Q. 148 - Intermediate Algebra

Question

A 40% antifreeze solution is to be mixed with a 70% antifreeze solution to get 240 liters of a 50% solution. How many liters of the 40% and how many liters of the 70% solutions will be used?

Step-by-Step Solution

Answer

liters of the 40% solution =160

liters of the 70% solution =80

Step 1. Given Information

Total liters required= 240

240 liters of a 50% solution of an antifreeze solution is required.

We have 40%  and 70%  solutions available

Step 2. Formation of two linear Equations

Let number of liters of  40%  solutions required=$x$ 

Let number of liters of  70%  solutions requires =$y$

As we want  240 liters  in total so,

$x+y=240$

The trial mix is 

$$\begin{gathered} \left(0.40\right)x+\left(0.70\right)y=\left(0.50\right)\times 240 \\ \left(0.40\right)x+\left(0.70\right)y=120 \end{gathered}$$

Step 3. Solving the equations

From the first equation,

$x=240-y$

Now we will substitute this value in the second equation.

$$\begin{gathered} \left(0.40\right)\left(240-y\right)+\left(0.70\right)y=120 \\ 96-\left(0.40\right)y+\left(0.70\right)y=120 \\ 0.30y=24 \\ y=80 \end{gathered}$$

No. of  liters of the 70% solution 80

Now put this value in the equation 

$$\begin{gathered} x=240-y \\ x=240-80 \\ x=160 \end{gathered}$$

No. of  liters of the 40% solutions =160

Step 4. Check the solution

Substitute $160$ for $x$ and $80$ for $y$ in the first equation formed.

$\begin{array}{rcl}x+y& =& 240\\ 160+80& =& 240\\ 240& =& 240\end{array}$

It is a true statement.

Again, substitute the values in the second equation formed.

$\begin{array}{rcl}0.40x+0.70y& =& 120\\ 0.40·160+0.70·80& =& 120\\ 64+56& =& 120\\ 120& =& 120\end{array}$

This is also a true statement.

So the point satisfies both the equations.