There are two containers X and Y. X contains 100ml of milk and y contains 100ml of water. 20ml of milk from X is transferred to Y. After mixing well,…

Correct Option (1)

To determine the relationship between 'm' (proportion of milk in X) and 'n' (proportion of water in Y), we analyze the contents of the containers through the two transfer steps:

• Initial State:
  • Container X: 100 ml Milk
  • Container Y: 100 ml Water
• Step 1: 20 ml of milk from X is transferred to Y.
  • Container X now contains: 100 ml - 20 ml = 80 ml Milk.
  • Container Y now contains: 100 ml Water + 20 ml Milk = 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.
• Step 2: 20 ml of the mixture from Y is transferred back to X.
  • The 20 ml mixture transferred from Y to X will contain:
    • Milk: (1/6) * 20 ml = 20/6 ml
    • Water: (5/6) * 20 ml = 100/6 ml
• Final Composition of Container X:
  • Total milk in X = (Milk remaining in X from Step 1) + (Milk transferred from Y)
  • Total milk in X = 80 ml + 20/6 ml = (480 + 20)/6 ml = 500/6 ml.
  • Total volume in X = 80 ml (remaining) + 20 ml (transferred) = 100 ml.
  • The proportion of milk in X (m) = (500/6 ml) / 100 ml = 500/600 = 5/6.
• Final Composition of Container Y:
  • Total water in Y = (Water initially in Y) - (Water transferred from Y)
  • Total water in Y = 100 ml - 100/6 ml = (600 - 100)/6 ml = 500/6 ml.
  • Total volume in Y = 120 ml (after Step 1) - 20 ml (transferred out) = 100 ml.
  • The proportion of water in Y (n) = (500/6 ml) / 100 ml = 500/600 = 5/6.

Since m = 5/6 and n = 5/6, it is concluded that m = n.

Incorrect Options:

Options 2 (m > n), 3 (m < n), and 4 (Cannot be determined due to insufficient data) are incorrect because, as demonstrated by the step-by-step calculation, the proportion of milk in container X (m) is precisely equal to the proportion of water in container Y (n) after both transfers are completed. The quantities are definitively determinable and found to be equal.