knight247 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
Don't have an OA on this. Detailed explanations would be highly appreciated.
The number of ways to choose 5 cards from 10 choices = 10C5 = 252.
We need to subtract from 252 the number of combinations in which the average = the median:
Median and average of 3:
{1,2,3,4,5}
Number of options = 1.
Median and average of 4:
{1,2,4,5,8} {1,2,4,6,7} {1,3,4,5,7} {2,3,4,5,6}
Number of options = 4.
Median and average of 5:
{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}
Number of options = 10.
Median and average of 6:
{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}
Number of options = 10.
Median and average of 7:
{3,6,7,9,10} {4,5,7,9,10} {4,6,7,8,10} {5,6,7,8,9}
Number of options = 4.
Median and average of 8:
{6,7,8,9,10}
Number of options = 1.
Total combinations in which the average = the median: 1+4+10+10+4+1 = 30.
Thus, we have 252-30 = 222 combinations in which the average ≠median.
In half of these combinations, the average will be less than the median; in the other half, the average will be greater than the median.
Thus, the number of combinations in which the average is greater than the median: 222/2 = 111.
The correct answer is
A.
Please note that this question would NOT be asked on the GMAT. WAY too time-consuming.
