Bunuel wrote:
Official Solution:
Statement 1: \(xy = 6\) is not 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 \(y = - 10\), \(x = -\frac{3}{5}\). In this case, \(x - y = -\frac{3}{5} - (-10) = \frac{47}{5}\). Yes.
If \(y = 3\), \(x = 2\). In this case, \(x - y = 2 - 3 = -1\). No.
In each case \((x-y) \gt 0\) and \(\lt 0\). 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 \(x = y+1\)
\(5x = y + 7\)
\(5(y+1) = y + 7\)
\(4y = 2\)
\(y = \frac{1}{2}\). Then \(x = \frac{3}{2}\). However this is not possible because \(x\) 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.
Answer: B
Hi, Thank you for the detailed explanation.
The question is asking : is (x-y)>0?
Let us go straight to the last line of your explanation, when y
=x+1, x-y= -1, which is <0. Therefore (2) alone is not sufficient
Please advise.
Thanks