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mclaren7
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Post subject: If w, x, y, and z are integers Posted: Sun Apr 13, 2008 10:12 am |
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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
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RonPurewal
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Post subject: Posted: Mon Apr 14, 2008 4:48 am |
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| ManhattanGMAT Staff |
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Posts: 7146
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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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mclaren7
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Post subject: Posted: Thu Apr 17, 2008 12:50 am |
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My goodness Ron, you are superb.
Thanks
KH
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rfernandez
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Post subject: Posted: Thu Apr 24, 2008 2:58 pm |
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Posts: 386
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teenup124
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Post subject: Re: Posted: Fri Aug 12, 2011 7:37 am |
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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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RonPurewal
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Post subject: Re: Re: Posted: Mon Aug 15, 2011 2:40 am |
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Posts: 7146
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teenup124, your solution looks legitimate to me.
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sannamalai1
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Post subject: Re: Posted: Mon Sep 12, 2011 12:04 am |
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Posts: 1
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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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RonPurewal
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Post subject: Re: Re: Posted: Mon Sep 12, 2011 4:39 am |
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| ManhattanGMAT Staff |
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Posts: 7146
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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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Post subject: Re: Posted: Sat Feb 11, 2012 5:26 pm |
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Posts: 47
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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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RonPurewal
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Post subject: Re: Re: Posted: Tue Feb 14, 2012 8:21 am |
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Posts: 7146
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no prob, thanks for editing.
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