Mixture Problem... An auto tech needs a radiator to
have a 40% antifreeze solution. The radiator currently
is filled with 4 gallons of a 25% antifreeze solution.
How much of the antifreeze mixture should be
drained from the car if the mechanic replaces it with
pure antifreeze?
If he drains x gallons, then
.25(4-x) + 1.00x = .40(4)
1 - .25x + x = 1.6
.75x = .6
x = .8
So, after draining .8 gals, he has 3.2 gals of 25% antifreeze. That means he has .8 gals of antifreeze in the radiator.
Now he adds .8 gals of pure antifreeze. Then he has 1.6 gals of antifreeze in 4 gals of mixture. That's .16/4 = 40% antifreeze.
To solve this mixture problem, we can use the concept of the equation: (amount of pure substance in new mixture) = (amount of pure substance in original mixture).
Let's break down the problem into steps:
Step 1: Determine the original amount of antifreeze in the radiator.
The original mixture in the radiator is 4 gallons of a 25% antifreeze solution. We can calculate the original amount of antifreeze in the mixture by multiplying the volume of the mixture (4 gallons) by the percentage of antifreeze (25%).
Original amount of antifreeze = 4 gallons * 0.25 = 1 gallon of antifreeze
Step 2: Determine the desired amount of antifreeze in the new mixture.
The mechanic wants the final mixture to contain a 40% antifreeze solution. Since the pure antifreeze has a concentration of 100%, the desired amount of antifreeze in the new mixture will be the same as the volume of the mixture (the amount to be drained).
Desired amount of antifreeze = Volume of mixture to be drained
Step 3: Set up the equation using the concept mentioned before.
(amount of pure substance in new mixture) = (amount of pure substance in original mixture)
Desired amount of antifreeze = Original amount of antifreeze
Step 4: Solve the equation to find the volume of the mixture to be drained.
Desired amount of antifreeze = Original amount of antifreeze
Volume of mixture to be drained = 1 gallon
Therefore, the mechanic should drain 1 gallon of the antifreeze mixture from the car and replace it with 1 gallon of pure antifreeze.