There are two containers X and Y. X contains 100 ml of milk and Y contains 100 ml of water. 20 ml of milk from Xis transferred to Y. After mixing well, 20 ml of the mixture in Yis transferred back to X. If mdenotes the proportion of milk in Xand ndenotes the proportion of water in Y, then which one of the following is correct?

Here's a step-by-step breakdown:

1. Initial State:

   • Container X: 100 ml Milk, 0 ml Water.
   • Container Y: 0 ml Milk, 100 ml Water.

2. First Transfer (20 ml from X to Y):

   • Container X now has: 100 - 20 = 80 ml Milk.
   • Container Y now has: 20 ml Milk + 100 ml Water = 120 ml total mixture.
   • The proportion of milk in Y is 20/120 = 1/6.
   • The proportion of water in Y is 100/120 = 5/6.

3. Second Transfer (20 ml from Y back to X):

   • 20 ml of the mixture from Y is transferred.
   • Amount of milk transferred from Y to X = 20 ml * (1/6) = 20/6 ml.
   • Amount of water transferred from Y to X = 20 ml * (5/6) = 100/6 ml.

4. Final Composition:

   • Container X:
     • Milk in X = 80 ml (remaining) + 20/6 ml (transferred from Y) = (480 + 20)/6 = 500/6 ml.
     • Water in X = 0 ml + 100/6 ml (transferred from Y) = 100/6 ml.
     • Total volume in X = 500/6 + 100/6 = 600/6 = 100 ml.
   • Container Y:
     • Milk in Y = 20 ml (initial transfer) - 20/6 ml (transferred back) = (120 - 20)/6 = 100/6 ml.
     • Water in Y = 100 ml (initial) - 100/6 ml (transferred back) = (600 - 100)/6 = 500/6 ml.
     • Total volume in Y = 100/6 + 500/6 = 600/6 = 100 ml.

5. Calculate m and n:

   • m = proportion of milk in X = (Milk in X) / (Total in X) = (500/6) / 100 = 500/600 = 5/6.
   • n = proportion of water in Y = (Water in Y) / (Total in Y) = (500/6) / 100 = 500/600 = 5/6.

6. Comparison:

   • Since m = 5/6 and n = 5/6, we have m = n.

The correct answer is A) m = n.