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A contractor builds an over tank with 3 pipes. First pipe can fill the
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15 May 2021, 23:22
15 May 2021, 23:22
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Question Stats:
45% (03:08) correct
54% (03:04) wrong based on 11 sessions
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A contractor builds an over tank with 3 pipes. First pipe can fill the tank in 12 hours, second one takes 1.5 times longer and the third one empties it in 36 hours. One day, contractor opens the first pipe and after 2 hours, the second pipe. At the end of 6 hours, he realized that the third pipe is open too and now closes it. What is the total time required to fill the tank?
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9 hours 12 mins or 552 minutes
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
A contractor builds an over tank with 3 pipes. First pipe can fill the
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16 May 2021, 23:25
16 May 2021, 23:25
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Let us name the pipes as A, B and C
\(T_A = 12\) hours
\(T_B = 18\) hours
\(T_C = 36\) hours
We know Rate, \(R = \frac{1}{T}\)
Therefore,
\(R_A = \frac{1}{12}\)
\(R_B = \frac{1}{18}\)
\(R_C = \frac{1}{36}\)
contractor opens the first pipe and after 2 hours, the second pipe. At the end of 6 hours, he realized that the third pipe is open too
i.e. A worked for 6 hours, B worked for 4 hours and C worked for 6 hours too
So, their combined work = \(\frac{6}{12} + \frac{4}{18} - \frac{6}{36} = \frac{5}{9}\)
(NOTE: -ve sign indicates that pipe C is doing opposite work)
Since, these pipes have filled \(\frac{5}{9}^{th}\) of the tank in 6 hours, the remaining work is \(1 - \frac{5}{9} = \frac{4}{9}\)
i.e. \(\frac{4}{9}^{th}\) of the work is to be completed by A and B together
\(R_{A+B} = \frac{1}{12} + \frac{1}{18} = \frac{5}{36}\)
\(Rate = \frac{Work}{Time} = (\frac{4}{9})(\frac{36}{5}) = \frac{16}{5}\) hours
TOTAL time required to fill the tank = 6 hours + \(\frac{16}{5}\) hours = \(\frac{46}{5}\) hours = \(\frac{46}{5}(60)\) minutes = \(552\) minutes = \(9\) hours \(12\) minutes
\(T_A = 12\) hours
\(T_B = 18\) hours
\(T_C = 36\) hours
We know Rate, \(R = \frac{1}{T}\)
Therefore,
\(R_A = \frac{1}{12}\)
\(R_B = \frac{1}{18}\)
\(R_C = \frac{1}{36}\)
contractor opens the first pipe and after 2 hours, the second pipe. At the end of 6 hours, he realized that the third pipe is open too
i.e. A worked for 6 hours, B worked for 4 hours and C worked for 6 hours too
So, their combined work = \(\frac{6}{12} + \frac{4}{18} - \frac{6}{36} = \frac{5}{9}\)
(NOTE: -ve sign indicates that pipe C is doing opposite work)
Since, these pipes have filled \(\frac{5}{9}^{th}\) of the tank in 6 hours, the remaining work is \(1 - \frac{5}{9} = \frac{4}{9}\)
i.e. \(\frac{4}{9}^{th}\) of the work is to be completed by A and B together
\(R_{A+B} = \frac{1}{12} + \frac{1}{18} = \frac{5}{36}\)
\(Rate = \frac{Work}{Time} = (\frac{4}{9})(\frac{36}{5}) = \frac{16}{5}\) hours
TOTAL time required to fill the tank = 6 hours + \(\frac{16}{5}\) hours = \(\frac{46}{5}\) hours = \(\frac{46}{5}(60)\) minutes = \(552\) minutes = \(9\) hours \(12\) minutes
Re: A contractor builds an over tank with 3 pipes. First pipe can fill the
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17 May 2021, 23:49
17 May 2021, 23:49
we are assuming that pipe c is open for 6 hr?
because que don't tell properly when pipe c was open
because que don't tell properly when pipe c was open
