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The average (arithmetic mean) of the positive integers x, y, and z is 3. If x < y < z, what is the greatest possible val

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by BTGModeratorVI » Wed Dec 16, 2020 12:37 pm
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The average (arithmetic mean) of the positive integers x, y, and z is 3. If x < y < z, what is the greatest possible value of z ?

A. 5
B. 6
C. 7
D. 8
E. 9

Answer: B
Source: official guide
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Re: The average (arithmetic mean) of the positive integers x, y, and z is 3. If x < y < z, what is the greatest possible

by Brent@GMATPrepNow » Mon Dec 28, 2020 8:17 am
BTGModeratorVI wrote: ↑
Wed Dec 16, 2020 12:37 pm
 The average (arithmetic mean) of the positive integers x, y, and z is 3. If x < y < z, what is the greatest possible value of z ?

A. 5
B. 6
C. 7
D. 8
E. 9

Answer: B
Source: official guide
The average (arithmetic mean) of the positive integers x, y, and z is 3
We can write: (x+y+z)/3 = 3

Multiply both sides of the equation by [m]3[/m] to get: x + y + z = 9

At this point we need to find the greatest possible value of z
Important: Keep in mind that the three numbers are DIFFERENT POSITIVE INTEGERS

Since we know that x+y+z=9, we can MAXIMIZE the value of z by MINIMIZING the values of x and y
We know that x is the smallest value.
Since x must be a positive integer, the smallest possible value of x is 1
Since y must be different from x, the smallest possible value of y is 2
At this point we have maximized the value of z

If x=1 and y=2, then z=6

Answer: B

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Brent
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Re: The average (arithmetic mean) of the positive integers x, y, and z is 3. If x < y < z, what is the greatest possible

by swerve » Mon Dec 28, 2020 4:11 pm
BTGModeratorVI wrote: ↑
Wed Dec 16, 2020 12:37 pm
 The average (arithmetic mean) of the positive integers x, y, and z is 3. If x < y < z, what is the greatest possible value of z ?

A. 5
B. 6
C. 7
D. 8
E. 9

Answer: B
Source: official guide
The smallest positive integer is 1 and the 2nd smallest 2.

So, \(3\cdot 3 = 3+x \Longrightarrow x= 6\).

Therefore, B is the correct answer.
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