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19 Jun 2018, 23:43
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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:
25%
(medium)
Question Stats:
81% (01:40) correct
19%
(01:52)
wrong
based on 104
sessions
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Time
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Not Attempted Yet
If x and y are positive integers, is x^2 + y^2 odd?
(1) xy is even.
(2) (x + 2)(y + 4) is even.
(1) xy is even.
(2) (x + 2)(y + 4) is even.
20 Jun 2018, 00:06
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1. \(xy\) is even.
Case 1: Both x and y are even. Here, \(x^2 + y^2\) is even.
Case 2: Either of x and y is odd, the other variable is even. Here,
\(x^2 + y^2\) is odd. We can't arrive at a unique solution. (Insufficient)
2. \((x + 2)(y + 4)\) is even.
This is pretty much the same thing as Statement 1. So, the same
cases apply here. Again, we can't arrive at a unique solution. (Insufficient)
When we combine the information in both the statements, we cannot
arrive at a unique solution for \(x^2 + y^2\), which can be both odd/even. (Insufficient - Option E)
Case 1: Both x and y are even. Here, \(x^2 + y^2\) is even.
Case 2: Either of x and y is odd, the other variable is even. Here,
\(x^2 + y^2\) is odd. We can't arrive at a unique solution. (Insufficient)
2. \((x + 2)(y + 4)\) is even.
This is pretty much the same thing as Statement 1. So, the same
cases apply here. Again, we can't arrive at a unique solution. (Insufficient)
When we combine the information in both the statements, we cannot
arrive at a unique solution for \(x^2 + y^2\), which can be both odd/even. (Insufficient - Option E)
20 Jun 2018, 00:37
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Bunuel
we are asked to find out whether x^2+y^2 is odd or not.
Statement 1: xy is even. Not sufficient.
both x any could be even or x is even , y is odd, or vice versa. Thus we are failed to determine the ultimate result of the x^2+y^2
Statement 2: The same as statement 1. there is a multiplication relation between 2 properties. we are sure that either x+2 or y+4 is even. so, x or y both can be even or either x or y could be even.
Thus nit sufficient.
(1+2):
Still impossible to determine both x or y is even or odd. If and only if both x and y are even , x^2+y^2 will be even.
Thus the best answer is E.
