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:

I think question is asking w/x + y/z odd and not w/x + w/z odd or not? If this is the case then here is my explanation. Rephrased question is wz+xy/xz is odd or not?

Statement 1: wx + yz is odd, this implies one pair is odd and other pair is even. As even + odd = odd. But we are not sure which pair is even or odd. So this is not sufficient and answer cannot be A or D.

Statement 2; wz + xy is odd, this numerator of question is odd. As given in question w/x + y/z is integer so wz+xy/xz is not a fraction and this implies wz+xy is divisible by xz. Only way a odd number divisible by another number is that divisor has to be odd as well. So odd divided by odd will yield odd. So question is answered.

Re: If w, x, y, and z are integers such that w/x and y/z are [#permalink]

Show Tags

13 Mar 2012, 12:06

2

This post received KUDOS

japped187 wrote:

If w, x, y, and z are integers such that w/x and y/z are integers, is w/x + w/z odd?

(1) wx + yz is odd (2) wz + xy is odd

IMO A is also satisfactory .... wx + yz is odd

Case 1 wx is odd & yz is even wx is odd ==> w & x are odd (odd*odd = odd) ==> w/x is odd yz is even ==> There can be 3 cases y is even and z is odd = Not possible as we are given y/z is integer y is odd and z is even = Not possible as we are given y/z is integer y is even and z is even = Possible ==> y/z is even w/x (odd) + y/z (even) = Odd

Case 2 wx is even & yz is odd can be proved in a similar manner.

If w, x, y and z are integers such that w/x and y/z are integers, is w/x + y/z odd?

Given: \(\frac{w}{x}=integer\) and \(\frac{y}{z}=integer\). Hence, \(\frac{w}{x}+\frac{y}{z}=\frac{wz+yx}{xz}=integer\) and the question is whether this integer is odd.

(1) wx + yz is odd --> if \(w=x=1\) and \(y=z=2\) then \(\frac{w}{x}+\frac{y}{z}=2=even\) but if \(w=x=1\) and \(y=2\), \(z=1\) then \(\frac{w}{x}+\frac{y}{z}=3=odd\). Not sufficient.

(2) wz + yx is odd --> \(\frac{wz+yx}{xz}=\frac{odd}{xz}=integer\) --> \(odd=(xz)*integer\) --> all multiple must be odd in order the product to be odd, hence \(integer =odd\). Sufficient.

Re: If w, x, y and z are integers such that w/x and y/z are [#permalink]

Show Tags

20 Sep 2012, 06:33

5

This post received KUDOS

Other way to look at the problem

As w/x is integer we can say w=xa+0 , where a= any integer------> xa/x = a Same way y/z is integer we can say y=zb+0 , where b= any integer------->zb/z = b

So the basically the question is, "Is a+b Odd"

1) wx + yz ----> x^2a + z^2b = odd------> a+b may be or may not be odd ---> Insufficient 2) wz + yx -----> xza + xzb ----> xz(a+b) = odd--->it means that both xz & (a+b) are odd ---->Sufficient

Answer B

Hope it helps.
_________________

If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS. Kudos always maximizes GMATCLUB worth-Game Theory

If you have any question regarding my post, kindly pm me or else I won't be able to reply

Re: If w, x, y and z are integers such that w/x and y/z are [#permalink]

Show Tags

20 Sep 2012, 07:16

1

This post received KUDOS

rohitgoel15 wrote:

IMO A is also satisfactory .... wx + yz is odd

Case 1 wx is odd & yz is even wx is odd ==> w & x are odd (odd*odd = odd) ==> w/x is odd yz is even ==> There can be 3 cases y is even and z is odd = Not possible as we are given y/z is integer y is odd and z is even = Not possible as we are given y/z is integer y is even and z is even = Possible ==> y/z is even w/x (odd) + y/z (even) = Odd

Case 2 wx is even & yz is odd can be proved in a similar manner.

Please advice if i am wrong.

rohitgoel15 wrote:

Hi Bunuel,

I agree with ur exp .. but is there a problem with my algebraic method ?

