you need to use 3 solutions of drug H containing respectively 50%, 20% and 5% to make 600ml of 10% drug H. How much of the 50% will you use?
try 400 ml of 5% and 200 ml of 20% for example then you need NONE of the 50%.
Unless you have another criterion, the problem has many solutions.
If the amounts of the three solutions are x,y,z, respectively, then all you have given is
x+y+z = 600
.50x + .20y + .05z = .10*600
There are many solutions (as it were), such as
5% 20% 50%
400 200 0
440 140 60
500 50 50
520 20 60
533 0 67
To find out how much of the 50% drug H solution you need to use, we first need to understand the concept of weighted averages. We can use a weighted average formula to solve this problem.
Let's assume you will use x mL of the 50% drug H solution.
The first step is to calculate the concentration of the drug H in each of the three solutions. We have:
- 50% drug H solution: concentration = 50%
- 20% drug H solution: concentration = 20%
- 5% drug H solution: concentration = 5%
Now, we can calculate the weighted average concentration of the three solutions:
(50% * x) + (20% * y) + (5% * z) = 10% * 600
Where x, y, and z represent the volumes (in mL) of the three solutions.
Since we are looking for the volume of the 50% drug H solution, we have:
(50% * x) = (10% * 600) - (20% * y) - (5% * z)
Simplifying:
0.5x = 0.1 * 600 - 0.2y - 0.05z
0.5x = 60 - 0.2y - 0.05z
Now, we need to consider another constraint: the total volume of the solution. We know that the total volume is 600 mL. Hence, we have:
x + y + z = 600
Now, we have a system of equations:
0.5x = 60 - 0.2y - 0.05z (equation 1)
x + y + z = 600 (equation 2)
We can solve this system of equations using substitution or elimination methods to find the value of x, which represents the volume of the 50% drug H solution you should use.