Asked by kay
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?
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?
Answers
Answered by
Steve
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.
.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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.