Algebra Mixture problems?
A chemist has one solution which is 30% pure acid and another solution which is 60% pure acid. how many liters of each solution must be used to produce 60 liters of a solution which is 50% pure acid?
when I did the equation (.30x + .60(60)=.5(x+60)) I got that x equals 30 but I didn't know how to find the amount of the other pure acid
Comments
Your equation is set up wrong.
Suppose that you have x liters of 30% pure acid. In order for the resulting solution to have 60 liters, the rest of the solution must consist of 60% acid, so there are 60 - x liters of 60% acid.
The total solution has 60 liters of solution, so your equation should be:
0.30x + 0.60(60 - x) = 0.5(60),
which solves to yield x = 20. Therefore, x = 20 liters of 30% pure acid and 60 - x = 40 liters of 60% issue should be used.
I hope this helps!
30x + 60y = 50 * 60
x + y = 60
30 * (x + 2y) = 50 * 30 * 2
x + 2y = 100
x + 2y - x - y = 100 - 60
y = 40
x + y = 60
x + 40 = 60
x = 20
20 liters of 30%
40 liters of 60%
Use a table.
type ----- % as dec ----- amount ----- pure
30% ------- .30 ------------- x ------------- .30x
60% ------- .60 ----------- 60 - x --------- .60(60 - x)
50% ------- .50 ------------ 60 ------------ .50 * 60
.30x + .60(60 - x) = .50 * 60
3x + 360 - 6x = 300
3x = 60
x = 20
20L of 30% solution plus 40L of 60% solution gives 60L of 50% solution
Note-
Your equation was using 60L of the 60% solution, not 60L of the 50% solution. The total in your problem was 90L of the 50% solution, with 30L of the 30% solution.
Using a table can help to avoid mistakes like that in the future. It helps to keep things straight.