SOLUTION: How much water should be added to 20 ounces of a 15% solution of alcohol to dilute it to a 10% solution?

Question 10331: How much water should be added to 20 ounces of a 15% solution of alcohol to dilute it to a 10% solution?
Found 3 solutions by Earlsdon, amalm06, ikleyn:
Answer by Earlsdon(6294)   (Show Source): You can put this solution on YOUR website!
Here's one approach.
20 ounces (oz) of a 15% alcohol solution has 85% water.
The 10% alcohol solution has 90% water. You need to add x ounces (oz) of 100% water.
Change percents to decimals.
20 oz(.85) + x oz(1.0) = (20+x)oz(0.9) Simplify and solve for x.
17 + x = 18 + 0.9x Subtract 0.9x from both sides.
17 + 0.1x = 18 Subtract 17 from both sides.
0.1x = 1 Divide both sides by 0.1
x = 10
Add 10 oz of water.
Answer by amalm06(224)   (Show Source): You can put this solution on YOUR website!
In this problem, you are replacing 5 parts of solution with 5 parts of water. You also want the ratio of water to solution to be 5/10.

If there are 20 parts of solution, then (5/10)(20)=10 oz water (Answer)
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.

Let W be the required volume of water to add (in fluid ounces). Then you have the resulting liquid volume of (20+w) ounces with the alcohol volume content of 0.15*20 ounces. The percentage equation is = 0.1. (<<<---=== 10% of alcohol in the resulting mixture) You can solve it by multiplying both sides by (20+w) and dividing by 0.1. You will get = 20+w, or 30 = 20 + w, which implies w = 10. Answer. 10 ounces of water must be added.

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