Question 1109268
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The algebraic solution provided by the other tutor is fine.<br>
Here is another way to solve these "mixture" problems.<br>
The desired 50% is "3 times as close" to 60% as it is to 20%. (60-50 = 10; 50-20 = 30).<br>
That means the amount of the 60% solution much be 3 times as much as the amount of the 20% solution.<br>
Since the chemist is using 8 liters of the 60% solution, the amount of the 20% solution required is 1/3 of 8 liters, or 8/3 liters, or 2.67 liters (approximately).