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?
OA 111
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Five cards are to be selected at random from 10 cards number
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- cans
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if median=3, we have to select 1,2,3. then selecting any 2 numbers other than 4,5 will result in mean > median. thus 7C2 - 1 = 20
if median=8, mean max=8. thus 0 way.
if median=7, mean max=8: sum of other 4=40-7 = 33. sum max can be: 5+6+9+10=30 0 way.
if median=6, mean 7: sum of other 4 numbers = 29. max = 4+5+9+10 = 28. not possible.
if median=6, mean =8: sum of other 4 numbers = 34. not possible
if median=5, mean=8: sum of other 4 = 35. not possible.
if median=5, mean=7: sum of other 4 = 30. not possible.
if median=5, mean=6: sum of other 4 = 25. max = 26. possible.
if sum of lowest 2=7, we have to choose 8 and 10.
if sum of lowest 2=6, we have to choose 9 and 10. 2 ways
if median=4, mean=8,7,6 not possible. mean=5 possible.
mean=5 and median=4, sum of other 4=21.
sum of lowest2 = 5, larger: (6,10)(7,9)
sum of lowest2 = 4, larger: (7,10)(8,9)
sum of lowest2 = 3, larger: (8,10)
5 ways. thus total = 20+2+5 = 27..
if median=8, mean max=8. thus 0 way.
if median=7, mean max=8: sum of other 4=40-7 = 33. sum max can be: 5+6+9+10=30 0 way.
if median=6, mean 7: sum of other 4 numbers = 29. max = 4+5+9+10 = 28. not possible.
if median=6, mean =8: sum of other 4 numbers = 34. not possible
if median=5, mean=8: sum of other 4 = 35. not possible.
if median=5, mean=7: sum of other 4 = 30. not possible.
if median=5, mean=6: sum of other 4 = 25. max = 26. possible.
if sum of lowest 2=7, we have to choose 8 and 10.
if sum of lowest 2=6, we have to choose 9 and 10. 2 ways
if median=4, mean=8,7,6 not possible. mean=5 possible.
mean=5 and median=4, sum of other 4=21.
sum of lowest2 = 5, larger: (6,10)(7,9)
sum of lowest2 = 4, larger: (7,10)(8,9)
sum of lowest2 = 3, larger: (8,10)
5 ways. thus total = 20+2+5 = 27..
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Cans!!
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Cans!!
- krishnasty
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Cans, i m not sure if i understand your solution. It would be helpful if you can please elaborate on your explanation?
cans wrote:if median=3, we have to select 1,2,3. then selecting any 2 numbers other than 4,5 will result in mean > median. thus 7C2 - 1 = 20
if median=8, mean max=8. thus 0 way.
if median=7, mean max=8: sum of other 4=40-7 = 33. sum max can be: 5+6+9+10=30 0 way.
if median=6, mean 7: sum of other 4 numbers = 29. max = 4+5+9+10 = 28. not possible.
if median=6, mean =8: sum of other 4 numbers = 34. not possible
if median=5, mean=8: sum of other 4 = 35. not possible.
if median=5, mean=7: sum of other 4 = 30. not possible.
if median=5, mean=6: sum of other 4 = 25. max = 26. possible.
if sum of lowest 2=7, we have to choose 8 and 10.
if sum of lowest 2=6, we have to choose 9 and 10. 2 ways
if median=4, mean=8,7,6 not possible. mean=5 possible.
mean=5 and median=4, sum of other 4=21.
sum of lowest2 = 5, larger: (6,10)(7,9)
sum of lowest2 = 4, larger: (7,10)(8,9)
sum of lowest2 = 3, larger: (8,10)
5 ways. thus total = 20+2+5 = 27..
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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