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:

50% (02:23) correct
50% (01:54) wrong based on 34 sessions

HideShow timer Statistics

Five cards are to be selected from 10 cards numbered 1 to 10. In how many ways can the cards be selected so that the average of the five numbers selected is greater than their median?

I could figure out that the number of selections has to be even and less than half the total possibilities = 10c5/2 = 252/2 = 126 so B is the only possibility.
_________________

Five cards are to be selected from 10 cards numbered 1 to 10. In how many ways can the cards be selected so that the average of the five numbers selected is greater than their median?

(A) 111 (B) 116 (C) 222 (D) 232 (E) 252

Number of ways to select 5 cards out of 10 cards = 10C5 = 10!/5!5! = 252

These 252 ways will have 3 possibilities: Average = Median Average > Median Average < Median

These are the combinations (30 in all) where Average = Median: {1,2,3,4,5} {1,2,4,5,8} {1,2,4,6,7} {1,3,4,5,7} {2,3,4,5,6} {1,2,5,7,10} {1,2,5,8,9} {1,3,5,6,10} {1,3,5,7,9} {1,4,5,6,9} {1,4,5,7,8} {2,3,5,6,9} {2,3,5,7,8} {2,4,5,6,8} {3,4,5,6,7} {1,4,6,9,10} {1,5,6,8,10} {2,3,6,9,10} {2,4,6,8,10} {2,5,6,7,10} {2,5,6,8,9} {3,4,6,7,10} {3,4,6,8,9} {3,5,6,7,9} {4,5,6,7,8} {3,6,7,9,10} {4,5,7,9,10} {4,6,7,8,10} {5,6,7,8,9} {6,7,8,9,10}

This means there are 252 - 30 = 222 combinations where Average ≠ Median

Now (the difficult part to explain!) - for each of the combinations where Average<Median, there will be one where Average>Median.

Consider the format {1,2,3,x,x} => there are 7!/2!5! i.e. 21 ways of selecting the rest two (i.e. 21 combinations for five cards in the format 1,2,3,x,x). Out of 21 such combinations, except the one listed above where average=median, the rest 20 will have Average>Median.

Now, conversely, consider the format {x,x,8,9,10} => except the one listed above where average=median, the rest 20 combinations will have Average<Median.

What I am trying to say (in a long-winded way ) is that the answer should be 222/2 = 111.

pls confirm the OA. Does anyone have a shorter method of solving this question? Took me a lot of time to even think of an approach on this one...

definitely a tricky question but if mean does not equal median, data is either skewed left or right and you can assume this is 50/50 based on randomness of selection

however, determining the 30 mean=median seems like it will take a lot of time not sure if theres a shortcut for this?

thanks
_________________

Kudos are greatly appreciated and I'll always return the favor on one of your posts.

I know for sure these numbers cannot be consecutive because it they were, mean = media.

I know the number of ways to select 5 cards out of 10 cards = 10C5 = 10!/5!5! = 252

Now figure how many we can get teh average of these numbers greater than the median is the hard part. I guess B because I do some magic that give me a number close to B. So, B is my final answer.

Hi Bunuel... do we have a shorter way for such problems?
_________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Hi Bunuel... do we have a shorter way for such problems?

Frankly speaking I don't know much shorter way to determine the # of sets when mean=median. But don't worry, you won't see such problems on GMAT as they are really very time consuming.
_________________

Five cards are to be selected at random from 10 cards [#permalink]

Show Tags

25 Sep 2012, 23:33

Five cards are to be selected at random from 10 cards numbered 1 to 10. How many ways are possible that the average of five numbers selected will be greater than the median?

Five cards are to be selected at random from 10 cards numbered 1 to 10. How many ways are possible that the average of five numbers selected will be greater than the median?

a.99 b.54 c.111 d.102 e.79

Merging similar topics. Please refer to the solutions above. Also note that this question is out of the scope of GMAT.
_________________

Five cards are to be selected at random from 10 cards numbered 1 to 10. How many ways are possible that the average of five numbers selected will be greater than the median?

No. of ways of selecting 5 cards out of 10 = 10C5 = 252 ways

Now in these 252 ways there are 3 categories

1. mean = median 2. mean < median 3. mean > median

Now no. of case 2 = no . case 3

With case 1 i.e. mean = median

the sum of deviation of the nos. less than median should be equal to those greater than median

for example in case of median 3, there is 1 probability 12345 (deviations -2, -1, +1, +2) median 4, 4 probabilities 12467, 12458, 13457, 23456 (deviations [-3, -2, +2, +3], [-3, -2, +1, +4], [-3, -1, +1, +3], [-2, -1, +1, +2] similarly for median 5, there are 10 possibilities median 6, 10 ways (ssame as median 5) median 7, 4 ways (same as median 4) median 8, 1 way (same as median 3)

Five cards are to be selected at random from 10 cards numbered 1 to 10. How many ways are possible that the average of five numbers selected will be greater than the median?

--- Ans will be provided later.

By my logic, there is no set in which mena is lesser than the median. Either they have to be equal or the mean has to be greater. There are 10C5 ways in total. We have to subtract the number of combinations for which mean and median are equal. I'm jus stuck here though. No idea on how to get that. If answer choices were given, I'd just choose the option which is just a bit lesser than 10C5 since mode and median being equal is quite a constrained requirement and I dont think there would be that many of those.

Kudos Please if my post helped.
_________________

Did you find this post helpful?... Please let me know through the Kudos button.

Five cards are to be selected at random from 10 cards numbered 1 to 10. How many ways are possible that the average of five numbers selected will be greater than the median?

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...

The words of John O’Donohue ring in my head every time I reflect on the transformative, euphoric, life-changing, demanding, emotional, and great year that 2016 was! The fourth to...