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10 Jul 2018, 20:59
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Difficulty:
75%
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Question Stats:
50% (01:49) correct
50%
(02:09)
wrong
based on 475
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If x and y are integers, is y an even integer?
(1) \(2y - x = x^2 - y^2\)
(2) x is an odd integer.
(1) \(2y - x = x^2 - y^2\)
(2) x is an odd integer.
10 Jul 2018, 21:59
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Bunuel
Statement 1: \(2y - x = x^2 - y^2\). \(2y\) will always be Even
\(=>y^2=x^2+x+Even\). Now irrespective of the value of \(x\), \(x^2+x\) will always be Even, because \(x\) is an integer.
Hence \(y^2=Even+Even=Even\). Sufficient
Statement 2: nothing mentioned about \(y\). Insufficient
Option A
General Discussion
10 Jul 2018, 22:05
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GIven X and Y are integers
We need to find if Y is an even integer
Statement 1
\(2y - x = x^{2}-y^{2}\)
=> \(2y = x^{2}-y^{2} + x\)
The left hand side of above equation is always even
=> \(x^{2}-y^{2} + x\) is even
Lets see the result for different combinations of X and Y
X | Y | \(x^{2}-y^{2} + x\)
Even | Even | Even
Odd | Even | Even
Odd | Odd | Odd
Even | Odd | Odd
We can see that Y is an even integer whenever \(x^{2}-y^{2} + x\) is even
Statement 1 is sufficient
Statement 2
X is an odd integer => doesn't tell anything about Y
Statement 2 not sufficient
Analysis of statement 1 and 2 together is not required
Hence option A
We need to find if Y is an even integer
Statement 1
\(2y - x = x^{2}-y^{2}\)
=> \(2y = x^{2}-y^{2} + x\)
The left hand side of above equation is always even
=> \(x^{2}-y^{2} + x\) is even
Lets see the result for different combinations of X and Y
X | Y | \(x^{2}-y^{2} + x\)
Even | Even | Even
Odd | Even | Even
Odd | Odd | Odd
Even | Odd | Odd
We can see that Y is an even integer whenever \(x^{2}-y^{2} + x\) is even
Statement 1 is sufficient
Statement 2
X is an odd integer => doesn't tell anything about Y
Statement 2 not sufficient
Analysis of statement 1 and 2 together is not required
Hence option A
