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The average (arithmetic mean) of the 5 positive integers [#permalink]
28 Dec 2010, 16:54

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Difficulty:

35% (medium)

Question Stats:

60% (02:43) correct
39% (01:29) wrong based on 163 sessions

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.

Re: The average (arithmetic mean) of the 5 positive integers k, [#permalink]
28 Dec 2010, 17:15

1

This post received KUDOS

Expert's post

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

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
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Last edited by Bunuel on 12 Jan 2012, 06:24, edited 1 time in total.

NOTE: this question has the different average value from that of in the initial post, though the questions are basically the same. Thus merged.

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

Given , mean is 22

total sum = 5 * mean = 5* 22 = 110

e is 40 so a+b+c+d = 70 and a<b<c<d<e

Median will be the middle term so value of "c" will be median

let a = 1 and b = 2 as we need to find the maximum value of median

c+d = 70 -(a+b) = 70 -3 = 67

so c < 33.5 and d >33.5 (on dividing 67 by 2) . As c and d are integers c = 33 and d =34 Hence median is 33

To find the greatest median, you need have smallest values for the numbers below the median.

That is a=1, b=2. Now, subtract from the sum of numbers = (22 x 5) - (1 + 2 + 40) = 68.

So, we need to fit in c and d within a sum of 68 such that c is the largest it can be. We get c = 33 and d = 35.
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Re: The average (arithmetic mean) of the 5 positive integers [#permalink]
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.

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.

Re: The average (arithmetic mean) of the 5 positive integers [#permalink]
04 Apr 2014, 01:50

1

This post received KUDOS

Expert's post

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

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