SOLUTION: A liter of fluid is 50% alcohol. How much water must be added to dilute it to a 20% solution?
my work: so x is the amount of water to be added.
so something + x shou
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my work: so x is the amount of water to be added.
so something + x shou
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Question 154167This question is from textbook college algebra
: A liter of fluid is 50% alcohol. How much water must be added to dilute it to a 20% solution?
my work: so x is the amount of water to be added.
so something + x should = .20?? This question is from textbook college algebra
You can put this solution on YOUR website! The following fact is used to set up the equation:
The amount of the pure alcohol before mixing must be equal to the amount of the pure alcohol after mixing.
So
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The solution before mixing: 1L of 50% solution
The amount of alcohol before mixing is 50%*1
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The solution after mixing: 1+x liters of 20% solution
The amount of alcohol after mixing is 20%*(1+x)
*****************************************************
So the equation is:
50%*1 = 20%(1+x)
Solving for x, we have
50 = 20(1+x) (multiply both sides by 100)
50 = 20 + 20x
30 = 20x
30/20 = x
1.5 = x
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