Mixture problems occur in many different situations. For example, a store owner may wish to combine two goods in order to sell a new blend at a given price. A chemist may wish to obtain a solution of a desired strength by combining other solutions. In any case, mixture problems may all be solved by using the bucket method.

• The key to the bucket method is setting up the buckets correctly.

• Generally, the buckets will be set up as follows:

• Each bucket must contain two values: •

An amount (liters, tons, pounds, ounces, grams, etc.)

A type (usually either a percent or a price)

Example 1: How many pounds of coffee worth $1.00 per pound must be mixed with 15 pounds of coffee worth $1.60 per pound to obtain a blend worth $1.20 per pound?

•HOW MANY LITERS OF 20% ALCOHOL SOLUTION SHOULD BE ADDED TO 40 LITERS OF A 50% ALCOHOL SOLUTION TO MAKE A 30% SOLUTION?

• LET X BE THE QUANTITY OF THE 20% ALCOHOL SOLUTION TO BE ADDED TO THE 40 LITERS OF A 50% ALCOHOL. LET Y BE THE QUANTITY OF THE FINAL 30% SOLUTION. HENCE 

X + 40 = Y 

• WE SHALL NOW EXPRESS MATHEMATICALLY THAT THE QUANTITY OF ALCOHOL IN X LITERS PLUS THE QUANTITY OF ALCOHOL IN THE 40 LITERS IS EQUAL TO THE QUANTITY OF ALCOHOL IN Y LITERS. BUT REMEMBER THE ALCOHOL IS MEASURED IN PERCENTAGE TERM. 

20% X + 50% * 40 = 30% Y 

• SUBSTITUTE Y BY X + 40 IN THE LAST EQUATION TO OBTAIN. 

20% X + 50% * 40 = 30% (X + 40) 

• CHANGE PERCENTAGES INTO FRACTIONS. 

20 X / 100 + 50 * 40 / 100= 30 X / 100 + 30 * 40 / 100 

• MULTIPLY ALL TERMS BY 100 TO SIMPLIFY. 

20 X + 50 * 40 = 30 X + 30 * 40

•STERLING SILVER IS 92.5% PURE SILVER. HOW MANY GRAMS OF STERLING SILVER MUST BE MIXED TO A 90% SILVER ALLOY TO OBTAIN A 500G OF A 91% SILVER ALLOY?

•Let x and y be the weights, in grams, of sterling silver and of the 90% alloy to make the 500 grams at 91%. Hence 

x + y =500 

•The number of grams of pure silver in x plus the number of grams of pure silver in y is equal to the number of grams of pure silver in the 500 grams. The pure silver is given in percentage forms. Hence 

92.5% x + 90% y = 91% 500

•Substitute y by 500 - x in the last equation to write 

92.5% x + 90% (500 - x) = 91% 500 

•Simplify and solve 

92.5 x + 45000 - 90 x = 45500 

x = grams

• HOW MANY KILOGRAMS OF PURE WATER IS TO BE ADDED TO 100 KILOGRAMS OF A 30% SALINE SOLUTION TO MAKE IT A 10% SALINE SOLUTION.

•Let x be the weights, in Kilograms, of pure water to be added. Let y be the weight, in Kilograms, of the 10% solution. Hence x + 100 = y 

•Let us now express the fact that the amount of salt in the pure water (which 0) plus the amount of salt in the 30% solution is equal to the amount of salt in the final saline solution at 10%. 

0 + 30% 100 = 10% y 

•Substitute y by x + 100 in the last equation and solve. 

30% 100 = 10% (x + 100) 

•Solve for x. 

x = Kilograms

• A 50 ML AFTER-SHAVE LOTION AT 30% ALCOHOL IS MIXED WITH 30 ML OF PURE WATER. WHAT IS THE PERCENTAGE OF ALCOHOL IN THE NEW SOLUTION?

•The amount of the final mixture is given by 

50 ml + 30 ml = 80 ml 

•The amount of alcohol is equal to the amount of alcohol in pure water ( which is 0) plus the amount of alcohol in the 30% solution. Let x be the percentage of alcohol in the final solution. Hence 

0 + 30% 50 ml = x (80) 

•Solve for x x = 0.1817 = 18.75%