A laboratory assistant is asked to mix a 30% alcohol solution with 21 liters of an 80% alcohol solution to make a 60% alcohol solution. How many liters of the 30% alcohol solution should be used?
need help do not understand this question how would i set the equation up
To solve this problem, you can set up an equation based on the information provided.
Let's assume that the laboratory assistant needs to mix x liters of the 30% alcohol solution.
The amount of alcohol in the 30% solution can be calculated as 0.3x.
The amount of alcohol in the 80% solution is given as 21 liters, so the amount of alcohol in this solution is 0.8 * 21 = 16.8 liters.
In the final mixture, the total volume of the solution will be x + 21 liters.
The amount of alcohol in the final mixture can be calculated as 0.6 times the total volume of the solution. So, the amount of alcohol in the final mixture is 0.6 * (x + 21) liters.
Since the goal is to have a 60% alcohol solution, the amount of alcohol in the final mixture should be equal to the sum of the amounts of alcohol in the two initial solutions.
Setting up the equation:
0.3x + 16.8 = 0.6(x + 21)
Now you can solve this equation for x, the number of liters of the 30% alcohol solution that should be used.