Bunuel wrote:
An animal shelter began the day Tuesday with a ratio of 5 cats for every 11 dogs. If no new animals arrived at the shelter, and the only animals that left the shelter were those that were adopted, what was the ratio of cats to dogs at the end of the day Tuesday?
(1) No cats were adopted on Tuesday.
(2) 4 dogs were adopted on Tuesday.
\(\begin{array}{*{20}{c}}
{{\text{cats}} = \,\,5\,k} \\
{{\text{dogs}} = 11\,k}
\end{array}\,\,\,\,\,\,\left( {k > 0\,\,\,\operatorname{int} } \right)\,\,\,\,\,\,\,\left( * \right)\)
\(\left( * \right)\,\,{\mkern 1mu} k\,{\mkern 1mu} {\mkern 1mu} = \,\,{\mkern 1mu} {\mkern 1mu} 11k - 2\left( {5k} \right){\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\text{ = }}{\mkern 1mu} {\mkern 1mu} \operatorname{int} - \operatorname{int} \,\,\, = \,\,\,\operatorname{int}\)
That´s the
k technique, a "killer" tool of our method (when dealing with ratios)!
\(\left( {1 + 2} \right)\,\,\,\,\,{\text{?}}\,\,\, = \,\,\,\frac{{5k}}{{\,11k - 4\,}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,k = 1\,\,\,\, \Rightarrow \,\,\,? = \frac{5}{7} \hfill \\
{\text{Take}}\,\,k = 2\,\,\,\, \Rightarrow \,\,\,? = \frac{{10}}{{18}} \ne \frac{5}{7} \hfill \\
\end{gathered} \right.\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
fskilnik.
_________________
Fabio Skilnik ::
GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here:
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