The average (arithmetic mean) of the 5 positive integers

Why cant we take the smallest number as zero

Because stem says that k, m, r, s, and t are positive integers and zero is not a positive number.

Since last digit t is 40 and arithmatic mean of all 5 digits is 16
k+m+r+s = 16*5 - 40 = 40

Questions is asking maximum possible value of median which is nothing but maximum possible value of r.
median is middle number if there are odd number of numbers. here 5 are there, so middle one is median.

possible numbers, since they are in ascending positive integers, giving least value to k
start with k=1, then m = 2(should be greater than k)
sum of r+s = 40-3 = 37

maximum value of r is 18 and s is 19.

so the numbers are 1,2,18,19,40

median(r) is 18.

Please give kudos if you like my answer

Using the logical approach:

We need to find the median which is the third value when the numbers are in increasing order. Since k<m<r<s<t, the median would be r.

The average of the positive integers is 16 which means that in effect, all numbers are equal to 16. If the largest number is 40, it is 24 more than 16. We need r to be maximum so k and m should be as small as possible to get the average of 16. Since all the numbers are positive integers, k and m cannot be less than 1 and 2 respectively. 1 is 15 less than 16 and 2 is 14 less than 16 which means k and m combined are 29 less than the average. 40 is already 24 more than 16 and hence we only have 29 - 24 = 5 extra to distribute between r and s. Since s must be greater than r, r can be 16+2 = 18 and s can be 16+3 = 19.

So r is 18.

Answer (B)

Given that k<m<r<s<t and t =40, we are required to maximize the median which is r. R can be maximized if k, m and s is least. As all are +ive integers the least k and m are 1 & 2 respectively. 's' is greater than r so minimum value of s should 1 more than r, therefore s = r + 1. Sum of all no. = mean * 5 = 16*5 = 80.
It means k + m + r + S + t = 80, we can write 1 + 2 + r + r + 1 + 40 = 80 , it gives 2r=36, therefore r=18.

Answer = B = 18

t = 40, so

k + m + r + s = 40

Median of k,m,r,s and t is r

Require to find MAX value of r

s should be less than \(\frac{40}{2} = 20\)

If s is 19, then r can be kept 18 (Max value)

Series would come up as

1 , 2 , 18 , 19 , 40

Same as Bunuel's..

\(\frac{(k+m+r+s+40)}{5} = 16\)

\((k+m+r+s) = 40\)

The only way by which the median can be maximised is by minimising the values lesser than the median and keeping the values greater than the median in such a way that their differences are minimum.

\(1+2+r+s+40 = 80\)

\(r+s = 80-40-2-1 = 37\)

Maximum value of \(r\) can be 18 (such that \(s = 19\)). So Ans (B).

This one is such a great Question.
One thing we should notice here is that it is full of restrictions.
The set in increasing order => k,m,r,s,t
Now it given give that they are all integers >0
Mean = 16

Using the Basic Mean theory => \(Mean = \frac{Sum}{#}\)


Hence Sum(5) = 16*5 = 80

k+m+r+s+t=80
t is 40
So => k+m+r+s=40

Now Since the number of terms is odd => Median = 5+1/2 Term => r

Hence median = r
Now to maximise any term in the set we must minimise all the other terms in the set keeping in mind =>

Maximum value of any data element to the left of median = Median
Minimum value of any data element to the left of median = Smallest element in the set
Maximum value of any data element to the right of the median = Largest element in set
Minimum value of any data element to the right of the median =Median


Hence minimum value of k=1
m= 2
r=r
s= r+1

Hence 1+2+r+r+1=40
2r=36
r=18

Hence B

Given:
1. The average (arithmetic mean) of the 5 positive integers k, m, r, s, and t is 16, and k < m < r < s < t.
2. t is 40

Asked; What is the greatest possible value of the median of the 5 integers?
Q. Max(r)=?

k + m + r + s + 40 = 5*16 = 80
k + m + r + s = 40

For r to be maximum, all other positive integers k,m & s should be minimum.
For Max (r) => s = r+1, k =1, m = 2
1 + 2 + r + r+1 = 40
2r + 4 = 40
r = 18

IMO B

Hi Bunuel, the question said positive integers, didn't really mention positive disctinct integers. Why are we assuming that, and not thinking that the integers can be 1+1+x+x+40. If we assume this then the largest medium is 19, not 18.

Check the highlighted part.

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