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Mathematics 19 Online
OpenStudy (anonymous):

How much pure antifreeze must be added to 12 gallons of 20% antifreeze to make a 40% antifreeze solution?

OpenStudy (amistre64):

12(.40-.20)/(1-.20) ... maybe?

OpenStudy (anonymous):

these are the options : A. 2 gallons B. 8 gallons C. 4 gallons D. 6 gallons

OpenStudy (amistre64):

gonna have to review my thoughts on that one :)

OpenStudy (amistre64):

a(1) + 12(.20) = (12+a)(.40) a(1) + 12(.20) = 12(.40)+a(.40) a(1-.40) = 12(.40-.20) a = 12(.40-.20)/(1-.40) ... I was thiiiisssss close lol

OpenStudy (anonymous):

so whats the answer?

OpenStudy (amistre64):

im gonna assume your smart enough to actually pick up at least a calculator and enter the numbers ...

OpenStudy (anonymous):

4 gallons

OpenStudy (anonymous):

lol. ok thanks

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