Mean and Median

[nakulbatra]

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 after some posts.

[hardik.jadeja]

I am getting 49. Is 49 the answer? I am not very confident. But I can explain how i got 49 if its the correct answer.

[cjb]

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 after some posts.

My first attempt at this is:

10C5 = 252

For some of those picks of 5 cards, the mean will equal the median. The rest will be equally split between having a higher mean or a higher median. So we can solve if we know all the way in which mean and median will be the same.

Possible card distributions with equal mean and median is the problem. Can't think of a quick way of enumerating them.

For cards a,b,c,d,e, such that 1 <= a < b < c < d < e <=10
mean = (a+b+c+d+e)/5
median = c

(a+b+c+d+e)/5 = c
a+b+c+d+e = 5c

=> a+b+d+e =4c

(c-a)+(c-b) = (d-c)+(e-c)

Consider c=5. Max LHS of eqn above =
(5-1)+(5-2) = 7
This can be balanced on the RHS either with 7 and 10, or with 8 and 9.

I could probably get somewhere with this if I had infinite time, but I'd be way over the two minute mark now. Must be a trick I'm not seeing.

[logitech]

For some of those picks of 5 cards, the mean will equal the median. The rest will be equally split between having a higher mean or a higher median. So we can solve if we know all the way in which mean and median will be the same.

Great start.

Median will be between 3 and 8 inclusive.

for 3 and 8 we only have one way to choose the rest of the 4 numbers

1 2 3 4 5

6 7 8 9 10

Between 3 and 8

4, 5, 6, and 7


For example FOR Median = Average = 4

2 3 4 5 6

We will have only (n-1) number on the left to play with 2 numbers and every time you make a change you have to make another change on the other side and you can choose 2 of them

For 4 = REST of the sum will be 16 ( you can use 1,2 or 3) C(3:2) = 3
For 5= REST of the sum will be 20 (you can use 1, 2, 3 or 4) C(4:2) = 6
For 6= REST of the sum will be 24 ( you can use 7,8,9 or 10 ) C(4:2) = 6
For 7= REST of the sum will be 28 ( you can use 8,9 or 10 ) C(3:2)=3


So AV=MEDIAN= 1+3+6+6+3+1 = 20 ways ( If I am not mistaken)

C(10:5)= 252

252 - 20 = 232 way for > Median and < Median

So answer should be 232/2 = 116



:roll:

I realized that I have missed couple of numbers!

median=3: only one way: 12345
median=4: four ways: 12467, 12458, 13457, 23456
median=5: ten ways: 12589, 1257T, 13579, 1356T, 14569, 14578, 23578, 23569, 24568, 34567
(I listed the numbers lower than the median systematically, then looked for pairs of numbers higher than the median that added to the right sum)
median=6 ten ways (symmetric with median=5)
median=7,four ways (symmetric with median=4),
median=8, 1 way 6789T
Total ways = 1+4+10+10+4+1 = 30

This makes the correct answer: 111

252-30 =222

and 222/2 = 111

[x2suresh]

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 after some posts.

Is this GMAT question..

If yes.. can you post 5 options too...

[nakulbatra]

the OA is 111

[cjb]

Now - what's the quick way of doing this?