mikemcgarry wrote:

If x and y are consecutive odd integers such that x < y, what is the value of y + x?

(1) The product of xy is negative.

(2) The sum x + y is the square of an integer.See the answer and a full video explanation here.

http://gmat.magoosh.com/questions/880 (1) Obviously one integer must be negative and the other one positive. Being both odd, the only possibility is -1 and 1.

Sufficient.

(2) There are infinitely many pairs of consecutive integers for which their sum is a square of an integer.

If \(x\) and \(y\) are consecutive odd integers, \(x<y\), then between them there is a unique even integer \(2k\), where \(2k=(x+y)/2.\)

\(x=2k-1\) and \(y=2k+1.\) Their sum is \(4k\) and necessarily \(4k=n^2\), or \(2k=n^2/2.\)

So, take any even integer \(n\), then \(\frac{n^2}{2}-1\) and \(\frac{n^2}{2}+1\) are two consecutive odd integers.

Tor example, 1 and 3, 7 and 9, 17 and 19, 31 and 33,...

Not sufficient.

Answer A

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