The integers v,w,x,y and z are such that 0<v<w<x<y<z. The average of these integers is 36 and median of these 5 integers is 28. What is the greatest possible value of Z?
a) 128
b) 130
c) 140
d) 132
e) 120
Please assist with above problem.
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The integers v,w,x,y and z are such that 0<v<w<x<
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alanforde800Maximus
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The average of these integers is 36alanforde800Maximus wrote:The integers v, w, x, y and z are such that 0 < v < w < x < y < z. The average of these integers is 36 and median of these 5 integers is 28. What is the greatest possible value of Z?
a) 128
b) 130
c) 140
d) 132
e) 120
So, (v + w + x + y + z)/5 = 36
So, v + w + x + y + z = 180
The median of these 5 integers is 28
Since x is the middlemost value (in ascending order), we know that x = 28
So, we have v, w, 28, y, z
If we want to MAXIMIZE the value of z, we must MINIMIZE the remaining values.
Since v is a positive integer, the smallest value of v is 1
1, w, 28, y, z
Since v < w, the smallest value of w is 2
1, 2, 28, y, z
Since x < y, the smallest value of y is 29
1, 2, 28, 29, z
Since v + w + x + y + z = 180, we know that 1 + 2 + 28 + 29 + z = 180
Simplify: 60 + z = 180
z = 120
Answer: E
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The GMAT loves its min-max problems. For another good one, see here: https://www.beatthegmat.com/gmat-prep-av ... 75647.html
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Let's make it easy on ourselves. x is the middle number of the five, so x = median = 28.
The sum of the five numbers = (Average * # of #'s) = 5*36 = 180.
We'll make v, w, and y as small as possible:
v = 1 (it has to be greater than 0, but we want it to be as close to 0 as possible)
w = 2 (it has to be greater than v, which is 1, but we want it to be as close to 1 as possible)
y = 29 (it has to be greater than x, which is 28, but we want it to be as close to 28 as possible)
From there, we've got
Sum = v + w + x + y + z
180 = 1 + 2 + 28 + 29 + z
and we're left with the max of z, or z = 120.
The sum of the five numbers = (Average * # of #'s) = 5*36 = 180.
We'll make v, w, and y as small as possible:
v = 1 (it has to be greater than 0, but we want it to be as close to 0 as possible)
w = 2 (it has to be greater than v, which is 1, but we want it to be as close to 1 as possible)
y = 29 (it has to be greater than x, which is 28, but we want it to be as close to 28 as possible)
From there, we've got
Sum = v + w + x + y + z
180 = 1 + 2 + 28 + 29 + z
and we're left with the max of z, or z = 120.
