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

tagged by: swerve

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A

B

C

D

E

## Global Stats

Difficult

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

The OA is E.

Source: Veritas Prep

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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

The OA is E.

Source: Veritas Prep
Say the rate of each of the two wide pipes = 1 unit

Thus, the rate of each of the three narrow pipes = 1 * 1/2 = 1/2 unit

Combined rate of the three narrow pipes = 3 * 1/2 = 3/2
Combined rate of the two wide pipes = 2 * 1 = 2

Combined rate of all the five pipes = 3/2 + 2 = 7/2

Thus, the fraction of time taken by the two wide pipes working together to fill the tank = 2 / (7/2) = 4/7

Hope this helps!

-Jay
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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.

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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.

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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
We can let x = the rate of each small inlet pipe and thus 2x = the rate of each large inlet pipe.

Let’s now assume that it will take 10 hours to fill the pool if all the pipes work together. Thus, the capacity of the pool is 10(3x + 2(2x)) = 10(7x) = 70x.

If the pool is filled only by the 2 large pipes, then it would take 70x/(2(2x)) = 70x/(4x) = 35/2 = 17.5 hours. Therefore, the ratio of the time if all 5 pipes work together to the time if only the 2 large pipes work together is 10/17.5 = 20/35 = 4/7.

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