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05 Jun 2016, 01:07
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C
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Is \(x^2 - y^2\) an even integer?
(1) x-y is an even integer.
(2) x is an odd integer.
(1) x-y is an even integer.
(2) x is an odd integer.
Chetan's questions
05 Jun 2016, 04:51
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fantaisie
Pl recheck the highlighted portion..
If x-y is already given as EVEN, how can you take that as ODD
It is a bit tricky as Divyadisha has mentioned below....
The following point is very important from point of view of GMAT
Solution -
We are looking at whether x^2-y^2 is EVEN integer..
lets see the statements-
(1) x-y is an even integer.
Most of us will correctly factorize \(x^2-y^2\)as (x-y)*(x+y), but will say further since x-y is an EVEN integer... (x-y)(x+y) should also be even integer...
But NO.... are we told that x and y are integers...
a) \(x = 5, y = 1\)........ YES \(x^2-y^2\)is an even integer.....
b) \(x=\frac{25}{3} , y=\frac{1}{3}\).........\(.x-y =\frac{25}{3}-\frac{1}{3}= \frac{24}{3} = 8............\).But \(x+y = \frac{25}{3}+\frac{1}{3} =\frac{26}{3.}.............\) so\(x^2-y^2 = 8*\frac{26}{3}..........\)NO,\(x^2-y^2\) is not an even integer..
Different answers....
Insuff
(2) x is an odd integer.
Nothing about y..
Insuff...
Combined..
We know x-y is an EVEN integer and x is an integer, so y will also be an integer ..
and \(x^2-y^2\) will be EVEN integer
Suff
C
General Discussion
05 Jun 2016, 01:25
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\(x^2\) - \(y^2\) = (x-y)*(x+y)
1) We know that (x - y) is even, but don't know if (x+y) is even or odd
If (x-y) & (x+y) is odd: (x+y)(x-y) = O*O = O
If (x-y) is odd & (x+y) is even: O*E = E
NOT SUFFICIENT
2) x is an odd integer
If x is odd, & y is odd, then:
(x-y) = O - O = E or 0
(x+y) = O + O = E or 0
(x+y)(x-y) = E * E = E
If x is odd, & y is even, then:
(x-y) = O - E = O
(x+y) = O + E = O
(x+y)(x-y) = O * O = O
NOT SUFFICIENT
1 & 2)
We know that (x-y) is even & that x is odd.
Given those circumstances, we can conclude that y is also odd:
(x-y) = O - O = E (In the formulas above we can see, that if y is even, then (x-y) would be odd)
If x is odd and y is odd, then:
(x-y)*(x+y) = E * E = E
Answer: C
1) We know that (x - y) is even, but don't know if (x+y) is even or odd
If (x-y) & (x+y) is odd: (x+y)(x-y) = O*O = O
If (x-y) is odd & (x+y) is even: O*E = E
NOT SUFFICIENT
2) x is an odd integer
If x is odd, & y is odd, then:
(x-y) = O - O = E or 0
(x+y) = O + O = E or 0
(x+y)(x-y) = E * E = E
If x is odd, & y is even, then:
(x-y) = O - E = O
(x+y) = O + E = O
(x+y)(x-y) = O * O = O
NOT SUFFICIENT
1 & 2)
We know that (x-y) is even & that x is odd.
Given those circumstances, we can conclude that y is also odd:
(x-y) = O - O = E (In the formulas above we can see, that if y is even, then (x-y) would be odd)
If x is odd and y is odd, then:
(x-y)*(x+y) = E * E = E
Answer: C
