UGC NET CSE | September 2013 | Part 3 | Question: 26

Question

Let A be a set of comfortable houses given as $A = \big\{ \frac{x_1}{0.8}, \frac{x_2}{0.9}, \frac{x_3}{0.1}, \frac{x_4}{0.7} \big \}$ and be the set of affordable houses $B = \big\{ \frac{x_1}{0.9}, \frac{x_2}{0.8}, \frac{x_3}{0.6}, \frac{x_4}{0.2} \big \}$ Then the set of comfortable and affordable houses is

• A. $ \big\{ \frac{x_1}{0.8}, \frac{x_2}{0.8}, \frac{x_3}{0.1}, \frac{x_4}{0.2} \big \}$
• B. $\big\{ \frac{x_1}{0.9}, \frac{x_2}{0.9}, \frac{x_3}{0.6}, \frac{x_4}{0.7} \big \}$
• C. $\big\{ \frac{x_1}{0.8}, \frac{x_2}{0.8}, \frac{x_3}{0.6}, \frac{x_4}{0.7} \big \}$
• D. $\big\{ \frac{x_1}{0.7}, \frac{x_2}{0.7}, \frac{x_3}{0.7}, \frac{x_4}{0.9} \big \}$

Answer 1

here we have to find A intersection B in fuzzy logic intersection takes min of 2 value so ans is A

x10.8,x20.8,x30.1,x40.2}