SOLUTION: Adjusting Antifreeze: Angela needs 20 quarts of 50% antifreeze solution in her radiator. She plans to obtain this by mixing some pure antifreeze with an appropriate amount of a 4

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Question 190731: Adjusting Antifreeze: Angela needs 20 quarts of 50% antifreeze solution in her radiator. She plans to obtain this by mixing some pure antifreeze with an appropriate amount of a 40% antifreeze solution. How many quarts of each should she use?
Found 2 solutions by nerdybill, ankor@dixie-net.com:
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
Adjusting Antifreeze: Angela needs 20 quarts of 50% antifreeze solution in her radiator. She plans to obtain this by mixing some pure antifreeze with an appropriate amount of a 40% antifreeze solution. How many quarts of each should she use?
.
Let x = quarts of pure antifreeze
20-x = quarts of 40% antifreeze
.
"amt of antifreeze from pure" + "amt from 40%" = "50% of total mixture"
x + .40(20-x) = .50(20)
x + 8 - .40x = 10
x - .40x = 2
.60x = 2
x = 2/.60
x = 20/6
x = 10/3 quarts (3.333)
.
Quarts of 40% antifreeze:
20-x = 20-10/3 = 60/3 - 10/3 = 50/3 quarts (16.667)

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Adjusting Antifreeze: Angela needs 20 quarts of 50% antifreeze solution
in her radiator. She plans to obtain this by mixing some pure antifreeze
with an appropriate amount of a 40% antifreeze solution.
How many quarts of each should she use?
:
Let x = amt of 40% solution to be removed
and
Let x = amt of pure antifreeze to be added
:
.4(20-x) + 1.0x = .50(20)
:
8 - .4x + 1x = 10
:
-.4x + 1x = 10 - 8
:
.6x = 2
x = 2%2F.6
x = 31%2F3 qts removed and added
;
:
Check:
.4(16.67) + 1(3.33) = .5(20)
6.67 + 3.33 = 10