Question

A full 275 L tank contains a 20% saline solution. How many litres must be replaced with a 100% saline solution to produce a full tank with a 45% saline solution? Round your final answer to 1 decimal place if necessary.

Answer 1

Let x be the volume of 20% solution in the tank after the given process

Let y be the volume of 100% solution used.

The sum of x and y needs to be equal to the final volume 275L:

$x+y=275$

The amount of substance (salt) in each solution is calculated by multipliying the volume by the concentration (in decimals); then, the amount of salt in 20% solution is 0.2x, in 100% solution is 1y and in the final solution (45%) is 0.45(275).

Sum amount in 20% solution with amount in 100% solution to get the amount in final solution:

$\begin{gathered} 0.2x+y=0.45\left(275\right) \\ 0.2x+y=123.75 \end{gathered}$

Use the next system of equations to answer the question:

$\begin{gathered} x+y=275 \\ 0.2x+y=123.75 \end{gathered}$

1. Solve x in the first equation:

$x=275-y$

2. Use the value of x (step 1) in the second equation:

$0.2\left(275-y\right)+y=123.75$

3. Solve y:

$\begin{gathered} 55-0.2y+y=123.75 \\ 55+0.8y=123.75 \\ 0.8y=123.75-55 \\ 0.8y=68.75 \\ y=\frac{68.75}{0.8} \\ \\ y=85.93 \end{gathered}$

The volume of 100% solution that needs to be used is 85.9 Litres.

Then, the litres that must be replaced with 100% solution to produce a full tank with 45% saline solution is 85.9