It is currently 26 Jun 2017, 16:14

GMAT Club Daily Prep

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

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Five cards are to be selected from 10 cards numbered 1 to 10

Author Message
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 39702
Five cards are to be selected from 10 cards numbered 1 to 10 [#permalink]

Show Tags

19 Nov 2009, 18:08
Expert's post
3
This post was
BOOKMARKED
00:00

Difficulty:

(N/A)

Question Stats:

50% (02:23) correct 50% (01:54) wrong based on 37 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?

(A) 111
(B) 116
(C) 222
(D) 232
(E) 252
_________________
Manager
Joined: 25 Aug 2009
Posts: 168
Location: Streamwood IL
Schools: Kellogg(Evening),Booth (Evening)
WE 1: 5 Years
Re: Really hard 800+ problem. [#permalink]

Show Tags

19 Nov 2009, 21:44

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.
_________________

Rock On

Intern
Joined: 14 Aug 2009
Posts: 8
Re: Really hard 800+ problem. [#permalink]

Show Tags

19 Nov 2009, 22:18
2
KUDOS
Bunuel wrote:
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...
Intern
Affiliations: University of Florida Alumni
Joined: 25 Oct 2009
Posts: 36
Schools: Wharton, Booth, Stanford, HBS
Re: Really hard 800+ problem. [#permalink]

Show Tags

19 Nov 2009, 22:40
ill agree with A 111

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.

Thanks!

Intern
Joined: 12 Nov 2009
Posts: 13
Re: Really hard 800+ problem. [#permalink]

Show Tags

20 Nov 2009, 07:29
This is hard!

B?

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.
Intern
Joined: 10 Feb 2010
Posts: 34
Re: Really hard 800+ problem. [#permalink]

Show Tags

17 Feb 2010, 11:38
very interested if anyone found a way to determine the number of sets in which mean=median.

The steps to solve this problem appear to be::

1 - calculate maximum # of combinations
2 - subtract the number of combinations where mean=median
3 - divide the remaining number by 2
Senior Manager
Joined: 01 Feb 2010
Posts: 264
Re: Really hard 800+ problem. [#permalink]

Show Tags

17 Feb 2010, 23:21
masland wrote:
very interested if anyone found a way to determine the number of sets in which mean=median.

The steps to solve this problem appear to be::

1 - calculate maximum # of combinations
2 - subtract the number of combinations where mean=median
3 - divide the remaining number by 2

mean = median in case when Difference between consecutive numbers is same i.e
1,3,5,7,9 or 1,2,3,4,5 cases.

What is the OA for the question
Math Expert
Joined: 02 Sep 2009
Posts: 39702
Re: Really hard 800+ problem. [#permalink]

Show Tags

18 Feb 2010, 07:34
OA: A (111). Gnet's explanation is correct. +1.
_________________
Senior Manager
Joined: 22 Dec 2009
Posts: 359
Re: Really hard 800+ problem. [#permalink]

Show Tags

20 Feb 2010, 11:37
Bunuel wrote:
OA: A (111). Gnet's explanation is correct. +1.

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

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

|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

~~Better Burn Out... Than Fade Away~~

Math Expert
Joined: 02 Sep 2009
Posts: 39702
Re: Really hard 800+ problem. [#permalink]

Show Tags

20 Feb 2010, 12:31
jeeteshsingh wrote:
Bunuel wrote:
OA: A (111). Gnet's explanation is correct. +1.

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.
_________________
Senior Manager
Joined: 19 Apr 2011
Posts: 277
Schools: Booth,NUS,St.Gallon
Five cards are to be selected at random from 10 cards [#permalink]

Show Tags

26 Sep 2012, 00:33
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?

a.99
b.54
c.111
d.102
e.79
Math Expert
Joined: 02 Sep 2009
Posts: 39702
Re: Five cards are to be selected at random from 10 cards [#permalink]

Show Tags

26 Sep 2012, 01:06
saikarthikreddy wrote:
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.
_________________
Manager
Joined: 19 Nov 2010
Posts: 90

Show Tags

13 Nov 2012, 21:50
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.
Intern
Joined: 10 Nov 2012
Posts: 21
GMAT Date: 01-16-2013
GPA: 3.37
WE: Management Consulting (Consulting)

Show Tags

13 Nov 2012, 22:48
Ans. 111

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)

Hence total 30 ways

Thus the answer = (10C5-30)/2 = (252-30)/2 = 222/2 = 111

Last edited by suryanshg on 14 Nov 2012, 01:00, edited 1 time in total.
Moderator
Joined: 02 Jul 2012
Posts: 1223
Location: India
Concentration: Strategy
GMAT 1: 740 Q49 V42
GPA: 3.8
WE: Engineering (Energy and Utilities)

Show Tags

13 Nov 2012, 23:35
bellcurve wrote:
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.

Thanks To The Almighty - My GMAT Debrief

GMAT Reading Comprehension: 7 Most Common Passage Types

Math Expert
Joined: 02 Sep 2009
Posts: 39702

Show Tags

14 Nov 2012, 03:04
bellcurve wrote:
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.

Merging similar topics.
_________________
Re: Tough Combinatorics!   [#permalink] 14 Nov 2012, 03:04
Similar topics Replies Last post
Similar
Topics:
In a deck of cards there are 52 cards numbered from 1 to 13. There are 4 05 Dec 2016, 23:19
4 A certain box has 10 cards written integers from 1 to 10 inclusive and 3 05 Sep 2016, 22:26
If 2 cards are selected at random from the deck of 52 cards then What 1 27 Jul 2015, 05:53
10 If you select two cards from a pile of cards numbered 1 to 10, what is 6 25 Dec 2015, 11:53
2 A deck of cards contains 6 cards numbered from 1 to 6. If 15 03 Apr 2013, 03:44
Display posts from previous: Sort by