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21 Feb 2012, 14:45
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
56% (02:00) correct
44%
(02:10)
wrong
based on 844
sessions
History
Date
Time
Result
Not Attempted Yet
If x and y are positive integers and \(\sqrt{x}^y = y\), then x-y could be equal to which of the following:
A. -2
B. -1
C. 1
D. 2
E. 4
MGMAT 1 (Q.9)
The explanation given is unclear to me. Can someone help!
A. -2
B. -1
C. 1
D. 2
E. 4
MGMAT 1 (Q.9)
The explanation given is unclear to me. Can someone help!
21 Feb 2012, 15:09
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docabuzar
Square \(\sqrt{x}^y = y\) --> \(x^y=y^2\). Now, use plug-in method: since both \(x\) and \(y\) are positive integers start from the lowest value:
If \(y=1\) then \(x^1=1^2\) --> \(x=1\) --> \(x-y=0\), 0 is not among answer choices (it's better to plug for y as it'll give the value of x right away);
If \(y=2\) then \(x^2=2^2\) --> \(x=2\) --> \(x-y=0\), 0 is not among answer choices;
If \(y=3\) then \(x^3=3^2\) --> \(x\neq{integer}\);
If \(y=4\) then \(x^4=4^2\) --> \(x=2\) --> \(x-y=-2\), we have answer choice A and since we cannot have two correct answers we can stop at this point.
Answer: A.
Hope it's clear.
General Discussion
08 Oct 2014, 04:19
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ok well square both the sides to get x^y = y^2.
Now to get option A as an answer, we can plug in x=2, and y=4.
we get 2^4= 4^2 (where the LHS = RHS) and therefore option A is correct.
Now to get option A as an answer, we can plug in x=2, and y=4.
we get 2^4= 4^2 (where the LHS = RHS) and therefore option A is correct.
