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mclaren7
 
 

If w, x, y, and z are integers

by mclaren7 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
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by RonPurewal Mon Apr 14, 2008 4:48 am

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
 
 

by mclaren7 Thu Apr 17, 2008 12:50 am

My goodness Ron, you are superb.

Thanks
KH
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by rfernandez Thu Apr 24, 2008 2:58 pm

teenup124
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Re:

by teenup124 Fri Aug 12, 2011 7:37 am

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
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Re: Re:

by RonPurewal Mon Aug 15, 2011 2:40 am

teenup124, your solution looks legitimate to me.
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Re:

by sannamalai1 Mon Sep 12, 2011 12:04 am

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
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Re: Re:

by RonPurewal Mon Sep 12, 2011 4:39 am

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.
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rachelhong2012
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Re:

by rachelhong2012 Sat Feb 11, 2012 5:26 pm

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 :)
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Re: Re:

by RonPurewal Tue Feb 14, 2012 8:21 am

no prob, thanks for editing.
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Re: If w, x, y, and z are integers

by supratim7 Thu Apr 11, 2013 12:54 pm

Made a flash card for such problems..
Sharing it.. hope it helps..

Odd / Odd = could be odd; could be a non-integer
e.g. 3/1 = 3 (odd); 1/3 = (non integer)

Odd / Even = could be a non-integer; could be "undefined"
e.g. 3/2 (non integer); 3/0 (undefined)

Even / Odd = could be even; could be a non-integer
e.g. 6/3 = 2 (even); 2/3 (non-integer)

Even / Even = could be even; could be odd; could be a non-integer; could be "undefined"
e.g. 4/2 = 2 (even); 2/2 = 1 (odd); 2/4 (non-integer); 2/0 (undefined)

Regards | Supratim
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Re: If w, x, y, and z are integers

by RonPurewal Fri Apr 12, 2013 4:35 am

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.
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Re: If w, x, y, and z are integers

by supratim7 Fri Apr 12, 2013 5:15 am

Noted Ron. Appreciate it :)
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Re: If w, x, y, and z are integers

by jlucero Sat Apr 13, 2013 1:32 pm

Glad it helped!
Joe Lucero
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