In a ‘10Y’ litres of mixture of milk and water, the

Solution

In a ‘10Y’ litres of mixture of milk and water, the quantity of milk is 60%. Quantity of milk in the initial mixture is 60%. Then the quantity of water in the initial mixture will be (100-60)% = 40%. Ratio of milk and water initially ⇒ 60% : 40% ⇒ 3 : 2 So the initial quantity of milk = 10Y of (3/5) = 6Y initial quantity of water = 10Y of (2/5) = 4Y (i) If (Y-54) litres of milk and (Y-90) litres of water are added into the mixture, then the ratio between the quantity of milk and water in the new mixture will be 11:7 respectively. [6Y+(Y-54)]/[4Y+(Y-90)] = 11/7 [7Y-54]/[5Y-90] = 11/7 49Y-378 = 55Y-990 55Y-49Y = 990-378 6Y = 612 Y = 102 Here the value of ‘Y’ will be the multiple of 3. (ii) If 25 litres of mixture is taken out from the mixture and 3 and 2 litres of milk and water is added into the mixture, then the quantity of milk and water in the new mixture will be (6Y-12) and (4Y-8) respectively. [6Y-25 of 60%+3]/[4Y-25 of 40%+2] = (6Y-12)/(4Y-8) [6Y-15+3]/[4Y-10+2] = (6Y-12)/(4Y-8) [6Y-12]/[4Y-8] = (6Y-12)/(4Y-8) Here both of the sides are equal. So the value of ‘Y’ cannot be determined. So we canot say that the value is the multiple of three or not. (iii) If (Y-40) litres of water is added into the mixture, then the quantity of milk will be 27.5% more than the quantity of water in the new mixture. [6Y]/[4Y+(Y-40)] = 127.5/100 [6Y]/[4Y+(Y-40)] = 127.5/100 [6Y]/[5Y-40] = 51/40 [6Y]/[Y-8] = 51/8 48Y = 51Y-408 51Y-48Y = 408 3Y = 408 Here the value of ‘Y’ will be the multiple of 3.