At the end of Summer, Puss discovers that his radiator antifreeze solution has dropped below the safe level. If the radiator contains 4 gallons of a 25% antifreeze solution, how many gallons of pure antifreeze must he add to bring it up to a desired 50% solution?

Question

is the answer 4/3 OR 1 1/3 0R 1.2

Answer 1

None of those.

the 4 gal of 25% antifreeze contains
3 gal water
1 gal antifreeze

So, you need 2 more gal of antifreeze to make the result 50%

algebraically,

.25*4 + 1.00 x = .50(4+x)

Answer 2

Yesterday, I gave you this solution, but for some reason you didn't like it:
See Related Question #1 below.

"Let the amount of old stuff drained be x gallons
so you have 4-x gallons left of 25% solution

.25(4-x) + x = .5(4)
1 - .25x + x = 2
.75x = 1
(3/4)x = 1
x = 4/3

so drain 4/3 gallon of the current solution, then top it up with pure antifreeze"

check:
after draining 4/3 gallons of the current fluid, you would have 4-4/3 or 8/3 gallons left
amount of antifreeze contained in that = .25(8/3) = 2/3 gallons
we are adding 4/3 of 100% antifreeze (if there is such a thing)
so the rad contains 2/3 + 4/3 or 2 gallons of antifreeze
at a full capacity of 50% of 4 gallons, that would be 2 gallons of antifreeze.

I stand by my answer.

Steve, I was assuming that the rad was full at 4 gallons, so you can't just add more, you would have to drain some out first.

Answer 3

The problem as stated above does not mention draining.

However, your answer is one of the choices, so I guess his posting today was garbled.

Answer 4

Its 2 gallons 100 percent sure