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An operation Ф is defined by the equation x Ф y = x²/4

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An operation Ф is defined by the equation x Ф y = x²/4 [#permalink]
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GMATPrepNow wrote:
An operation Ф is defined by the equation $$x Ф y = \frac{x^2}{4} - xy + y^2$$ for all numbers x and y. What is the value of 186 Ф 91?

A) 1
B) 4
C) 8
D) 16
E) 24

*kudos for all correct solutions

$$x$$ $$Ф$$ $$y = \frac{x^2}{4} - xy + y^2$$ $$=> \frac{(x^2 - 4xy + 4y^2)}{4}$$

$$x$$ $$Ф$$ $$y = \frac{(x - 2y)^2}{4}$$

$$186$$ $$Ф$$ $$91 = \frac{(186 - 2*91)^2}{4}$$

$$186$$ $$Ф$$ $$91 = \frac{(186 - 182)^2}{4}$$

$$186$$ $$Ф$$ $$91 = \frac{4^2}{4}$$

$$186$$ $$Ф$$ $$91 = 4$$

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An operation Ф is defined by the equation x Ф y = x²/4 [#permalink]
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Good question to be done mentally

xФy=(x/2−y)^2=(93−91)^2=4

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An operation Ф is defined by the equation x Ф y = x²/4 [#permalink]
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GMATPrepNow wrote:
An operation Ф is defined by the equation $$x Ф y = \frac{x^2}{4} - xy + y^2$$ for all numbers x and y. What is the value of 186 Ф 91?

A) 1
B) 4
C) 8
D) 16
E) 24

*kudos for all correct solutions

When we see that the expression x²/4 - xy + y² has two squares (x²/4 = (x/2)² and y² is obviously a square), we should start wondering whether the expression can be factored into the form (something)²

To make this factorization easier to see, let's factor out 1/4 first.
We get: x²/4 - xy + y² = (1/4)(x² - 4xy + 4y²)
= (1/4)(x - 2y)²

Perfect, we can now say that x Ф y = (1/4)(x - 2y)²
So, 186 Ф 91 = (1/4)[186 - 2(91)]²
= (1/4)[186 - 182]²
= (1/4)[4]²
= (1/4)[16]
= 4

Originally posted by BrentGMATPrepNow on 03 Aug 2017, 07:40.
Last edited by BrentGMATPrepNow on 04 Jul 2020, 06:49, edited 1 time in total.
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Re: An operation is defined by the equation x y = x²/4 [#permalink]
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Re: An operation is defined by the equation x y = x²/4 [#permalink]
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