Question 263512: If a pharmacist wants to make 150 milliliters of a 20 percent alcohol solution, but only has a 30 percent alcohol solution and a 15 percent alcohol solution, how much of each solution should she mix together to get what she needs?
Could you please take me step by step?
Found 2 solutions by oberobic, josmiceli:
Answer by oberobic(2304)   (Show Source):
You can put this solution on YOUR website! When solving a solution problem, you have to determine how much of the 'pure' stuff you need in the solution.
In this case, you need 150 ml of 20% alcohol.
That means there will be 30 ml of alcohol and 120 ml of water.
Using the 30% solution, how much would you need to have 30 ml of pure 'alcohol'?
30 ml = .3x
.3x = 30
x = 100 ml
That is, if you have 100 ml of 30% alcohol solution, then you have 30 ml of alcohol and 70 ml of water.
So, measure our 100 ml of the 30% alcohol solution and add 50 ml of pure water.
You now will have a total volume of 150 ml, of which, 30 ml are pure alcohol.
30/150 = .2
which means you have your required 150 ml of 20% alcohol solution.
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Unless you're required to use the 15% alcohol solution, why bother with it at all? Just dilute the 30% solution and you're done.
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But teachers can be picky...
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So another way to look at the problem is to think in terms any other word problem.
You need to mix a 30% solution and a 15% solution to arrive at a 20% solution.
We know the final result is supposed to be 150 ml of 20% solution, which we can show as .2*150.
If we use 'x' amount of the 30% solution, then we are forced to use 150-x of the 15% solution.
That defines an equation:
.3x + .15(150-x) = .2(150)
.3x + 22.5 - .15x = 30
.3x - .15x = 30 - 22.5
.15x = 7.5
x = 50
So you use 50 ml of the 30% solution and 150-50 = 100 ml of the 15% solution.
Check this by determining how much 'pure stuff' is in the final solution.
.15*100 = 15
.3*50 = 15
total = 30 ml of pure alcohol in the 150 ml
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Answer by josmiceli(19441)  (Show Source):
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