SOLUTION: Suppose 44 pints of aa 13​% alcohol solution is mixed with 66 pints of aa 60​% alcohol solution. What is the concentration of alcohol in the new 10​-pint ​m

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Question 1069802: Suppose 44 pints of aa 13​% alcohol solution is mixed with 66 pints of aa 60​% alcohol solution. What is the concentration of alcohol in the new 10​-pint ​mixture?
The concentration of alcohol in the new 10​-pint mixture is?
How many ounces of a 15% alcohol solution must be mixed with 44 ounces of a 20 % alcohol solution to make a 19 %19% alcohol​ solution?
The amount of 15 % alcohol solution is
?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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Suppose 44 pints of aa 13​% alcohol solution is mixed with 66 pints of aa 60​% alcohol solution.
What is the concentration of alcohol in the new 10​-pint ​mixture?
The concentration of alcohol in the new 10​-pint mixture is?
:
let x = the percent concentration of the new mixture in decimal form
.13(44) + .60(66) = x(44 + 66)
5.72 + 39.6 = 110x
45.32 = 110x
x = 45.32/110
x = .416 which is 41.6%
:
:
How many ounces of a 15% alcohol solution must be mixed with 44 ounces of a 20 % alcohol solution to make a 19 %19% alcohol​ solution?
The amount of 15 % alcohol solution is
.15x + .20(44) = .19(44+x)
.15x + 8.8 = 8.36 + .19x
8.8 - 8.36 = .19x - .15x
.44 = .04x
x = .44/.04
x = 11 oz of 15% solution