SOLUTION: We have 20% alcohol solution and 50% acid solution, how many pints must be added from each to obtain 8 pints of 30%?
-Because it mixes alcohol and acid solutions, I'm not sure i
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-Because it mixes alcohol and acid solutions, I'm not sure i
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Question 731016: We have 20% alcohol solution and 50% acid solution, how many pints must be added from each to obtain 8 pints of 30%?
-Because it mixes alcohol and acid solutions, I'm not sure if the same equation for mixtures works for this problem. Found 2 solutions by mananth, josgarithmetic:Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! Percent ---------------- quantity
Solution I 20.00% ---------------- x pints
Solution II 50.00% ------ 8 - x pints
Mixture 30.00% ---------------- 8
8
20.00% x + 50.00% ( 8 - x ) = 30.00% * 8
20 x + 50 ( 8 - x ) = 240
20 x + 400 - 50 x = 240
20 x - 50 x = 240 - -400
-30 x = -160
/ -30
x = 5.33 pints 20.00% Solution I
2.67 pints 50.00% Solution II
You can put this solution on YOUR website! Bad question. No specification for the ratio of the alcohol to the acid in the 8 pints mixture, and you cannot expect to INCREASE the amount of alcohol from 20% to 30%. If you want to accept this as an open-ended question, then you could pick any ratio between the alcohol and the acid that is possible.
One way to approach is to make a total percent concentration rational equation and give one of the conditions that the volumes of each quantity of the available solutions must sum to 8 pints. The thought is like this:
x for the alcohol volume, y for the acid volume, and . You would not care what the ratio is between x and y; only that you care that the value of p is 30, for 30%.
According to that, you are back to a typical two component mixture problem without regard to the identity of the pure material in the available solutions or the resulting solution.
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