Q. 144 - Intermediate Algebra

Question

Jotham needs 70 liters of a 50% solution of an alcohol solution. He has a 30% and an 80% solution available. How many liters of the 30% and how many liters of the 80% solutions should he mix to make the 50% solution?

Step-by-Step Solution

Answer

No. of  liters of the 30% solutions =42

No. of  liters of the 80% solutions =28

Step 1. Given Information

Total liters required= 70 liters.

70 liters of a 50% solution of an alcohol solution is required.

We have 80%  and 30%  solutions available.

Step 2. Formation of two linear Equations

Let number of liters of  30%  solutions required= $x$.

Let number of liters of  80%  solutions requires =$y$.

As we want  70 liters  in total so,

$x+y=70$

The trial mix is 

$$\begin{gathered} \left(0.30\right)x+\left(0.80\right)y=\left(0.50\right)\times 70 \\ \left(0.30\right)x+\left(0.80\right)y=35 \end{gathered}$$

Step 3. Solving the equations

From the first equation,

$x=70-y$

Now we will substitute this value in the second equation.

$$\begin{gathered} \left(0.30\right)\left(70-y\right)+\left(0.80\right)y=35 \\ 21-\left(0.3\right)y+\left(0.80\right)y=35 \\ \left(0.5\right)y=14 \\ y=28 \end{gathered}$$

No. of  liters of the 80% solutions =28

Now put this value in the equation 

$$\begin{gathered} x=70-y \\ x=70-28 \\ x=42 \end{gathered}$$

No. of  liters of the 30% solutions =42

Step 4. Check the solution

Substitute $42$ for $x$ and $28$ for $y$ in the first equation formed.

$\begin{array}{rcl}x+y& =& 70\\ 42+28& =& 70\\ 70& =& 70\end{array}$

It is a true statement.

Again, substitute the values in the second equation formed.

$\begin{array}{rcl}0.30x+0.80y& =& 35\\ 0.30·42+0.80·28& =& 35\\ 12.6+22.4& =& 35\\ 35& =& 35\end{array}$

This is also a true statement.

So the point satisfies both the equations.