Final OA/OE after a couple of changes and updates.
Let me know if there is still an issue.
Explanation[rating1]yellow/793752[/rating1]
Official Answer: BStatement 1: \(xy = 6\) is sufficient to answer the question because solving the equation gives \(y = 3\) when \(x\) is 2 and \(y\) is -10 when \(x\) is \(-\frac{3}{5}\).
\(5x = y + 7\)
\(x = \frac{y + 7}{5}\) ................i
\(xy = 6\) ........................ ii
Replacing the value of \(x\) on eq ii: \(\frac{y(y + 7)}{5} = 6\)
\(y^2 + 7y = 30\)
\(y^2 + 10y - 3y - 30= 0\)
\(y (y + 10) - 3 (y + 10) = 0\)
\((y + 10) (y - 3) = 0\)
\(y = - 10\) or \(3\)
If \(y = - 10\), \(x = -3/5\). In this case, \(x - y = -3/5 - (-10) = 47/5\). Yes.
If \(y = 3\), \(x = 2\). In this case, \(x - y = 2 - 3 = -1\). No.
In each case (x-y) > 0 and < 0. Hence statement 1 is not sufficient.
Statement 2: If \(x\) and \(y\) are consecutive integers, \(x = 2\) and \(y = 3\).
When x and y are consecutive integers, either x = y+1 or y = x+1 is possible.
(i) If x = y+1 \(5x = y + 7\)
\(5(y+1) = y + 7\)
\(4y = 2\)
\(y = 1/2\). Then \(x = 3/2\). However this is not possible becase x and y are even not integers. So this option is not valid.
(ii) If y = x+1 \(5x = y + 7\)
\(5x = x + 1 + 7\)
\(4x = 8\)
\(x = 2\)
Then, \(y = 3\). This is valid because x and y are consecutive integers.
So (x-y) = 2-3 = -1. Hence statement 2 is sufficient.
Therefore B is correct.