Salt Water Concentration Problems ③ Applications and Problems Involving Adding Salt
In this lesson, we will cover problems involving adding salt itself, as well as compound problems such as removing some salt water and adding water.
Problems Involving Adding Salt
When adding salt, think of the salt as “salt water with a concentration of 100%.”
Example 1
If 20 g of salt is added to 400 g of 5% salt water, what will the concentration be?
Solution:
Original amount of salt: 400×5÷100＝20g
Salt after adding: 20＋20＝40g
Amount of salt water after adding: 400＋20＝420g
Concentration: 40÷420×100＝approximately 9.5%
If 20 g of salt is added to 400 g of 5% salt water, what will the concentration be?
Solution:
Original amount of salt: 400×5÷100＝20g
Salt after adding: 20＋20＝40g
Amount of salt water after adding: 400＋20＝420g
Concentration: 40÷420×100＝approximately 9.5%
Problems Involving Removing Salt Water and Adding Water
Problems in which some salt water is removed and then replaced with water are also common. Even after some salt water is removed, the concentration of the remaining salt water does not change.
Example 2
100 g is removed from 500 g of 10% salt water, and 100 g of water is added in its place. What is the concentration of the resulting salt water?
Solution:
Original amount of salt: 500×10÷100＝50g
Salt removed: 100×10÷100＝10g
Salt remaining: 50－10＝40g
Even after adding water, the amount of salt water is 500g
Concentration: 40÷500×100＝8%
100 g is removed from 500 g of 10% salt water, and 100 g of water is added in its place. What is the concentration of the resulting salt water?
Solution:
Original amount of salt: 500×10÷100＝50g
Salt removed: 100×10÷100＝10g
Salt remaining: 50－10＝40g
Even after adding water, the amount of salt water is 500g
Concentration: 40÷500×100＝8%
Adding Salt Until Reaching a Target Concentration
Example 3
How many grams of salt must be added to 300 g of 6% salt water to make it 15%?
Solution:
Original amount of salt: 300×6÷100＝18g
Let the amount of salt to add be xg:
(18＋x)÷(300＋x)×100＝15
18＋x＝15×(300＋x)÷100
1800＋100x＝4500＋15x
85x＝2700
x＝approximately 31.8g
How many grams of salt must be added to 300 g of 6% salt water to make it 15%?
Solution:
Original amount of salt: 300×6÷100＝18g
Let the amount of salt to add be xg:
(18＋x)÷(300＋x)×100＝15
18＋x＝15×(300＋x)÷100
1800＋100x＝4500＋15x
85x＝2700
x＝approximately 31.8g
Key Points
- Adding salt → both the amount of salt and the amount of salt water increase
- Adding water → the amount of salt does not increase; only the amount of salt water increases
- Removing some salt water → the concentration does not change; only the amount decreases
- When solving with an equation, focus on the “amount of salt” when setting up the equation
Practice Problems
- If 30 g of salt is added to 500 g of 4% salt water, what will the concentration be?
- 200 g is removed from 600 g of 8% salt water, and 200 g of water is added. What is the concentration?
Answers and Explanations
- Answer: Approximately 9.4%
Explanation: Original amount of salt: 500×4÷100＝20g. Salt after adding: 20＋30＝50g. Amount of salt water after adding: 500＋30＝530g. Concentration: 50÷530×100≈9.43%. (Precisely, 50/530×100≒9.4%.) - Answer: 5.33…% (approximately 5.3%)
Explanation: Original amount of salt: 600×8÷100＝48g. Salt removed: 200×8÷100＝16g. Salt remaining: 48－16＝32g. Even after adding water, the amount of salt water is 600g. Concentration: 32÷600×100＝16/3≒5.3%.