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For all positive numbers x and y, the operation x ♦ y is defined by x

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Bunuel
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For all positive numbers x and y, the operation x ♦ y is defined by x [#permalink] New post   19 Mar 2020, 09:07
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For all positive numbers x and y, the operation \(x ♦ y\) is defined by \(x ♦ y = \frac{x^2 - x}{2xy}\). If \(x ♦ 5 = 2\), then \(x =\)

A. 1
B. 5
C. 11
D. 20
E. 21
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E
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hiranmay
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Re: For all positive numbers x and y, the operation x ♦ y is defined by x [#permalink] New post   19 Mar 2020, 09:46
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For all positive numbers x and y, the operation \(x ♦ y\) is defined by \(x ♦ y = \frac{x^2 - x}{2xy}\). If \(x ♦ 5 = 2\), then \(x =\)

A. 1
B. 5
C. 11
D. 20
E. 21 --> correct: \(x ♦ 5 = (x^2 -x)/2*x*5= 2\) => x^2 -x = 20x => x^2-21x = 0 => x(x-21)=0 => x >0, so x =21
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For all positive numbers x and y, the operation x ♦ y is defined by x [#permalink] New post   19 Mar 2020, 09:15
Bunuel wrote:
For all positive numbers x and y, the operation \(x ♦ y\) is defined by \(x ♦ y = \frac{x^2 - x}{2xy}\). If \(x ♦ 5 = 2\), then \(x =\)

A. 1
B. 5
C. 11
D. 20
E. 21


Asked; For all positive numbers x and y, the operation \(x ♦ y\) is defined by \(x ♦ y = \frac{x^2 - x}{2xy}\). If \(x ♦ 5 = 2\), then \(x =\)

m]x ♦ y = \frac{x^2 - x}{2xy}[/m]
\(x ♦ 5 = \frac{x^2 - x}{10x} = 2\)
\(x^2 - x = 20x\)
\(x^2 - 21x = 0\)
x(x-21) = 0
x = 21 ; Since x <>0

IMO E
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Re: For all positive numbers x and y, the operation x ♦ y is defined by x [#permalink] New post   20 Mar 2020, 13:39
21
x♦5=2
x^2-x/2.x.5=2
x^2-x=20x
x^2=21x
Hence x=21
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Re: For all positive numbers x and y, the operation x ♦ y is defined by x [#permalink] New post   21 Mar 2020, 15:56
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Bunuel wrote:
For all positive numbers x and y, the operation \(x ♦ y\) is defined by \(x ♦ y = \frac{x^2 - x}{2xy}\). If \(x ♦ 5 = 2\), then \(x =\)

A. 1
B. 5
C. 11
D. 20
E. 21


Since x * 5 = 2:

(x^2 - x) / [(2)(x)(5)] = 2

(x^2 - x) / (10x) = 2

x^2 - x = 20x

x^2 - 21x = 0

x(x - 21) = 0

x = 0 or x = 21

Since x is positive, x = 21.

Answer: E
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gmatclubot
Re: For all positive numbers x and y, the operation x ♦ y is defined by x [#permalink]
21 Mar 2020, 15:56
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