 0 numbers less than 50 40 + 60 = 100 and 40 + 100 = 140 => 2 numbers are less than 50[/strike] Eliminated because we need distinct 4 numbers 40 + 60 = 100 and 50 + 90 = 140 => 1 number is less than 50 40 + 60 = 100 and 30 + 110 = 140 => 2 numbers are less than 50

Where am I thinking wrong? _________________

--------------------------------------------------------------- Check-out the following: --------------------------------------------------------------- 1. Math Divisibility Test

Re: average of four distinct positive integers [#permalink]
04 Sep 2009, 22:12

Excellent point. Now I got it. Many thanks for your help. Now, I see there will always 2 numbers less than 50. Wow!!! _________________

--------------------------------------------------------------- Check-out the following: --------------------------------------------------------------- 1. Math Divisibility Test

Re: If the average of four distinct positive integers is 60, how [#permalink]
30 Mar 2013, 01:59

i believe that (A) should be the answer. the problem with secoand statement is that 50, 50, 50, 90 can be a case where median is sum of the middle two terms that is 50. so, none of the terms are less than 50 here.

Re: If the average of four distinct positive integers is 60, how [#permalink]
30 Mar 2013, 02:37

Expert's post

cyberjadugar wrote:

i believe that (A) should be the answer. the problem with secoand statement is that 50, 50, 50, 90 can be a case where median is sum of the middle two terms that is 50. so, none of the terms are less than 50 here.

regards,

We are given that: "the average of four distinct positive integers is..." _________________

How the growth of emerging markets will strain global finance : Emerging economies need access to capital (i.e., finance) in order to fund the projects necessary for...

Good news for globetrotting MBAs: travel can make you a better leader. A recent article I read espoused the benefits of traveling from a managerial perspective, stating that it...