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x,x2,xy,xy1,x4,x6 [#permalink]
22 Nov 2017, 19:00
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Question Stats:
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50% (00:48) wrong based on 4 sessions
\(x\),\(x_{2}\),\(x_{y}\),\(x_{y1}\),\(x_{4}\),\(x_{6}\)
Quantity A 
Quantity B 
The mode of the numbers above when y = 4 
The median of the numbers above when y =5 
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. Drill 2 Question: 1 Page: 524
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Re: x,x2,xy,xy1,x4,x6 [#permalink]
23 Nov 2017, 05:46
Can someone explain this question please?



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Re: x,x2,xy,xy1,x4,x6 [#permalink]
26 Nov 2017, 17:06
ExplanationIn Quantity A, if y = 4, then the numbers (arranged in increasing order) become \(x, x_{2}, x_{3}, x_{4}, x_{4}, x_{6}\); the mode is \(x_{2}\). In Quantity B, if y = 5, then the numbers become \(x, x_{2}, x_{4}, x_{4}, x_{5}, x_{6}\). Usually, you’d need to take the average of the middle two numbers to find the median because there is an even number of values, but in this case they’re both \(x_{4}\). The median, then, is \(x_{4}\), so the quantities are equal. Because x > 1, you don’t have to worry about special cases such as 0, 1, negatives, or fractions, and the correct answer is choice (C).
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Re: x,x2,xy,xy1,x4,x6 [#permalink]
10 Dec 2017, 17:17
ExplanationIn Quantity A, if y = 4, then the numbers (arranged in increasing order) become \(x, x_2, x_3, x_4, x_4, x_6\); the mode is \(x_4\). In Quantity B, if y = 5, then the numbers become \(x, x_2, x_3, x_4, x_4, x_6\). Usually, you’d need to take the average of the middle two numbers to find the median because there is an even number of values, but in this case they’re both \(x_4\). The median, then, is \(x_4\), so the quantities are equal. Because x > 1, you don’t have to worry about special cases such as 0, 1, negatives, or fractions, and the correct answer is choice C.
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Re: x,x2,xy,xy1,x4,x6 [#permalink]
14 Dec 2017, 12:26
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How do you know that x > 1. And what does the subscript mean?




Re: x,x2,xy,xy1,x4,x6
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14 Dec 2017, 12:26





