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  • Typical Day of a UCLA MBA Student - Recording of Webinar with UCLA Adcom and Student

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If pipe B fills a container at a constant rate in 100 minutes, how man

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If pipe B fills a container at a constant rate in 100 minutes, how man  [#permalink]

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New post 13 Jul 2018, 03:42
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If pipe B fills a container at a constant rate in 100 minutes, how many minutes does it take pipe A, working at is own constant rate, to fill the same container?

(1) Pipe A and pipe B together fill the container at (1/4) the time it takes pipe A alone.
(2) Pipe A and pipe B together fill the container at (3/4) the time it takes pipe B alone.

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If pipe B fills a container at a constant rate in 100 minutes, how man  [#permalink]

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New post 13 Jul 2018, 04:23
Bunuel wrote:
If pipe B fills a container at a constant rate in 100 minutes, how many minutes does it take pipe A, working at is own constant rate, to fill the same container?

(1) Pipe A and pipe B together fill the container at (1/4) the time it takes pipe A alone.
(2) Pipe A and pipe B together fill the container at (3/4) the time it takes pipe B alone.


If pipe B fill the container in 100 minutes, it will fill \(\frac{1}{100}\)th of the container in a minute.

1. If x is the time taken by both the pipes to fill the container, then \(\frac{x}{4}\) is the
time taken for Pipe A to fill the container. We will have one equation and one unknown
Therefore, we can find the time taken for the pipe A along to fill the container (Sufficient)

2. If x is the time taken by both the pipes to fill the container, then \(\frac{3x}{4}\) is the
time taken for Pipe B to fill the container. Now that we can find the time taken by Pipe B to fill
the container, we can find the value of x(which is the time taken by both the pipes to fill the
container). Using this we can calculate the time taken by pipe A to fill the container (Sufficient - Option D)
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If pipe B fills a container at a constant rate in 100 minutes, how man  [#permalink]

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New post 13 Jul 2018, 04:29
Bunuel wrote:
If pipe B fills a container at a constant rate in 100 minutes, how many minutes does it take pipe A, working at is own constant rate, to fill the same container?

(1) Pipe A and pipe B together fill the container at (1/4) the time it takes pipe A alone.
(2) Pipe A and pipe B together fill the container at (3/4) the time it takes pipe B alone.


Given\(r_{B}=\frac{1}{100}, t_{B}=100

Question stem:- [m]t_{A}=?\)

St1:- Pipe A and pipe B together fill the container at (1/4) the time it takes pipe A alone.
\(t_{A+B}=\frac{1}{4}*t_{A}\)
Combined rate, \(r_{A+B}=\frac{1}{100}+\frac{1}{t_{A}}\)
So, \(\frac{4}{t_{A}}=\frac{1}{100}+\frac{1}{t_{A}}\)
Hence \(t_{A}\) can be determined definitely.
Sufficient.

St2:-Pipe A and pipe B together fill the container at (3/4) the time it takes pipe B alone

Similarly, Given \(t_{A+B}=\frac{3}{4}*t_{B}\)
So, \((\frac{4}{3})*t_{B}=\frac{1}{100}+\frac{1}{t_{B}}\)
Hence, \(t_{B}\) can be determined definitely & subsequently value of t_{A} can be determined.
Sufficient.

Ans. (D)
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If pipe B fills a container at a constant rate in 100 minutes, how man &nbs [#permalink] 13 Jul 2018, 04:29
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