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This question makes you realize that you have to have an "even integer". If the question were merely, "is this even?" you would not need the statements to answer it. If you have \(2 * x * 5 * y\), then you have 10xy. Generally, you need to know if you have odd * odd or odd * even, or even * even in order to know if the product is even, but when you have 10 * any number, the result will always be even because 2 of the 3 scenarios results in an even number. {1) odd*odd=odd 2) even*odd = even 3) even*even=even} If you have even just 1 even number, the product will always be even. So, since we know that there is a 2 involved, the product will always be even.

But the key is "integer", so we don't know if x = 1.15 or any other decimal which may or may not give an even integer.

Statement 1) Insuffucient. 2 + 2.5 + 5 + 2.5 = even integer of 12. But if you take 2 * 2.5 * 5 * 2.5, that gives you 62.5, not an integer at all, so it could not possibly be an even integer, but the sum is even, therefore, #1 insufficient.

2) Insufficient because if x = 3.5 and y = 0.5, then the different is 3, an odd integer, but if you take the product of 2 * 3.5 * .5 * 5, you don't get an integer at all.

Together) Insufficient. If, from statement 1, you add 2 + 5 =7, then, in order to make Statement 1 true, that the sum is an even integer, the sum of x + y must be odd, so you have odd + odd = even.

Any numbers that will sum an odd integer will never have a difference of an odd integer. 4+2=6, but 4-2=2. Take this principle to the bank. So, these statements cannot both be true at the same time.

I don't think this is a good question, because GMAT will not write questions that have contradictory statements. While I believe the answer to be E, I think the question is flawed.

prinits wrote:

Is product 2*x*5*y an even integer?

1.2 + x + 5 + y is an even integer 2.x - y is an odd integer

Please explain.I could not follow the explanation given in GMAT Club test.

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

1) 2 + x + 5 + y is an even integer 2) x - y is an odd integer

Please explain.I could not follow the explanation given in GMAT Club test.

Is product 2*x*5*y (or 10xy) an even integer? Agree with Allen.

1) 2+x+5+y is an even integer. 2+x+5+y=7+ (x+y), where x and y could be integers or fractions but (x+y) must be odd. NSF. 2) x - y is an odd integer = > here also x and y could be integers or fractions. NSF.

From 1 and 2, it is not possible to have (x+y) and (x-y) must be odd. So the question is not properly designed.

Why is not possible to have x and y which satisfy this?

what if x is any odd number and y is 0? or x=2 and y= 1

This is why it is not possible to have X and Y satisfy this.

For this to be the type of question that the GMAT would actually use, it must be possible for the statements to both be true with a single set of numbers. There exists sets of numbers for which both statements will not be true, but for us to answer the question either "yes" or "no" requires that we be able to say "ALWAYS YES" or "ALWAYS NO". We cannot do that here.

In your example of x =2 and 1=1:

Statement 1: 2 + 2 + 5 + 1 = 10, yes that's an even number.

Statement 2:

2 - 1 = 1, and yes this is an odd number.

The test in these types of questions is not can we find 1 set that WILL work, but can we find a set that WILL NOT. If we know there is a set that will not work, then we cannot answer the question, becasue there is an implied "always" to the questions. Is product 2 * x * 5 * y (always) an even integer?

what if x is any odd number and y is 0? or x=2 and y= 1

hemantsood wrote:

GMAT TIGER wrote:

prinits wrote:

Is product 2*x*5*y an even integer?

1) 2 + x + 5 + y is an even integer 2) x - y is an odd integer

Please explain.I could not follow the explanation given in GMAT Club test.

Is product 2*x*5*y (or 10xy) an even integer? Agree with Allen.

1) 2+x+5+y is an even integer. 2+x+5+y=7+ (x+y), where x and y could be integers or fractions but (x+y) must be odd. NSF. 2) x - y is an odd integer = > here also x and y could be integers or fractions. NSF.

From 1 and 2, it is not possible to have (x+y) and (x-y) must be odd. So the question is not properly designed.

Why is not possible to have x and y which satisfy this?

what if x is any odd number and y is 0? or x=2 and y= 1

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

1) 2 + x + 5 + y is an even integer 2) x - y is an odd integer

Can someone post OA for this Q?

