The average (arithmetic mean) of the 5 positive integers

Question

The average (arithmetic mean) of the 5 positive integers k, m, r, s, and t is 16, and k < m < r < s < t. If t is 40, what is the greatest possible value of the median of the 5 integers?

A. 16
B. 18
C. 19
D. 20
E. 22

Bunuel · Accepted Answer

Note that k, m, r, s, and t are positive integers, thus neither of them can be zero. Given: 0 k+m+r+s+40=16*5=80 --> k+m+r+s=40 . Now, we want to maximize r : General rule for such kind of problems: to maximize one quantity, minimize the others; to minimize one quantity, maximize the others. So to maximize r we should minimize k and m , as k and m must be distinct positive integers then the least values for them are 1 and 2 respectively --> 1+2+r+s=40 --> r+s=37 --> r_{max}=18 and s=19 (as r and s also must be distinct positive integers and r

Bunuel · Answer

Min/max questions: search.php?search_id=tag&tag_id=63

Eshaninan · Answer

Why cant we take the smallest number as zero

Bunuel · Answer

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

GMatAspirerCA · Answer

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

KarishmaB · 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)

nitin1negi · Answer

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.

PareshGmat · Answer

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

TimeTraveller · Answer

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).

stonecold · Answer

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

Kinshook · Answer

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

shubhs76 · Answer

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.

Bunuel · Answer

Check the highlighted part.

aditsatt · Answer

Hi  Bunuel

Please help me understand my mistake.

To maximise median, I took k & m as 1 & 2 but took s & t as closer to the median, so as to increase value of r (median)

Using hit & trial, I found - 1+2+22+27+28 = 80 = 16*5 (1+2+24+26+27 also sum to 80, but not in the options - also, any value of s & t lesser than 26 & 27 would result in r being greater)

Shouldn't the answer hence, be 22?

Bunuel · Answer

Please check the question:

The average (arithmetic mean) of the 5 positive integers k, m, r, s, and t is 16, and k < m < r < s < t.   If t is 40  , what is the greatest possible value of the median of the 5 integers?

Demonicgod · Answer

Given: 5 integers: k 2r + 1 = 37 > r = 18r+(r+1)=37⇒2r+1=37⇒r=18 ✅ That’s the maximum possible median

The average (arithmetic mean) of the 5 positive integers

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The average (arithmetic mean) of the 5 positive integers

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