A pharmacist has an 18% alcohol solution. How much of this solution and how much water must be mixed together to make 10 liters of a 12% alcohol solution?
To make 10 L in total you will need x L of the 18% solution and 10-x L of water.
.18x + 0(10-x) = .12(10)
.18x = 1.2
x = 1.2/.18 = 120/18 = 20/3
He needs 20/3 L or 6 2/3 L of water.
(btw, a Russian would never dilute his alcohol)
plz help how much solution does he need?
To find out how much of the 18% alcohol solution and how much water needs to be mixed, we can use a proportion based on the concentration of alcohol.
Let's assume that x liters of the 18% alcohol solution need to be mixed with y liters of water to make 10 liters of a 12% alcohol solution.
The proportion can be set up as follows:
((18% alcohol concentration in x liters) + (0% alcohol concentration in y liters)) / 10 liters = 12% alcohol concentration
To solve the equation, we need to convert the percentages into decimal form:
(0.18x + 0) / 10 = 0.12
Now we can solve for x:
0.18x = 0.12 * 10
0.18x = 1.2
x = 1.2 / 0.18
x ≈ 6.67 liters
This means that approximately 6.67 liters of the 18% alcohol solution need to be mixed.
Next, we need to find out how much water (y liters) should be mixed with the alcohol solution. Since the total volume needs to be 10 liters, we can subtract the amount of alcohol solution from 10 liters:
y = 10 - 6.67
y ≈ 3.33 liters
Therefore, approximately 3.33 liters of water should be mixed with the 6.67 liters of the 18% alcohol solution to make 10 liters of a 12% alcohol solution.
Current work:
.18x+0(10-x)=1.2
@Reiny: THAT WAS A PHARMACIST... And also I go to a stupid math school. they make me do this.
help a Russian out, how much liters of water does he have?
Spasibo