A tank has 5 inlet pipes. Three pipes are narrow and two are

swerve wrote:A tank has 5 inlet pipes. Three pipes are narrow and two are wide. Each of the three narrow pipes works at 1/2 the rate of each of the wide pipes. All the pipes working together will take what fraction of time taken by the two wide pipes working together to fill the tank?

A. 1/2
B. 2/3
C. 3/4
D. 3/7
E. 4/7

Source: Veritas Prep

\[5\,\,{\text{pipes}}\,\,\,\left\{ \begin{gathered}
\,3\,\,{\text{narrow}}\,\,\,\, \to \,\,\,{\text{each}}\,\,\,1\,\,{\text{gallons}}/\min \,\,\, \hfill \\
\,2\,\,{\text{wide}}\,\,\,\,\,\,\,\,\, \to \,\,\,{\text{each}}\,\,\,2\,\,{\text{gallons}}/\min \hfill \\
\end{gathered} \right.\,\,\,\,\,\left( {{\text{particular}}\,\,{\text{case}}!} \right)\]
\[{\text{A}}\,\,\,{\text{ = }}\,\,\,{\text{2}}\,\,{\text{wide}}\,\,{\text{together}}\,\,\,{\text{:}}\,\,\,\,2 \cdot 2 = 4\,\,{\text{gallons/min}}\]
\[{\text{B}}\,\,\,{\text{ = }}\,\,\,{\text{all}}\,\,{\text{5}}\,\,{\text{together}}\,\,\,{\text{:}}\,\,\,\,3 \cdot 1 + 2 \cdot 2 = 7\,\,{\text{gallons/min}}\]
\[{\text{B:A}}\,\,\underline {{\text{work}}} \,\,{\text{ratio}}\,\,\left( {{\text{per}}\,\,{\text{any}}\,\,{\text{time}}} \right)\,\,\,{\text{ = }}\,\,\,\,\frac{7}{4}\,\,\,\,\,\]
\[?\,\,\, = \,\,\,B:A\,\,\underline {{\text{time}}} \,\,{\text{ratio}}\,\,\,\left( {{\text{per}}\,\,{\text{any}}\,\,{\text{work}}} \right)\,\,\, = \,\,\,{\left( {\frac{7}{4}} \right)^{ - 1}} = \,\,\,\frac{4}{7}\]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

swerve wrote:A tank has 5 inlet pipes. Three pipes are narrow and two are wide. Each of the three narrow pipes works at 1/2 the rate of each of the wide pipes. All the pipes working together will take what fraction of time taken by the two wide pipes working together to fill the tank?

A. 1/2
B. 2/3
C. 3/4
D. 3/7
E. 4/7

Let the rate of each wide pipe = 2 liters per hour.
Combined rate for 2 wide pipes = 2*2 = 4 liters per hour.

Since each narrow pipe works at 1/2 the rate of each wide pipe, the rate of each narrow pipe = (1/2)(2) = 1 liter per hour.
Combined rate for 3 narrow pipes and 2 wide pipes = (3*1) + (2*2) = 7 liters per hour.

Let the tank 28 liters.
Since the rate for all 5 pipes = 7 liters per hour, the time for all 5 pipes to fill the 28-liter tank = 28/7 = 4 hours.
Since the rate for 2 wide pipes = 4 liters per hour, the time for 2 wide pipes to fill the 28-liter tank = 28/4 = 7 hours.
(time for all 5 pipes)/(time for 2 wide pipes) = 4/7.

The correct answer is E.