Q. 147 - Intermediate Algebra

Question

A scientist needs 120 milliliters of a 20% acid solution for an experiment. The lab has available a 25% and a 10% solution. How many liters of the

25% and how many liters of the 10% solutions should the scientist mix to make the 20% solution?

Step-by-Step Solution

Answer

 Liters of the 25% solution = 80
 Liters of the 10% solution = 40

Step 1. Given Information

Total liters required= 120 milliliters.

120 milliliters of a 20% solution of an acid solution is required.

We have 25%  and 10%  solutions available.

Step 2. Formation of two linear Equations

Let the number of liters of  25%  solutions required = x

Let the number of liters of  10%  solutions requires = y

As we want  120 milliliters  in total so,

$x+y=120$

The trial mix is 

$$\begin{gathered} \left(0.25\right)x+\left(0.10\right)y=\left(0.20\right)\times 120 \\ \left(0.25\right)x+\left(0.10\right)y=24 \end{gathered}$$

Step 3. Solving the equations

From the first equation,

$x=120-y$

Now we will substitute this value in the second equation.

$$\begin{gathered} \left(0.25\right)\left(120-y\right)+\left(0.10\right)y=24 \\ 30-\left(0.25\right)y+\left(0.10\right)y=24 \\ -\left(0.15\right)y=-6 \\ 0.15y=6 \\ y=40 \end{gathered}$$

No. of  liters of the 10% solution = 40

Now put this value in the equation 

$$\begin{gathered} x=120-y \\ x=120-40 \\ x=80 \end{gathered}$$

No. of  liters of the 25% solutions = 80

Step 4. Check the solution

Substitute $80$ for $x$ and $40$ for $y$ in the first equation formed.

$\begin{array}{rcl}x+y& =& 120\\ 80+40& =& 120\\ 120& =& 120\end{array}$

It is a true statement.

Again, substitute the values in the second equation formed.

$\begin{array}{rcl}0.25x+0.10y& =& 24\\ 0.25·80+0.10·40& =& 24\\ 20+4& =& 24\\ 24& =& 24\end{array}$

This is also a true statement.

So the point satisfies both the equations.