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

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Five cards are to be selected from 10 cards numbered 1 to 10 [#permalink] New post 19 Nov 2009, 17:08
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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

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Re: Really hard 800+ problem. [#permalink] New post 19 Nov 2009, 20:44
Is the answer B-116?

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.

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Re: Really hard 800+ problem. [#permalink] New post 19 Nov 2009, 21:18
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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...
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Re: Really hard 800+ problem. [#permalink] New post 19 Nov 2009, 21:40
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

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Re: Really hard 800+ problem. [#permalink] New post 20 Nov 2009, 06:29
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.
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Re: Really hard 800+ problem. [#permalink] New post 17 Feb 2010, 10:38
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
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Re: Really hard 800+ problem. [#permalink] New post 17 Feb 2010, 22:21
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
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Re: Really hard 800+ problem. [#permalink] New post 18 Feb 2010, 06:34
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Re: Really hard 800+ problem. [#permalink] New post 20 Feb 2010, 10:37
Bunuel wrote:
OA: A (111). Gnet's explanation is correct. +1.


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

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Re: Really hard 800+ problem. [#permalink] New post 20 Feb 2010, 11:31
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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.

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

a.99
b.54
c.111
d.102
e.79
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Re: Five cards are to be selected at random from 10 cards [#permalink] New post 26 Sep 2012, 00:06
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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.

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Tough Combinatorics! [#permalink] New post 13 Nov 2012, 20:50
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.
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Re: Tough Combinatorics! [#permalink] New post 13 Nov 2012, 21:48
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, 00:00, edited 1 time in total.
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Re: Tough Combinatorics! [#permalink] New post 13 Nov 2012, 22:35
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.

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Re: Tough Combinatorics! [#permalink] New post 14 Nov 2012, 02:04
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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.

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Re: Tough Combinatorics!   [#permalink] 14 Nov 2012, 02:04
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