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.)

Question

anonymous · Answer

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

anonymous · Answer

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 
$$25(5-x)+100x=300$$

anonymous · Answer

Awesome! Thank you!