Let N denote the set of all natural numbers. Define two binary relations on N as R1 = {(x,y) ∈ N × N: 2x + y = 10} and R2 = {(x,y) ∈ N × N: x + 2y = 10}. Then:

Define two binary relations on N as R1 = {(x,y) ∈ N × N: 2x + y = 10} and R2 = {(x,y) ∈ N × N: x + 2y = 10}
From R1, 2x + y = 10 and x, y ∈ N
So, possible values for x and y are:
x = 1, y = 8 i.e (1,8)
x = 2, y = 6 i.e (2,6)
x = 3, y = 4 i.e (3,4)
x = 4, y = 2 i.e (4,2)
R1 = {(1,8), (2,6), (3,4), (4,2)}
Therefore, Range of R1 is {2,4,6,8}
R1 is not symmetric
Also, R1 is not transitive because (3,4), (4,2) ∈ R1 but (3,2) ∉ R1
Thus, options A, B and D are incorrect.
From R2, x + 2y = 10 and x, y ∈ N
So, possible values for x and y are:
x = 8, y = 1 i.e (8,1)
x = 6, y = 2 i.e (6,2)
x = 4, y = 3 i.e (4,3)
x = 2, y = 4 i.e (2,4)
R2 = {(8,1), (6,2), (4,3), (2,4)}
Therefore, Range of R2 is {1,2,3,4}
R2 is not symmetric
Hence, option C is correct.