Final answer:
20 fluid ounces of water were added to dilute an initial 80 fluid ounces of lemonade from a 3:1 lemon-to-water ratio to a 3:2 ratio.
Step-by-step explanation:
The question involves solving a problem related to ratios and proportions, where we initially have a mixture of lemon and water in a 3:1 ratio in 80 fluid ounces of lemonade. We want to determine how much water was added to dilute the lemonade to a new ratio of 3:2. To find the answer, we need to work through a step-by-step process, first by understanding how much lemon and water are in the initial mixture and then determining the amount of water needed to achieve the new ratio.
Initial lemon to water ratio: 3:1, which means out of 4 parts total (3 parts lemon + 1 part water), 3 are lemon and 1 is water.
The total initial volume of lemonade: 80 fluid ounces. So, 60 fluid ounces are lemon and 20 fluid ounces are water (since 80/4 = 20).
We need the final ratio to be 3:2, which means out of 5 parts total, 3 are lemon (which is still 60 fluid ounces) and 2 are water.
Since the lemon portion remains constant, we find the new volume for the water to maintain the 3:2 ratio by setting up a proportion: 3/2 = 60/W.
W = (2/3) * 60 = 40 fluid ounces of water.
To find out how much water was added, we subtract the original water amount from the new amount: 40 - 20 = 20 fluid ounces.
Therefore, 20 fluid ounces of water were added to dilute the lemonade to the desired ratio.