SOLUTION: Hello, My name is Eric and I have been having a simple conundrum of being unable to understand this particular question: "The radiator in a particular car needs 40 gallons of a l

Question 528687: Hello,
My name is Eric and I have been having a simple conundrum of being unable to understand this particular question: "The radiator in a particular car needs 40 gallons of a liquid that is composed of 40% antifreeze. Only 20% and 100% antifreeze are availible. How much of each should be mixed to get a 40% solution?" I have attempted to solve this simple, but strangly(please excuse my spelling if this is incorrect) worded problem. I have found that the answer should be near 30 gallons or so parts of 20% and 10 gallons of 100%. Please help me understand the concept as I am confused.
Thank You,
Eric
Answer by Maths68(1474) (Show Source): You can put this solution on YOUR website!
Solutoin A
==========
Amount = x gallons
Antifreeze = 20% = 20/100 = 0.2

Solutoin B
==========
Amount = 40-x gallons
Antifreeze = 100% = 100/100 = 1

Final Solution
==============
Amount = 40 gallons
Antifreeze = 40% = 40/100=0.4

(Amount of Solution A)*(Antifreeze)+(Amount of Solution B)*(Antifreeze)= (Amount of Mixture)*(Antifreeze)
x*0.2+(40-x)*1=40*0.4
0.2x+40-x=40*0.4
0.2x+40-x=16
0.2x-x=16-40
-0.8x=24
-0.8x/-0.8=24/-0.8
x=30

Solutoin A
==========
Amount = x = 30 gallons
Antifreeze = 20% = 20/100 = 0.2

Solutoin B
==========
Amount = 40-x = 40-30 = 10gallons
Antifreeze = 100% = 100/100 = 1
30 gallons of 20% Antifreeze and 10 gallons of 100% Antifreeze should be mixed to get a 40 gallons of 40% solution.

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