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The average (arithmetic mean) of the positive integers x, y,
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20 Aug 2020, 10:02
20 Aug 2020, 10:02
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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
A. 5
B. 6
C. 7
D. 8
E. 9
Re: The average (arithmetic mean) of the positive integers x, y,
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20 Aug 2020, 13:31
20 Aug 2020, 13:31
We are told x,y, and z are positive integers, and x<y<z
(x+y+z)/3 = 3
Therefore x+y+z=9
We want to find the greatest possible value of z.
The greatest possible value of z, is when z makes up the largest part possible of the sum.
That is, when x and y make up the smallest part of the sum.
When will x and y make up the smallest part of the sum?
When they attain their lowest possible values.
Since x and y are positive integers, the smallest possible value of x is 1.
Since y>x, the smallest possible value of y is 2.
We get
1+2+z=9
3+z=9
z=6
Final Answer: B
(x+y+z)/3 = 3
Therefore x+y+z=9
We want to find the greatest possible value of z.
The greatest possible value of z, is when z makes up the largest part possible of the sum.
That is, when x and y make up the smallest part of the sum.
When will x and y make up the smallest part of the sum?
When they attain their lowest possible values.
Since x and y are positive integers, the smallest possible value of x is 1.
Since y>x, the smallest possible value of y is 2.
We get
1+2+z=9
3+z=9
z=6
Final Answer: B
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Re: The average (arithmetic mean) of the positive integers x, y,
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04 Aug 2021, 06:40
04 Aug 2021, 06:40
1
Carcass wrote:
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
A. 5
B. 6
C. 7
D. 8
E. 9
The average (arithmetic mean) of the positive integers x, y, and z is 3
We can write: \(\frac{x+y+z}{3}=3\)
Multiply both sides of the equation by \(3\) 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
Cheers,
Brent
