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If the operation ? is defined by x?y = yx^(1/2) for all positive

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exc4libur
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Re: If the operation ? is defined by x?y = yx^(1/2) for all positive [#permalink] New post  19 Aug 2019, 12:53
Quote:
If the operation ? is defined by x?y=y√x for all positive integers x and y, then 9?(16?25)=

A. 60
B. 180
C. 240
D. 300
E. 1200


16?25=25√16=25•4=100
9?100=100√9=300

Answer (D).
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Mohammadmo
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Re: If the operation ? is defined by x?y = yx^(1/2) for all positive [#permalink] New post  19 Aug 2019, 14:07
First we should indicate the term in the parenthesis.
X=16,Y=25
25×4=100
9?100 ==>100×3=300
Option D

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vaibhav1221
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Re: If the operation ? is defined by x?y = yx^(1/2) for all positive [#permalink] New post  19 Aug 2019, 21:44
\(x?y = y\sqrt{x}\)

\(? = \sqrt{x}/x\)

9?(16?25) =

Solve the bracket first.

\((\sqrt{16}*25)\) \(= 4*25 = 100\)

\(\sqrt{9}*100 = 3*100 = 300\)

Answer: 300
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lacktutor
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Re: If the operation ? is defined by x?y = yx^(1/2) for all positive [#permalink] New post  19 Aug 2019, 22:02
If the operation ? is defined by x?y=y√x for all positive integers x and y, then 9?(16?25)=

x?y=y√x
x?y=\(yx^{\frac{1}{2}}\)
?=1/2 - the square root of a positive number

--> 9?(16?25)=\(9^{\frac{1}{2}}(16^{\frac{1}{2}}*25)\)=3*4*25=300

The answer is D.
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Hazel1134
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Re: If the operation ? is defined by x?y = yx^(1/2) for all positive [#permalink] New post  19 Aug 2019, 22:28
Lets not forget about PEMDAS for this question! :O
Although i am pretty sure most of us here will not!

Now, the ones in the Para first -- 16?25 = 25*sqrt16 = 25 * 4 = 100.

Next, it becomes 9?100. Again, 100 * sqrt9 = 100 * 3 = 300.

Hence, D is and should be the answer.
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Re: If the operation ? is defined by x?y = yx^(1/2) for all positive [#permalink]
19 Aug 2019, 22:28
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