Register    Login    Search    Rss Feeds

 Page 1 of 1 [ 10 posts ] 



 
Author Message
 Post subject: If w, x, y, and z are integers
 Post 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


Top 
 Post subject:
 Post Posted: Mon Apr 14, 2008 4:48 am 
Offline
ManhattanGMAT Staff


Posts: 7146
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


Top 
 Post subject:
 Post Posted: Thu Apr 17, 2008 12:50 am 
My goodness Ron, you are superb.

Thanks
KH


Top 
 Post subject:
 Post Posted: Thu Apr 24, 2008 2:58 pm 
Offline
ManhattanGMAT Staff


Posts: 386


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


Top 
 Post subject: Re: Re:
 Post Posted: Mon Aug 15, 2011 2:40 am 
Offline
ManhattanGMAT Staff


Posts: 7146
teenup124, your solution looks legitimate to me.


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


Top 
 Post subject: Re: Re:
 Post Posted: Mon Sep 12, 2011 4:39 am 
Offline
ManhattanGMAT Staff


Posts: 7146
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.


Top 
 Post subject: Re:
 Post Posted: Sat Feb 11, 2012 5:26 pm 
Offline
Course Students


Posts: 47
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 :)


Top 
 Post subject: Re: Re:
 Post Posted: Tue Feb 14, 2012 8:21 am 
Offline
ManhattanGMAT Staff


Posts: 7146
no prob, thanks for editing.


Top 
Display posts from previous:  Sort by  
 
 Page 1 of 1 [ 10 posts ] 





Who is online

Users browsing this forum: No registered users and 1 guest

 
 

 
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to: