Chris needs to make 500 L of a 35% acidic solution. He has only two of the acidic solutions available, a 25% solution and a 50% solution. How many litres of each acidic solution should he mix?
amount needed of the 25% stuff --- x L
amount needed of the 50% stuff ---- 500-x L
.25x + .50(500-x) = .35(500)
times 100
25x + 50(500-x) = 35(500)
25x + 25000 - 50x = 17500
-25x = -7500
x = 300 L
state your conclusion
To determine how many liters of each acidic solution Chris should mix, we can set up a system of equations based on the given information.
Let's assume Chris mixes x liters of the 25% acidic solution and y liters of the 50% acidic solution.
According to the problem, Chris needs to make 500 L of a 35% acidic solution. So, the sum of the volumes of the two solutions is equal to 500:
x + y = 500 (equation 1)
We also know that the percentage of acid in the resulting mixture will be 35%. Thus, the total amount of acid in the mixture will be:
0.25x + 0.50y = 0.35(500) (equation 2)
Simplifying equation 2:
0.25x + 0.50y = 175
Now we can solve the system of equations (equations 1 and 2) simultaneously to find the values of x and y.
Let's multiply equation 1 by 0.25 to align the coefficients of the x term:
0.25(x + y) = 0.25(500)
0.25x + 0.25y = 125 (equation 3)
Now we can subtract equation 3 from equation 2 to eliminate the x term:
(0.25x + 0.50y) - (0.25x + 0.25y) = 175 - 125
0.50y - 0.25y = 50
0.25y = 50
y = 50 / 0.25
y = 200
Substituting the value of y back into equation 1 to solve for x:
x + 200 = 500
x = 500 - 200
x = 300
Therefore, Chris should mix 300 liters of the 25% acidic solution and 200 liters of the 50% acidic solution to make 500 liters of a 35% acidic solution.
To find out how many liters of each acidic solution Chris should mix, we can set up a system of equations based on the given information.
Let's assume Chris needs to mix x liters of the 25% solution and y liters of the 50% solution.
Since he needs to make a total of 500 liters of the 35% solution, we can write the first equation:
x + y = 500 (equation 1)
The second equation relates to the acidic content or concentration of the solution:
0.25x + 0.50y = 0.35(500) (equation 2)
Let's solve this system of equations:
First, multiply equation 1 by -0.25, so that when we add it to equation 2, the x coefficients will cancel each other out:
-0.25(x + y) = -0.25(500)
-0.25x - 0.25y = -125 (equation 3)
Now, add equation 3 to equation 2:
(-0.25x + 0.25x) + (-0.25y + 0.50y) = -125 + 0.35(500)
0.25y = 175
Divide both sides of the equation by 0.25:
y = 175 / 0.25
y = 700
Now substitute the value of y back into equation 1 to solve for x:
x + 700 = 500
x = 500 - 700
x = -200
Since it doesn't make sense to have a negative quantity of a solution, we can conclude that Chris cannot achieve the desired 35% acidic solution using the given 25% and 50% solutions.
Therefore, it is not possible for Chris to mix the two available acidic solutions to make a 35% acidic solution.