1. ## 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

2. 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 $\displaystyle 0.2\cdot 125 = 25$ gallons of water.

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

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

3. Solve for x.

Spoiler:
I've got $\displaystyle x = \dfrac{25}3\ gallons\ of\ water$

3. Hello, a69356!

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

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

. . We 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.

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

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]

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 in two ways.

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

Got it?

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# 125 gallons of a mixture contains 20 water

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