Concentration

• August 5th 2008, 03:37 PM
Kitty216
Concentration
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 How much of this mixture must be withdrawn and replaced by a 100% concentrate to bring the mixture up to 50% concentration?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...
• August 5th 2008, 04:01 PM
galactus
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)

New 50% mixture: .5(55)

So, we have this to solve for x:

$.3(55-x)+x=.5(55)$
• August 5th 2008, 05:27 PM
Kitty216
Hmm.. How did you know to multiply the .3 to the 55?
• August 5th 2008, 08:33 PM
Soroban
Hello, Kitty!

I'll show you my favorite approach to "Mixture Problems."
For reference, assume the solution is water and acid.

Quote:

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How much of this mixture must be withdrawn and replaced by 100% acid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: . $0.30 \times 55 \:=\: 16.5$ gallons of acid.
Write that in the first row of our chart . . .

$\begin{array}{c|c|c|c|}
& \text{sol'n} & \% & \text{acid} \\ \hline
\text{Start} & 55 & 30\% & 16.5 \end{array}$

We remove $x$ gallons of mixture.
. . This contains: . $30\%\times x \:=\:0.3x$ gallons of acid.
Write this in the second row of our chart . . .

$\begin{array}{c|c|c|c|}
& \text{sol'n} & \% & \text{acid} \\ \hline
\text{Start} & 55 & 30\% & 16.5 \\
\text{Remove} & \text{-}x & 30\% & \text{-}0.3x \\\end{array}$

We add $x$ gallons of pure (100%) acid.
. . This contains: . $100\% \times x \:=\:x$ gallons of acid.
Write this in the third row of our chart . . .

$\begin{array}{c|c|c|c|}
& \text{sol'n} & \% & \text{acid} \\ \hline
\text{Start} & 55 & 30\% & 16.5 \\
\text{Remove} & \text{-}x & 30\% & \text{-}0.3x \\
\text{Add} & +x & 100\% & +x \end{array}$

We know that the final mixture is 55 gallons which is 50% acid.
. . This contains: . $50\% \times 55 \:=\:27.5$ gallons of acid.
Write this in the last row of our chart . . .

$\begin{array}{c|c|c|c|}
& \text{sol'n} & \% & \text{acid} \\ \hline
\text{Start} & 55 & 30\% & 16.5 \\
\text{Remove} & \text{-}x & 30\% & \text{-}0.3x \\
\text{Add} & +x & 100\% & +x \\ \hline
\text{Mixture} & 55 & 50\% & 27.5 \end{array}$

Look at the last (rightmost) column.
It describes the amount of acid in the various stages.

. . we remove $0.3x$ gallons of acid,
. . we add $x$ gallons of acid,
and we end up with $27.5$ gallons of acid.
And there is our equation! . . . . . ${\color{blue}16.5 - 0.3x + x \:=\:27.5}$