Yes there is a problem in statement: y is even and z is odd = Not possible as we are given y/z is integer Even/Odd can be integer, Consider eg Y=6 , z=3.. Hence that solution is incorrect.
_________________

Re: If w, x, y and z are integers such that w/x and y/z are [#permalink]

Show Tags

02 Aug 2013, 08:33

2

This post received KUDOS

Hi All,

Problem helps us to understand the following: For \(\frac{w}{x}\) to be an integer, both w and x need to be either even or odd. Same is the case with y and z, both either need to be even or odd. We need, w/x + y/z to be of the following format:

ODD + EVEN, so of the two terms, one of them has to be even and the other to be odd.

Now, lets look the statements..

Statement (1): \(wx+yz\) is odd

From this statement we can deduce that either wx is odd and yz is even OR wx is even and yz is odd. Lets assume that wx is odd and yz is even. For the term wx to be odd and w/x to be an integer, both w and x needs to be odd. However, even though yz is even since both y and z are even, it doesn't guarantee that y/z will be even. (6/2 = 3)

So from statement 1, what we get is ODD + EVEN/ODD, hence not sufficient.

Statement (2): \(wz+yx\) is odd

We can simply multiply and divide by xz as below:

(w/x + y/z) * xz = ODD

Now since we know that only ODD * ODD = ODD, we can be certain that w/x + y/z is odd, hence sufficient.

if AB is odd, is A/B also odd? What is the rule on this type of construction?

Thanks

You could try some examples to answer your own question.

For integers a and b, ab is odd only if both are odd. Now, odd/odd can be odd or not an integer at all. For example, 3/1=3=odd but 1/3 is not an integer at all.
_________________

If w, x, y and z are integers such that w/x and y/z are integers, is w/x + y/z odd?

(1) wx + yz is odd (2) wz + yx is odd

Plug in approach requiring minimal thinking:

Take the case that is to be proved i.e, w/x + y/z is odd and then take the contrary case i.e, w/x + y/z is even for both the statements

(i) Case to be proved : wx+ yz is odd and w/x + y/z is odd.

The former is satisfied by odd + even and the latter again by odd+ even

Thus this will be satisfied say, if the following is satisfied: wx and w/x are both odd and y/z and yz are both even. an example is w=15 , x=3 and y=4 and z=2

How about the contrary case for (i). i.e., wx+ yz is odd but w/x + y/z is even.

The former is satisfied by odd+ even and the latter by odd+odd or even +even

This will be satisfied say, if wx and w/x are both odd and yz is even and y/z is odd. An example is w=9 x=3 and y=6 z=2

since the case to be proved and the contrary are both satisfied (i) is not sufficient

(ii) Case to be proved: wz+yx is odd and w/x and y/z is odd

The former is satisfied by odd+ even and the latter again by odd+ even

Thus this will be satisfied say, if the following is satisfied: wz and w/x are both odd and y/z and yx are both even. an example is w=15 , x=3 and y=12 and z=1

How about the contrary case for (i). i.e., wx+ yz is odd but w/x + y/z is even.

The former is satisfied by odd+ even and the latter by odd+odd or even +even

This will be satisfied say, if wz and w/x are both odd and yx is even and y/z is odd.

We find this cannot be satisfied

Since the contrary case cannot be proved for (ii) , this alone is sufficient and the answer is B.
_________________

Re: If w, x, y and z are integers such that w/x and y/z are [#permalink]

Show Tags

07 Feb 2015, 00:08

1

This post received KUDOS

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: If w, x, y and z are integers such that w/x and y/z are [#permalink]

Show Tags

18 Mar 2016, 22:37

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 w, x, y and z are integers such that w/x and y/z are integers, is w/x + y/z odd?

(1) wx + yz is odd (2) wz + yx is odd

Please find the solution as attached

Attachments

File comment: www.GMATinsight.com

1115.jpg [ 115.29 KiB | Viewed 4489 times ]

_________________

Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html

Re: If w, x, y and z are integers such that w/x and y/z are [#permalink]

Show Tags

16 Sep 2017, 10:26

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.
_________________