Alligation or Mixture problems
Two solution of 90% and 97% purity and mixed resulting in 21 liters of mixture of 94% purity. How much is the quantity of the first solution in the resulting mixture?
Method 1 to solve the equation.Let us assume the number of liters of the 90% purity solution = Aand the number of liters of the 97% purity solution = B.According to question,Since there are 21 liters of the solution,A + B = 21 ...................... (1)Since after mixing the two solutions the new mixture has 94% purity,Concentrate of A + Concentrate of B = Concentrate of (A + B)A x 90% + B x 94% = (A+ B) x 97%? A x 90/100 + B x 97/100 = (A + B) x 94/100 ? 90A + 97B = (A + B) x 94? 90A + 97B = 94A + 94B ? 94A + 94B - 90A- 97B = 0? 4A - 3B = 0 ........................(2)Multiply the 3 with Equation (1) and add with Equation (2),3A + 3B + 4A - 3B = 63 + 0? 7A = 63 ? A = 63/7 = 9Put the value of A in Equation (1) , we will get 9 + B = 21B = 21 - 9B = 12The first solution would be A = 9 liters.Method 2 to solve the equation.Hit and trail method.94% is closer to 97% but barely meaning the mixtures will not be equal parts but will be slightly more of the higher purity. Quickly eliminate A and B. Out of the others 9 is the easy choice. If the other choices were closer to half this wouldn't work.