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Sagot :
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:
[tex]x+y=275[/tex]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:
[tex]\begin{gathered} 0.2x+y=0.45\left(275\right) \\ 0.2x+y=123.75 \end{gathered}[/tex]Use the next system of equations to answer the question:
[tex]\begin{gathered} x+y=275 \\ 0.2x+y=123.75 \end{gathered}[/tex]1. Solve x in the first equation:
[tex]x=275-y[/tex]2. Use the value of x (step 1) in the second equation:
[tex]0.2\left(275-y\right)+y=123.75[/tex]3. Solve y:
[tex]\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}[/tex]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
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