Problem Solving

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

$$rate\ of\ \ x+y=\frac{1}{4}\ where\ x\ is\ the\ time\ pump$$
$$x\ needs\ to\ fill\ out\ the\ pool.\ $$
$$rate\ of\ pump\ x\ is\ \frac{1}{x}$$
$$rate\ of\ pump\ y\ is\ \frac{1}{48}$$
$$pump\ y\ drains\ water\ ou\ of\ the\ container\ $$
$$\frac{1}{x}-\frac{1}{48}=\frac{1}{4x}$$

$$-\frac{1}{48}=\frac{1}{4x}-\frac{1}{x}$$
$$-\frac{1}{48}=-\frac{3}{4x}$$
cross multiply
$$-4x=-3\cdot48$$
$$-\frac{4x}{-4}=-\frac{144}{-4}$$
$$x=36$$
$$answer\ is\ option\ B$$

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.