How many gallons of a 70% acid solution must be mixed
with 30 gallons of a 16% solution to obtain a solution that is
60% acid?
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We are mixing a stronger 70% acid solution with a weaker 16%
acid solution to obtain a medium strength 60% acid solution.
Let x = the number of gallons of the stronger
We mix the following:
x gallons of stronger
30 gallons of weaker
and therefore we get
x+30 gallons of medium strength solution
Now we figure the number of gallons of pure alcohol contained
in all three of these:
The x gallons of stronger solutions contains 0.70x gallons of pure
alcohol (and the rest is water, which we ignore.)
The 30 gallons of weaker solution contains 0.16(30) or 4.8 gallons
of pure alcohol (and the rest is water, which we ignore).
The x+30 gallons of the final medium-strength solution contains 0.60(x+30) gallons of pure alsohol (and the rest is water, which we ignore).
Now
The number of gallons of pure alcohol in the stronger +
The number of gallons of pure alcohol in the weaker solution =
The number of gallons of pure alcohol in the medium-strength
solution.
or 0.70x + 4.8 = 0.60(x+30)
Can you solve that? Answer: x = 132 gallons.
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Check:
The 132 gallons of 70% contains 92.4 gallons of pure alcohol and
39.6 gallons of water.
The 30 gallons of 16% contains 4.8 gallons of pure alcohol and
25.2 gallons of water.
When they are mixed:
The 162 gallons of 60% contains 97.2 gallons of pure alcohol and
64.8 gallons of water.
Adding up the liquid: 132 gallons + 30 gallons = 162 gallons
Adding up the pure acid: 92.4 gallons + 4.8 gallons = 97.2 gallons
Adding up the water: 39.6 gallons + 25.2 gallons = 64.8 gallons
It all adds up!
Edwin