Median - PS

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by GMATGuruNY » Mon Dec 19, 2011 7:07 pm
karthikpandian19 wrote:Set consists of five integers that sum to 89. If the median is distinct from the other integers, what is the smallest possible value of the range of ?

A. 0
B. 1
C. 2
D. 3
E. 4
Let m=median.

Since m must be distinct from the other integers, all 5 integers cannot be the same.
To minimize the range, the other integers must differ from m as little as possible.
The smallest possible difference between m and the other integers is 1.
Thus, the smallest possible range will be achieved if the 5 integers are:
m-1, m-1, m, m+1, m+1.

Given that the sum=89, we get:
(m-1) + (m-1) + m + (m+1) + (m+1) = 89
5m = 89.
m = 89/5.
Doesn't work, since m must be an integer.

To minimize the range, we want to change the values in our list as little as possible.
Thus, we need 5m to be equal to the nearest multiple of 5, which is 90.
For the right-hand side of 5m=89 to INCREASE by 1, the left-hand side must DECREASE by 1.
Thus, the smallest value in our list must decrease to m-2:
m-2, m-1, m, m+1, m+1.

With these values, we get:
(m-2) + (m-1) + m + (m+1) + (m+1) = 89
5m - 1 = 89
5m = 90
m = 18.

Thus:
The least value = m-2 = 18-2 = 16.
The greatest value = m+1 = 18+1 = 19.
Range = 19-16 = 3.

The correct answer is D.
Last edited by GMATGuruNY on Mon Dec 19, 2011 7:47 pm, edited 1 time in total.
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by karthikpandian19 » Mon Dec 19, 2011 7:38 pm
@GMATGuruNY

How did you assume the integers to be

"m-1, m-1, m, m+1, m+1."

I can understand "m" to be distinct. But the others can be anything right????
GMATGuruNY wrote:
karthikpandian19 wrote:Set consists of five integers that sum to 89. If the median is distinct from the other integers, what is the smallest possible value of the range of ?

A. 0
B. 1
C. 2
D. 3
E. 4
Let m=median.

Since m must be distinct from the other integers, the smallest possible range will be achieved if the 5 integers are:
m-1, m-1, m, m+1, m+1.

Given that the sum=89, we get:
(m-1) + (m-1) + m + (m+1) + (m+1) = 89
5m = 89.
m = 89/5.
Doesn't work, since m must be an integer.

To minimize the range, we want to change the values in our list as little as possible.
Thus, we need 5m to be equal to the nearest multiple of 5, which is 90.
For the right-hand side of 5m=89 to INCREASE by 1, the left-hand side must DECREASE by 1.
Thus, the smallest value in our list must decrease to m-2:
m-2, m-1, m, m+1, m+1.

With these values, we get:
(m-2) + (m-1) + m + (m+1) + (m+1) = 89
5m - 1 = 89
5m = 90
m = 18.

Thus:
The least value = m-2 = 18-2 = 16.
The greatest value = m+1 = 18+1 = 19.
Range = 19-16 = 3.

The correct answer is D.

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by GMATGuruNY » Mon Dec 19, 2011 7:51 pm
karthikpandian19 wrote:@GMATGuruNY

How did you assume the integers to be

"m-1, m-1, m, m+1, m+1."

I can understand "m" to be distinct. But the others can be anything right????
Please revisit my post above. I've fleshed out the reasoning behind representing the 5 integers as m-1, m-1, m, m+1, m+1.
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by karthikpandian19 » Mon Dec 19, 2011 9:14 pm
Ok...sure...now i understood.....thanks very much

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by ankush123251 » Mon Dec 19, 2011 10:34 pm
Nice explanation..thanks

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by sk8legend408 » Tue Dec 20, 2011 10:41 am
I did it a slightly different way:

89/5 is approximately 18. You know that the average is approximately 18. 18 can also be the median since it is an odd set of integer.

Try picking numbers keeping 18 in the middle. 17+17+18+19+19 = 90.

We want to get to 89 so make the first 17 a 16. 19-16=3.

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by karthikpandian19 » Wed Dec 28, 2011 9:56 pm
OA is D

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by jonathan123456 » Mon Jan 02, 2012 12:33 pm
This is a set of 5 (odd number) integers. Median would be the middle number but the question states that median is distinct from the numbers. Hows that possible?

karthikpandian19 wrote:Set consists of five integers that sum to 89. If the median is distinct from the other integers, what is the smallest possible value of the range of ?

A. 0
B. 1
C. 2
D. 3
E. 4
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