Question 171063: I'm trying to work this problem and I think that it actually will be a linear equation, but I'm having problems trying to set it up to make it come out right.
An auto service center has ten fifty-five gallon drums of 90% ethylene glycol, a common antifreeze fluid used in car radiators. The service center also has eight fifty-five gallon drums of 40% ethylene glycol solution as well. The manager wants to mix appropriate amounts of these two solutions to make 100 gallons of a 55% ethylene glycol concentration. How many gallons of the 90% solution will she need?
Can you help me with this one? I'm studying to take my final exam this coming Friday and I believe there will be questions of this type on my final.
Thank you
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! The problem description contains a lot of extraneous data that will not be needed in solving this problem, so, beware of this on your test.
You'll need to sort out the wheat from the chaff, as the saying goes.
The fact that the Auto Center has ten fifty-five gallon drums of 90% and eight fifty-five gallon drums of 40% ethylene glycol is interesting but not required to solve the problem.
Let x = the number of gallons of 90% ethylene glycol solution needed ((90%)x).
Then the manager will need to mix (100-x) gallons of the 40% ethylene glycol solution ((40%(100-x)) with that to get the 100 gallons of 55% ethylene glycol solution (55%(100)).
After changing the percentages to their decimal equivalents: (90% = 0.9, 40% = 0.4, and 55% = 0.55), you can express this as follows:
0.9x+0.4(100-x) = 0.55(100) Simplify and solve for x.
0.9x+40-0.4x = 55 Combine like-terms.
0.5x+40 = 55 Subtract 40 from both sides.
0.5x = 15 Divide both sides by 0.5
x = 30 gallons of 90% ethylene glycol solution.
Suffice it to say that the manager has enough ethylene glycol solution on hand to do the job.
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