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If x and y are positive integers, is x + y even?
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19 Nov 2014, 08:15
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Tough and Tricky questions: Number Properties. If x and y are positive integers, is x + y even? (1) xy is even (2) x/y is even Kudos for a correct solution.
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Re: If x and y are positive integers, is x + y even?
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19 Nov 2014, 09:25
statement 1: not sufficient If xy is even that means that at least one of them is even. This is not sufficient because if x=2 and y=3 then xy=6 and x+y=5, odd. But if x=4 and y = 4 then xy=16 and x+y=8, even.
statement 2: not sufficient X=14 and y=7, x/y=2, even. x+y=21, odd. x=8 and y=4, x/y=2, even. x+y=12, even.
If we combine statements, we still do not have enough information. The most we know is that one of them is even but we need to know for both x and y.
Answer E!



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Re: If x and y are positive integers, is x + y even?
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19 Nov 2014, 09:37
Take x=2 y=1; x+y=3=odd; statement 1xy=even correct Take x=2 y=4; x+y=6=even; statement 1xy= even correct hence statement 1 not sufficient Take x=2 y=1; x+y=3=odd;statement 1xy=even correct Take x=4 y=2; x+y=6=even; statement 1xy= even correct Hence statement 2 not sufficient Take x=2 and y=1 ;x+y=3=odd and take x=4 y=2; x+y=6=evenn Not sufficient hence E.



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Re: If x and y are positive integers, is x + y even?
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19 Nov 2014, 12:58
For x + y to be even, either both x and y has to be even or both has to be odd. stmt1 : xy is even This means that either both x and y are even OR one is even and other is odd. why so?? because even * even = even AND even * odd = even. Since we do not have suff information thus INSUFFICIENT. stmt2: x/y is even x/y = even> x = even * y which means that x is even. But we do not know is y is even or not thus INSUFFICIENT. combining both statements also do not give sufficient information about x and y. E.
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Re: If x and y are positive integers, is x + y even?
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19 Nov 2014, 20:32
Statement1: Even x Even = Even => Even + Even = Even OR Even x Odd = Even => Even + Odd = Odd Hence Insufficient.
Statement2: Even / Even = Even => Even + Even = Even OR Even / Odd = Even => Even + Odd = Odd Hence Insufficient.
Statement 1 and 2 combined also will give similar combination. So can't be determined. Hence answer is E.



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Re: If x and y are positive integers, is x + y even?
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19 Nov 2014, 21:14
Question : If x and y are positive integers, is x + y even?
Statement 1: xy is even If xy = 8, Valid combination of x and y are , x= 4 and y =2 . x+y = 6 (true) Another combination of x and y are , x= 8 and y =1 . x+y = 9 (false)
Insufficient
(2) x/y is even If x/y = 8, Valid combination of x and y are , x= 16 and y =2 . x+y = 18 (true) Valid combination of x and y are , x= 8 and y =1 . x + y = 9 (false)
Insufficient
Combining both statements as well, we will not be able to solve the problem. Hence E) is the correct answer.



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Re: If x and y are positive integers, is x + y even?
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19 Nov 2014, 21:21
x,y > 0; x+y = even if x,y = even or x,y = odd.
1) xy = even  x or y or x & y  even. So x+y = odd or even. Not Sufficient 2) x/y = even  x  even and y could be even or odd. So x+y = odd or even. Not Sufficient 1+2  xy = even & x/y = even  we can eliminate x and y both odd. So x = even for sure. Y = even or odd.  x+y = Even or Odd. So 1+2 not sufficient.
Answer. E. OA?



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Re: If x and y are positive integers, is x + y even?
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19 Nov 2014, 21:38
statement 1: If xy is even that means that at least one of them is even. This is not sufficient because if x=4 and y=5 then xy=20 and x+y=9, odd. But if x=2 and y = 6 then xy=12 and x+y=8, even. Hence Not Sufficient. Option A & D are out statement 2: X=12 and y=3, x/y=4, even. x+y=15, odd. x=8 and y=4, x/y=2, even. x+y=12, even. Hence Not Sufficient. Option B is out If we combine statements, we still do not have enough information. The most we know is that one of them is even but we need to know for both x and y. Answer E
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Re: If x and y are positive integers, is x + y even?
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20 Nov 2014, 03:09
from statement 1: xy is even as we know xy can be even in two cases 1) both x & y are even = x+y=EVEN 2) x is odd and y is even or vice versa. =x+y=ODD SO its conflicting and no definite result hence INSUFFICIENT.
From statement 2: x/y is even again x/y is even in following two scenarios 1) both x & y are even =x/y cane both odd and even 2) x is even and y is odd=x/y is even =x+y=ODD This statement also contradicts and INSUFFICIENT.
Even combination of both statements is also not sufficient.
Answer is E



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Re: If x and y are positive integers, is x + y even?
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20 Nov 2014, 08:56
Bunuel wrote: Tough and Tricky questions: Number Properties. If x and y are positive integers, is x + y even? (1) xy is even (2) x/y is even Kudos for a correct solution. Official Solution:If x and y are positive integers, is x + y even? For the sum of \(x\) and \(y\) to be even, the two variables must both be even or both be odd. Statement (1): INSUFFICIENT. We know that at least one of the variables is even, but we do not know whether they are both even. Statement (2): INSUFFICIENT. We know that \(x = y \times \text{(some even integer)}\). Since an even integer multiplied by any other integer is also even, we know that \(x\) is even. However, we do not know whether \(y\) is even. Statements (1) & (2): INSUFFICIENT. Using both statements, we only know that \(x\) is even. Meanwhile, y could be even, but it does not have to be. As a result, we cannot determine whether the sum of \(x\) and \(y\) is even or odd. All of these results can be confirmed by picking numbers. Answer: E.
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Re: If x and y are positive integers, is x + y even?
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19 Nov 2016, 08:19
There are a lot of ways to this one. Given info => x,y are integers (This is a very important piece of info and should be always checked ) we need the even/odd nature of x+y Statement 1=> xy=even => (x,y)=> even,even => sum = even even,odd => sum odd odd,even => sum = odd statement 2 x/y=even hence x=y*even so x must be even but y can be even or odd hence not sufficient combining them two cases are possible (x,y)=> even,even => sum=even even,odd => sum =odd hence not sufficient Hence E
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Re: If x and y are positive integers, is x + y even?
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Re: If x and y are positive integers, is x + y even?
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