Mixture Algebra Word Problem Game

Mixture Algebra Word Problem Quiz/Game
This game focuses on solving word problems involving Mixtures. Mixture problems involve combining two or more substances with different concentrations to achieve a specific final concentration. The key to solving them is understanding that the total amount of the specific substance (salt, acid, sugar, etc.) remains the same before and after mixing. Scroll down the page for a more detailed explanation.

 

 

How to Play the Mixture Algebra Explorer Game

1. Look at the Problem: Read the problem carefully. Solve it and select one of the answers.
   
2. Check Your Work: If you selected the right answer, it will be highlighted in green. If you are wrong, it will be highlighted in red and the correct answer will be highlighted in green. A hint will be given to help you find the correct answer.
   
3. Get a New Problem: Click “Next Question” for a new problem.
   Your score is tracked, showing how many you’ve gotten right.
   
4. Finish Game When you have completed 10 questions, your final score will be displayed.
    
   

How to Solve Mixture Algebra Word Problems
The Golden Formula
The fundamental equation for almost every mixture problem is:
\(V_1C_1 + V_2C_2 = V_{total}C_{final}\)
Where:
\(V_1, V_2\): The volumes (or weights) of the individual solutions.
\(C_1, C_2\): The concentrations (percentages) of those solutions.
\(V_{total}\): The sum of the volumes (\(V_1 + V_2\)).
\(C_{final}\): The concentration of the resulting mixture.

Example:
A chemist has one solution that is 20% acid and another that is 50% acid. How much of each should be mixed to create 10 liters of a solution that is 32% acid?
Step 1: Define Variables
Let x be the number of liters of the 20% solution.
Since the total volume must be 10 liters, the remaining volume must be (10 - x) liters for the 50% solution.
Step 2: Organize the Data
Organize your data in a table to visualize the equation:

Solution	Volume (V)	Concentration (C)	Pure Acid (V × C)
Weak Solution	x	0.20	0.20x
Strong Solution	10 - x	0.50	0.50(10 - x)
Final Mixture	10	0.32	10(0.32)

 

Step 3: Create the Equation
The acid from the weak solution plus the acid from the strong solution must equal the acid in the final mixture:
0.20x + 0.50(10 - x) = 10(0.32)
Step 4: Solve the Equation
Distribute and multiply:
0.20x + 5 - 0.50x = 3.2
Combine like terms (x):
-0.30x + 5 = 3.2
Subtract 5 from both sides:
-0.30x = -1.8
Divide by -0.30:
\(x = \frac{-1.8}{-0.3} = 6\)
Step 5: Final Answer & Check
Weak Solution (20%): x = 6 liters
Strong Solution (50%): 10 - 6 = 4 liters
Quick Check: Does 0.20(6) + 0.50(4) = 3.2?
1.2 + 2.0 = 3.2.

Special Scenarios
Adding Pure Water (Dilution)
When you add pure water, the concentration of the “substance” is 0%.

Adding Pure Substance (Strengthening)
When you add pure acid, salt, or sugar, the concentration is 100%.

Helpful Tips
Estimate First: If you mix 10% and 40% acid, your answer must be between 10% and 40%.
Check Units: Ensure all volumes are in the same unit (liters, gallons, ounces) before starting.
Evaporation: If a problem mentions evaporation, the volume is subtracted, but the amount of substance usually stays the same because only water evaporates.
 

This video gives a clear, step-by-step approach to explain how to solve mixture algebra word problems.

• Show Video

 

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