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16 Nov 2009, 22:19
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Is \((x - y)*(x + y)\) an even integer?
1. \(x\) and \(y\) are integers
2. \(x + y\) is even
Chose incorrect B and after looking at the solution thought of adding a little to OE (don't know if it is a superstition, but I'm getting #13 wrong in so many tests :evil: )...
x+y is even
in case both are integers, then they are either both odd or both even => x-y will be even.
in case both are fraction (if only 1 is fraction, then "x+y is even" does not holds true). Check it with few substitutions for odd/even combination on denominator so as to keep x+y even:
Even denominator... 1/2+3/2 => subtraction result odd. But overall result even.
Odd denominator... 5/3+1/3 => subtraction result is fraction making B false
So they have to be integers OA
1. \(x\) and \(y\) are integers
2. \(x + y\) is even
Chose incorrect B and after looking at the solution thought of adding a little to OE (don't know if it is a superstition, but I'm getting #13 wrong in so many tests :evil: )...
x+y is even
in case both are integers, then they are either both odd or both even => x-y will be even.
in case both are fraction (if only 1 is fraction, then "x+y is even" does not holds true). Check it with few substitutions for odd/even combination on denominator so as to keep x+y even:
Even denominator... 1/2+3/2 => subtraction result odd. But overall result even.
Odd denominator... 5/3+1/3 => subtraction result is fraction making B false
So they have to be integers OA
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22 Nov 2009, 08:48
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S1 is insufficient
s2 is sufficient because
(x-y) is either even or odd integer, but multiplication with even (x+y) will always be even. So it is sufficient to tell us the given statement result.
Answer is B
s2 is sufficient because
(x-y) is either even or odd integer, but multiplication with even (x+y) will always be even. So it is sufficient to tell us the given statement result.
Answer is B
