Question 152357
Some times, with mixture problems such as this one, it might be a little easier if you think in terms of just the antifreeze rather than the mixture.
You are told that Samantha needs 20 quarts of 50% antifreeze. First, change the percentage to its decimal equavalent (50% = 50/100 = 0.5).
Now 50% of 20 quarts of antifreeze mixture is really 0.5 (20) = 10 quarts of pure antifreeze.
The amount of 40% antifreeze mixture to be added is (20-x) quarts, that is this is the amount, added to the x quarts of pure antifreeze to get the 20 quarts.
Now let x = the amount of pure (100% = 1) antifreeze to be added to (20-x) quarts of 40% antifreeze mixture.
The amount of 40% antifreeze mixture is 0.4(20-x) quarts of antifreeze.
Now you can write an equation to solve for x, the number of quarts of pure antifreeze that Samantha needs.
{{{x + 0.4(20-x) = 0.5(20)}}} Perform the indicated multiplication.
{{{x + 8 - 0.4x = 10}}} Combine the x's
{{{0.6x + 8 = 10}}} Subtract 8 from both sides.
{{{0.6x = 2}}} Finally, didvide both sides by 0.6 to get x by itself.
{{{x = 3.33}}} or x = 3 1/3
So Samantha needs to mix 3 1/3 quarts of pure antifreeze with 20-x = 20-3 1/3 = 16 2/3 quarts of 40% antifreeze mixture to obtain 20 quarts of 50% antifreeze mixture..