Question

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

(1) 8 

(2) 16 

(3) 12 

(4) 10

Answer 1

Correct option is (4) 10 

x = {1, 2, 3, .......20}

R₁ = {(x, y) : 2x – 3y = 2} 

R₂ = {(x, y) : –5x + 4y = 0}

R₁ = {(4, 2), (7, 4), (10, 6), (13, 8), (16, 10), (19, 12)}

R₂ = {(4, 5), (8, 10), (12, 15), (16, 20)}

in R₁ 6 element needed

in R₂ 4 element needed

So, total 6+4 = 10 element