Question

A radiator contains 5 quarts of fluid, 25% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the new mixture is 60% antifreeze? (Round your answer to two decimal places.)

Answer 1

\[.25(5-x)+x=.6\times 5\] should solve it

Answer 2

you want the result to have 60% of 5 which is 3 liters of antifreeze
if you drain out \(x\) liters, you have \(5-x\) left, which is 25% antifreeze, i.e. contains \(.25(5-x) \) liters of anti freeze. then you add \(x\) liters of anti freeze, so your total is \(.25(5-x)+x\) and you want that to be equal to 3, so solve
\[.25(5-x)+x=3\]
probably easier to multiply by 100 and solve
\[25(5-x)+100x=300\]

Answer 3

Awesome! Thank you!