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 use twice as much of the 5% solution as the 15% solution. How much of each solution should she use?
I've been trying to figure out the equation but I'm so stumped :(
If the amounts of 5%,15% and 25% are x,y,z then we have
x+y+z = 50
.05x + .15y + .25z = .20(50)
x = 2y
Replace x with 2y and you have
2y+y+z = 50
.05(2y) + .15y + .25z = .20(50)
or,
3y+z = 50
.25y + .25z = 10
3y+z = 50
y+z = 40
So, subtracting, we get
2y=10
y=5
so, x=2y=10
10+5+z=50, so z=35
To solve this problem, let's break it down step by step:
Let's assume that the biologist uses x liters of the 5% solution.
According to the problem, the biologist wants to use twice as much of the 5% solution as the 15% solution. So, she will use 2x liters of the 15% solution.
Now let's find the amount of the 25% solution she will need. Since she needs a total of 50 liters of the 20% solution, and she has already used x liters of the 5% solution and 2x liters of the 15% solution, the amount of the 25% solution can be represented as (50 - x - 2x).
Next, we need to set up the equation based on the salt content. Remember, the salt content of the final solution should be 20%.
The equation is:
(0.05x + 0.15(2x) + 0.25(50 - x - 2x)) / 50 = 0.20
Simplifying the equation:
(0.05x + 0.30x + 0.25(50 - 3x)) / 50 = 0.20
Now, solve the equation for x:
0.05x + 0.30x + 12.5 - 0.75x = 10
0.60x + 12.5 - 0.75x = 10
-0.15x = -2.5
x = -2.5 / -0.15
x ≈ 16.67
Since we can't have a fraction of a liter, we can round it to 17 liters for the 5% solution.
Now, we can find the amount of the 15% and 25% solutions:
The amount of the 15% solution = 2x = 2 * 17 = 34 liters
The amount of the 25% solution = 50 - x - 2x = 50 - 17 - 34 = 50 - 51 = -1
Since we can't have a negative amount, it means that we can't use the 25% solution in this case.
Therefore, the biologist should use 17 liters of the 5% solution and 34 liters of the 15% solution to obtain 50 liters of the desired 20% solution.