# Mixture problem

• Aug 24th 2009, 03:55 AM
a69356
Mixture problem
Q) A mixture of 125 galllons of wine and water contains 20% of water.How much water must be added to the mixture in order to increase the percentage of the water to 25% of the new mixture?

Thanks
• Aug 24th 2009, 06:56 AM
earboth
Quote:

Originally Posted by a69356
Q) A mixture of 125 galllons of wine and water contains 20% of water.How much water must be added to the mixture in order to increase the percentage of the water to 25% of the new mixture?

Thanks

1. The actual mixture contains $0.2\cdot 125 = 25$ gallons of water.

2. Let x denote the additional water then you know:

$\dfrac{25+x}{125+x}=0.25$

3. Solve for x.

Spoiler:
I've got $x = \dfrac{25}3\ gallons\ of\ water$
• Aug 24th 2009, 03:00 PM
Soroban
Hello, a69356!

Quote:

A mixture of 125 galllons of wine and water contains 20% water.
How much water must be added to increase the percentage of the water to 25%?

When I first learned "Mixture Problems", I made a chart . . .

The chart has the formula across the top:
. . $\text{[Gallons of mixture]} \times \text{[Percent water]} \:=\:\text{[Gallons of water]}$

. . $\begin{array}{c||ccccc}
\qquad & \text{Gallons} & \times & \text{Percent} & = & \text{Water} \\ \hline\hline
& &|& &|& \\ \hline
& &|& &|& \\ \hline \hline
& &|& &|& \end{array}$

. . We have: . $20\% \times 125 \:=\:25$ gallons of water.
Write this in the first row:

. . $\begin{array}{c||ccccc}
& \text{Gallons} & \times & \text{Percent} & = & \text{Water} \\ \hline\hline
\text{Start} & 125 &|& 20\% &|& 25 \\ \hline
& &|& &|& \\ \hline \hline
& &|& &|& \end{array}$

We add $x$ gallons of water (which is 100% water).
. . We add: . $100\% \times x \:=\:x$ gallons of water.
Write this in the second row.

. . $\begin{array}{c||ccccc}
& \text{Gallons} & \times & \text{Percent} & = & \text{Water} \\ \hline\hline
\text{Start} & 125 &|& 20\% &|& 25 \\ \hline
\text{Add} & x &|& 100\% &|& x \\ \hline \hline
& &|& &|& \end{array}$

The final mixture has $x+125$ gallons which is 25% water.
. . It contains: . $25\% \times (x+125) \:=\:0.25(x+125)$ gallons of water.
Write this in the third row.

. . $\begin{array}{c||ccccc}
& \text{Gallons} & \times & \text{Percent} & = & \text{Water} \\ \hline\hline
\text{Start} & 125 &|& 20\% &|& 25 \\ \hline
\text{Add} & x &|& 100\% &|& x \\ \hline \hline
\text{Mixture} & x+125 &|& 25\% &|& 0.25(x+125) \end{array}$

We start with 25 gallons of water, then add $x$ gallons of water.
. . Hence, the mixture contains $x+25$ gallons of water. [1]

We have $x+125$ gallons which is 25% water.
. . Hence, the mixture contains $0.25(x+125)$ gallons of water. [2]
There is our equation! $\hdots\quad x + 25 \:=\:0.25(x+125)$