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Updated on: 09 Mar 2012, 23:39
Originally posted by pradeepparihar on 09 Mar 2012, 21:22.
Last edited by Bunuel on 09 Mar 2012, 23:39, edited 2 times in total.
Last edited by Bunuel on 09 Mar 2012, 23:39, edited 2 times in total.
Edited the question
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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79% (01:45) correct
21%
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If x and y are positive, what is the value of x?
(1) 200% of x equals to 400% of y.
(2) xy is the square of a positive integer.
(1) 200% of x equals to 400% of y.
(2) xy is the square of a positive integer.
09 Mar 2012, 23:38
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If x and y are positive, what is the value of x?
Notice that we are not told that \(x\) and \(y\) are integers only, we are just told that they are both positive.
(1) 200% of x equals to 400% of y --> \(\frac{200}{100}*x=\frac{400}{100}*y\) --> \(x=2y\). Not sufficient to get the single numerical value of \(x\).
(2) xy is the square of a positive integer --> \(xy=n^2\), for some positive integer \(n\). Not sufficient to get the single numerical value of \(x\).
(1)+(2) Since from (1) \(x=2y\) then from (2) \(x*\frac{x}{2}=n^2\) --> \(x^2=2n^2\) --> the value of \(x\) (x^2) is determined by the value of integer \(n\), so we still cannot get the single numerical value of \(x\). For example: if \(n=1\) then \(x=\sqrt{2}\) but if \(n=2\) then \(x=2\sqrt{2}\). Not sufficient.
Answer: E.
Hope it's clear.
Since we are not told that \(x\) and \(y\) are integers only, then from \(x=2y\) we cannot say whether \(x\) even:
\(x\) will be even if \(y=integer\);
\(x\) will be odd if \(y=\frac{odd}{2}\), for example if \(y=\frac{3}{2}\) then \(x=3=odd\);
\(x\) may not be an integer at all, for example if \(y=\frac{1}{3}\) then \(x=\frac{2}{3}\neq{integer}\).
In fact if we knew that \(x=even\) then \(x^2=2n^2\), from (1)+(2), would have only one integer solution \(x=0\) and \(n=0\), but this would contradict the given fact that \(x\) is positive.
Hope it's clear.
Notice that we are not told that \(x\) and \(y\) are integers only, we are just told that they are both positive.
(1) 200% of x equals to 400% of y --> \(\frac{200}{100}*x=\frac{400}{100}*y\) --> \(x=2y\). Not sufficient to get the single numerical value of \(x\).
(2) xy is the square of a positive integer --> \(xy=n^2\), for some positive integer \(n\). Not sufficient to get the single numerical value of \(x\).
(1)+(2) Since from (1) \(x=2y\) then from (2) \(x*\frac{x}{2}=n^2\) --> \(x^2=2n^2\) --> the value of \(x\) (x^2) is determined by the value of integer \(n\), so we still cannot get the single numerical value of \(x\). For example: if \(n=1\) then \(x=\sqrt{2}\) but if \(n=2\) then \(x=2\sqrt{2}\). Not sufficient.
Answer: E.
Hope it's clear.
subhashghosh
Since we are not told that \(x\) and \(y\) are integers only, then from \(x=2y\) we cannot say whether \(x\) even:
\(x\) will be even if \(y=integer\);
\(x\) will be odd if \(y=\frac{odd}{2}\), for example if \(y=\frac{3}{2}\) then \(x=3=odd\);
\(x\) may not be an integer at all, for example if \(y=\frac{1}{3}\) then \(x=\frac{2}{3}\neq{integer}\).
In fact if we knew that \(x=even\) then \(x^2=2n^2\), from (1)+(2), would have only one integer solution \(x=0\) and \(n=0\), but this would contradict the given fact that \(x\) is positive.
Hope it's clear.
General Discussion
09 Mar 2012, 23:17
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(1)
200/100 * x = 400/100 * y
2x = 4y
X = 2y
=> X is even
(2)
Xy = 1,4,9,16,…
Not Sufficient.
(1) + (2)
Suppose:
2y^2 = 16
Y = 2root(2)
X = 4root(2)
Or
2y^2 = 4
Y = root(2)
X = 2root(2)
Not Sufficient.
So Answer is E.
200/100 * x = 400/100 * y
2x = 4y
X = 2y
=> X is even
(2)
Xy = 1,4,9,16,…
Not Sufficient.
(1) + (2)
Suppose:
2y^2 = 16
Y = 2root(2)
X = 4root(2)
Or
2y^2 = 4
Y = root(2)
X = 2root(2)
Not Sufficient.
So Answer is E.
