Median

[shantascherla]

Set S contains seven distinct integers. The median of set S is the integer m, and all values in set S are equal to or less than 2m. What is the highest possible average (arithmetic mean) of all values in set S ?

m
10m/7
10m/7 - 9/7
5m/7 + 3/7
5m

Correct answer is c

I couldnt understand the following explanation

We must choose the highest possible value for every member of the set. The maximum value of any member of the set is 2m, so the 7th value is 2m

How can we assume that the 7th value would be 2m - when the question states that all values are equal to or < 2m. Can anybody explain

[jnelson0612]

Set S contains seven distinct integers. The median of set S is the integer m, and all values in set S are equal to or less than 2m. What is the highest possible average (arithmetic mean) of all values in set S ?

m
10m/7
10m/7 - 9/7
5m/7 + 3/7
5m

Correct answer is c

I couldnt understand the following explanation

We must choose the highest possible value for every member of the set. The maximum value of any member of the set is 2m, so the 7th value is 2m

How can we assume that the 7th value would be 2m - when the question states that all values are equal to or < 2m. Can anybody explain

shanta, keep in mind that our goal is to obtain the highest possible average for the seven numbers in the set. To do this, I must make each number in the set as large as possible.

Since I know that the largest value in the set is equal to or less than 2m, I will simply choose to set my largest value at 2m to maximize this value.

I hope this helps, thanks.

[niitsm]

Hi Is it possible to solve this one by assuming m some integer .

Can anyone please guide me with this one.

Suppose I assume m= 3 then 2m = 6 and my nos will be

0123456 & as these no's are consecutive no's my mean will be 3 .

[tim]

no. when they ask for the maximum possible, that means for all possible values. picking one value will not get you anywhere on a problem like this unfortunately..

[krishnan.anju1987]

Hi,

This is how I solved this question.

Since m must be the median, there must be three values equal to m or less than m. Also there must be three values greater than m or equal to m. Now every value in the set S can be 2m or lesser than 2m. To find the maximum possible average for this set, I consider the last three variables to be 2m and the variables to the left of median to be equal to m. Hence the set S becomes m,m,m,m,2m,2m,2m. Hence sum =10m and average =1m/7 which uis answer B. However, the answer mentioned in this question is C. Am I going wrong somewhere?

[jnelson0612]

Hi,

This is how I solved this question.

Since m must be the median, there must be three values equal to m or less than m. Also there must be three values greater than m or equal to m. Now every value in the set S can be 2m or lesser than 2m. To find the maximum possible average for this set, I consider the last three variables to be 2m and the variables to the left of median to be equal to m. Hence the set S becomes m,m,m,m,2m,2m,2m. Hence sum =10m and average =1m/7 which uis answer B. However, the answer mentioned in this question is C. Am I going wrong somewhere?

Krishnan, overall good thinking, but you missed one word in the question that makes all the difference: "distinct". We have seven "distinct" or different integers. Thus we can't repeately use m and 2m; you have to make each integer different from the others. I'll bet that you can solve now. :-)

[krishnan.anju1987]

Oops, yes.. Thanks for clarification. Got the answer now :)

[tim]

cool

[NL]

It would be interesting if the answer choice E would be 6m instead of 5m. If that, E could be true and C is always true.
(I read the explanation and wondered why the author thinks that 2m-1>m+2, then picked some numbers)

[RonPurewal]

It would be interesting if the answer choice E would be 6m instead of 5m. If that, E could be true and C is always true.

how can you get E to be true?

[NL]

how can you get E to be true?

Yup, I miscalculated.
It should be edited: “if E were 8m/7-3/7 instead of 5m, then E could be true”.

7 numbers can be represented by 2 groups: (in case m=3 or 2m=m+3)
(1) m-3; m-2; m-1; m; m+1; m+2; 2m
(2) m-3; m-2; m-1; m; 2m-2; 2m-1; 2m

[RonPurewal]

how can you get E to be true?

Yup, I miscalculated.
It should be edited: “if E were 8m/7-3/7 instead of 5m, then E could be true”.

7 numbers can be represented by 2 groups: (in case m=3 or 2m=m+3)
(1) m-3; m-2; m-1; m; m+1; m+2; 2m
(2) m-3; m-2; m-1; m; 2m-2; 2m-1; 2m

your #1 here works ONLY for m = 3. if m is anything else, it won't work.
you're trying to maximize everything in the whole set (in order to maximize the average). if the biggest possible value is 2m, then the second- and third-biggest possible values are, respectively, (2m – 1) and (2m – 2).

perhaps ironically, if m is the median, (m + 1) and (m + 2) are actually the smallest possible values for those two spots.

[RonPurewal]

if E were 8m/7-3/7 instead of 5m, then E could be true

in other words, that ^^ wouldn't be a correct answer, even if it were in the choices.
sure, it works for m = 3, but that's the risk that's always inherent in "plugging in" numbers: you might, by pure coincidence, get the desired value from more than one choice. in that case, you just have to try the surviving choices again with a different value. the correct answer must work for ALL values of the unknown(s).