Question

Pure acid is to be added to a 20% acid solution to obtain 44 L of a 40% acid solution.What amounts of each should be used?

Answer 1

Let:

x be the liters of 20% acid solution.

y be the liters of pure acid solution.

To find x and y, follow the steps below.

Step 01: Write an equation for the number of liters.

The number of liters of the solution is 44, which is the sum of x and y.

$x+y=44$

Step 02: Isolate y in the equation above.

To do it, subtract x from both sides.

$\begin{gathered} x+y-x=44-x \\ y=44-x \end{gathered}$

Step 03: Write an equation that expresses the total of acid.

The total of acid is 0.4*44, which is equal to 1y plus 0.2x.

$\begin{gathered} 0.4\cdot44=1\cdot y+0.2\cdot x \\ 17.6=y+0.2x \end{gathered}$

Step 04: Substitute y by 44 - x in the equation above and isolate x.

$\begin{gathered} 17.6=44-x+0.2x \\ 17.6=44-0.8x \end{gathered}$

To isolate x, subtract 44 from both sides.

$\begin{gathered} 17.6-44=44-0.8x-44 \\ -26.4=-0.8x \\ \end{gathered}$

Now, divide both sides by -0.8.

$\begin{gathered} \frac{-26.4}{-0.8}=\frac{-0.8}{-0.8}x \\ 33=x \end{gathered}$

Step 05: Use the equation from step 2 to find y.

$\begin{gathered} y=44-x \\ y=44-33 \\ y=11 \end{gathered}$

Answer:

x is the liters of 20% acid solution = 33 L.

y is the liters of pure acid solution = 11 L.