Question 106793This question is from textbook
: Angela needs 20 quarts of 50% antifreeze solution in her radiator. She plans to obtain this by mixing some pure antifreeze with an appropriate amount of 40% antifreeze solution. How many quarts of each should she use?
This question is from textbook
Answer by oberobic(2304)   (Show Source):
You can put this solution on YOUR website! The notion of mixtures always involves some amount of 'pure' stuff mixed with some amount of another 'not pure' stuff to produce some mixture that is not pure. But in all cases, you will have to figure out how much pure stuff there is. Also, water is consider the universal solvent, so adding water is a common way to reduce a pure substance to a more dilute version of the same.
In this case, you want 20 quarts of 50% pure antifreeze. That means you want to use .5(20) = 10 qt of pure antifreeze and 10 qt of water. So, 10/20 = 50% pure.
But the given problem is trickier because have to start with 40% mixture and add enough pure antifreeze to get your solution.
You stil want to end up with 10 qt of pure stuff (antifreeze) and 10 qt of water, so that is something you can search to find.
So, setting up the equation based on what you know:
.4*(x) + 1.0*(20-x) = .5(20), where x is the qt of 40% solution you need, and 20-x is the number of pure qt to add
So,
.4x + 20 - x = 10
Collecting like terms and subtracting 20 from both sides:
-.6x = -10
Dividing by -.6, we have
x = 16.67 qt of pure, to which you add 3.33 qt of pure antifreeze to obtain 20qt of 50% antifreeze.
ALWAYS check!
.4*16.67+1*3.33 = .5*20
6.67+3.33=10
Check!
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