SOLUTION: a car radiator has a capacity of 16 quartz and is filled with a 25% antifreeze solution. how much must be drained off and replaced with pure antifreeze to obtain a 40% antifreeze s
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Question 559646: a car radiator has a capacity of 16 quartz and is filled with a 25% antifreeze solution. how much must be drained off and replaced with pure antifreeze to obtain a 40% antifreeze solution? Found 3 solutions by mananth, josgarithmetic, greenestamps:Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! a car radiator has a capacity of 16 quartz and is filled with a 25% antifreeze solution. how much must be drained off and replaced with pure antifreeze to obtain a 40% antifreeze solution?
Initial conc = 25%Final conc = 40%
100% antifreeze x qts
Antifreeze to be removed =x
(16-x)*0.25+1*x = 0.4(16)
4-0.25x +x = 6.4
0.75x = 2.4
x = 3.2 qts
The "draining off" and "replacing" in the problem only complicates the process of solving the problem. Essentially, you are simply mixing a 25% antifreeze solution with a 100% antifreeze solution to obtain 16 quarts of 40% antifreeze.
The question asks for the amount of pure antifreeze to be used. So
x = quarts of 100% antifreeze
16-x = quarts of 25% antifreeze
The mixture is 40% antifreeze:
I leave it to you to solve the problem by that formal algebraic method. As shown the the other tutors, the answer is 3.2 quarts.
Here is an informal method that can be used to solve any 2-part mixture problem like this.
The three percentages in the problem are 25, 40, and 100. Picture (in your mind, or on paper if needed) those three percentages on a number line. Then observe/calculate that 40 is 15/75 = 1/5 of the way from 25 to 100. That means 1/5 of the mixture is the 100% antifreeze.