Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: Is product 2*x*5*y an even integer? [#permalink]

Show Tags

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?
_________________

"Nowadays, people know the price of everything, and the value of nothing."Oscar Wilde

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? [#permalink]

Show Tags

22 Feb 2012, 08:48

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

1. 7 + x + y = even (7+1+2 satisfy the statement, 7+2.5+0.5 satisfy the statement as well --> not sufficient) 2. x - y = odd (5-2 = odd and 5.5-2.5 = odd --> not sufficient)

Together: (x+y)+(x-y) = odd + odd --> 2x = even 2x can only be even if x is an integer, thus y must be an integer too

Re: Is product 2*x*5*y an even integer? [#permalink]

Show Tags

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? [#permalink]

Show Tags

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!!!

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? [#permalink]

Show Tags

17 Mar 2013, 00:05

No matter what the value of x and y is (as long as it is an integer) the product will always be an Even number , therefore what this question is asking essentially is whether x and y are integers..

Statement (1) : Simplified we get, x+y = Odd Integer... This is satisfied by x being 2.5 and y being 0.5 , therefore we are not certain whether x and y are integers .. Not Suff.

Statement (2) : x - y = odd integer, this again is satisfied with 3.5 (x) - 0.5 (y) .. therefore is also insuff.

Combining 1 and 2, and lining them up we get, 2x = Odd int + Odd Int , or 2x = even integer (odd Integer + Odd integer is always an even integer) . Solving further we now know that X is an integer (Even integer divided by 2 is always an integer) .. Similarly substituting this information in any one of the 2 equations we can verify that y is also an integer.

Therefore the Answer is C...

Hope this helps..
_________________

"When you want to succeed as bad as you want to breathe, then you’ll be successful.” - Eric Thomas

Re: Is product 2*x*5*y an even integer? [#permalink]

Show Tags

18 Mar 2013, 13:10

vomhorizon wrote:

No matter what the value of x and y is (as long as it is an integer) the product will always be an Even number , therefore what this question is asking essentially is whether x and y are integers..

Statement (1) : Simplified we get, x+y = Odd Integer... This is satisfied by x being 2.5 and y being 0.5 , therefore we are not certain whether x and y are integers .. Not Suff.

Statement (2) : x - y = odd integer, this again is satisfied with 3.5 (x) - 0.5 (y) .. therefore is also insuff.

Combining 1 and 2, and lining them up we get, 2x = Odd int + Odd Int , or 2x = even integer (odd Integer + Odd integer is always an even integer) . Solving further we now know that X is an integer (Even integer divided by 2 is always an integer) .. Similarly substituting this information in any one of the 2 equations we can verify that y is also an integer.

Re: Is product 2*x*5*y an even integer? [#permalink]

Show Tags

30 May 2015, 08:26

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: Is product 2*x*5*y an even integer? [#permalink]

Show Tags

07 Jun 2017, 09:18

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Version 8.1 of the WordPress for Android app is now available, with some great enhancements to publishing: background media uploading. Adding images to a post or page? Now...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...