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:

1. xy + xz is an even integer - SUFFICIENT Given: xy + z is odd ...(i) xy + xz is even ...(ii)

subtracting (ii) from (i) we get xz - z, which should be odd (* since odd - even = odd) => z(x-1) is odd => both z and (x-1) is odd => since (x-1) is odd, x must be even.

2. y + xz is an odd integer -INSUFFICIENT Given: xy + z is odd ...(i) y + xz is odd ...(ii)

subtracting (ii) from (i) we get xy + z - y - xz = (x-1)(y-z) , which should be even => either (x-1) is even or (y-z) is even ....insufficient to determine
_________________

You can look at Statement 1 conceptually: when we add z to xy, we get something odd. However, when we add xz to xy, we get something even. So certainly one of z or xz is odd, the other even. Now if xz is different from z, then multiplying by x must have changed z, and that could only happen if x is even and z odd. That's a bit tricky to explain, but I hope that's clear.

Or you can proceed algebraically - notice the similarity between the expression in the question and in Statement 1. We know that xy + xz is even, and xy + z is odd. When you subtract this second expression from the first, you're subtracting an odd from an even, so must get an odd: xy + xz - (xy + z) = xz - z = z(x-1) is odd. Since this is a product, z must be odd, and x-1 must be odd, so x is even. Sufficient.

For Statement 2, all the letters could be odd, so not sufficient.
_________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Re: GMAT Prep...How much time did u take to solve this one ?? [#permalink]

Show Tags

27 Sep 2009, 09:58

1

This post received KUDOS

IanStewart wrote:

You can look at Statement 1 conceptually: when we add z to xy, we get something odd. However, when we add xz to xy, we get something even. So certainly one of z or xz is odd, the other even. Now if xz is different from z, then multiplying by x must have changed z, and that could only happen if x is even and z odd. That's a bit tricky to explain, but I hope that's clear.

Or you can proceed algebraically - notice the similarity between the expression in the question and in Statement 1. We know that xy + xz is even, and xy + z is odd. When you subtract this second expression from the first, you're subtracting an odd from an even, so must get an odd: xy + xz - (xy + z) = xz - z = z(x-1) is odd. Since this is a product, z must be odd, and x-1 must be odd, so x is even. Sufficient.

For Statement 2, all the letters could be odd, so not sufficient.

Fr St2, y+xz odd xy+z odd => y+zx+xy+z even => y(x+1)+z(x+1) even => (y+z)(x+1) event x+1 can be odd or even means that x can be even or odd, insuff

Re: If x, y and z are integers and xy + z is an odd integer, is [#permalink]

Show Tags

19 Mar 2014, 19:41

Given condition: xy + z = odd implies either xy = odd (x =odd and y = odd) and z = even or xy = even (x or y can be odd and even respectively and vice versa) and z = odd

condition 1:

xy + xz = even; Implies x(y+z) = even which again implies the following:

i) x even and y+z = odd - where again y or z can be odd and even respectively and vice versa ii) x odd and y +z = even - where again y and z has to be both odd or both even

inconclusive

condition 2:

y + xz = odd

again inconclusive 1 + 2: Add xy + z + y + xz = odd + odd implies: (x + 1)(y+z) = even and x (y+z) is also even according to 2.. so y + z = even <y and z both even or y + z both odd>, x can be odd or even but by 1 xy + z = odd which means y and z both odd, so x is even.

If x, y and z are integers and xy + z is an odd integer, is x an even integer?

(1) xy + xz is an even integer (2) y + xz is an odd integer

Odd/Even questions can be usually solved quite easily if one tries some operations with the statements

We want to know if x is even integer

We are given that xy+z is odd

Statement 1

xq + xz is even

Subtracting

z(x+1) is odd

Therefore, x+1 should be odd and x should be even

Sufficient

Statement 2

Not sufficient

Answer is A

Just my 2c

Cheers J

I am not clear with the red part.

When you subtract \(xy + z=odd\) from \(xy+xz=even\) you'll get: \(xz-z=even-odd=odd\) --> \(z(x-1)=odd\). For the product of two integers to be odd, both of them must be odd --> \(z\) and \(x-1\) are odd. If \(x-1=odd\), then x must be even: \(x-1=x-odd=odd\) --> \(x=odd+odd=even\).

Re: If x, y and z are integers and xy + z is an odd integer, is [#permalink]

Show Tags

21 Mar 2014, 20:46

1

This post received KUDOS

1

This post was BOOKMARKED

Odd(O) Even (E) given: x,y,z integers xy+z=O so only the following scenarios can fulfill the constraints a) EO+O b) EE+O c) OE+O d) OO+E

question: x=E?

1) x(y+z)=E i. (E)(O+O) --> fits scenario a -->yes, x can be even ii. (O)(E+E) --> n/a - doesn't fit any scenarios iii. (O)(O+O) --> n/a - doesn't fit any scenarios

stop testing, x can't be odd, sufficient

2) y+xz = O i. E+(O)(O) --> fits scenario a -->yes, x can be even ii. O+(E)(E) --> n/a - doesn't fit any scenarios iii. O+(O)(E) --> fits scenario d -->no, x can be odd

Re: If x, y and z are integers and xy + z is an odd integer, is [#permalink]

Show Tags

08 Mar 2016, 16:27

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.
_________________

If x, y, z are integers and xy+z is an odd integer, is x an even integer.

1. xy +xz is an even integer

2. y+ xz is an odd integer

Statement 1

xy+xz is even xy+z is odd

Subtract the two. xy+xz-xy-z = xz-z

xz-z will be odd (even - odd will always result in odd)

z(x-1) will be odd

This is only possible when both z and (x-1) are odd.

x-1 is odd. This means that x is even.

Sufficient.

Statement 2

y+xz is odd xy+z is odd

Add the two. y+xz+xy+z = x(y+z)+1 (y+z) = (x+1) (y+z).

(x+1)(y+z) will be even (odd+odd is even)

Now here x+1 can be even and y+z can be odd. (Even * odd will result in an even result) or x+1 can be odd and y+z can be even (Even * odd will result in an even result) or both (x+1) and (y+z) can be even (Even * even will result in an even result)

So x+1 can be both odd and even. In other words, x can be both even and odd.

This is only possible when both z and (x-1) are odd.

Hey, guys, So, I’ve decided to run a contest in hopes of getting the word about the site out to as many applicants as possible this application season...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...

Term 1 has begun. If you're confused, wondering what my post on the last 2 official weeks was, that was pre-term. What that means is that the school...