Solving Mixture Problems
Mixture problems involve combining two or more things and determining some characteristic of either the ingredients or the resulting mixture. For example, we might want to know how much water to add to dilute a saline solution, or we might want to determine the percentage of concentrate in a jug of orange juice.
Introduction to Mixtures
We can use fractions, ratios, or percentages to describe quantities in mixtures.
If 200 grams of a saline solution has 40 grams of salt, what percentage of the solution is salt?
\(\frac{40}{200} = 0.20 = 20\%\) of the solution is salt.
If each can of orange juice concentrate contains the same amount of concentrate, which recipe will make the drink that is most orangey?
Using a Table to Problem Solve
There is a general strategy for solving these mixture problems that uses simple algebra organized with a chart.
How much 40% rubbing alcohol do we need to add to 90% rubbing alcohol to make a 50% solution of rubbing alcohol?
We could organize the data we are given in the following chart:
Solution Type Concentration Amount of Solution Amount of Pure Alcohol 40% rubbing alcohol 0.4 ? liters ? liters 90% rubbing alcohol 0.9 ? liters ? liters 50% rubbing alcohol 0.5 10 liters \(0.5(10) = 5\) liters In general, the rows of the chart are the mixture types that you have. The columns describe the amount of each compound you have and the concentration of that component in each mixture (represented as a decimal).
When you don't have some of the information needed to fill in the cart, use a variable instead.
Solution Type Concentration Amount of Solution Amount of Pure Compound 40% rubbing alcohol 0.4 \(x\) \(0.4(x)\) 90% rubbing alcohol 0.9 \(10-x\) \(0.9(10-x)\) 50% rubbing alcohol 0.5 10 \(0.5(10) = 5\) The amount of 40% solution that we'll need is unknown (so make it \(x\)). The amount of 90% solution that we'll need is also unknown, but must be \(10-x\) liters so that we'll, in total, make 10 liters of the final solution.
The amount of alcohol that each part of the mixture adds to the final result is equal to the amount of each solution mixed in, times the fraction of alcohol that that solution is made from.
To use this chart to solve the problem, we will use the fourth column as an equation to solve for \(x.\)
The 10 liters of our final mixture must have a total volume of 5 liters of alcohol in it in order to be 50% alcohol. Those 5 liters must come from a combination of the amount of 40% solution we mix in and the amount of 90% solution. If the volume (in liters) of 40% solution that we mix in is \(x,\) then \(0.4x\) will be the volume (in liters) of the amount of alcohol contributed. Similarly, \(0.9(10-x)\) will be the amount of alcohol contributed by the \(10-x\) liters of 90% alcohol solution that we add. Therefore, in total, \(0.4x + 0.9(10-x)\) must be equal to the 5 total liters we'll need in the final solution to make it 50% alcohol.
Solving the equation, \[\begin{align} 0.4x + (9 - 0.9x) &= 5\\ -0.5x + 9 &= 5\\ 0.5x &= 4\\ \Rightarrow x &= \frac{4}{0.5}= 8. \end{align}\] Therefore, we need \(x = 8\) liters of the 40% solution.
How many grams of pure water must be added to 40 grams of a 10% saline solution to make a saline solution that is 5% salt?
Let's set up a table to solve this problem, using \(x\) to represent the number of grams of water that must be added.
Solution Type Concentration Amount of Solution Total Amount of Salt water 0 \(x\) 0 10% solution 0.1 40 \((0.1)(40)\) 5% solution 0.05 40+\(x\) \((0.05)(40+x)\) Because the total amount of salt remains the same after we add the water, we can set up and solve the following equation: \[\begin{align} (0.1)(40)&=(0.05)(40+x) \\ 4 &= 2 + 0.05x \\ x&=40.\end{align}\]
We need to add \(40 \) grams of pure water to create the solution that is \(5\%\) salt.
I have a 100 ml mixture that is 20% isopropyl alcohol (80 ml of water and 20 ml of isopropyl alcohol).
How much more alcohol do I need to add to make the mixture 25% alcohol?
There is a 40 litre solution of milk and water in which the concentration of milk is 72%. How much water must be added to this solution to make it a solution in which the concentration of milk is 60% ?
Practice Problems
Let's practice using the strategies from above on a variety of problems.
Jill mixes 100 liters of A-beverage that contains 45% juice with 200 liters of B-beverage. The resulting C-beverage is 30% fruit juice. What is the percent of fruit juice in the 200 liters of the B-beverage?
Let's begin by making a table to show what we know.
Beverage Liters of Beverage Concentration Liters of Juice Beverage A 100 0.45 \((0.45)(100) = 45\) Beverage B 200 \(x\) \(200x\) Beverage C 300 0.30 \((0.30)(300)=90\) The total number of liters of juice in Beverages A and B must equal 90, so \[\begin{align} 45 + 200x &= 90 \\ 200x &= 45 \\ 0.225 &= x.\end{align}\]
The concentration of juice in Beverage B is 22.5%.
Unequal amounts of 40% and 10% acid solutions were mixed and the resulting mixture was 30% concentrated. However, the required concentration is 25%, so the Chemist added 300 cubic meters of 20% acid solution in order to get the required concentration. What was the original amount of 40% acid solution?
Write only the quantity without the units (cubic meters).
Legend:
wt = weight
Strawberries contain about 15 wt% solids and 85 wt% water. To make strawberry jam, crushed strawberries and sugar are mixed in a 45:55 mass ratio, and the mixture is heated to evaporate water until the residue contains one-third water by mass.
Question: 1.) Calculate how many pounds of strawberries are needed to make a pound of jam.
In a mixture of 60 litres, the ratio of milk and water 2 : 1. If this ratio is to be 1 : 2, then the quanity of water to be further added is: