Bunuel
If \(x>y>0\), is \(y<2\)?
(1) \(\frac{1}{x} = \frac{1}{2}\)
(2) \((\frac{1}{x})+(\frac{1}{y}) =1\)
\(x>y>0\)
On a number line, \(y\) lies to the left of \(x\)
------- \(0\) ------------ \(y\) ----------- \(x\) --------
Statement 1(1) \(\frac{1}{x} = \frac{1}{2}\)
\(x = 2\)
As y lies to the left of x on the number line, we can infer that \(y < 2\).
The statement alone is sufficient.
Statement 2(2) \((\frac{1}{x})+(\frac{1}{y}) =1\)
Assume, x = y. If that were the case, \(x = y = 2\) would satisfy this equation. However, \(x > y\).
Let's assume that \(x < 2\)
- the value of \(\frac{1}{x}\) will be greater than \(\frac{1}{2}\)
- as \(y\) is less than \(x\), the value of \(\frac{1}{y}\) will be greater than the value of \(\frac{1}{x}\)
- This situation will result in the sum of \(\frac{1}{y}\) and \(\frac{1}{x}\) greater than 1
\((\frac{1}{x})+(\frac{1}{y}) > 1\)
However, the statement states that \((\frac{1}{x})+(\frac{1}{y}) =1\). Therefore the value of \(x\) must be greater than 2. In such a case, the value of \(\frac{1}{x}\) will be less than \(\frac{1}{2}\). To compensate for the decrease in the value of \(\frac{1}{x }\) and to maintain the sum of \((\frac{1}{x})+(\frac{1}{y}) =1\), the value of \(y\) must now be less than \(2\), making \(\frac{1}{y} > \frac{1}{x}\).
Hence, we can conclude that \(y < 2\)
The statement is also sufficient.
Option D