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:

Of the 5 numbers, the largest number is 4 greater than the median. Is the mean greater than the median? (1) The largest number plus the median is 34. (2) The median minus the smallest number is 10

The method i used was too tedious and turned out to be wrong at the end Can someone help me with a easy method / explanation please ?

Hi, and welcome to Gmat Club. Very good question, so +1 for it. Below is a solution:

Le these 5 numbers in ascending order be \(a\), \(b\), \(c\), \(d\), \(e\): \(a\leq{b}\leq{c}\leq{d}\leq{e}\).

Median would be the middle number - \(c\) and the \(mean=\frac{a+b+c+d+e}{5}\).

Given: \(e=c+4\). Question: is \(\frac{a+b+c+d+e}{5}>c\) --> is \(\frac{a+b+c+d+(c+4)}{5}>c\) --> is \(a+b+d+4>3c\)

(1) The largest number plus the median is 34 --> \(e+c=34\) --> \(c=15\) and \(e=19\) --> question becomes: is \(a+b+d+4>45\). Now, if \(a=b=15\) (max values possible for \(a\) and \(b\)) and \(d=19\) (max value possible for \(d\)) then answer would be YES but as min values of \(a\) and \(b\) are not limited at all then the answer could be NO as well. Not sufficient.

(2) The median minus the smallest number is 10 --> \(a=c-10\) --> question becomes: is \(c-10+b+d+4>3c\) --> \(b+d>2c+6\). Now, max value of \(b\) is \(c\) and max value of \(d\) is \(e=c+4\), so max value of LHS (left hand side) would be \(LHS=b+d=c+c+4=2c+4\), which is always less than right hand side: \(RHS=2c+6\). Hence the answer to the question is NO. Sufficient.

Re: Of the 5 numbers, the largest number is 4 greater than the median. Is [#permalink]

Show Tags

21 Feb 2014, 05:46

1

This post received KUDOS

Nice question +1

Just to add a conceptual approach to this one

Statement 1 is insufficient since we can only find the numerical values of the median and the largest number (M=15, L=19) But we still don’t know about the other three numbers. For instance, smallest numbers could be 11 and other two numbers equally spaced giving an evenly spaced set with Median= Mean. We have extra flexibility to place the other two numbers since we are told that the numbers are not necessarily integers.

Statement 2 is a bit more tricky. It says that the smallest number is 10 less than the median. So let's say that the median is 10 and the largest and smallest numbers are 14 and 0 respectively. Now, we will have 2 numbers to the left of 10 and two to the right. Even if we have 10 and 10 to the left and 10 and 10 to the right, the mean will ALWAYS be larger than the median. Therefore B is the correct answer

Re: Of the 5 numbers, the largest number is 4 greater than the median. Is [#permalink]

Show Tags

18 May 2014, 10:35

Because St. 1 tells us : Median is 15. So the five numbers can be 19, (one number between 19-15), 15, (2 numbers less than 15)

St. 2 gives us a range between the largest and the smallest number. Now visualize the numbers on the numberline, the largest number is 4 greater than the median, and median in 10 greater than the smallest number. Now the Mean will be in the exact middle of this range. Therefore, it has to be less than the Median. IMO B

Re: Of the 5 numbers, the largest number is 4 greater than the median. Is [#permalink]

Show Tags

11 Jul 2017, 11:52

jlgdr wrote:

Nice question +1

Just to add a conceptual approach to this one

Statement 1 is insufficient since we can only find the numerical values of the median and the largest number (M=15, L=19) But we still don’t know about the other three numbers. For instance, smallest numbers could be 11 and other two numbers equally spaced giving an evenly spaced set with Median= Mean. We have extra flexibility to place the other two numbers since we are told that the numbers are not necessarily integers.

Statement 2 is a bit more tricky. It says that the smallest number is 10 less than the median. So let's say that the median is 10 and the largest and smallest numbers are 14 and 0 respectively. Now, we will have 2 numbers to the left of 10 and two to the right. Even if we have 10 and 10 to the left and 10 and 10 to the right, the mean will ALWAYS be larger than the median. Therefore B is the correct answer

Hope this helps Cheers J

Your conclusion for statement 2 should be the opposite. In fact, intuitively, the average is being "pulled" down more by the first number (median-10) than it is being pulled up by the last (median+4). This conclusively tells us that the mean will always be smaller than the median. Let's try extreme examples: