Q. 145 - Intermediate Algebra

Question

Joy is preparing 15 liters of a 25% saline solution. She only has 40% and 10% solution in her lab. How many liters of the 40% and how many liters of the 10% should she mix to make the 25% solution?

Step-by-Step Solution

Answer

 liters of the 10% = 7.5

 liters of the 40%=  7.5

Step 1. Given Information

Total liters required= 15 liters.

15 liters of a 25% solution of an saline solution is required.

We have 40%  and 10%  solutions available.

Step 2. Formation of two linear Equations

Let number of liters of  40%  solutions required= .

Let number of liters of  10%  solutions requires =.

As we want  15 liters  in total so,

$x+y=15$

The trial mix is 

$$\begin{gathered} \left(0.40\right)x+\left(0.10\right)y=\left(0.25\right)\times 15 \\ \left(0.40\right)x+\left(0.10\right)y=3.75 \end{gathered}$$

Step 3. Solving the equations

From the first equation,

$x=15-y$

Now we will substitute this value in the second equation.

$$\begin{gathered} \left(0.40\right)\left(15-y\right)+\left(0.10\right)y=3.75 \\ 6-\left(0.4\right)y+\left(0.10\right)y=3.75 \\ -\left(0.30\right)y=-2.25 \\ y=7.5 \end{gathered}$$

No. of  liters of the 10% solutions =7.5

Now put this value in the equation 

$$\begin{gathered} x=15-y \\ x=15-7.5 \\ x=7.5 \end{gathered}$$

No. of  liters of the 40% solutions =7.5

Step 4. Check the solution

Substitute $7.5$ for $x$ and $7.5$ for $y$ in the first equation formed.

$\begin{array}{rcl}x+y& =& 15\\ 7.5+7.5& =& 15\\ 15& =& 15\end{array}$

It is a true statement.

Again, substitute the values in the second equation formed.

$\begin{array}{rcl}0.40x+0.10y& =& 3.75\\ 0.40·7.5+0.10·7.5& =& 3.75\\ 3+0.75& =& 3.75\\ 3.75& =& 3.75\end{array}$

This is also a true statement.

So the point satisfies both the equations.