Let x=the number of gallons to be drained.
Since there are 55 gallons in the original 30%, we can set it up like so:
Original 30% less amount drained: .3(55-x)
Pure amount added back: x
New 50% mixture: .5(55)
So, we have this to solve for x:
A 55 gallon barrel contains a mixture with a concentration of 30%. How much of this mixture must be withdrawn and replaced by a 100% concentrate to bring the mixture up to 50% concentration?
I know that in the 55 gallon barrel with 30% concentration there is 16.5 gallons of solute and 38.5 gallons of solvent. Is that info useful in this problem? I’m not sure what to do from there...
Hello, Kitty!
I'll show you my favorite approach to "Mixture Problems."
For reference, assume the solution is water and acid.
We start with 55 gallons which is 30% acid.A 55-gallon barrel contains a mixture which is 30% acid.
How much of this mixture must be withdrawn and replaced by 100% acid
to bring the mixture up to 50% concentration?
. . The barrel contains: . gallons of acid.
Write that in the first row of our chart . . .
We remove gallons of mixture.
. . This contains: . gallons of acid.
Write this in the second row of our chart . . .
We add gallons of pure (100%) acid.
. . This contains: . gallons of acid.
Write this in the third row of our chart . . .
We know that the final mixture is 55 gallons which is 50% acid.
. . This contains: . gallons of acid.
Write this in the last row of our chart . . .
Look at the last (rightmost) column.
It describes the amount of acid in the various stages.
We start with 16.5 gallons of acid,
. . we remove gallons of acid,
. . we add gallons of acid,
and we end up with gallons of acid.
And there is our equation! . . . . .