BTGmoderatorDC wrote:A mixture of oil, water, and vinegar contains 10% oil. After all of the water evaporates, what percent of the mixture is oil?
(1) Before the water evaporated, the mixture exactly filled a one liter bottle.
(2) Before the water evaporated, vinegar accounted for forty percent of the mixture.
Source: Manhattan Prep
$${\text{Before: x}}\,\,{\text{liters}}\,\,\,\,\left\{ \begin{gathered}
{\text{oil}}\,\,:0.1x\,\,{\text{liters}} \hfill \\
water\,\,:\,\,k\left( {0.9x} \right)\,\,{\text{liters}} \hfill \\
vinegar\,\,:\,\,\,\left( {1 - k} \right)\left( {0.9x} \right)\,\,{\text{liters}}\,\, \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\left( {0 < k < 1} \right)$$
$${\text{After}}:\,\,\,{\text{x}}\left( {1 - 0.9k} \right)\,\,\,{\text{liters}}\,\,\,\left\{ \begin{gathered}
{\text{oil}}\,\,:0.1x\,\,{\text{liters}} \hfill \\
water\,\,:\,\,0\,\,{\text{liters}} \hfill \\
vinegar\,\,:\,\,\,\left( {1 - k} \right)\left( {0.9x} \right)\,\,{\text{liters}}\,\, \hfill \\
\end{gathered} \right.$$
$$? = \frac{{0.1x}}{{x\left( {1 - 0.9k} \right)}} = \frac{{0.1}}{{1 - 0.9k}}\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\boxed{? = k}\,$$
$$\left( 1 \right)\,\,\,x = 1\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,\,k = 0.1 \hfill \\
\,{\text{Take}}\,\,\,k = 0.2 \hfill \\
\end{gathered} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{INSUFF}}.$$
$$\left( 2 \right)\,\,\,\,\,\frac{{\left( {1 - k} \right)\left( {0.9x} \right)\,\,}}{x} = \frac{2}{5}\,\,\,\,\, \Rightarrow \,\,\,\,k\,\,{\text{unique}}\,\,\,\,\, \Rightarrow \,\,\,\,{\text{SUFF}}.\,$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
