Let the relations R1 and R2 on the set X = \1, 2, 3, , 20\ be given by R1 = \(x, y) : 2x - 3y = 2\ and R2 = \(x, y) : -5x + 4y = 0\. If M and N be the minimum number of elements required to be added in R1 and R2, respectively, in order to make the relations symmetric, then M + N equals
- 16
- 12
- 8
- 10
Answer
(4) 10
Solution
R1: x = 3y + 22 y even, x 20:
R1 = \(4, 2), (7, 4), (10, 6), (13, 8), (16, 10), (19, 12)\
no reverse pair (y, x) is in R1 add all 6 M = 6
R2: 4y = 5x x a multiple of 4, y 20:
R2 = \(4, 5), (8, 10), (12, 15), (16, 20)\ N = 4
M + N = 6 + 4 = 10
Answer: (4)