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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.
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.
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.
Re: Is product 2*x*5*y an even integer?
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22 Feb 2012, 08:04
I dont get it. \(2\) is already being multiplied to the original number. so unless \(x\) and \(y\) is a fraction, let's say \(\frac{1}{2}\) then it could be odd, but as long as \(x\) and \(y\) are integers, there is no way this could be an odd number since it is being multiplied by \(2\). Now if we look at Statement A, it still does not tell us that \(x\) and \(y\) are integers or not. Since \(x\) could be \(\frac{1}{2}\) and \(y\) again could be \(\frac{1}{2}\) so A is obviously Insufficient. We cannot even establish if \(2*x*5*y\) is an integer, let alone it is even or not. And the presence of \(2\) is extremely misleading.
Now B says that \((x - y)\) is an odd integer. Let's suppose \(x=\frac{4}{3}\) and \(y=\frac{1}{3}\), then \((x-y)=1\) which is an odd integer as it is supposed to be but that does not make \(2*x*5*y\) an even integer an integer at all. On the other hand let x=4 and y=3 than \((x-y)=1\) which is again an odd integer so yes \(2*x*5*y\) is an even integer. Two different answers, Hence Insufficient.
Now if we combine A & B:
Statement A: \(2+x+5+y\)is \(even\) so \((x+y)+7\) is \(even\) so \((x+y)\) has to be odd Statement B: \(x-y=odd\) which is basically just restating Statement A.
There is something wrong with the question. Do you have a source for this one?
Re: Is product 2*x*5*y an even integer?
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22 Feb 2012, 08:26
10
3
omerrauf wrote:
I dont get it. \(2\) is already being multiplied to the original number. so unless \(x\) and \(y\) is a fraction, let's say \(\frac{1}{2}\) then it could be odd, but as long as \(x\) and \(y\) are integers, there is no way this could be an odd number since it is being multiplied by \(2\). Now if we look at Statement A, it still does not tell us that \(x\) and \(y\) are integers or not. Since \(x\) could be \(\frac{1}{2}\) and \(y\) again could be \(\frac{1}{2}\) so A is obviously Insufficient. We cannot even establish if \(2*x*5*y\) is an integer, let alone it is even or not. And the presence of \(2\) is extremely misleading.
Now B says that \((x - y)\) is an odd integer. Let's suppose \(x=\frac{4}{3}\) and \(y=\frac{1}{3}\), then \((x-y)=1\) which is an odd integer as it is supposed to be but that does not make \(2*x*5*y\) an even integer an integer at all. On the other hand let x=4 and y=3 than \((x-y)=1\) which is again an odd integer so yes \(2*x*5*y\) is an even integer. Two different answers, Hence Insufficient.
Now if we combine A & B:
Statement A: \(2+x+5+y\)is \(even\) so \((x+y)+7\) is \(even\) so \((x+y)\) has to be odd Statement B: \(x-y=odd\) which is basically just restating Statement A.
There is something wrong with the question. Do you have a source for this one?
Is product 2*x*5*y an even integer?
Notice that we are not told that x and y are integers.
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.
Re: Is product 2*x*5*y an even integer?
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24 Feb 2012, 11:18
Bunuel wrote:
omerrauf wrote:
I dont get it. \(2\) is already being multiplied to the original number. so unless \(x\) and \(y\) is a fraction, let's say \(\frac{1}{2}\) then it could be odd, but as long as \(x\) and \(y\) are integers, there is no way this could be an odd number since it is being multiplied by \(2\). Now if we look at Statement A, it still does not tell us that \(x\) and \(y\) are integers or not. Since \(x\) could be \(\frac{1}{2}\) and \(y\) again could be \(\frac{1}{2}\) so A is obviously Insufficient. We cannot even establish if \(2*x*5*y\) is an integer, let alone it is even or not. And the presence of \(2\) is extremely misleading.
Now B says that \((x - y)\) is an odd integer. Let's suppose \(x=\frac{4}{3}\) and \(y=\frac{1}{3}\), then \((x-y)=1\) which is an odd integer as it is supposed to be but that does not make \(2*x*5*y\) an even integer an integer at all. On the other hand let x=4 and y=3 than \((x-y)=1\) which is again an odd integer so yes \(2*x*5*y\) is an even integer. Two different answers, Hence Insufficient.
Now if we combine A & B:
Statement A: \(2+x+5+y\)is \(even\) so \((x+y)+7\) is \(even\) so \((x+y)\) has to be odd Statement B: \(x-y=odd\) which is basically just restating Statement A.
There is something wrong with the question. Do you have a source for this one?
Is product 2*x*5*y an even integer?
Notice that we are not told that x and y are integers.
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.
Answer: C.
Hope it's clear.
A real tricky question and an awesome explanation. Thanks
Re: Is product 2*x*5*y an even integer?
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27 Feb 2013, 00:18
i knw its a silly question to ask but can anybody pls explain: 2x can only be even if x is an integer,?? what if x is a value like .7.. then 2x is 1.4.. which is even i assume.. or is it not?? any help would be highly appreciated!!!
Re: Is product 2*x*5*y an even integer?
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27 Feb 2013, 02:01
swarman wrote:
i knw its a silly question to ask but can anybody pls explain: 2x can only be even if x is an integer,?? what if x is a value like .7.. then 2x is 1.4.. which is even i assume.. or is it not?? any help would be highly appreciated!!!
No, 0.7 is not even. Only integers can be even or odd.
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.
An odd number is an integer that is not evenly divisible by 2. An odd number is an integer of the form \(n=2k+1\), where \(k\) is an integer.
Re: Is product 2*x*5*y an even integer?
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20 Mar 2017, 03:51
Bunuel 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
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.
Answer: C.
Hi..I did it exactly as you did it. However, I chose (E). Please help..while its true that both x and y are integers..there is nothing in the stem that says that they cant be NEGATIVE INTEGERS. And the case of negative integers can result into a non integer value for the expression, this made me choose E.
Re: Is product 2*x*5*y an even integer?
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20 Mar 2017, 04:34
ShashankDave wrote:
Bunuel 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
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.
Answer: C.
Hi..I did it exactly as you did it. However, I chose (E). Please help..while its true that both x and y are integers..there is nothing in the stem that says that they cant be NEGATIVE INTEGERS. And the case of negative integers can result into a non integer value for the expression, this made me choose E.
When we consider the statements together we get that both x and y are integers. It does not matter whether they are positive or negative, still in this case 2*x*5*y=10*x*y=10*integer=even.
_________________
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29% (02:21) correct 
any help would be highly appreciated!!!
