Question

How many gallons of a 90%solution must be mixed with 90 gallons of 30% antifreeze to get a mixture that is 80% antifreeze?

Answer 1

I got 432 is that correct?

Answer 2

lets call the amount of 90% solution \(x\) so it will contain \(.9x\) antifreeze
you have 90 gallons of 30% solution, so it contains \(.3\times 90=27\) gallons of antifreeze
when you add \(x\) gallons to the 90 gallons you will have \(x+90\) gallons that you want to be 80% antifreeze, so set
\[.9x+27=.8(90+x\] and solve for \(x\)

Answer 3

that is not what i get, no

Answer 4

\[.9x+27=.8(90+x)\] get rid of annoying decimal by multplying by 10 and get
\[9x+270=8(90+x)\]

Answer 5

ugh Im lost ive done this prob a million times

Answer 6

\[9x+270=720+8x\]
\[x+270=720\]
\[x=450\] is what i get

Answer 7

yeah they are a pain
any question about any steps?

Answer 8

No I think i WAS JUST MISCALCULATING