How many gallons of a 60% alcohol solution and a 30% alcohol solution must be mixed to get 12 gallons of a 50% alcohol solution?
in Word Problem Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

If there are a gallons of the 60% solution and b of the 30% solution, a+b=12.

The amount of alcohol in the mixture is 50% of 12=6 gallons.

So 0.6a+0.3b=6. Since b=12-a, we can substitute:

0.6a+0.3(12-a)=6, 0.6a+3.6-0.3a=6, 0.3a=6-3.6=2.4.

a=2.4/0.3=24/3=8 gallons, making b=4 gallons.

So we have 8 gallons of 60% solution and 4 gallons of 30% solution.

by Top Rated User (839k points)

Related questions

Welcome to, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
86,303 questions
92,362 answers
23,927 users