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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
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GPA: 3.39
The operation # is defined for all nonzero x and y by x # y
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13 Feb 2023, 09:21
13 Feb 2023, 09:21
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Question Stats:
66% (01:24) correct
33% (00:57) wrong based on 3 sessions
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The operation # is defined for all nonzero x and y by \(x\#y = x + \frac{x}{y}\). If \(a>0\), then \(1\#(1\#a) =\)
A. \(a\)
B. \(a+1\)
C. \(\frac{a}{a+1}\)
D. \(\frac{a+2}{a+1}\)
E. \(\frac{2a+1}{a+1}\)
A. \(a\)
B. \(a+1\)
C. \(\frac{a}{a+1}\)
D. \(\frac{a+2}{a+1}\)
E. \(\frac{2a+1}{a+1}\)
Re: The operation # is defined for all nonzero x and y by x # y
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14 Feb 2023, 07:45
14 Feb 2023, 07:45
1
Expert Reply
x#y = x + x/y. If a>0, then 1#(1#a) =
You can see that a stands for y and 1 stands for x. I like to work with numbers so I did x=1 a=2 and a=y so y=2
1#a= 1+(1/2)=1.5 so now a is 1.5 --> 1+(1/1.5)=1 2/3 = 5/3
So now you can check by filling in a, which is 2:
A. a --> 2
B. a+1 --> 2+1=3
C. a/(a+1)-->2/(2+1)=2/3
D. (a+2)/(a+1)-->(2+2)/(2+1)=4/3
E. (2a+1)/(a+1)-->(2*2+1)/(2+1)=5/3
So E is the answer.
You can see that a stands for y and 1 stands for x. I like to work with numbers so I did x=1 a=2 and a=y so y=2
1#a= 1+(1/2)=1.5 so now a is 1.5 --> 1+(1/1.5)=1 2/3 = 5/3
So now you can check by filling in a, which is 2:
A. a --> 2
B. a+1 --> 2+1=3
C. a/(a+1)-->2/(2+1)=2/3
D. (a+2)/(a+1)-->(2+2)/(2+1)=4/3
E. (2a+1)/(a+1)-->(2*2+1)/(2+1)=5/3
So E is the answer.
gmatclubot
Re: The operation # is defined for all nonzero x and y by x # y [#permalink]
14 Feb 2023, 07:45
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