A three-star medicine is prepared by mixing two type of fluids (A & B) in ratio of 1 : 4. Fluid A is prepared by mixing two type of water, water M and water N in ratio of 4 : 5. Fluid B is prepared by mixing two type of water, water X and water N in ratio of 3 : 7. The final mixture is prepared by 675 units of three-star medicine with water Y. If the concentration of the water N in the final mixture is 60%, then how much water Y had been added to three-star medicine? Correct Answer 302 In three-star medicine 1/5 is fluid A and 4/5 is fluid B Fluid A has water M and water N in ratio of 4 : 5 So, 5/9 of fluid A is water N The fraction of water N in three-star medicine from Fluid A = 1/5 × 5/9 = 1/9 Similarly, Fraction of water N in three-star medicine from Fluid B = 7/10 × 4/5 = 14/25 Fraction of water N in three-star medicine = 1/9 + 14/25 = 151/225 Units of water N in three-star medicine = 151/225 × 675 = 453 units Water N in the final mixture is 60% = 453 units Total units of water Y in final mixture (40%) = 453 × 40/60 = 302 units