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The average (arithmetic mean) of the 5 positive integers

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The average (arithmetic mean) of the 5 positive integers [#permalink]

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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
[Reveal] Spoiler: OA

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Re: The average (arithmetic mean) of the 5 positive integers [#permalink]

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ajit257 wrote:
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

Please could someone explain why we cant take the smallest value as 0.


Note that k, m, r, s, and t are positive integers, thus neither of them can be zero.

Given: \(0<k<m<r<s<t=40\) and \(average=\frac{k+m+r+s+40}{5}=16\). Question: \(median_{max}=?\) As median of 5 (odd) numbers is the middle number (when arranged in ascending or descending order) then the question basically asks to find the value of \(r_{max}\).

\(average=\frac{k+m+r+s+40}{5}=16\) --> \(k+m+r+s+40=16*5=80\) --> \(k+m+r+s=40\). 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 \(r<s\)). So, \(r_{max}=18\)

Answer: B.

Hope it's clear.
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Why cant we take the smallest number as zero

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Re: The average (arithmetic mean) of the 5 positive integers [#permalink]

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New post 04 Apr 2014, 01:43
ajit257 wrote:
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

Please could someone explain why we cant take the smallest value as 0.



Average of 5 integers = 16

So total = 80

Given that t = 40; so k+m+r+s = 80-40 = 40

Make two groups so that k+s = 20 & m+r = 20

We require to find the highest possible value of r which is the median, so take values of k & m the least possible & value of s maximum possible

1 < 2 < r < 20

20-2 = 18

So 18 is the maximum possible value to r which satisfies all the conditions given
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Last edited by PareshGmat on 04 Apr 2014, 01:56, edited 2 times in total.

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Re: The average (arithmetic mean) of the 5 positive integers [#permalink]

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New post 04 Apr 2014, 01:47
sairajesh063 wrote:
Please help me out in this question.


The arithmetic mean of 5 positive integers a,b,c,d and e is 22 , and a<b<c<d<e.if e is 40,what is the greatest possible value of the median of the 5 integers?

Ans is : 33



a+b+c+d+e = 110

e = 40

a+b+c+d = 70

1<2<c<34

70/2 - 1 = 34 = Max possible value of d

34-1 = 33 = Maximum possible value of c = Answer

Details of calculation as provided in the post above
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Last edited by PareshGmat on 04 Apr 2014, 01:57, edited 1 time in total.

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Re: The average (arithmetic mean) of the 5 positive integers [#permalink]

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PareshGmat wrote:
ajit257 wrote:
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

Please could someone explain why we cant take the smallest value as 0.



Average of 5 integers = 16

So total = 80

Given that t = 40; so k+m+r+s = 80-40 = 40

Make two groups so that k+s = 20 & m+r = 20

We require to find the highest possible value of r which is the median, so take value of k the least possible & value of s maximum possible

1 < m < r < 20

20-1 = 19

So 19 is the maximum possible value to r which satisfies all the conditions given


Te correct answer is B, not C. Check here: the-average-arithmetic-mean-of-the-5-positive-integers-k-107059.html#p842762
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Re: The average (arithmetic mean) of the 5 positive integers [#permalink]

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New post 04 Apr 2014, 01:58
Bunuel wrote:
PareshGmat wrote:
ajit257 wrote:
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

Please could someone explain why we cant take the smallest value as 0.



Average of 5 integers = 16

So total = 80

Given that t = 40; so k+m+r+s = 80-40 = 40

Make two groups so that k+s = 20 & m+r = 20

We require to find the highest possible value of r which is the median, so take value of k the least possible & value of s maximum possible

1 < m < r < 20

20-1 = 19

So 19 is the maximum possible value to r which satisfies all the conditions given


Te correct answer is B, not C. Check here: the-average-arithmetic-mean-of-the-5-positive-integers-k-107059.html#p842762



I've corrected the post. Thanks for pointing it out

Can you kindly suggest if this approach is ok?
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Re: The average (arithmetic mean) of the 5 positive integers [#permalink]

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New post 18 Jun 2014, 19:51
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.

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Re: The average (arithmetic mean) of the 5 positive integers [#permalink]

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

A. 16
B. 18
c. 19
d. 20
e. 22


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)
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Re: The average (arithmetic mean) of the 5 positive integers [#permalink]

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New post 18 Jun 2014, 20:16
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.

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Re: The average (arithmetic mean) of the 5 positive integers [#permalink]

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

A. 16
B. 18
c. 19
d. 20
e. 22



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
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Re: The average (arithmetic mean) of the 5 positive integers [#permalink]

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Re: The average (arithmetic mean) of the 5 positive integers [#permalink]

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New post 19 Jul 2015, 06:20
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).
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The average (arithmetic mean) of the 5 positive integers [#permalink]

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New post 18 Dec 2016, 10:50
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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

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