Salt Water Concentration Problems ② Mixture Problems
In this lesson, you will learn about “mixture problems,” in which two types of salt water are mixed. We use the fact that the total amount of salt does not change.
How Mixing Works
When two types of salt water are mixed:
(Amount of salt after mixing) = (Salt in salt water A) + (Salt in salt water B)
Example 1
200 g of 5% salt water and 100 g of 15% salt water are mixed. What is the concentration after mixing?
Solution:
Salt in the 5% salt water: 200×5÷100＝10 g
Salt in the 15% salt water: 100×15÷100＝15 g
Salt water after mixing: 200＋100＝300 g
Salt after mixing: 10＋15＝25 g
Concentration: 25÷300×100＝8.33…≒8.3%
200 g of 5% salt water and 100 g of 15% salt water are mixed. What is the concentration after mixing?
Solution:
Salt in the 5% salt water: 200×5÷100＝10 g
Salt in the 15% salt water: 100×15÷100＝15 g
Salt water after mixing: 200＋100＝300 g
Salt after mixing: 10＋15＝25 g
Concentration: 25÷300×100＝8.33…≒8.3%
Alligation Method
Using the alligation method, you can quickly find the ratio of the amounts to be mixed.
How the alligation method works:
Using the concentration after mixing as the fulcrum, the ratio of the amounts to be mixed is the inverse ratio of the differences between each salt water concentration and the final concentration.
Example 2 (Alligation Method)
You want to make 7% salt water by mixing 3% salt water and 12% salt water. What is the ratio of the amounts of the two types of salt water?
Solution:
Difference on the 12% side: 12－7＝5
Difference on the 3% side: 7－3＝4
Ratio of amounts to be mixed (3%：12%)＝5：4
You want to make 7% salt water by mixing 3% salt water and 12% salt water. What is the ratio of the amounts of the two types of salt water?
Solution:
Difference on the 12% side: 12－7＝5
Difference on the 3% side: 7－3＝4
Ratio of amounts to be mixed (3%：12%)＝5：4
Key Points
- The total amount of salt is the key to mixture problems
- Alligation method: the inverse ratio of the differences gives the mixing ratio
- When adding salt, treat it as “100% salt water”
Practice Problems
- What concentration of salt water results from mixing 300 g of 8% salt water and 200 g of 2% salt water?
- You want to make 500 g of 10% salt water by mixing 6% and 14% salt water. How many grams of each should be mixed?
Answers and Explanations
- Answer: 5.6%
Explanation: Salt in the 8% salt water: 300×8÷100＝24 g. Salt in the 2% salt water: 200×2÷100＝4 g. Salt water after mixing: 300＋200＝500 g. Salt after mixing: 24＋4＝28 g. Concentration: 28÷500×100＝5.6%. - Answer: 250 g of 6% salt water and 250 g of 14% salt water
Explanation: Alligation method: 14－10＝4, 10－6＝4. The ratio is 6%：14%＝4：4＝1：1. Dividing 500 g in a 1：1 ratio gives 250 g of each.