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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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13 Sep 2022, 06:43
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.
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
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]
13 Sep 2022, 16:29
13 Sep 2022, 16:29
2
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.
\(\frac{(8+x+y+10)}{(4)} = 12\)
\(x+y= 30\)
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.
Re: K = {8, x, y, 10} The mean of set K is 12, there is no mode, and x > y
[#permalink]
29 Jan 2026, 07:32
29 Jan 2026, 07:32
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