A chemist has three different acid solutions. The first acid solution contains
25
%
acid, the second contains
35
%
and the third contains
55
%
. He wants to use all three solutions to obtain a mixture of
50
liters containing
45
%
acid, using
3
times as much of the
55
%
solution as the
35
%
solution. How many liters of each solution should be used?
Well, it seems like we have quite the colorful chemistry problem here! Let's break it down.
Let's call the amount of the 25% acid solution that the chemist needs to use "x" liters. Since the total mixture should be 50 liters, we can say that the amount of the 35% acid solution needed is 3x liters (3 times the amount of the 35% solution).
Now, the total amount of acid in the mixture should be 45% of 50 liters (0.45 * 50 = 22.5 liters). We can express this as an equation:
0.25x + 0.35(3x) + 0.55(3x) = 22.5
After some math glitter, we find that x = 5 liters.
So, the chemist should use 5 liters of the 25% acid solution, 15 liters of the 35% acid solution, and 15 liters of the 55% acid solution to achieve a mixture of 50 liters containing 45% acid. Ta-da!
Let's assume:
x = liters of the 25% acid solution
y = liters of the 35% acid solution
z = liters of the 55% acid solution
Given:
Total volume of the mixture = 50 liters
Target acid concentration in the mixture = 45%
We know that:
The amount of acid in the mixture is the sum of the amounts of acid in each solution.
Based on this information, we can form two equations:
Equation 1: The total volume of the mixture is the sum of the volumes of each solution.
x + y + z = 50
Equation 2: The total amount of acid in the mixture is the sum of the amounts of acid in each solution.
0.25x + 0.35y + 0.55z = 0.45 * 50
We also know that the volume of the 55% acid solution should be three times the volume of the 35% acid solution.
z = 3y
Now we can substitute the value of z in terms of y into equation 1:
x + y + 3y = 50
x + 4y = 50
Solving equation 3 for x:
x = 50 - 4y
Now we can substitute the values of x and z into equation 2:
0.25(50 - 4y) + 0.35y + 0.55(3y) = 0.45 * 50
Simplifying the equation:
12.5 - y + 0.35y + 1.65y = 22.5
Combining like terms:
-22.5 + 12.5 + 22.5 = -y + 0.35y + 1.65y
Solving for y:
12.5 = 2y
Dividing both sides by 2:
6.25 = y
Now we can substitute the value of y into equation 3 to find z:
z = 3(6.25)
z = 18.75
Finally, we can substitute the values of y and z into equation 1 to find x:
x + 6.25 + 18.75 = 50
x = 50 - 25
x = 25
Therefore, the chemist should use 25 liters of the 25% acid solution, 6.25 liters of the 35% acid solution, and 18.75 liters of the 55% acid solution to obtain a mixture of 50 liters containing 45% acid.
To solve this problem, we need to set up a system of equations based on the information provided.
Let x represent the number of liters of the 25% acid solution.
Let y represent the number of liters of the 35% acid solution.
Let z represent the number of liters of the 55% acid solution.
We are given that the total volume of the final mixture is 50 liters, so we have the equation:
x + y + z = 50
We are also given that the final mixture should contain 45% acid, so we have the equation:
(0.25x + 0.35y + 0.55z) / 50 = 0.45
Finally, we are told that the amount of the 55% acid solution used should be three times the amount of the 35% acid solution used, so we have the equation:
z = 3y
Now, we can solve these equations simultaneously to find the values of x, y, and z.
why all those annoying newlines?
x+y+z = 50
z = 3y
.25x + .35y + .55z = .45*50
Now solve for x,y,z