Chapter 8, Problem 15CT

To determine

To find liter’s of each solution required to obtain desired mixture

Answer to Problem 15CT

The 25 liters of 20% solution is mixed with 75 litres of 60% solution to obtain 100 liters of a 50% solution to get the desired mixture.

Explanation of Solution

Given:

One hundred litres of a 50% solutions is obtained by mixing 60% solution with a 20% solution.

Let x litres of 20% solution is mixed with y liters of 60% solution to obtain 100 liters of a 50% solution.

Then, the amount of solute as:

  $\begin{array}{l}\left[\begin{array}{cc}x& y\end{array}\right]\left[\begin{array}{c}0.2\\ 0.6\end{array}\right]=100\left[0.5\right]\hfill \\ \left[0.2x+0.6y\right]=\left[50\right]\hfill \\ 0.2x+0.6y=50\hfill \end{array}$

This equation can also be rewritten as:

  $x+3y=250$

Now, the total volume of the mixture is 100 liters.

So, $x+y=100$

Now, the system of linear equations is:

  $\begin{array}{l}x+y=100\hfill \\ x+3y=250\hfill \end{array}$

Here, the augmented matrix is:

  $\left[\begin{array}{cccc}1& 1& ⋮& 100\\ 1& 3& ⋮& 250\end{array}\right]$

The system can be solved by Gauss-Jordan elimination.

The reduced row-echelon form is obtained by the row transformations as:

  $\left[\begin{array}{cccc}1& 1& ⋮& 100\\ 0& 2& ⋮& 150\end{array}\right]$

  $R_2\to R_2-R_1$

  $\left[\begin{array}{cccc}1& 1& ⋮& 100\\ 0& 1& ⋮& 75\end{array}\right]$

  $R_2\to \frac{1}{2}{R}_2$

  $\left[\begin{array}{cccc}1& 0& ⋮& 25\\ 0& 1& ⋮& 75\end{array}\right]$

  $R_1\to R_1-R_2$

So,

  $\begin{array}{l}x=25\\ y=75\end{array}$

So, the solution of the system of linear equations is $x=25andy=75$ .

Hence, 25 liters of 20% solution is mixed with 75 litres of 60% solution to obtain 100 liters of a 50% solution.