Question

How much of a 60% saline solution would you need to add to pure water to create 10 L of a 50% saline solution ?

Answer 1

so say, we add "x" of the 60% saline solution, now, how much salt is in that solution? now 60% is salt but the rest is something else, but only salt is 60% of "x", or (60/100) * x, or 0.6x.

likewise, for the 10 Liters, if only 50% is salt, then is just (50/100) * 10 or 5.

now... let's check the water whilst we add "y" amount, the water has no salt at all, freshwater for that matter, so, salinity is 0%. or (0/100) * y or 0.


`\bf \begin{array}{lccclll} &\stackrel{liters}{amount}&\stackrel{\textit{\% of salt}}{quantity}&\stackrel{\textit{total salt}}{quantity}\\ &------&------&------\\ \textit{60\% sol'n}&x&0.60&0.6x\\ \textit{pure water}&y&0.00&0y\\ ------&------&------&------\\ mixture&10&0.50&5 \end{array}`


`\bf \begin{cases} x+y=10\\ 0.6x+0y=5 \end{cases} \\\\\\ 0.6x=5\implies x=\cfrac{5}{0.6}\implies x=\cfrac{25}{3}\implies x=8(1)/(3)`