Bunuel wrote:
If \(x+2y≠0\), is \(\frac{x}{x+2y}<1\)?
(1) \(10|x|=|y|\)
(2) \(x>0\)
Let's simplify the question asked -
\(\frac{x}{x+2y}<1\)
\(\frac{x}{x+2y} - 1 < 0\)
\(\frac{x - (x+2y) }{(x+2y)} < 0\)
\(\frac{-2y }{(x+2y)} < 0\)
\(\frac{2y }{(x+2y)} > 0\)
Case 1: y > 0
If \(y > 0\), \(x+2y > 0\)
\(x > -2y\)
Case 2: y > 0
If \(y < 0\), \(x+2y < 0\)
\(x < -2y\)
Statement 1(1) \(10|x|=|y|\)Inference: The distance of \(y\) from zero is \(10\) times the distance of \(x\) from zero. Hence, we can conclude that \(y\) lies farther from zero than \(x\).
The placements are shown below -
Attachment:
Screenshot 2023-09-27 121245.png [ 41.3 KiB | Viewed 503 times ]
Case 1: y > 0
\(x\) can lie only in the positions shown below. Hence, in both scenarios \(x\) lies to the right of \(-2y\). Hence, we can say that \(x > -2y\)
Attachment:
Screenshot 2023-09-27 121733.png [ 16.18 KiB | Viewed 499 times ]
Case 2: y < 0
\(x\) can lie only in the positions shown below. Hence, in both scenarios \(x\) lies to the left of \(-2y\). Hence, we can say that \(x < -2y\)
Attachment:
Screenshot 2023-09-27 121957.png [ 15.05 KiB | Viewed 506 times ]
Hence, we can conclude that statement 1 is sufficient. We can eliminate B, C, and E.
Statement 2(2) \(x>0\)This statement tells us that \(x\) is positive. However, we do not have any information on \(y\). Hence, the statement alone is not sufficient.
Option A