Mixture problems and their solutions are presented along with their solutions. Percentages are also used to solve these types of problems.
Problem 1:
How many liters of 20% alcohol solution should be added to 40 liters of a 50% alcohol solution to make a 30% solution?
Solution to Problem 1:

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

Solve for x.
x = 80 liters

80 liters of 20% alcohol is be added to 40 liters of a 50% alcohol solution to make a 30% solution.
Problem 2:
John wants to make a 100 ml of 5% alcohol solution mixing a quantity of a 2% alcohol solution with a 7% alcohol solution. What are the quantities of each of the two solutions (2% and 7%) he has to use?
Solution to Problem 2:

Let x and y be the quantities of the 2% and 7% alcohol solutions to be used to make 100 ml. Hence
x + y = 100

We now write mathematically that the quantity of alcohol in x ml plus the quantity of alcohol in y ml is equal to the quantity of alcohol in 100 ml.
2% x + 7% y = 5% 100

The first equation gives y = 100  x. Substitute in the last equation to obtain
2% x + 7% (100  x) = 5% 100

Multiply by 100 and simplify
2 x + 700  7 x = 5 * 100

Solve for x
x = 40 ml

Substitute x by 40 in the first equation to find y
y = 100  x = 60 ml
Problem 3:
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?
Solution to Problem 3:

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 = 200 grams.

200 grams of Sterling Silver is needed to make the 91% alloy.
Problem 4:
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.
Solution to Problem 4:

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 = 200 Kilograms.
Problem 5:
A 50 ml aftershave lotion at 30% alcohol is mixed with 30 ml of pure water. What is the percentage of alcohol in the new solution?
Solution to Problem 5:

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%
Problem 6:
You add x ml of a 25% alcohol solution to a 200 ml of a 10% alcohol solution to obtain another solution. Find the amount of alcohol in the final solution in terms of x. Find the ratio, in terms of x, of the alcohol in the final solution to the total amount of the solution. What do you think will happen if x is very large? Find x so that the final solution has a percentage of 15%.
Solution to Problem 6:

Let us first find the amount of alcohol in the 10% solution of 200 ml.
200 * 10% = 20 ml

The amount of alcohol in the x ml of 25% solution is given by
25% x = 0.25 x

The total amount of alcohol in the final solution is given by
20 + 0.25 x

The ratio of alcohol in the final solution to the total amount of the solution is given by
[ ( 20 + 0.25 x ) / (x + 200)]

If x becomes very large in the above formula for the ratio, then the ratio becomes close to 0.25 or 25% (The above function is a rational function and 0.25 is its horizontal asymptote). This means that if you increase the amount x of the 25% solution, this will dominate and the final solution will be very close to a 25% solution.

To have a percentage of 15%, we need to have
[ ( 20 + 0.25 x ) / (x + 200)] = 15% = 0.15

Solve the above equation for x
20 + 0.25 x = 0.15 * (x + 200)
x = 100 ml
More math problems with detailed solutions in this site.