x, y, and z are positive integers such that x ≥ y ≥ z. If the average

Question

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

BrentGMATPrepNow · Accepted Answer

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: A

obliraj · Answer

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.

GMATinsight · Answer

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

guptarahul · Answer

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.

ScottTargetTestPrep · Answer

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

energetics · Answer

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.

Aswathraj · Answer

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

thyagi · Answer

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

KarishmaB · Answer

Consider this: For 3 numbers, z, y and x, the median will be the middle number. 
Hence y = x - 13. So maximum value of z is also x - 13 since z <= y.

Assuming this,  (x - 13) + (x - 13) + x = 40 * 3 = 120 
 x = \frac{146}{3} 
But this is not an integer. 146 is 1 less than 147, a multiple of 3. 
Hence, if we reduce z by 1 to make it (x - 14), we get
 (x - 14) + (x - 13) + x = 120

x = \frac{147}{3} = 49  (an integer)

So z must be  x - 14 = 49 - 14 = 35

Answer (A)

Purnank · Answer

Best way is going through option elimination.
Now we know y = x-13
so write the x+x-13+z = 120, 2x+z = 133
Now check with options
se here 2x is multiple of 2 means z has to be odd number here that makes our life bit easy eliminate option B and E.
now lets check for max value from remaining options.
lets pick D.
z=39 gives us x = 47 now y = x-13 = 34 this is violating y ≥ z So Eliminate D.
even z=37 wont work if you want you can check. 
So A has to be our answer.
lets check z=35 gives us x = 49 and y = x-13 = 36 which is y ≥ z following the given constraints.
  So our answer is A

NuttyIan · Answer

Oddly enough I made a mistake here but still got the correct answer. I didn't realize that z could be equal to y, I did not see the equal to symbol. Anyway I worked out the formula to be 3(40)= x + (x-13) + (x-14), or 120= 3x -27. The actual problem should be 120= 3x - 26 like the others worked out. Oddly enough easier to solve as 146/3 is more annoying than 147/3.

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