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We are asked to find how many gallons of a 20% salt solution must be mixed with 4 gallons of a 30% salt solution to create a 25% salt solution.

Let:
- \( x \) = the number of gallons of the 20% salt solution that we need to find.

### Step-by-step solution:

1. **Set up the salt contribution equation**:

   - The amount of salt in the 20% solution is \( 0.20x \) (since 20% of \( x \) gallons is salt).
   - The amount of salt in the 4 gallons of 30% solution is \( 0.30 \times 4 = 1.2 \) gallons of salt.

   After mixing the two solutions, the total volume is \( x + 4 \) gallons, and we want this mixture to be 25% salt. Thus, the total amount of salt in the final solution will be \( 0.25(x + 4) \).

2. **Set up the equation**:

   The total salt from both solutions must equal the total salt in the final mixture:
   \[
   0.20x + 1.2 = 0.25(x + 4)
   \]

3. **Solve the equation**:

   \[
   0.20x + 1.2 = 0.25(x + 4)
   \]

   First, distribute the \( 0.25 \) on the right-hand side:
   \[
   0.20x + 1.2 = 0.25x + 1
   \]

   Now, subtract \( 0.20x \) from both sides:
   \[
   1.2 = 0.05x + 1
   \]

   Subtract 1 from both sides:
   \[
   0.2 = 0.05x
   \]

   Now, divide both sides by \( 0.05 \):
   \[
   x = \frac{0.2}{0.05} = 4
   \]

So, you need **4 gallons** of the 20% salt solution.

### Would you like further details, or do you have any questions?

Here are 5 related questions to consider:
1. What would happen if you used a 10% solution instead of a 20% solution?
2. How would you adjust the formula if the final solution needed to be 30% salt?
3. What is the total volume of the final solution after mixing?
4. Can this approach be applied to solutions with other types of solutes?
5. How would the equation change if you wanted to create a solution of a different volume?

**Tip:** When solving mixture problems, setting up an equation based on the concentration of the components is a powerful approach for finding the unknowns!

How many gallons of a 20% salt solution must be mixed with 4 gallons of a 30% salt solution to make a 25% salt solution?

Learn how to solve a mixture problem where a 20% salt solution is mixed with a 30% salt solution to create a 25% salt solution. This example includes detailed steps and algebraic methods, perfect for Grades 8-10 students.

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