Problema Solution
A tank 100L contains 60% saline solution. How much of the solution must be drained an replaced with water in order to decrease the concentration to 50%?
Answer provided by our tutors
let
x = the water that needs to be added (it has 0%concentration of salt)
x = the amount that we will drain so we will be left with (100 - x) L from the 60% solution
60% of (100 - x) L + 0% of x = 50% of 100
0.6*(100 - x) + 0*x = 0.5*100
60 - 0.6*x = 50
x = (60 - 50) / 0.6
x = 16.67 L
we need to drain 16.67 L and then replace that amount with pure water.