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.
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)
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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