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collection of tough problems from G PREP - 16

Expert replies
by abhasjha » Sat Feb 06, 2010 5:34 am
If x, y, and z are integers and xy+z is an odd integer, is x an even integer?

1. xy+xz is an even integer

2. y+xz is an odd integer
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Source: — Data Sufficiency |

by harsh.champ » Sat Feb 06, 2010 6:14 am
abhasjha wrote:If x, y, and z are integers and xy+z is an odd integer, is x an even integer?

1. xy+xz is an even integer

2. y+xz is an odd integer
Now,if xy + z is odd.(either xy-->odd and z-->even or xy-->even and z-->odd)

Statement 1: x(y+z) is even [that means x-->even or y+z-->even]
(y+ z-->even means y and z both should either be odd or both should be even)


Statement 2:y +xz is odd.(either y=even;xz odd or y=odd;xz even)


IMO E.Any suggestions??
Last edited by harsh.champ on Sat Feb 06, 2010 6:48 am, edited 1 time in total.
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by shashank.ism » Sat Feb 06, 2010 6:29 am
"xy + z is odd"
two numbers can add to give an odd sum only if they have opposite parity. hence:
case 1: xy is odd, z is even
there's only one way this can happen:
x = odd, y = odd, z = even. ............(1)
case 2: xy is even, z is odd
there are 3 ways in which this can happen:
x = even, y = even, z = odd .........(2a)
x = odd, y = even, z = odd ..........(2b)
x = even, y = odd, z = odd .........(2c)


statement (1)

pull out x:
xy+xz = x(y + z) is even.
this means that at least one of x and (y + z) is even.
--> if x is even, regardless of the parity of (y + z), then the answer to the prompt question is "yes" and we're done.
--> the other possibility would be x = odd and (y + z) = even. this is impossible, though, as it doesn't satisfy any of the cases above.
therefore, the answer must be "yes".
sufficient.

-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --

statement (2)
this means that y and xz have opposite parity.
--> y = even, xz = odd --> this means x = odd, y = even, z = odd. that's case (2b), which gives "no" to the question.
at this point you're done, because STATEMENTS CAN'T CONTRADICT EACH OTHER, se you know that "yes" MUST be a possibility with this statement (as statement #1 gives exclusively "yes" answers).
if you use this statement first, you'll have to keep going through the cases.
insufficient.

ans = A..
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by ajith » Sat Feb 06, 2010 8:38 am
abhasjha wrote:If x, y, and z are integers and xy+z is an odd integer, is x an even integer?

1. xy+xz is an even integer

2. y+xz is an odd integer
xy+z is odd => either xy or z is odd but not both

1. x(y+z) is even

xy+xz is even
xy+xz - xy+z is odd

xz-z is odd

=>z is not even z is odd
x is even

sufficient

2.) y+xz is odd
xy+z is also odd

now if x is odd and z is odd y has to be even
if y is odd x has to be even z has to be odd

both the conditions are possible
insufficient

A IMO
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by abhasjha » Mon Feb 08, 2010 12:29 am
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