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Re: x,x2,xy,xy-1,x4,x6 [#permalink]
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Explanation

In 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,xy-1,x4,x6 [#permalink]
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How do you know that x > 1. And what does the subscript mean?
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Re: x,x2,xy,xy-1,x4,x6 [#permalink]
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Re: x,x2,xy,xy-1,x4,x6 [#permalink]
Can someone explain this question again?
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Re: x,x2,xy,xy-1,x4,x6 [#permalink]
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OE

In Quantity A, if y = 4, then the numbers (arranged in increasing order) become \(x,x^2,x^3,x^4,x^4,x^6\), and the mode is \(x^4\)

In Quantity B, if y = 5, then the numbers become \(x,x^2,x^3,x^4,x^4,x^5,x^6\)

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,xy-1,x4,x6 [#permalink]
The explanation is not still clear to me, I mean after plugging the value of y = 4 and 5 there is a list given, but I don't understand how this list is inferred. Can anyone share a more detailed answer.
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x,x2,xy,xy-1,x4,x6 [#permalink]
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sandy wrote:
\(x>1\)

\(x,x^2,xy,xy^{-1},x^4,x^6\)

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

I don't like this question...

Given set: \({x,x^2,xy,xy^{-1},x^4,x^6}\)

If \(y = 4\), then the set becomes: \({x,x^2,4x,\frac{1}{4}x,x^4,x^6}\)
So, if x = 2, then the set becomes \({2,4,8,\frac{1}{2},16,64}\), in which case the mode is \(2,4,8,\frac{1}{2},16,64\)

Similarly, if \(y = 5\), then the set becomes: \(x,x^2,5x,\frac{1}{5}x,x^4,x^6\)
So, if x = 2, then the set becomes \(2,4,20,\frac{2}{5},16,64\), in which case the median is \(10\)

So we have:
QUANTITY A: \(2,4,8,\frac{1}{2},16,64\)
QUANTITY B: \(10\)

There's no way to compare these two quantities.

Answer: Bad question
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Re: x,x2,xy,xy-1,x4,x6 [#permalink]
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Re: x,x2,xy,xy-1,x4,x6 [#permalink]
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