
Author 
Message 
mclaren7

Post subject: If w, x, y, and z are integers Posted: Sun Apr 13, 2008 10:12 am 


Dear moderators and friends,
If w, x, y, and z are integers such that w/x and y/z are integers, is w/x + y/z odd?
a. wx + yz = odd
b. wz + xy = odd
I will post the OA later.
Thanks
KH





RonPurewal

Post subject: Posted: Mon Apr 14, 2008 4:48 am 


ManhattanGMAT Staff 

Posts: 14602

this problem involves two fractions that are added together. for no other reason than that 'it's the normal thing to do with two fractions added together', let's find the common denominator:
w/x + y/z = wz/xz + xy/xz = (wz + xy)/xz
therefore
the question can be rearranged to:
is (wz + xy)/xz  which is the same thing as w/x + y/z  odd?
 (2) alone 
if wz + xy is an odd integer, then all of its factors are odd. this means that (wz + xy)/xz, which is guaranteed to be an integer**, must also be odd  because it's a factor of an odd number.
sufficient
**we know this is an integer because it's equal to w/x + y/z, which, according to the information given in the problem statement, is integer + integer.
 (1) alone 
try to come up with contradictory examples**:
w=2, x=1, y=3, z=1 (so that wx + yz = 5 = odd, per the requirement):
w/x + y/z = 2 + 3 = 5 = odd
w=2, x=2, y=3, z=1 (so that wx + yz = 7 = odd, per the requirement):
w/x + y/z = 1 + 3 = 4 = even
insufficient
**of course, if you're at a loss for the theory, you should try this for statement (1) too ... but you'll find that all the examples you get are odd.

answer = b





mclaren7

Post subject: Posted: Thu Apr 17, 2008 12:50 am 


My goodness Ron, you are superb.
Thanks
KH





rfernandez

Post subject: Posted: Thu Apr 24, 2008 2:58 pm 


ManhattanGMAT Staff 

Posts: 383





teenup124

Post subject: Re: Posted: Fri Aug 12, 2011 7:37 am 


Students 

Posts: 2

RonPurewal wrote: this problem involves two fractions that are added together. for no other reason than that 'it's the normal thing to do with two fractions added together', let's find the common denominator: w/x + y/z = wz/xz + xy/xz = (wz + xy)/xz
therefore the question can be rearranged to: is (wz + xy)/xz  which is the same thing as w/x + y/z  odd?
 (2) alone 
if wz + xy is an odd integer, then all of its factors are odd. this means that (wz + xy)/xz, which is guaranteed to be an integer**, must also be odd  because it's a factor of an odd number.
sufficient
**we know this is an integer because it's equal to w/x + y/z, which, according to the information given in the problem statement, is integer + integer.
 (1) alone 
try to come up with contradictory examples**:
w=2, x=1, y=3, z=1 (so that wx + yz = 5 = odd, per the requirement): w/x + y/z = 2 + 3 = 5 = odd
w=2, x=2, y=3, z=1 (so that wx + yz = 7 = odd, per the requirement): w/x + y/z = 1 + 3 = 4 = even
insufficient
**of course, if you're at a loss for the theory, you should try this for statement (1) too ... but you'll find that all the examples you get are odd.

answer = b Dear ron please correct me i am trying to make it short and simple Given w/x is an integer so w=x * a y/z is an integer so y=z * b is w/x + y/z odd? bt putting values w/x + y/z = (x*a)/x+(z*b)/z= (a+b)= odd? now the given statements a. wx + yz = odd putting these values (x *a) x+ (z*b)z is odd so this is a x^2 +bz^2 dont have any info about individual terms so not sufficient b. wz + xy = odd (x *a) z+ (z*b)x is odd so this is xza+xzb = odd xz(a+b)= odd now we know that Odd x Odd = Odd so (a+b) = odd hence sufficient





RonPurewal

Post subject: Re: Re: Posted: Mon Aug 15, 2011 2:40 am 


ManhattanGMAT Staff 

Posts: 14602

teenup124, your solution looks legitimate to me.
_________________ Pueden hacerle preguntas a Ron en castellano Potete fare domande a Ron in italiano On peut poser des questions ã Ron en français Voit esittää kysymyksiä Ron:lle myös suomeksi
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sannamalai1

Post subject: Re: Posted: Mon Sep 12, 2011 12:04 am 


Forum Guests 

Posts: 1

In the example for statement 1 (w=2, x=2, y=3, z=1), is it okay to take the same value of 2 for both w and x ? When W and X are given as two different variables, how can we assign the same value. Thanks. RonPurewal wrote: this problem involves two fractions that are added together. for no other reason than that 'it's the normal thing to do with two fractions added together', let's find the common denominator: w/x + y/z = wz/xz + xy/xz = (wz + xy)/xz
therefore the question can be rearranged to: is (wz + xy)/xz  which is the same thing as w/x + y/z  odd?
 (2) alone 
if wz + xy is an odd integer, then all of its factors are odd. this means that (wz + xy)/xz, which is guaranteed to be an integer**, must also be odd  because it's a factor of an odd number.
sufficient
**we know this is an integer because it's equal to w/x + y/z, which, according to the information given in the problem statement, is integer + integer.
 (1) alone 
try to come up with contradictory examples**:
w=2, x=1, y=3, z=1 (so that wx + yz = 5 = odd, per the requirement): w/x + y/z = 2 + 3 = 5 = odd
w=2, x=2, y=3, z=1 (so that wx + yz = 7 = odd, per the requirement): w/x + y/z = 1 + 3 = 4 = even
insufficient
**of course, if you're at a loss for the theory, you should try this for statement (1) too ... but you'll find that all the examples you get are odd.

