A chemist has 10% acid solution and 60% acid solution. HOW many liters of each solution does the chemist need to make 200 L of a solution that is 50% acid?
amount of 10% solution ---- x L
amount of 60% solution ---- 200-x L
.1x + .6(200-x) = .5(200)
times 10
x + 6(200-x) = 5(200)
x + 1200 - 6x = 1000
-5x = -200
x = 40
check:
.1(40) + .6(160)
= 4+96
= 100 which is 50% of 200L
To solve this problem, we need to set up an equation based on the amount of acid in each solution.
Let's denote the amount of 10% acid solution as x liters and the amount of 60% acid solution as y liters.
Based on the given information, the equation for the amount of acid in the final solution can be written as:
0.10x + 0.60y = 0.50(200)
Simplifying the equation, we have:
0.10x + 0.60y = 100
To determine the values of x and y, we need to solve this system of equations. There are multiple methods to solve this. Let's use the substitution method:
1. Rearrange the first equation to solve for x:
0.10x = 100 - 0.60y
x = (100 - 0.60y) / 0.10
2. Substitute the value of x in the second equation:
0.10[(100 - 0.60y) / 0.10] + 0.60y = 100
Simplifying further:
(100 - 0.60y) + 0.60y = 100
100 - 0.60y + 0.60y = 100
100 = 100
This equation shows that y can be any value since both terms cancel out. Therefore, there is no unique solution to this problem.
In other words, there are infinitely many combinations of the 10% and 60% acid solutions that can be mixed to obtain a 50% acid solution.
To find out how many liters of each solution the chemist needs to make 200 L of a solution that is 50% acid, we can set up an equation based on the concept of mixtures.
Let's assume the chemist needs x liters of the 10% acid solution and (200 - x) liters of the 60% acid solution.
Now, let's consider the acid content in each solution. The 10% acid solution contains 10% acid or 0.10 acid per liter, and the 60% acid solution contains 60% acid or 0.60 acid per liter.
To determine the acid content in the final mixture, we need to find the average of the acid content in the two solutions. Since we want a 50% acid solution, we multiply the acid content by the amount of solution.
For the 10% acid solution, the acid content would be 0.10x, and for the 60% acid solution, the acid content would be 0.60(200 - x).
The sum of these two acid contents should be equal to the acid content in the final mixture, which is 0.50(200):
0.10x + 0.60(200 - x) = 0.50(200)
Now, we can solve this equation to find the value of x.
0.10x + 120 - 0.60x = 100
0.10x - 0.60x = 100 - 120
-0.50x = -20
x = -20 / -0.50
x = 40
Therefore, the chemist needs 40 liters of the 10% acid solution and (200 - 40) = 160 liters of the 60% acid solution to make 200 liters of a solution that is 50% acid.