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

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- Aug 24th 2009, 02:55 AMa69356Mixture 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, 05:56 AMearboth
- Aug 24th 2009, 02:00 PMSoroban
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%?

The chart has the formula across the top:

. . $\displaystyle \text{[Gallons of mixture]} \times \text{[Percent water]} \:=\:\text{[Gallons of water]}$

. . $\displaystyle \begin{array}{c||ccccc}

\qquad & \text{Gallons} & \times & \text{Percent} & = & \text{Water} \\ \hline\hline

& &|& &|& \\ \hline

& &|& &|& \\ \hline \hline

& &|& &|& \end{array}$

We start with 125 gallons of mixture which is 20% water.

. . We have: .$\displaystyle 20\% \times 125 \:=\:25$ gallons of water.

Write this in the first row:

. . $\displaystyle \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 $\displaystyle x$ gallons of water (which is 100% water).

. . We add: .$\displaystyle 100\% \times x \:=\:x$ gallons of water.

Write this in the second row.

. . $\displaystyle \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 $\displaystyle x+125$ gallons which is 25% water.

. . It contains: .$\displaystyle 25\% \times (x+125) \:=\:0.25(x+125)$ gallons of water.

Write this in the third row.

. . $\displaystyle \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}$

Read down the last column.

We start with 25 gallons of water, then add $\displaystyle x$ gallons of water.

. . Hence, the mixture contains $\displaystyle x+25$ gallons of water. [1]

Read across the last row.

We have $\displaystyle x+125$ gallons which is 25% water.

. . Hence, the mixture contains $\displaystyle 0.25(x+125)$ gallons of water. [2]

We just described the final amount of water inways.*two*

There is our equation! $\displaystyle \hdots\quad x + 25 \:=\:0.25(x+125)$

Got it?