how many liters of a 5% solution of salt should be added to a 25% solution in order to obtain 1120 liters of a 22% solution
volume of the 5% solution needed is x L
volume of the 25% solution needed is 1120 - x L
.05x + .25(1120-x) = .22(1120)
times 100
5x + 25(1120-x) = 22(1120)
5x + 28000 - 25x = 24640
-20x = -3360
x = 168
They need 168 L of the 5% solution and 952 L of the stronger stuff
To solve this problem, we need to determine the amount of each solution that should be mixed to obtain the desired concentration.
Let's break down the problem step-by-step:
Let's assume x liters of the 5% solution should be added.
So, the amount of the 25% solution would be 1120 - x liters.
Now, let's calculate the amount of salt in both solutions:
Amount of salt in the 5% solution = 5% * x liters
Amount of salt in the 25% solution = 25% * (1120 - x) liters
Since the final solution should be 1120 liters with a concentration of 22%, we can write the equation:
(5% * x) + (25% * (1120 - x)) = 22% * 1120
Now, let's solve the equation:
(0.05x) + (0.25 * (1120 - x)) = 0.22 * 1120
0.05x + 280 - 0.25x = 246.4
-0.20x = -33.6
x = 168 liters
So, 168 liters of the 5% solution should be added to obtain 1120 liters of a 22% solution.