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# How many ounces of water must be added to 10 ounces of 3 % alcohol

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Re: How many ounces of water must be added to 10 ounces of 3 % alcohol [#permalink]
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Abhi077 wrote:
How many ounces of water must be added to 10 ounces of 3 % alcohol and water solution to reduce the solution to 1 % alcohol.
A)5
B)10
C)15
D)20
E)30

The original solution contains 10 ounces
3% of that is alcohol
3% of 10 = 0.3
So, the original solution contains 0.3 ounces of alcohol

Let x = the amount of water (in ounces) that must be added
So, the volume of the RESULTING solution = 10 + x ounces

IMPORTANT: Since we're adding water only, the volume of alcohol does not change.
In other words, there are 0.3 ounces of alcohol in the RESULTING solution

We want the RESULTING solution to be 1% alcohol
In other words: (volume of alcohol in solution)/(total volume of solution) = 1/100
Rewrite as: 0.3 /(10 + x) = 1/100
Multiply both sides by (10 + x) to get: 0.3 = (10 + x)/100
Multiply both sides by 100 to get: 30 = 10 + x
Solve: x = 20

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Re: How many ounces of water must be added to 10 ounces of 3 % alcohol [#permalink]
Abhi077 wrote:
How many ounces of water must be added to 10 ounces of 3 % alcohol and water solution to reduce the solution to 1 % alcohol.
A)5
B)10
C)15
D)20
E)30

Let´s do it with another technique: alligation!

$$? = x$$

$${{10} \over {10 + x}} = {{1 - 0} \over {3 - 0}}\,\,\left( { = {{10} \over {30}}} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 20$$

This solution follows the notations and rationale taught in the GMATH method.

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Re: How many ounces of water must be added to 10 ounces of 3 % alcohol [#permalink]
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Re: How many ounces of water must be added to 10 ounces of 3 % alcohol [#permalink]
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