Mixture A is 15 percent alcohol, and mixture B is 50 percent alcohol. If the two are poured together to create a 4-gallon mixture that contains 30 percent alcohol, approximately how many gallons of mixture A are in the mixture?[/color]
All the good methods are already outlined above: Another way of doing this problem, without the need to write down anything:
Assume that both Mixture A and Mixture B are in equal proportion \to
The total amount of alcohol would be 50 % of 2 gallons+15% of 2 gallons \to
1+0.3 = 1.3 gallons. As per the problem, we need only 1.2 gallons. Thus, to account for this extra 0.1 gallons, and we can safely assume that the mixture which has more percentage of alcohol would be lesser, and Mixture A will be a little more than 2 gallons
Now, if we are confused between 2.3 or 2.5, we can always back calculate :
For eg, assuming that 2.5 is the correct answer, amount of Mixture B = 4-2.5 = 1.5 gallons \to
50% of this is 0.75 gallons.
Now, we know that 10% of 2.5 gallons can be calculated in a jiffy and it equals 0.25. Thus, 15% would be 10%+5% \to
0.25+0.125 = 0.375 and the net alcohol content stands at 1.125 which is less than 1.2 gallons.Hence, we have to increase the alcohol content a little, and this can be done by increasing amount of Mixture B. Thus, the correct answer is C.
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