How many liters of a 25% antifreeze solution must be mixed with 4liters of a 60%antifreeze solution to make a mixture that is 45%antifreeze?
I'm not sure of the correct formula to use.
keep track of the amount of antifreeze in each part, and the whole:
.25x + .60*4 = .45(x+4)
To solve this problem, you can use the following formula:
(Amount of antifreeze in solution 1) + (Amount of antifreeze in solution 2) = (Amount of antifreeze in the final mixture)
In this case, let's assume that you need to mix x liters of the 25% antifreeze solution.
The amount of antifreeze in the 25% solution would be 25% of x liters, which is 0.25x liters.
The amount of antifreeze in the 60% solution is 60% of the 4 liters you already have, which is 0.6*4 = 2.4 liters.
The amount of antifreeze in the final mixture would be 45% of (x + 4) liters, which is 0.45(x + 4) liters.
Now, we can set up the equation:
0.25x + 2.4 = 0.45(x + 4)
Solving this equation will give you the value of x, which represents the number of liters of the 25% solution you need to mix.