Let the relations R₁ and R₂ on the set X={1,2,3, …, 20} be given by R₁={(x, y): 2 x-3 y=2} and R₂={(x, y):-5 x+4 y=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

Solution

Maths sir · 4 steps

List every pair in each relation; each pair needs its mirror image added, so just count the pairs.

1. Pairs in R1. From 2x - 3y = 2 we get x = (3y+2)/2, so y must be even. Taking y = 2, 4, …, 12 gives x = 4, 7, 10, 13, 16, 19. At y = 14, x = 22 falls outside X.
2. Pairs in R2. From -5x + 4y = 0 we get y = 5x/4, so x must be a multiple of 4. Taking x = 4, 8, 12, 16 gives y = 5, 10, 15, 20. At x = 20, y = 25 falls outside X.

   Relation	Pairs (x, y)	Reverse pairs to add
   R₁	(4,2), (7,4), (10,6), (13,8), (16,10), (19,12)	M = 6
   R₂	(4,5), (8,10), (12,15), (16,20)	N = 4

3. Check the reverses. No pair has x = y, and no reversed pair satisfies its own equation. For example, (2,4) gives 2(2) - 3(4) = -8 ≠ 2. So every reversed pair is genuinely missing and must be added.
4. So. M + N = 6 + 4 = 10, which is option (d).

The idea to keep. To make a relation symmetric, add the mirror (y, x) of every pair (x, y) that does not already have one. Count the solutions in the set carefully, then check none are already mirrored.

Answer from the NTA final answer key. Solution by PrepWiser.