Question

a radiator contains 15 gal of 20% antifreeze solution. how many gallons must be drained from the radiator and replaced by pure antifreeze so that the radiator will contain 15 gal of a 40% antifreeze solution?

Explanation:

Step1: Determine the initial amount of antifreeze

The radiator has 15 gal of 20% antifreeze solution. So the initial amount of antifreeze is $0.2\times15 = 3$ gal.

Step2: Let the amount drained and replaced be $x$ gallons

After draining $x$ gallons of the 20% solution, the amount of antifreeze drained is $0.2x$ gallons. The remaining amount of solution is $15 - x$ gallons and the remaining amount of antifreeze is $3-0.2x$ gallons. Then $x$ gallons of pure (100% or 1 in decimal - form) antifreeze is added.

Step3: Set up an equation for the final 40% solution

The final volume of the solution is still 15 gallons and the final amount of antifreeze is $0.4\times15 = 6$ gallons. So we have the equation $(3 - 0.2x)+x=6$.

Step4: Solve the equation

First, simplify the left - hand side: $3 - 0.2x+x=3 + 0.8x$. Then the equation becomes $3+0.8x = 6$. Subtract 3 from both sides: $0.8x=6 - 3=3$. Divide both sides by 0.8: $x=\frac{3}{0.8}=\frac{30}{8}=3.75$ gallons.

Answer:

$3.75$