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Bunuel
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Five cards are to be selected from 10 cards numbered 1 to 10 19 Nov 2009, 18:08
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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95% (hard)

Question Stats:

21% (02:06) correct 79% (02:47) wrong based on 50 sessions

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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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gnet
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Five cards are to be selected from 10 cards numbered 1 to 10 19 Nov 2009, 22:18
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Bunuel
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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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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Five cards are to be selected from 10 cards numbered 1 to 10 19 Nov 2009, 22:40
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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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Five cards are to be selected from 10 cards numbered 1 to 10 17 Feb 2010, 11:38
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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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Five cards are to be selected from 10 cards numbered 1 to 10 20 Feb 2010, 11:37
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Bunuel
OA: A (111). Gnet's explanation is correct. +1.
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Hi Bunuel... do we have a shorter way for such problems?
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Five cards are to be selected from 10 cards numbered 1 to 10 20 Feb 2010, 12:31
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jeeteshsingh
Bunuel
OA: A (111). Gnet's explanation is correct. +1.

Hi Bunuel... do we have a shorter way for such problems?
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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 from 10 cards numbered 1 to 10 Updated on: 14 Nov 2012, 01:00
Originally posted by suryanshg on 13 Nov 2012, 22:48.
Last edited by suryanshg on 14 Nov 2012, 01:00, edited 1 time in total.
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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
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Five cards are to be selected from 10 cards numbered 1 to 10 28 Jan 2022, 06:30
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