A laboratory technician wishes to mix a 70% saline solutio and a 20% solution to prepare 500ml of a 40% solution. What amount of each solution should be used?
.7 x + .2 y = .4*500
x + y = 500
To solve this problem, we need to use a mixture equation. Let's assume that the amount of 70% saline solution needed is x ml, and the amount of 20% saline solution needed is y ml.
According to the question, the total volume of the mixture should be 500ml, so we can write our first equation:
x + y = 500
Next, let's calculate the amount of saline in each solution. For the 70% solution, 70% of x ml is saline, which is 0.7x ml. Similarly, for the 20% solution, 20% of y ml is saline, which is 0.2y ml.
Now, we can write our second equation based on the saline content:
0.7x + 0.2y = 0.4 * 500 (since we want a 40% saline solution)
Simplifying the equation:
0.7x + 0.2y = 200
Now, we have a system of two equations:
Equation 1: x + y = 500
Equation 2: 0.7x + 0.2y = 200
We can solve this system of equations using various methods such as substitution or elimination. Let's use the substitution method:
From Equation 1, we can get x = 500 - y.
Substituting this value of x into Equation 2, we have:
0.7(500 - y) + 0.2y = 200
Now, we can solve this equation for y:
350 - 0.7y + 0.2y = 200
-0.5y = -150
y = 300
Substituting this value of y back into Equation 1, we can find x:
x + 300 = 500
x = 200
Therefore, 200ml of the 70% solution and 300ml of the 20% solution should be mixed to prepare 500ml of a 40% solution.