A barrel contains some wine. A man drank 15 litres from

Solution

We know quantity of wine left is = P × {1 - (R/P)}ⁿ {Where 'P' is the initial quantity of wine, 'R' is the quantity withdrawn each time and 'n' is number of times the process is done.} Let the initial quantity of wine in the barrel be 'I' litres. 48 = I × {(I - 15)/I}² 48I = (I – 15)² 48I = I² – 30I + 225 I² – 78I + 225 = 0 I² – 75I – 3I + 225 = 0 I(I – 75) – 3(I – 75) = 0 (I – 3)(I – 75) = 0 I = 75, 3 Since, quantity of wine left in the barrel is 48 litres. So, initial quantity of wine cannot be 3 litres. So, initial quantity of wine in the barrel = 75 litres