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?

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

Its 2 gallons 100 percent sure

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.

To find out how many gallons of pure antifreeze Puss must add to bring the radiator's antifreeze solution up to a 50% concentration, we need to calculate the difference between the current concentration and the desired concentration, and then use that to determine the amount of pure antifreeze needed.

Here's the calculation:

1. Calculate the amount of antifreeze in the current solution:
- The radiator contains 4 gallons of a 25% antifreeze solution.
- The amount of antifreeze in the current solution is 4 gallons × 25% = 1 gallon.

2. Calculate the desired amount of antifreeze for a 50% solution:
- The desired concentration is 50%.
- We need to determine how much antifreeze is required to achieve this concentration.
- Let's assume Puss needs to add x gallons of pure antifreeze.

3. Set up an equation to solve for x:
- The amount of antifreeze in the final solution is the sum of the antifreeze in the current solution and the amount of pure antifreeze added, which should equal the desired concentration.
- So, 1 gallon + x gallons = 4 gallons × 50%.
- Simplifying the equation gives us 1 + x = 4 × 0.50.

4. Solve for x:
- 1 + x = 2.
- Subtracting 1 from both sides gives us x = 1.

Therefore, Puss must add 1 gallon of pure antifreeze to bring the radiator's antifreeze solution up to a desired 50% concentration.

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)

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.