A chemist has three different acid solutions. The first acid solution contains $25 \%$ acid, the second contains $45 \%$ and the third contains $90 \%$. He wants to use all three solutions to obtain a mixture of 100 liters containing $40 \%$ acid, using 2 times as much of the $90 \%$ solution as the $45 \%$ solution. How many liters of each solution should be used?
The chemist should use liters of $25 \%$ solution, liters of $45 \%$ solution, and liters of $90 \%$ solution.

Step 1 ：Let's denote the amount of the $25\%$ solution as $x$, the amount of the $45\%$ solution as $y$, and the amount of the $90\%$ solution as $z$.

Step 2 ：We know that the total amount of solution is 100 liters, so we have the equation \(x + y + z = 100\).

Step 3 ：We also know that the chemist uses 2 times as much of the $90\%$ solution as the $45\%$ solution, so we have the equation \(z = 2y\).

Step 4 ：Finally, we know that the final solution contains $40\%$ acid, so we have the equation \(0.25x + 0.45y + 0.9z = 0.4 * 100\).

Step 5 ：We can solve this system of equations to find the values of $x$, $y$, and $z$.

Step 6 ：The solution to the system of equations is \(x = 70\), \(y = 10\), and \(z = 20\).

Step 7 ：Final Answer: The chemist should use \(\boxed{70}\) liters of $25 \%$ solution, \(\boxed{10}\) liters of $45 \%$ solution, and \(\boxed{20}\) liters of $90 \%$ solution.