SOLUTION: I'm really stuck on this word problem.word problem are my weakness: It is necessary to have a 40% antifreeze solution in the radiator of a certain car. The radiator now has 40 lite

Question 467328: I'm really stuck on this word problem.word problem are my weakness: It is necessary to have a 40% antifreeze solution in the radiator of a certain car. The radiator now has 40 liters of 20% solution. How many liters of this should be drained and replaced with 100% anitfreeze to get the desired strength?
Found 2 solutions by ankor@dixie-net.com, solver91311:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
It is necessary to have a 40% antifreeze solution in the radiator of a certain car.
The radiator now has 40 liters of 20% solution.
How many liters of this should be drained and replaced with 100% antifreeze to get the desired strength?
:
Let x = amt of original anti-freeze to be drained, also the amt of pure antifreeze to be added
:
Write a decimal amt of antifreeze equation
:
.20(40-x) + 1x = .40(40)
8 - .2x + 1x = 16
-.2x + 1x = 16 - 8
.8x = 8
x =
x = 10 liters removed and 10 liters of pure antifreeze to be added
:
:
Check this in the original equation
.20(40-10) + 10 = .4(40)
.2(30) + 10 = 16
6 + 10 = 16; confirms our solution of x = 10
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

Let represent the amount drained (and the amount of 100% anti-freeze). Then the amount of anti-freeze that remains after draining gallons is . 20% of that which remains is antifreeze, to which you want to add gallons of 100% antifreeze which is, by definition, gallons of antifreeze. The two add up to 40 gallons of 40% antifreeze which is 16 gallons of antifreeze, so:

Solve for

John

My calculator said it, I believe it, that settles it

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