SOLUTION: I have 500 hectolitres of 4.6% abv (alcohol by volume) beer in a tank. I need to increase the alcohol % up to 5%abv from 4.6%abv the max volume of tank 500hl so we need to remove s

Algebra ->  Volume -> SOLUTION: I have 500 hectolitres of 4.6% abv (alcohol by volume) beer in a tank. I need to increase the alcohol % up to 5%abv from 4.6%abv the max volume of tank 500hl so we need to remove s      Log On


   

Question 199597: I have 500 hectolitres of 4.6% abv (alcohol by volume) beer in a tank. I need to increase the alcohol % up to 5%abv from 4.6%abv the max volume of tank 500hl so we need to remove some of the 4.6% and replace it with 7.15% abv bringing up the abv percentage to 5%. I need to know how much 4.6% to remove and how much 7.15% to replace it with. Can you please show me the formula used to figure out this question please
Thank you
Mark
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
I have 500 hectolitres of 4.6% abv (alcohol by volume) beer in a tank.
I need to increase the alcohol % up to 5%abv from 4.6%abv the max volume
of tank 500hl so we need to remove some of the 4.6% and replace it with 7.15%
abv bringing up the abv percentage to 5%.
I need to know how much 4.6% to remove and how much 7.15% to replace it with.
:
Let x = amt to be removed, this is also the amt of 7.15% to be added
:
.046(500 - x) + .0715x = .05(500)
:
All you need now is some basic algebra
Multiply what's inside the brackets
23 - .046x + .0715x = 25
:
.046x + .0715x = 25 - 23
:
.0255x = 2
x = 2%2F.0255
x = 78.43 hl removed and 78.43 hl of 7.15% added
:
:
See if this is true
.046(500 - 78.43) + .0715(78.43) = .05(500)
.046(421.57) + .0715(78.43) = .05(500)
19.39 + 5.61 = 25; confirms our solution