# Thread: % problem or possibly volume

1. ## % problem or possibly volume

Q. A 5 liter container contains the following mix of ingredients expressed as percentages.
ingredient A 35%
ingredient B 15%
ingredient C 2.5%
ingredient D 1%
Water 46.5%

The solution is then diluted in a 250ml container which contains 15ml of the initial mix of ingredients and 235ml of water.
Please could somebody express the percentages of the individual ingredients which are in the 250ml container.
Thank you

2. Hello, stump!

This is an arithmetic problem . . .

A solution contains the following mix of ingredients expressed as percentages:

. . $\begin{array}{cccc}\text{ingredient A} & 35\% \\
\text{ingredient B} & 15\% \\ \text{ingredient C} & 2.5\% \\
\text{ingredient D} & 1\% \\ \text{Water} & 46.5\% \end{array}$

In a container, 15ml of the initial solution is diluted with 235ml of water.

Find the percentages of the individual ingredients in the final mixture.

In 15ml of solution, we have:

. . $\begin{array}{cc}
& \text{ml} \\ \hline
A & 5.250 \\ B & 2.250 \\ C & 0.375 \\ D & 0.150 \\ \text{Water} & 6.975 \\ \hline & 15.000\;\; \end{array}$

Afer 235ml of water is added, we have:

. . $\begin{array}{cc}
& \text{ml} \\ \hline
A & 5.250 \\ B & 2.250 \\ C & 0.375 \\ D & 0.1250 \\ \text{Water} & 241.975\;\; \\ \hline & 250.000\;\;\end{array}$

Therefore:

. . $\begin{array}{ccccc}
A &=& \dfrac{5.25}{250} &=& 2.10\% \\ \\[-2mm]
B &=& \dfrac{2.25}{250} &=& 0.90\% \\ \\[-2mm]
C &=& \dfrac{0.375}{250} &=& 0.15\% \\ \\[-2mm]
D &=& \dfrac{0.15}{250} &=& 0.06\% \\ \\[-2mm]
\text{Water} &=& \dfrac{241.975}{250} &=& 96.79\%
\end{array}$