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x, y, and z are positive integers such that x ≥ y ≥ z. If the average
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18 May 2017, 00:28

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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?

Re: x, y, and z are positive integers such that x ≥ y ≥ z. If the average
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18 May 2017, 07:49

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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

Re: x, y, and z are positive integers such that x ≥ y ≥ z. If the average
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18 May 2017, 01:12

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.

x, y, and z are positive integers such that x ≥ y ≥ z. If the average
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18 May 2017, 07:43

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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Re: x, y, and z are positive integers such that x ≥ y ≥ z. If the average
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19 May 2017, 03:18

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 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
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22 May 2017, 18:34

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.

Re: x, y, and z are positive integers such that x ≥ y ≥ z. If the average
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16 Mar 2019, 12:16

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.

Re: x, y, and z are positive integers such that x ≥ y ≥ z. If the average
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16 Mar 2019, 23:06

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

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gmatclubot

Re: x, y, and z are positive integers such that x ≥ y ≥ z. If the average
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16 Mar 2019, 23:06