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When x, y are positive integers, x@y=xy/(x+y). If a, b, and

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by Max@Math Revolution » Thu Jun 09, 2016 9:44 pm
When x, y are positive integers, x@y=xy/(x+y). If a, b, and c are positive numbers, what is the value of 1/a@(1/b@1/c)?
A. a+b+c
B. 1/abc
C. 1/(a+b+c)
D. 1/(ab+bc+ca)
E. 3/abc

*An answer will be posted in 2 days.
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by 800_or_bust » Fri Jun 10, 2016 6:07 am
Max@Math Revolution wrote:When x, y are positive integers, x@y=xy/(x+y). If a, b, and c are positive numbers, what is the value of 1/a@(1/b@1/c)?
A. a+b+c
B. 1/abc
C. 1/(a+b+c)
D. 1/(ab+bc+ca)
E. 3/abc

*An answer will be posted in 2 days.
First, perform the operation set aside in parentheses (following standard order of operation rules). So 1/b @ 1/c = ((1/b)(1/c))/((1/b)+(1/c)) = (1/bc)/((c/bc)+(b/bc)) = (1/bc) x (bc/b+c) = 1/(b+c)

Now, do the remaining operation using this expression for (1/b @ 1/c):

1/a @ 1/(b+c) = (1/a)(1/b+c)/((1/a) + (1/b+c)) = 1/(ba+bc) / (((b+c)/(ba+bc)) + a/(ba+bc))= 1/(a+b+c)
800 or bust!
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by Max@Math Revolution » Mon Jun 13, 2016 12:06 am
Since (1/b@1/c)=1/(b+c), we get 1/a@(1/b@1/c)=1/(a+b+c). Hence, the correct answer is C.
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