aishwaryav12 wrote:Two pumps are connected to a certain empty container at the same time. Pump X fills the container water at a constant rate, while pump Y drains water out of the container at a constant rate. The two pumps finish filling the container in four times the duration it would take pump X alone to fill the container. If pump Y alone can empty a whole container in 48 minutes, then how many minutes does it take pump X alone to fill the container?
A. 24
B. 36
C. 48
D. 50
E. 64
\[?\,\,\,:\,\,\,\# \,\,\,{\text{minutes}}\,\,X\,\,{\text{fills}}\,\,{\text{container}}\]
Excellent opportunity to use
UNITS CONTROL, one of the most powerful tools of our method!
\[X\,\,:\,\,\,\,\frac{{x\,\,{\text{gallons}}}}{{1\,\,{\text{minute}}}}\,\,\,\,\,\left( {{\text{filling}}} \right)\]
\[Y\,\,:\,\,\,\,\frac{{y\,\,{\text{gallons}}}}{{1\,\,{\text{minute}}}}\,\,\,\,\,\left( {{\text{draining}}} \right)\]
The filling TIME ratio 4:1 (for any given volume) of water is
inversely proportional to the filling VOLUME ratio (for any given time), hence:
\[{\text{Relative}}\,\,{\text{volume}}\,\,\left( {{\text{filling}}} \right)\,\,{\text{rate}}\,\,\,:\,\,\,\,\frac{1}{4} = \frac{{\frac{{\left( {x - y} \right)\,\,{\text{gallons}}}}{{1\,\,{\text{minute}}}}}}{{\frac{{x\,\,{\text{gallons}}}}{{1\,\,{\text{minute}}}}}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\, \ldots \,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\frac{x}{y} = \frac{4}{3}\,\,\,\]
Finally :
\[y = 3\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\{ \begin{gathered}
\,\,x = 4\, \hfill \\
48\,\,{\text{minutes}}\,\,\,\left( {\frac{{3\,\,{\text{gallons}}}}{{1\,\,{\text{minute}}}}} \right)\,\,\, = \,\,\,48 \cdot 3\,\,{\text{gallons}}\,\,\, = \,\,\,{\text{Volume}}\,\,{\text{container}} \hfill \\
\end{gathered} \right.\]
\[{\text{?}}\,\,\,{\text{ = }}\,\,\,48 \cdot 3\,\,{\text{gallons}}\,\,\left( {\frac{{1\,\,{\text{minute}}}}{{4\,\,{\text{gallons}}}}} \right)\,\,\,\, = \,\,\,\,36\,\,{\text{minutes}}\,\,\]
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
