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This problem is a typical mixture problem where we need to determine how much of each solution (30% and 60%) should be used to get a desired solution of 94 liters of 32% acid.

Let the amount of the 30% solution be \( x \) liters. Then, the amount of the 60% solution will be \( 94 - x \) liters (since the total volume of the final solution is 94 liters).

We can set up the following equation based on the percentage of acid in each solution:

\[
0.30x + 0.60(94 - x) = 0.32(94)
\]

### Solving the equation:
1. Expand the terms:
   \[
   0.30x + 0.60(94 - x) = 0.32 \times 94
   \]
   \[
   0.30x + 0.60 \times 94 - 0.60x = 0.32 \times 94
   \]
   \[
   0.30x + 56.4 - 0.60x = 30.08
   \]

2. Combine like terms:
   \[
   -0.30x + 56.4 = 30.08
   \]

3. Subtract 56.4 from both sides:
   \[
   -0.30x = 30.08 - 56.4
   \]
   \[
   -0.30x = -26.32
   \]

4. Divide by -0.30:
   \[
   x = \frac{-26.32}{-0.30} = 87.7
   \]

Thus, \( x = 87.7 \) liters of the 30% solution should be used.

### Final answer: 
The chemist should use **87.7 liters** of the 30% solution.

Let me know if you'd like further details on this solution! Here are some related questions to deepen your understanding:

1. What would happen if the desired solution was 40% acid instead of 32%?
2. How would the equation change if the chemist only had a 50% solution instead of 60%?
3. Can you determine how much of the 60% solution would be needed?
4. How do you verify if the solution makes sense with the given constraints?
5. How does changing the total volume to something like 150 liters affect the calculation?

**Tip:** When solving mixture problems, setting up the equation based on total percentages is key to avoiding confusion in larger problems.

A chemist has 30% and 60% solutions of acid available. How many liters of 30% solution should be used to obtain 94 liters of 32% acid solution? Give your answer with 1 decimal place (e.g., 4.2).

Solve the mixture problem of how much 30% acid solution is needed to make 94 liters of 32% acid solution. This problem involves setting up a linear equation based on percentages and solving for the unknown. Suitable for Grades 9-12.

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