SOLUTION: Give an example of a solution you could add to achieve the desired result. Then explain why that solution would work.: You have a 68% alcohol solution. You want to end up with a so
Algebra ->
Customizable Word Problem Solvers
-> Mixtures
-> SOLUTION: Give an example of a solution you could add to achieve the desired result. Then explain why that solution would work.: You have a 68% alcohol solution. You want to end up with a so
Log On
Question 370860: Give an example of a solution you could add to achieve the desired result. Then explain why that solution would work.: You have a 68% alcohol solution. You want to end up with a solution containing 50% alcohol. Found 2 solutions by ankor@dixie-net.com, NaturalBornNerd:Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! You have a 68% alcohol solution.
You want to end up with a solution containing 50% alcohol.
:
Add pure water, let w = amt of water, start with 10 liters of 68% solution
:
.68(10) = .50(w+10)
6.8 = .5w + 5
6.8 - .5 = .5w
1.8 = .5w
multiply both sides by 2
w = 3.6 liters of water added to 10 liters of 68% solution yields 13.6 liters of 50% solution
:
:
Check this, the amt of alcohol will remain the same; only the per cent changes
.68(10) = .5(13.6)
6.8 = 6.8
You can put this solution on YOUR website! You need to add a certain amount of water, which we'll call 'w', in order to dilute your mixture down to 50% alcohol.
You know that the current mixture is 32% water by subtracting 68% alcohol from 100%,where 100% represents the total mixture.
w(the unknown amount of water you need to add)
0.32w (the amount of water in the current mixture)
w + 0.32w = 0.50
w =0.50 - 0.32
w = 0.18
You will need to add 18% more water to the current mixture to get a mixture with 50% alcohol.
