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Official answer is that both these statements are independently sufficient (D), with the following explanations :- (1)- since x and y are consecutive numbers, so one of these would be even and thus xy is also even. (2) if the fraction is even, then it means x is even and hence xy is also even..

Here is my doubt...

I chose option (B), meaning (2) alone is sufficient but (1) is not. Explanation: x and y are integers which means x could be 0 as well, in that case y will be 1, or x could be -1 and then y would be 0. in both these cases the product xy will not be even.

Can someone please help me in clarifying this doubt!

In order the product of two integers to be even either (or both) of them must be even. So, the question basically asks whether either x or y is even.

(1) x = y + 1. If x is odd then y is even and vise-versa. Sufficient. (2) x/y is an even integer --> \(\frac{x}{y}=even\) --> \(x=y*even=even\). Sufficient.

Answer: D.

As for your doubt: if either x or y is zero, then xy=0=even, because zero is an even integer. Zero is nether positive nor negative, but zero is definitely an even number.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even (in fact zero is divisible by every integer except zero itself).

In order the product of two integers to be even either (or both) of them must be even. So, the question basically asks whether either x or y is even.

(1) x = y + 1. If x is odd then y is even and vise-versa. Sufficient. (2) x/y is an even integer --> \(\frac{x}{y}=even\) --> \(x=y*even=even\). Sufficient.

Answer: D.

As for your doubt: if either x or y is zero, then xy=0=even, because zero is an even integer. Zero is nether positive nor negative, but zero is definitely an even number.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even (in fact zero is divisible by every integer except zero itself).

Hope it helps.

For (1) if y=-1 and x=0, then xy=0=even.

Zero is an even integer.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even (in fact zero is divisible by every integer except zero itself).

You have to be careful with your 'generalizations' and Number Properties.

hersheykitts wrote:

Am I missing something? Part 2 says x/y is even. Odd / Odd is even. Even / Even is even. Even / Odd is also even (24/3=8). How can we be sure what x & y are?

First off, ODD/ODD is NOT an even.... it's either ODD or it's a non-integer (which means it's neither even nor odd)

Here are some examples:

3/3 = 1 9/3 = 3 7/5 = 1.4

In that same way, EVEN/EVEN is usually even or a non-integer....but COULD be odd (if the two evens are the SAME NUMBER)....

2/2 = 1 4/2 = 2 6/4 = 1.5

EVEN/ODD is either even or a non-integer....

2/1 = 2 12/3 = 4 4/3 = 1.33333

To answer your question, the prompt tells us that X and Y are integers and Fact 2 tells us that X/Y is an EVEN INTEGER. This means that AT LEAST one of the two variables is even....

4/1 = 4 6/3 = 2 4/2 = 2 Etc.

The question asks if XY is even. Since one or both of the variables will be even in this situation, the answer to the question is ALWAYS YES. Fact 2 is SUFFICIENT.

Re: If x and Y are integers, is xy even? [#permalink]

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19 Apr 2013, 11:35

Hi bunuel... what if y=-1 and x=0 in case 1 ?

Bunuel wrote:

If x and Y are integers, is xy even?

In order the product of two integers to be even either (or both) of them must be even. So, the question basically asks whether either x or y is even.

(1) x = y + 1. If x is odd then y is even and vise-versa. Sufficient. (2) x/y is an even integer --> \(\frac{x}{y}=even\) --> \(x=y*even=even\). Sufficient.

Answer: D.

As for your doubt: if either x or y is zero, then xy=0=even, because zero is an even integer. Zero is nether positive nor negative, but zero is definitely an even number.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even (in fact zero is divisible by every integer except zero itself).

Re: If x and y are integers, is xy even? [#permalink]

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19 Aug 2014, 03:13

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Re: If x and y are integers, is xy even? [#permalink]

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27 Feb 2015, 16:30

Am I missing something? Part 2 says x/y is even. Odd / Odd is even. Even / Even is even. Even / Odd is also even (24/3=8). How can we be sure what x & y are?

Statement 1: x=y+1 i.e. if y is odd then x is even OR if y is Even then x is odd but in each case xy will be even as one of them is even and other is odd. hence SUFFICIENT

Statement 2: x/y is even i.e. x must be an even Integers as both are Integers that is already given and also y is a factor of x SUFFICIENT

Answer: Option D
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Re: If x and y are integers, is xy even? [#permalink]

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03 Aug 2016, 10:25

Top Contributor

inderjeetdhillon wrote:

If x and y are integers, is xy even?

(1) x = y + 1. (2) x/y is an even integer.

Target question: Is xy even?

Aside: For xy to be even, we need x to be even, or y to be even (or both even).

Statement 1: x = y+1 This tells us that x is 1 greater than y. This means that x and y are consecutive integers. If x and y are consecutive integers, then one must be odd and the other must be even. As such, the product xy must be even. So, statement 1 is SUFFICIENT

Statement 2: x/y is an even integer. If x/y is an even integer, then we can write x/y = 2k (where k is an integer) Now take the equation and multiply both sides by y to get: x = 2ky If k and y are both integers, we can see that 2ky (also known as x) must be even. If x is even, then the product xy must be even. So, statement 2 is SUFFICIENT

Re: If x and y are integers, is xy even? [#permalink]

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22 Aug 2016, 02:25

Here we need to check whether xy is even or not Statement 1 => x=y+1 => x-y=1 so x and y must be consecutive Hence the product must be even as one out of them must be even. Statement 2 => x/y=even => x=even => sufficient Smash that D
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If x and y are integers, is xy even? (1) x=y + 1 (2) x/y is an even integer.

hi experts,

I picked up B, because I think that state 1 is not sufficient and that state 2 is sufficient I did not get the idea of OE for state 1, following reasoning: for example, X=1, Y=0, then XY = 0, which is not even for example ,X=2, Y=1, then XY = 2, which is even.

If x and y are integers, is xy even? (1) x=y + 1 (2) x/y is an even integer.

hi experts,

I picked up B, because I think that state 1 is not sufficient and that state 2 is sufficient I did not get the idea of OE for state 1, following reasoning: for example, X=1, Y=0, then XY = 0, which is not even for example ,X=2, Y=1, then XY = 2, which is even.

please point out my fault.

thank a lot have a nice day >_~

Merging topics. Please refer to the discussion above.

As for your doubt: 0 is even integer.
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