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Let's solve this problem step by step.

### Information provided:
- There are three acid solutions:
  - First solution: 25% acid.
  - Second solution: 40% acid.
  - Third solution: 80% acid.
- The total mixture should be 78 liters containing 55% acid.
- The amount of the 80% solution is 3 times the amount of the 40% solution.

Let the volume of each solution be:
- \( x \) liters of the 25% solution,
- \( y \) liters of the 40% solution,
- \( 3y \) liters of the 80% solution (since the amount of the 80% solution is 3 times the amount of the 40% solution).

### Step 1: Total volume equation
The total volume of the solution is 78 liters, so:
\[
x + y + 3y = 78
\]
This simplifies to:
\[
x + 4y = 78 \quad \text{(Equation 1)}
\]

### Step 2: Total acid equation
The total amount of acid in the mixture should be 55% of 78 liters, which is:
\[
0.55 \times 78 = 42.9 \text{ liters of acid.}
\]

The total amount of acid from each solution is:
- \( 0.25x \) liters from the 25% solution,
- \( 0.40y \) liters from the 40% solution,
- \( 0.80 \times 3y = 2.4y \) liters from the 80% solution.

Thus, the total acid content equation becomes:
\[
0.25x + 0.40y + 2.4y = 42.9
\]
Simplifying this:
\[
0.25x + 2.8y = 42.9 \quad \text{(Equation 2)}
\]

### Step 3: Solve the system of equations
We have the two equations:
1. \( x + 4y = 78 \)
2. \( 0.25x + 2.8y = 42.9 \)

From Equation 1, solve for \( x \):
\[
x = 78 - 4y
\]

Substitute this into Equation 2:
\[
0.25(78 - 4y) + 2.8y = 42.9
\]
Expanding:
\[
0.25 \times 78 - 0.25 \times 4y + 2.8y = 42.9
\]
\[
19.5 - y + 2.8y = 42.9
\]
\[
1.8y = 42.9 - 19.5
\]
\[
1.8y = 23.4
\]
\[
y = \frac{23.4}{1.8} = 13
\]

### Step 4: Find \( x \) and \( 3y \)
Now that \( y = 13 \):
- From Equation 1: \( x + 4 \times 13 = 78 \)
\[
x + 52 = 78
\]
\[
x = 78 - 52 = 26
\]
- \( 3y = 3 \times 13 = 39 \)

### Final answer:
- The chemist should use:
  - 26 liters of 25% solution,
  - 13 liters of 40% solution,
  - 39 liters of 80% solution.

Would you like further details or have any questions?

Here are 5 related questions to explore further:
1. What happens to the acid concentration if more 25% solution is used?
2. How would the equations change if the 80% solution were only twice the amount of the 40% solution?
3. Can you solve for the volume of each solution if the final concentration were 60%?
4. How does the ratio between the solutions affect the total volume?
5. What if the chemist had to create 100 liters of solution with the same concentration?

**Tip**: Always double-check your system of equations when solving for multiple variables to avoid calculation errors.

A chemist has three different acid solutions. The first acid solution contains 25% acid, the second contains 40%, and the third contains 80%. They want to use all three solutions to obtain a mixture of 78 liters containing 55% acid, using 3 times as much of the 80% solution as the 40% solution. How many liters of each solution should be used?

Learn how to solve a system of linear equations to find the amount of acid in three solutions (25%, 40%, 80%) needed to create 78 liters of a 55% acid mixture. This problem involves algebra, percentages, and system of equations, suitable for high school students.

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