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Math Expert V
Joined: 02 Sep 2009
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x, y, and z are positive integers such that x ≥ y ≥ z. If the average  [#permalink]

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Question Stats: 60% (02:58) correct 40% (03:00) wrong based on 148 sessions

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x, y, and z are positive integers such that x ≥ y ≥ z. If the average (arithmetic mean) of x,y, and z is 40, and the median is (x–13), what is the greatest possible value of z?

A. 35
B. 36
C. 37
D. 39
E. 40

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CEO  V
Joined: 12 Sep 2015
Posts: 3717
Location: Canada
Re: x, y, and z are positive integers such that x ≥ y ≥ z. If the average  [#permalink]

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Bunuel wrote:
x, y, and z are positive integers such that x ≥ y ≥ z. If the average (arithmetic mean) of x,y, and z is 40, and the median is (x–13), what is the greatest possible value of z?

A. 35
B. 36
C. 37
D. 39
E. 40

Since y is the middlemost value among the 3 numbers, we know that y is the median.
Since we're told that the median = x-13, we can conclude that y = x-13

The average (arithmetic mean) of x,y, and z is 40
So, (x+y+z)/3 = 40
Multiply both sides by 3 to get: x + y + z = 120
Replace y with x-13 to get: x + x-13 + z = 120
Simplify: 2x - 13 + z = 120
Add 13 to both sides: 2x + z = 133
Solve for z to get: z = 133 - 2x

We're told that y ≥ z
So, we can conclude that x-13 ≥ 133 - 2x
Add 2x to both sides: 3x - 13 ≥ 133
Add 13 to both sides: 3x ≥ 146
Divide both sides by 3 to get: x ≥ 48.666...

Since x is an INTEGER, the SMALLEST possible value of x is 49
We already concluded that z = 133 - 2x
We can see that we can MAXIMIZE the value of z by MINIMIZING the value of x
The SMALLEST possible value of x is 49
So, plug in x = 49 to get: z = 133 - 2(49) = 35

Answer:

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Intern  B
Joined: 26 Jul 2016
Posts: 41
Location: India
Re: x, y, and z are positive integers such that x ≥ y ≥ z. If the average  [#permalink]

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To maximize z; from the given condition we shall take z=y
Mean (z,y,x) = 40; i.e Sum = 120
Median= y= x-13
From A; If z=35, y=35 and x= 48 (35+13); sum= 70+48 = 118
From B; If z=36, y=36 and x=49 (36+13); sum=72+49 = 121 (which is more than sum)
For remaining options sum will be more than 120
Hence Answer A.
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x, y, and z are positive integers such that x ≥ y ≥ z. If the average  [#permalink]

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Bunuel wrote:
x, y, and z are positive integers such that x ≥ y ≥ z. If the average (arithmetic mean) of x,y, and z is 40, and the median is (x–13), what is the greatest possible value of z?

A. 35
B. 36
C. 37
D. 39
E. 40

average (arithmetic mean) of x,y, and z is 40
i.e. x+y+z = 40*3 = 120

median is (x–13)
but since x ≥ y ≥ z
so median of (x, y, z) = y = (x-13)

i.e. x+(x-13)+z = 120
i.e. 2x+z = 133

for z to be greatest, x must be smallest and z will be greatest when it is equal to y i.e. (x-13)

2x+(x-13) = 133
i.e. 3x = 146
i.e. x = 48.66

i.e. x min = 49
y and z max = ((120-49)/2 = 35.5
i.e. y = 36 and x max = 35

Answer: option A
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Manager  S
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Joined: 27 Mar 2017
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Re: x, y, and z are positive integers such that x ≥ y ≥ z. If the average  [#permalink]

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Bunuel wrote:
x, y, and z are positive integers such that x ≥ y ≥ z. If the average (arithmetic mean) of x,y, and z is 40, and the median is (x–13), what is the greatest possible value of z?

A. 35
B. 36
C. 37
D. 39
E. 40

we have to maximize z;
to do that take y=z
Mean (x,y,z) = 40;
(x+y+z)/3=40
x+y+z = 120
Median= y= x-13 given;
2x-13+z=120
2x+z=133
as we took y=z=x-13
we get 3x=146 but by this x cant be integer
instead of taking z= x-13 if we take x-14 we will have 3x= 147 which gives us x=49,y=36 and z=35.
Hence Answer A.
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Re: x, y, and z are positive integers such that x ≥ y ≥ z. If the average  [#permalink]

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1
Bunuel wrote:
x, y, and z are positive integers such that x ≥ y ≥ z. If the average (arithmetic mean) of x,y, and z is 40, and the median is (x–13), what is the greatest possible value of z?

A. 35
B. 36
C. 37
D. 39
E. 40

We are given that the average of x, y, and z is 40; thus:

(x + y + z)/3 = 40

x + y + z = 120

We know that x ≥ y ≥ z and the median is x - 13. The median of three numbers is the second largest number, so y is the median. That is, y = x - 13. However, if we want to determine the greatest possible value of z (i.e., the smallest number), we want z to be as close to y (i.e., the median) as possible. Thus, we can let z be x - 13 also and we have:

x + x - 13 + x - 13 = 120

3x = 146

x = 146/3 = 48.66

Since x must be an integer, the smallest value of x is 49, y = x - 13 = 36, and thus the greatest value of z is 120 - (49 + 36) = 35.

Answer: A
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Manager  G
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Joined: 05 Feb 2018
Posts: 249
Re: x, y, and z are positive integers such that x ≥ y ≥ z. If the average  [#permalink]

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Testing answers works well here. Noting that x≥y≥z, I looked at E) first, I had a hunch this would be wrong but it helped clear up the min-max relationship:

E) If z is 40, the only possibility is 40≥40≥40, obviously this doesn't satisfy that median = x-13.

A) If z is 35, then y could be 36. This means we can 'give' 5 from z and 4 from y to x. So we could have 49≥36≥35
49 - 13 = 36, which matches our median. Thus, it's A.
Intern  Joined: 10 Feb 2019
Posts: 17
Re: x, y, and z are positive integers such that x ≥ y ≥ z. If the average  [#permalink]

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Nice question we have to maximize z;
to do that take y=z
Mean (x,y,z) = 40;
(x+y+z)/3=40
x+y+z = 120
Median= y= x-13 given;
2x-13+z=120

Posted from my mobile device
Manager  S
Joined: 19 Jan 2019
Posts: 77
Re: x, y, and z are positive integers such that x ≥ y ≥ z. If the average  [#permalink]

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We have to maximize z in such a way that it satisfies the equality X>y>z... Remember that z is going to have the least value among all the three. .
(X+y+z)/3 = 40
X+y+z = 120..
X + x-13 + z = 120
2x + z = 133...
2x = 133 - z ...
2x is even, which means z has to be odd.... Delete choices B and E...

Plug in remaining choices in place of z...

We'll get value of X = 49 when z = 35
X = 48 , when z = 37
X = 47 , when z = 39....
Among the above 3 .. only first case will satisfy the condition of X>y>z...

For
X=48, y = 35, z = 37.. not true
X = 47, y = 34, z= 39.. not true..

Therefore value of z is 35.. option A

Posted from my mobile device Re: x, y, and z are positive integers such that x ≥ y ≥ z. If the average   [#permalink] 16 Mar 2019, 23:06
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