SOLUTION: A radiator holds 4 gallons of fluid. If it is fullwith a 20% solution how muchfluid should be drained and replaced with a 70% antifreeze mixture to result in a 50% mixture of antif

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Question 103624This question is from textbook
: A radiator holds 4 gallons of fluid. If it is fullwith a 20% solution how muchfluid should be drained and replaced with a 70% antifreeze mixture to result in a 50% mixture of antifreeze? This question is from textbook

Answer by Earlsdon(6294)   (Show Source): You can put this solution on YOUR website!
Start by assigning a variable to the unknown quantity.
Let x = the number of gallons that you need to remove and replace.
Now, when you remove x gallons from the radiator that presently holds 4 gallons of 20% solution, you can show this algebraically as: (After changing the percentages to their decimal equivalents)
(4-x)(0.2)
Following this, you want to add back the same amount but this time it will be x gallons of 70% solution, right? This can be shown as:
+x(0.7)
After the addition of the x gallons of 70% (0.7) solution, you expect to have 4 gallons of 50% (0.5) solution. So, putting this all together, you have:
(4-x)(0.2)+x(0.7) = 4(0.5) Now you can simplify this and solve for x.
0.8-0.2x+0.7x = 2 Combine like-terms.
0.8+0.5x = 2 Subtract 0.8 from both sides.
0.5x = 1.2 Finally, divide both sides by 0.5
x = 2.4
Solution (no pun intended!): You need to drain 2.4 gallons of the 20% solution and replace it with 2.4 gallons of 70% soltion to obtain 4 gallons of 50% solution.

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