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The average (arithmetic mean) of x and z is greater than y,
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15 Nov 2019, 16:39
15 Nov 2019, 16:39
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The average (arithmetic mean) of x and z is greater than y, and \(x < y < z\).
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Kudos for the right answer and solution.
Quantity A |
Quantity B |
The average of x, y, and z |
The median of x, y, and z |
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Kudos for the right answer and solution.
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The average (arithmetic mean) of x and z is greater than y,
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16 Nov 2019, 07:11
16 Nov 2019, 07:11
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Carcass wrote:
The average (arithmetic mean) of x and z is greater than y, and \(x < y < z\).
Quantity A |
Quantity B |
The average of x, y, and z |
The median of x, y, and z |
Since \(x < y < z\), we can see that the median of the set must be y. So Quantity B = y
We have:
Quantity A: \(\frac{x + y + z}{3}\)
Quantity B: \(y\)
Multiply both quantities by 3 to get:
Quantity A: \(x + y + z\)
Quantity B: \(3y\)
Subtract \(y\) from both quantities to get
Quantity A: \(x + z\)
Quantity B: \(2y\)
GIVEN: The average (arithmetic mean) of x and z is greater than y
We can write: \(\frac{x + y}{2} > y\)
Multiply both sides of the inequality by 2 to get: \(x + y > 2y\)
Perfect! This last bit of information tells us that Quantity A must be greater
Answer: A
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Re: The average (arithmetic mean) of x and z is greater than y,
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14 Oct 2025, 01:19
14 Oct 2025, 01:19
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