A pharmacist mixes 14 ounces of 68% saline solution with 55% saline solution. If the final mixture is 62% saline, how many ounces of the 55% solution were used?
55x+68(14-x)=62(14)
55x+143-68x=868
55x-68x=725
To determine the amount of the 55% saline solution used, we need to set up the equation based on the given information.
Let's assume that x ounces of the 55% saline solution were used.
Since the pharmacist mixed 14 ounces of the 68% saline solution with x ounces of the 55% saline solution, the total mixture will be (14 + x) ounces.
To calculate the amount of saline in the mixture, we multiply the amount of solution by the percentage of saline. For example, for the 68% saline solution, the amount of saline is 0.68 * 14 = 9.52 ounces.
Similarly, for the 55% saline solution, the amount of saline is 0.55 * x = 0.55x ounces.
Now, we can set up the equation:
(9.52 + 0.55x) / (14 + x) = 0.62
To solve this equation, we can cross-multiply:
9.52 + 0.55x = 0.62 * (14 + x)
Simplify:
9.52 + 0.55x = 8.68 + 0.62x
Subtract 0.55x from both sides:
9.52 - 0.55x = 8.68 + 0.07x
Subtract 8.68 from both sides:
9.52 - 8.68 - 0.07x = 0.07x
0.84 = 0.07x
Divide by 0.07:
0.84 / 0.07 = x
12 = x
Therefore, 12 ounces of the 55% saline solution were used.
let amount of 55% solution be x ounces
then amount of 68% solution is 14-x
solve for x:
.55x + .68(14-x) = .62(14)
multiply by 100
55x+ 68(14-x) = 62(14)
take over....