a chemist needs 10 liters of a 25% acid solution. the solution is to be mixed from three solutions whose acid concentrations are 10% 20% 50% how may liters of each solution should the chmist use to satisfy the following?
a use as little as possible of the 50% solution
b use as much as possible of the 50% solution
c use 2 liters of the 50% solution
let the amount of 50% solution be x L
let the amount of 20% solution by y L
let the amount of 10% solution be z or (10-x-y) L
then .1(10-x-y) + .2y + .5x = .25(10)
10-x-y + 2y + 5x = 25
y = -4x + 15 , relation (A)
also x+y ≤ 10 , relation (B)
so any (x,y) which satisfies both relations A and B will work
where do they intersect?
y = -4x + 15 with y = -x + 10
x = 5/3, then y = 25/3 , z = 0
notice the x intercept falls within (B)
let y=0, then x= 3.75 , z = 6.25
if x = 2 , then y = 7, and z = 1
50% . 20% . 10%
0 .. 15 .. not possible
5/3 . 25/3 ..0
15/4 . 0 ...25/4
2 ... 7 ... 1
a) for which of these is x as small as possible
b) for which of these is x as large as possible
c) when x = 2 , y = 7, z = 1
To calculate the liters of each solution needed to satisfy the conditions, we can follow these steps:
a) To use as little as possible of the 50% solution:
Let x be the liters of 10% solution, y be the liters of 20% solution, and z be the liters of 50% solution.
The equation for the total amount of acid in the final solution can be written as:
0.10x + 0.20y + 0.50z = 0.25(10)
Since we want to use as little as possible of the 50% solution, we can set z = 0. So the equation becomes:
0.10x + 0.20y = 0.25(10)
Simplifying the equation:
0.10x + 0.20y = 2.5
We also need to consider the constraint that the total volume of the solution is 10 liters:
x + y + z = 10
Since z = 0, the equation becomes:
x + y = 10
Solving the two equations simultaneously:
0.10x + 0.20y = 2.5
x + y = 10
We can solve this system of equations to find the values of x and y.
b) To use as much as possible of the 50% solution:
In this case, we want to maximize the amount of the 50% solution, which means setting z to the maximum value, which is 10 liters.
We can follow the same steps as in part a, but with z = 10.
0.10x + 0.20y + 0.50(10) = 0.25(10)
Simplifying the equation:
0.10x + 0.20y + 5 = 2.5
0.10x + 0.20y = -2.5
x + y + 10 = 10
Solving the two equations simultaneously will give us the values of x and y.
c) To use 2 liters of the 50% solution:
In this case, we can set z = 2 and follow the same steps as in part a.
0.10x + 0.20y + 0.50(2) = 0.25(10)
Simplifying the equation:
0.10x + 0.20y + 1 = 2.5
0.10x + 0.20y = 1.5
x + y + 2 = 10
Solving the two equations simultaneously will give us the values of x and y.
To determine how many liters of each solution the chemist should use to create a 10 liter, 25% acid solution, we can follow these steps:
Step 1: Define Variables
Let's denote the amount of the 10% solution as x₁ liters, the amount of the 20% solution as x₂ liters, and the amount of the 50% solution as x₃ liters.
Step 2: Write Equations
Since we want to create a 10 liter solution, we can write the following equation for the total volume:
x₁ + x₂ + x₃ = 10 --------- (Equation 1)
To achieve a 25% acid concentration, we need to calculate the total amount of acid in the solution. The equation for the total amount of acid can be written as:
(0.10 * x₁) + (0.20 * x₂) + (0.50 * x₃) = 0.25 * (10) --------- (Equation 2)
Step 3: Solve the Equations
We have two equations with three unknowns, so we will need to use some logical reasoning to find the solutions.
a) Use as little as possible of the 50% solution:
To use as little as possible of the 50% solution, let's assume x₃ = 0 (no 50% solution). Using Equation 1:
x₁ + x₂ + 0 = 10 implies x₁ + x₂ = 10 --------- (Equation 3)
We can solve Equation 3. Since x₃ is 0, the solution is simply to divide the 10 liters equally between the 10% and 20% solutions. So, the chemist should use 5 liters of each.
b) Use as much as possible of the 50% solution:
To use as much as possible of the 50% solution, let's assume x₃ = 10 (all 50% solution). Using Equation 1:
x₁ + x₂ + 10 = 10 implies x₁ + x₂ = 0 --------- (Equation 4)
Again, we can solve Equation 4. Since x₁ + x₂ = 0, there is no solution for obtaining a total volume of 10 liters while using all 50% solution. Therefore, this scenario is not possible.
c) Use 2 liters of the 50% solution:
To use 2 liters of the 50% solution, we substitute x₃ = 2 into Equation 1:
x₁ + x₂ + 2 = 10 implies x₁ + x₂ = 8 --------- (Equation 5)
We can solve Equation 5. Since x₁ + x₂ = 8, the chemist can divide the remaining 8 liters between the 10% and 20% solutions. The specific quantities depend on how the chemist wants to distribute these two solutions.
In summary:
a) The chemist should use 5 liters of the 10% solution and 5 liters of the 20% solution to use as little as possible of the 50% solution.
b) It is not possible to use as much as possible of the 50% solution.
c) The chemist can use 2 liters of the 50% solution along with a combination of the 10% and 20% solutions that add up to 8 liters. The exact amounts of the 10% and 20% solutions will depend on the chemist's preferred distribution.