I think the above reasoning is not correct and the answer is C.

Question: \(2*x*5*y=even\). As there is 2 as a multiple, then this expression will be even if \(5xy=integer\). Basically we are asked is \(5xy=integer\) true?

Note that \(x\) and \(y\) may not be integers for \(2*x*5*y\) to be even (example \(x=\frac{7}{9}\) and \(y=\frac{9}{7}\)) BUT if they are integers then \(2*x*5*y\) is even.

(1) \(2+x+5+y=even\) --> \(7+x+y=even\) --> \(x+y=odd\). Not sufficient. (x=1 and y=2 answer YES BUT x=1.3 and y=1.7 answer NO)

(2) \(x-y=odd\). Not sufficient. (x=1 and y=2 answer YES BUT x=1.3 and y=0.3 answer NO)

(1)+(2) Sum (1) and (2) \((x+y)+(x-y)=odd_1+odd_2\) --> \(2x=even\) --> \(x=integer\) --> \(y=integer\) --> Both \(x\) and \(y\) are integers. Hence sufficient.

I wud also go for C.... SI gives x+y to be an odd no..... SII gives x-y to be an odd no..... both these conditions are satisfied only when both x anf y are integer themselves.... if u add two fractions with denominator 2, u get only one of the two(sum or difference) as odd and other as even .... any other denominator generally dont give both sum and difference as integers... so both Statements together satisfy x and y to be integers....hence sufficient..

If I understand correctly, x+y=odd and x-y=odd are only both always true if x and y are integers. If fractions are involved, the statments contradict because 1/3+2/3=1=odd but 1/3-2/3=-1/3=neither odd nor even. However, I am confused by jallen's statement:

"Any numbers that will sum an odd integer will never have a difference of an odd integer. 4+2=6, but 4-2=2"

4+3 sum odd (7) and have an odd difference (1), right? And why are positive #s used in the example? I assume jallen knows what he is talking about and that I am missing something. _________________

If I understand correctly, x+y=odd and x-y=odd are only both always true if x and y are integers. If fractions are involved, the statments contradict because 1/3+2/3=1=odd but 1/3-2/3=-1/3=neither odd nor even. However, I am confused by jallen's statement:

"Any numbers that will sum an odd integer will never have a difference of an odd integer. 4+2=6, but 4-2=2"

4+3 sum odd (7) and have an odd difference (1), right? And why are positive #s used in the example? I assume jallen knows what he is talking about and that I am missing something.

jallenmorris's solution is not right:

jallenmorris wrote:

This question makes you realize that you have to have an "even integer". If the question were merely, "is this even?" you would not need the statements to answer it.

Only integers can be even or odd. There is no difference in asking "is x even integer" and "is x even".

jallenmorris wrote:

Any numbers that will sum an odd integer will never have a difference of an odd integer. 4+2=6, but 4-2=2. Take this principle to the bank. So, these statements cannot both be true at the same time.

\(x=3=odd\) and \(y=2=even\) --> \(x+y=5=odd\) and \(x-y=1=odd\) OR \(x=6=even\) and \(y=1=odd\) --> \(x+y=7=odd\) and \(x-y=5=odd\).

alphastrike wrote:

If I understand correctly, x+y=odd and x-y=odd are only both always true if x and y are integers.

You are absolutely right. Pleas refer to my post for solution for this problem. _________________

I disagree with the OA mentioned above in this problem.

The condition when either x or y = 0 is not considered at all.

look at this example,

consider x=1 and y=0

(1) 2+5+1+0 = 8 (Even) Hence (A) is satisfied (2) 1-0 =1 (Odd) Hence (B) is also satisfied

The product 2*5*1*0 = 0 (Not an even integer)

Similarly consider x=2 and y=1

(1) 2+5+2+1 = 10 (Even) A is satisfied (2) 2-1 = 1 (Odd) B is Satisfied Product 2*5*2*1 = 20 (Even integer)

Hence , OA is E and definitely not C .

OA for this question is C.

You should know that: 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. An even number is an integer of the form \(n=2k\), where \(k\) is an integer.

My last interview took place at the Johnson School of Management at Cornell University. Since it was my final interview, I had my answers to the general interview questions...