Five cards are to be selected at random from 1 to 10 cards numbered 1 to 10. How many ways are possible that the average of five numbers selected will be greater than the median ?
Greater than the median ?
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is the answer 85?
hmmm not Corrrect !!. Please reply with your logic.
i guess i wont be able to solve this one….whats the answer and solution
i was trying to counting method. There are possibly 252 combinations of 5 cards out of 10, i tried counting cases where median = average..
but then wasn’t able to proceed further…and my initial calculations were incorrect..
I think the answer is 10C5-6
Let, us consider the lowest possible value of avg. possible with the numbers from 1 to 10….now whichever 5 numbers u select their sum wl b divided by 5 to get the avg…….so,the lesser the numerator the lesser the avg.
the least avg tht we can get by selecting 5 numbers from 1 to 10 is by selecting the least numbers…….(1+2+3+4+5)/5=3
Now, we r selecting odd number of numbers i.e.,5 n whenever we select consecutive numbers the middle term wl b the avg and it wl b equal to the median…..
so, the cases for which median>avg is 0
the cases for which median=avg is-
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
5 6 7 8 9
6 7 8 9 10
i.e.,6 cases.
for all other cases, the median is > avg
Thus,
I think the answer is-
10C5-6= 252-6= 246
i think the above person has made the same mistake as i did…
because its not just 6 cases….
its consistent difference cases where average = median
1-3-5-7-9 and 2-4-6-8-10 come to my mind..so there are eight cases where median = average…
Take the gmat, comn out with the answer, releive us of our misery…
Hi 800 Aspirants,
Effort is not up to mark. Looking for genuine efforts.
sorry there r other cases possible……
my answer is wrong
Hi,
A very good question.
Median comes out to be 5.5 , and the question asks for “How many ways are possible that the average of five numbers selected will be greater than the median ? ”
it means minimum total of those 5 numbers must be equal to 28 and the max total goes upto 40 (6+7+8+9+10=40)
So we need to pick those five no., whose sum comes out to be >=28 and goes till 40
or other way arnd, we need those no. whose sum is greater thn 15 and less thn 28
Choosng out 5 no. from 10 numbers, total answer is 252
Problem arises in my case is, there are many cases in which different combinations of no. give me same total
for eg. 4+5+6+7+8=30
1+4+6+9+10=30
Am working on it and will get a possible solution
A.c.
even for numbers in Arithmetic progression, average = median
1 3 5 7 9
2 4 6 8 10
so: 10C5 – 6(Ap with d = 1) – 2( AP with d =2)
Answer is 244
@ Ankur
How can median be 5.5?
firstly, here each set consists of odd number of elements so median has to be the middlemost number
I worked on this and here’s another solution……
Since, Median is the middlemost number…….in this case, the possible values of median are 3,4,5,6,7,8…..
the condition is that the avg > median
So, median min total=avg(=median)*5
3 15
4 20
5 25
6 30
7 35
8 40
If median =3, the cases possible are:
_ _ _ _ _
(3)
1 2 10 4,5,6,7,8,9 =6sets
9 4,5,6,7,8 =5sets =18sets
8 4,5,6,7 =4sets
7 5,6 =2sets
6 5 =1set
_ _ _ _ _
(4)
1 2 10 9,8,7 =3sets
9 8 =1set
1 3 10 9,8,7,6 =4sets
9 8,7 =2sets =19sets
2 3 10 9,8,7,6,5 =5sets
9 8,7,6 =3sets
8 7 =1set
_ _ _ _ _
(5)
2 4 10 9 =1set
3 4 10 9,8 =2sets =4sets
9 8 =1set
So, total of 18+19+4=41 sets possible
Thus, answer is 10C5-41 =252-41=211
Anyone has better ideas??
Am a little confuse here……
Are we talking about the median of all the 10 numbers(thn it comes out to bed 5.5) or are we talking about the median of those 5 numbers which we will select out from 1-10
10C5 – 36 = 215
This is the solution provided by Prachi……im posting on her behalf-
With small set of numbers assume —- {1,2,3,4,5}
When Mean = Median: (1,2,3) (2,3,4) (3,4,5) (1,3,5)
When Mean > Median: (1,2,4) (1,2,5) (2,3,5)
When Mean Median) = P(MeanMedian) + P(Mean P(Mean > Median) = 1/2* {1 – P(Mean = Median)} *
*For the set {1,2,3,4,5,6, 7,8,9,10}
No of cases when Mean = Median : 30
which are as follows:
(1,2,3,4,5),
(2,3,4,5,6), (1,2,4,5,8), (1,2,4,6,7), (1,3,4,5,7),
(3,4,5,6,7), (1,2,5,7, 10), (1,2,5,8,9), (1,4,5,7,8), (1,4,5,6,9),
(1,3,5,6,10) , (1,3,5,7,9), (2,4,5,6,8), (2,3,5,6,9), (2,3,5,7,8),
(4,5,6,7,8), (2,3,6,9,10) , (3,4,6,8,9), (3,5,6,7,9), (2,4,6,8,10) ,
(2,5,6,8,9), (2,5,6,7,10) , (3,4,6,7,10) , (1,4,6,9,10) , (1,5,6,8,10)
(5,6,7,8,9), (4,5,7,9,10) , (4,6,7,8,10) , (3,6,7,9,10)
(6,7,8,9,10) ,
Total ways = 252
Hence the number of ways in which the average of five numbers selected
will be greater than the median = (252 – 30) /2 = 111 is the answer…
TakeGMAT team plz tell us the answer….
If you select 5 numbers, obviously the middle will be the median number (i think this point is well understood by now).
The numbers cannot be sequential, otherwise the mean will be equal to the median. So, of the two numbers in the set of five that are greater than the mean, they both must sum to at least one more than the sequential numbering of the last two would sum to
Now here are what I will call the base sets (or the sets in which the mean and median are the same, but can satisfy my above condition). If you replace the last bracketed numbers in the base sets with the following bracketed numbers, the mean/median condition is met
1 2 3 [4 5] … [4 6] [4 7] [4 8] [4 9] [4 10], etc (20 cases in all)
2 3 4 [5 6] … [5 7] [5 8] [5 9] [5 10], etc (14 cases in all)
3 4 5 [6 7] (9 cases in all)
4 5 6 [7 8] (5 cases in all)
5 6 7 [8 9] (2 cases in all)
Total of 50 cases.
Multiply by how many ways you can choose five numbers (since numbers are picked at random)
50*120=6000
6000 ways ladies and gents.
TakeGMAT, am I right? Am I an aspiring 800 member?
in my answer:
So, of the two numbers in the set of five that are greater than the mean, they both must sum to at least one more than the sequential numbering of the last two would sum to
should state that the last two numbers in a set of five must be at least sum to one more than the last two numbers in a set of five sequentially (but i can’t write in english apparently)
Really a good question.
First, the method of simple calculations for finding the answer is very time-confusing, so there must be a point, an idea leading to simple and quick solution. (GMAT style).
Here is an approach to solve this problem in alloted 2 minutes or so.
All possible combinations can be seperated into three different subgroups.
/ 2 = 122
1) When median = mean.
2) When median > mean.
3) When median 8
And so, The number we are looking for is (252 –
Hi webmaster!
Hi webmaster!
my answer is 239 . there r only 13 cases out of 10c5 whose total will be 25 or less than 25
sorry 14 cases so the answer is 238
sory final answer is 237