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If x ≠ 0 and the average (arithmetic mean) of set A is twice the avera
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13 Jan 2022, 09:08
13 Jan 2022, 09:08
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\(A = \{x, 2x, 3x, 4x\}\)
\(B = \{x^2, 2x^2, 3x^2, 4x^2, 5x^2\}\)
If x ≠ 0 and the average (arithmetic mean) of set A is twice the average of set B, then x =
A. 1/3
B. 5/12
C. 2/3
D. 5/6
E. 5/3
\(B = \{x^2, 2x^2, 3x^2, 4x^2, 5x^2\}\)
If x ≠ 0 and the average (arithmetic mean) of set A is twice the average of set B, then x =
A. 1/3
B. 5/12
C. 2/3
D. 5/6
E. 5/3
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Re: If x ≠ 0 and the average (arithmetic mean) of set A is twice the avera
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13 Jan 2022, 10:58
13 Jan 2022, 10:58
1
Carcass wrote:
\(A = \{x, 2x, 3x, 4x\}\)
\(B = \{x^2, 2x^2, 3x^2, 4x^2, 5x^2\}\)
If x ≠ 0 and the average (arithmetic mean) of set A is twice the average of set B, then x =
A. 1/3
B. 5/12
C. 2/3
D. 5/6
E. 5/3
\(B = \{x^2, 2x^2, 3x^2, 4x^2, 5x^2\}\)
If x ≠ 0 and the average (arithmetic mean) of set A is twice the average of set B, then x =
A. 1/3
B. 5/12
C. 2/3
D. 5/6
E. 5/3
There's a nice rule that says, "In a set where the numbers are equally spaced, the mean will equal the median."
For example, in each of the following sets, the mean and median are equal:
{7, 9, 11, 13, 15}
{-1, 4, 9, 14}
{3, 4, 5, 6}
This means the average of set A is 2.5x
And the average of set B is 3x²
If the average (arithmetic mean) of set A is twice the average of set B, we can write: 2.5x = 2(3x²)
Simplify: 2.5x = 6x²
To make things easier on ourselves, let's multiply both sides of the equation by 2 to get: 5x = 12x²
Subtract 5x from both sides: 0 = 12x² - 5x
Factor: 0 = x(12x - 5)
So, either x = 0 or 12x - 5 = 0
Since we are told x ≠ 0, it must be the case that 12x - 5 = 0
Add 5 to both sides: 12x = 5
Divide both sides by 12 to get: x = 5/12
Answer: B
gmatclubot
Re: If x ≠ 0 and the average (arithmetic mean) of set A is twice the avera [#permalink]
13 Jan 2022, 10:58
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