An industrial chemist will mix two acid solutions. One is 37% solution and the other is a 27% solution. How many liters of each should be mixed to get 50 liters of a 36% acid solution?
x = liters of 37% acid solution
Other = (50 - x) = liters of 27% solution
37x + 27*(50 - x) = 50*36
37x + 1350 - 27x = 1800
10x = 450
x = 45 liters
45 liters of 37% acid and 5 liters of 27% acid.
45 liters of 37% acid and 5 liters of 27% acid.
To find out how many liters of each solution should be mixed, we can use a mathematical approach called the "Mixture Problem."
Let's assign variables for the two unknown quantities:
Let x represent the number of liters of the 37% solution.
Let y represent the number of liters of the 27% solution.
According to the given information, we need to mix these two solutions to obtain 50 liters of a 36% acid solution.
So, we can write the equation for the mixture problem as follows:
0.37x + 0.27y = 0.36 * 50
Let's solve this equation to find the values of x and y.
0.37x + 0.27y = 18
To eliminate the decimals, we can multiply both sides of the equation by 100:
37x + 27y = 1800
Now, we need to solve a system of equations. We have two equations:
37x + 27y = 1800 (equation 1)
x + y = 50 (equation 2)
We can use either substitution or elimination to solve this system. Let's use the elimination method:
Multiply equation 2 by 27 to make the coefficients of y the same in both equations:
27x + 27y = 1350 (equation 3)
37x + 27y = 1800 (equation 1)
Now, subtract equation 3 from equation 1:
(37x + 27y) - (27x + 27y) = 1800 - 1350
10x = 450
x = 45
Substitute the value of x back into equation 2 to find y:
45 + y = 50
y = 50 - 45
y = 5
Therefore, you will need to mix 45 liters of the 37% solution and 5 liters of the 27% solution to obtain 50 liters of a 36% acid solution.