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This problem involves solving a system of equations based on the given conditions. Let's define the variables:

- Let \( x_1 \) be the number of liters of the 25% solution.
- Let \( x_2 \) be the number of liters of the 45% solution.
- Let \( x_3 \) be the number of liters of the 90% solution.

### Conditions:
1. The total mixture is 90 liters:
   \[
   x_1 + x_2 + x_3 = 90
   \]

2. The amount of 90% solution is twice the amount of 45% solution:
   \[
   x_3 = 2x_2
   \]

3. The final solution contains 55% acid, so the total acid from all solutions must equal 55% of 90 liters. Therefore:
   \[
   0.25x_1 + 0.45x_2 + 0.90x_3 = 0.55 \times 90
   \]
   Simplifying the right-hand side:
   \[
   0.25x_1 + 0.45x_2 + 0.90x_3 = 49.5
   \]

### System of Equations:
We now have the following system of equations:
1. \( x_1 + x_2 + x_3 = 90 \)
2. \( x_3 = 2x_2 \)
3. \( 0.25x_1 + 0.45x_2 + 0.90x_3 = 49.5 \)

#### Step-by-Step Solution:
1. Substitute \( x_3 = 2x_2 \) into the first and third equations.

   From the first equation:
   \[
   x_1 + x_2 + 2x_2 = 90 \implies x_1 + 3x_2 = 90 \quad \text{(Equation 4)}
   \]

   From the third equation:
   \[
   0.25x_1 + 0.45x_2 + 0.90(2x_2) = 49.5
   \]
   Simplifying:
   \[
   0.25x_1 + 0.45x_2 + 1.8x_2 = 49.5
   \]
   \[
   0.25x_1 + 2.25x_2 = 49.5 \quad \text{(Equation 5)}
   \]

2. Now solve the system of two equations (Equation 4 and Equation 5).

   From Equation 4:
   \[
   x_1 = 90 - 3x_2
   \]

   Substitute this into Equation 5:
   \[
   0.25(90 - 3x_2) + 2.25x_2 = 49.5
   \]
   Expanding:
   \[
   0.25 \times 90 - 0.75x_2 + 2.25x_2 = 49.5
   \]
   Simplifying:
   \[
   22.5 + 1.5x_2 = 49.5
   \]
   \[
   1.5x_2 = 49.5 - 22.5 = 27
   \]
   \[
   x_2 = \frac{27}{1.5} = 18
   \]

3. Now, use \( x_2 = 18 \) to find \( x_1 \) and \( x_3 \).

   From Equation 4:
   \[
   x_1 = 90 - 3 \times 18 = 90 - 54 = 36
   \]

   From \( x_3 = 2x_2 \):
   \[
   x_3 = 2 \times 18 = 36
   \]

### Final Answer:
- The chemist should use **36 liters** of 25% solution.
- The chemist should use **18 liters** of 45% solution.
- The chemist should use **36 liters** of 90% solution.

Would you like further details or explanations on any step?

---

Here are five related questions for further practice:
1. How would the solution change if the final mixture had 50% acid instead of 55%?
2. What if the chemist wanted to use three times as much 90% solution as 45% solution?
3. Can you solve this system if only two solutions were used instead of three?
4. How can you generalize the process for different percentages and total volume?
5. What would happen if the total mixture was 100 liters instead of 90?

**Tip:** When solving mixture problems, it's helpful to first express unknown quantities in terms of one variable to simplify the equations.

A chemist has three different acid solutions. The first acid solution contains 25% acid, the second contains 45% and the third contains 90%. They want to use all three solutions to obtain a mixture of 90 liters containing 55% acid, using 2 times as much of the 90% solution as the 45% solution. How many liters of each solution should be used?

Learn how to solve a system of equations to determine the volumes of different acid solutions required to create a 90-liter mixture with 55% acid concentration. This problem involves algebraic techniques and concentration calculations, suitable for high school students in Grades 10-12.

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