A biologist has three salt solutions: some 5% solution, some 15% solution, and some 25% solution. She needs to mix some of each to get 50 liters of 20% solution. She wants to us twice as much of the 5% solution as the 15% solution. How much of each solution should she use? How do I set it up using 3 varibles

Question

A biologist has three salt solutions: some 5% solution, some 15% solution, and some 25% solution. She needs to mix some of each to get 50 liters of 20% solution. She wants to us twice as much of the 5% solution as the 15% solution. How much of each solution should she use? How do I set it up using 3 varibles

Answer 1

x+y+z = 50

x = 2y
.05x + .15y + .25z = .20(50)

(x,y,z) = (10,5,35)

Answer 2

To solve this problem using three variables, let's call the amount of the 5% solution as x liters, the amount of the 15% solution as y liters, and the amount of the 25% solution as z liters.

Now, we'll set up the equation based on the given conditions:

1) The total volume of the solution is 50 liters:
x + y + z = 50

2) The concentration of the final solution is 20%, which means 20% of the 50 liters is salt:
0.20 * 50 = 0.20 * (x + y + z)

3) The biologist wants to use twice as much of the 5% solution as the 15% solution:
x = 2y

To solve these equations simultaneously, substitute the value of x in terms of y from equation 3) into equations 1) and 2).

Substituting x = 2y in equation 1):
2y + y + z = 50
3y + z = 50 ----(Equation 4)

Substituting x = 2y in equation 2):
0.20 * 50 = 0.20 * (2y + y + z)
10 = 0.20y + 0.20y + 0.20z
10 = 0.40y + 0.20z
5 = 0.2y + 0.1z
50 = 2y + z ----(Equation 5)

Now, we have a system of two linear equations with two variables (y and z), which can be solved to find their values.

By solving equations 4) and 5) simultaneously, you can find the values of y and z. Then, substitute them back into the equation x = 2y to find the value of x.