answer = b





RonPurewal

Post subject: Re: Re: Posted: Mon Sep 12, 2011 4:39 am 


ManhattanGMAT Staff 

Posts: 14602

sannamalai1 wrote: In the example for statement 1 (w=2, x=2, y=3, z=1), is it okay to take the same value of 2 for both w and x ? When W and X are given as two different variables, how can we assign the same value. given two variables, the default assumption is that the two quantities are allowed to take on the same value, if possible. if that situation is to be prohibited, then the prohibition must be explicitly imposed (by means of an expression such as "different integers" or "distinct integers"). by the way, even if you don't know this default, you can figure it out by observation: just note that lots of problems specify "distinct" or "different", but that no problem will ever explicitly specify "variables that are allowed to take on the same value". if the latter is never explicitly articulated, it must be the default.
_________________ Pueden hacerle preguntas a Ron en castellano Potete fare domande a Ron in italiano On peut poser des questions ã Ron en français Voit esittää kysymyksiä Ron:lle myös suomeksi
Un bon vêtement, c'est un passeport pour le bonheur. – Yves SaintLaurent





rachelhong2012

Post subject: Re: Posted: Sat Feb 11, 2012 5:26 pm 


Course Students 

Posts: 52

RonPurewal wrote: this problem involves two fractions that are added together. for no other reason than that 'it's the normal thing to do with two fractions added together', let's find the common denominator: w/x + y/z = wz/xz + xy/xz = (wz + xy)/xz
therefore the question can be rearranged to: is (wz + xy)/xz  which is the same thing as w/x + y/z  odd?
 (2) alone 
if wz + xy is an odd integer, then all of its factors are odd. this means that (wz + xy)/xz, which is guaranteed to be an integer**, must also be odd  because it's a factor of an odd number.
sufficient
**we know this is an integer because it's equal to w/x + y/z, which, according to the information given in the problem statement, is integer + integer.
 (1) alone 
try to come up with contradictory examples**:
w=2, x=1, y=3, z=1 (so that wx + yz = 5 = odd, per the requirement): w/x + y/z = 2 + 3 = 5 = odd
w=2, x=2, y=3, z=1 (so that wx + yz = 7 = odd, per the requirement): w/x + y/z = 1 + 3 = 4 = even
insufficient
**of course, if you're at a loss for the theory, you should try this for statement (1) too ... but you'll find that all the examples you get are odd.

answer = b I didn't quite understand the highlighted part but now I get it, please ignore my bumping up the post :)





RonPurewal

Post subject: Re: Re: Posted: Tue Feb 14, 2012 8:21 am 


ManhattanGMAT Staff 

Posts: 14602

no prob, thanks for editing.
_________________ Pueden hacerle preguntas a Ron en castellano Potete fare domande a Ron in italiano On peut poser des questions ã Ron en français Voit esittää kysymyksiä Ron:lle myös suomeksi
Un bon vêtement, c'est un passeport pour le bonheur. – Yves SaintLaurent





supratim7

Post subject: Re: If w, x, y, and z are integers Posted: Thu Apr 11, 2013 12:54 pm 


Forum Guests 

Posts: 149

Made a flash card for such problems.. Sharing it.. hope it helps..
Odd / Odd = could be odd; could be a noninteger e.g. 3/1 = 3 (odd); 1/3 = (non integer)
Odd / Even = could be a noninteger; could be "undefined" e.g. 3/2 (non integer); 3/0 (undefined)
Even / Odd = could be even; could be a noninteger e.g. 6/3 = 2 (even); 2/3 (noninteger)
Even / Even = could be even; could be odd; could be a noninteger; could be "undefined" e.g. 4/2 = 2 (even); 2/2 = 1 (odd); 2/4 (noninteger); 2/0 (undefined)
Regards  Supratim





RonPurewal

Post subject: Re: If w, x, y, and z are integers Posted: Fri Apr 12, 2013 4:35 am 


ManhattanGMAT Staff 

Posts: 14602

on the gmat, you won't have to worry about "undefined" quantities. otherwise, that looks ok.
to me, though, that seems like the kind of thing that would be awfully hard (and unnecessary) to memorize, and much easier just to figure out from scratch if you need to.
_________________ Pueden hacerle preguntas a Ron en castellano Potete fare domande a Ron in italiano On peut poser des questions ã Ron en français Voit esittää kysymyksiä Ron:lle myös suomeksi
Un bon vêtement, c'est un passeport pour le bonheur. – Yves SaintLaurent





supratim7

Post subject: Re: If w, x, y, and z are integers Posted: Fri Apr 12, 2013 5:15 am 


Forum Guests 

Posts: 149

Noted Ron. Appreciate it :)





jlucero

Post subject: Re: If w, x, y, and z are integers Posted: Sat Apr 13, 2013 1:32 pm 


ManhattanGMAT Staff 

Posts: 1119

Glad it helped!
_________________ Joe Lucero Manhattan GMAT Instructor






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