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K = {8, x, y, 10} The mean of set K is 12, there is no mode, and x > y

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K = {8, x, y, 10} The mean of set K is 12, there is no mode, and x > y [#permalink] New post   13 Sep 2022, 06:43
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K = {8, x, y, 10}
The mean of set K is 12, there is no mode, and x > y.

Quantity A
Quantity B
2x
30



A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

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K = {8, x, y, 10} The mean of set K is 12, there is no mode, and x > y [#permalink] New post   13 Sep 2022, 16:29
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Utilizing the information partially, we can deduce that:

\(\frac{(8+x+y+10)}{(4)} = 12\)
\(x+y= 30\)
manipulating the last euality:
\(y= 30 - x\)
Given that information, and knowing the fact that \(x>y\), replacing the ineuality with the last expression, we have \(x > 30 - x\), and doing some algebra, \(2x > 30\). Therefore, Option A.
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K = {8, x, y, 10} The mean of set K is 12, there is no mode, and x > y [#permalink]
13 Sep 2022, 16:29
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