How many quarts of water must be mixed with 8 quarts of a 25% salt solutions to obtain a 17.5% salt solution?
thats not algebra
It is a question on my algebra homework!
To solve this problem, we need to determine how many quarts of pure water should be added to 8 quarts of a 25% salt solution to obtain a 17.5% salt solution.
Let's break down the information we have:
1. The initial mixture consists of 8 quarts of a 25% salt solution.
2. We want to add a certain amount of pure water to this mixture to dilute the salt concentration to 17.5%.
Now, let's calculate the salt content in the initial mixture:
Salt content in initial mixture = 8 quarts * 25% = 2 quarts of salt
Next, we can set up an equation based on the salt content in the initial mixture and the target salt concentration in the final solution. Let's assume x quarts of water are added:
Salt content in final solution = (2 quarts of salt) / (8 quarts + x quarts) = 17.5%
Using this equation, we can solve for x:
2/(8+x) = 17.5/100
To simplify the equation, let's convert the percentages to decimals:
2/(8+x) = 0.175
Now, we can solve for x by cross-multiplying:
2 = 0.175(8+x)
2 = 1.4 + 0.175x
0.175x = 2 - 1.4
0.175x = 0.6
x = 0.6 / 0.175
x ≈ 3.428
Therefore, approximately 3.428 quarts of pure water must be mixed with 8 quarts of a 25% salt solution to obtain a 17.5% salt solution.