SOLUTION: a radiator contains 25 quartz of water and antifreeze solution, which of 60 percent (by volume) is antifreeze. how much of this solution should be drained and replaced with water f

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Question 427227: a radiator contains 25 quartz of water and antifreeze solution, which of 60 percent (by volume) is antifreeze. how much of this solution should be drained and replaced with water for the new solution to be 40 percent antifreeze.?
Found 4 solutions by mananth, ikleyn, greenestamps, timofer:
Answer by mananth(16949)   (Show Source):
You can put this solution on YOUR website!
.........Percent ---------------- quantity
Solution A 60 ---------------- 30
Water 0----------------x
Total 40----------------30+x
60*25+0* x=40(25+ x)
1500 + 0 x = 1000 + 40 x
0 x - 40 x = 1000 - 1500
-40 x = -500
/-40

x=12.5 quartz water
12.5 quartz of solution has to be drained.

Answer by ikleyn(53751)   (Show Source):
You can put this solution on YOUR website!
.
a radiator contains 25 quartz of water and antifreeze solution, which of 60 percent (by volume) is antifreeze.
how much of this solution should be drained and replaced with water for the new solution to be 40 percent antifreeze?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        The solution in the post by @mananth is incorrect.
        I came to bring a correct solution.

Let x be the volume of the original 60% antifreeze solution to partly drain and to replace by pure water to get the 40% antifreeze solution. After draining, we then have (25-x) quartz of the 60% antifreeze solution. It contains 0.6*(25-x) quartz of the pure antifreeze. Adding water does not change the amount of the antifreeze in the solution. At the end, the volume of the pure antifreeze in the radiator after adding x quartz of water is 0.4*25 quartz. So, we equate these two expressions for the pure antifreeze amount 0.6*(25-x) = 0.4*25 quartz. (1) Simplify and find x 15 - 0.6x = 10, 15 - 10 = 0.6x, 5 = 0.6x x = = = 8. ANSWER. 8 quartz of the original 60% solution should be drained and replaced by pure water to get the 40% antifreeze solution.

Solved correctly.

Answer by greenestamps(13327)   (Show Source):
You can put this solution on YOUR website!

Here are two informal solutions using logical reasoning instead of formal algebra.

(1)

At the start, the 25-quart radiator is full of 60% antifreeze, so it contains .60(25) = 15 quarts of antifreeze.

In the end, we want it to be full of 40% antifreeze, so it will contain .40(25) = 10 quarts of antifreeze.

The amount of antifreeze we want to finish with (10 quarts) is 2/3 of the 15 quarts we started with; since we are adding water which contains no antifreeze, we want the radiator to finish with 2/3 of the original antifreeze mixture, which means we want to drain 1/3 of the original mixture and replace it with water. 1/3 of 25 quarts is 8 1/3 quarts.

ANSWER: 8 1/3 quarts

(2)

We are mixing 60% antifreeze with 0% antifreeze to obtain a mixture that is 40% antifreeze. 40% is twice as close to 60% as it is to 0%, so the amount of the original antifreeze mixture must be twice as much as the added water -- i.e., 2/3 of the final mixture must be the original 60% antifreeze and 1/3 must be the added water. Again, 1/3 of 25 quarts is 8 1/3 quarts.

ANSWER (again, of course): 8 1/3 quarts

Answer by timofer(155)   (Show Source):
You can put this solution on YOUR website!
drain and replace some v quarts
volume to be unchanged