SOLUTION: We have similar problems to this in our textbook, but this is off a handwritten midterm review from our "teacher." It's an online course. A 2004 Pontiac Montana has an engine c

Algebra ->  Linear-equations -> SOLUTION: We have similar problems to this in our textbook, but this is off a handwritten midterm review from our "teacher." It's an online course. A 2004 Pontiac Montana has an engine c      Log On


   

Question 146093: We have similar problems to this in our textbook, but this is off a handwritten midterm review from our "teacher." It's an online course.
A 2004 Pontiac Montana has an engine coolant capacity of 10.1 liters. If the system is currently filled with a mixture that is 30% antifreeze, how much should be drained and replaced with pure antifreeze so the system is filled with a mix that is 50% pure antifreeze?
The answer we have is 2.89. I do not understand how to get to this answer! HELP! :) Thanks.
Found 2 solutions by nerdybill, stanbon:
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
A 2004 Pontiac Montana has an engine coolant capacity of 10.1 liters. If the system is currently filled with a mixture that is 30% antifreeze, how much should be drained and replaced with pure antifreeze so the system is filled with a mix that is 50% pure antifreeze?
.
Let x = amt in liters drained and replaced by 100% antifreeze
.
amt of 30% antifreeze = 10.1 - x
amt of 100% antifreeze = x
.
"half of radiator content" = amt of anti-freeze
.5(10.1) = "amt antifreeze from 30% mix" + "amt antifreeze from 100% mix"
.5(10.1) = .30(10.1 - x) + x
5.05 = 3.03 - .30x + x
5.05 = 3.03 + .70x
2.02 = .70x
2.89 liters = x

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A 2004 Pontiac Montana has an engine coolant capacity of 10.1 liters.
If the system is currently filled with a mixture that is 30% antifreeze, how much should be drained and replaced with pure antifreeze so the system is filled with a mix that is 50% pure antifreeze?
----------------
30% DATA:
Amt. = 10.1 liters ; Amt of active ingredient = 0.3(10.1) = 3.03 liters
-------------------------------
After removing x liters:
Amt = 10.1-x liters ; Amt of active ingredient = 0.3(10.1-x) = 3.03-0.3x liters
-------------------------------
After add x liters of 100% active ingredient:
Amt = 10.1 liters ; Amt of active ingredient = 0.5*10.1 = 5.05 liters
-------------------------------
EQUATION:
active + active = active
3.03-0.03x + x + 5.05
0.97x = 2.02
x = 2.0825 liters (amount that is subtracted and then added)
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Cheers,
Stan H.