# If two pipes function simultaneously, a reservoir will be filled in 12 hours. One pipe fills the reservoir 10 hours faster than the other. How many hours will be taken by the second pipe to fill the reservoir.

Last updated date: 26th Mar 2023

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Answer

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Hint- Here, we will simply use the approach to finding the parts filled by each pipe.

Let the first pipe be pipe A and the second pipe be pipe B.

Given, Time taken to fill the reservoir when both pipes A and B function simultaneously$ = 12{\text{ hours}}$

$ \Rightarrow $Part of the reservoir filled by pipes A and B per hour when functioning simultaneously$ = \dfrac{1}{{12}}$

Also, given that pipe A fills the reservoir 10 hours faster than pipe B so if the time taken by pipe A alone to fill the reservoir is $x$ hours (say), then the time taken by pipe B alone to fill the reservoir will be $\left( {x - 10} \right)$ hours.

Time taken by pipe A alone to fill the reservoir$ = x$ hours

$ \Rightarrow $Part of the reservoir filled by pipe A alone per hour$ = \dfrac{1}{x}$

Time taken by pipe B alone to fill the reservoir$ = \left( {x - 10} \right)$ hours

$ \Rightarrow $Part of the reservoir filled by pipe B alone per hour$ = \dfrac{1}{{\left( {x - 10} \right)}}$

As we know that the part of the reservoir filled per hour when both the pipes are functioning will be equal to the sum of the parts of the reservoir filled per hour when pipe A is functioning alone and when pipe B is functioning alone.

\[

\dfrac{1}{{12}} = \dfrac{1}{x} + \dfrac{1}{{\left( {x - 10} \right)}} \Rightarrow \dfrac{1}{{12}} = \dfrac{{\left( {x - 10} \right) + x}}{{x\left( {x - 10} \right)}} \Rightarrow \dfrac{1}{{12}} = \dfrac{{2x - 10}}{{x\left( {x - 10} \right)}} \Rightarrow x\left( {x - 10} \right) = 12\left( {2x - 10} \right) \\

\Rightarrow {x^2} - 10x = 24x - 120 \Rightarrow {x^2} - 34x + 120 = 0 \Rightarrow {x^2} - 4x - 30x + 120 = 0 \Rightarrow x\left( {x - 4} \right) - 30\left( {x - 4} \right) = 0 \\

\Rightarrow \left( {x - 4} \right)\left( {x - 30} \right) = 0 \\

\]

Either $x = 4$ or $x = 30$

When $x = 4$ then $\left( {x - 10} \right) = 4 - 10 = - 6$ which is not possible because time is always positive. So, $x = 4$ is neglected.

When $x = 30$ then $\left( {x - 10} \right) = 30 - 10 = 20$.

Therefore, time taken by pipe A alone to fill the reservoir is 30 hours and the time taken by pipe B alone to fill the reservoir is 20 hours.

Hence, 20 hours are taken by the second pipe to fill the reservoir.

Note- In these types of problems, we simply take the reciprocal of the respective time taken by the pipes to find the part of the reservoir filled by each pipe. Then, we finally equal the part of the reservoir filled per hour by both pipes acting together with the sum of the parts of the reservoir filled per hour by each pipe when acting individually.

Let the first pipe be pipe A and the second pipe be pipe B.

Given, Time taken to fill the reservoir when both pipes A and B function simultaneously$ = 12{\text{ hours}}$

$ \Rightarrow $Part of the reservoir filled by pipes A and B per hour when functioning simultaneously$ = \dfrac{1}{{12}}$

Also, given that pipe A fills the reservoir 10 hours faster than pipe B so if the time taken by pipe A alone to fill the reservoir is $x$ hours (say), then the time taken by pipe B alone to fill the reservoir will be $\left( {x - 10} \right)$ hours.

Time taken by pipe A alone to fill the reservoir$ = x$ hours

$ \Rightarrow $Part of the reservoir filled by pipe A alone per hour$ = \dfrac{1}{x}$

Time taken by pipe B alone to fill the reservoir$ = \left( {x - 10} \right)$ hours

$ \Rightarrow $Part of the reservoir filled by pipe B alone per hour$ = \dfrac{1}{{\left( {x - 10} \right)}}$

As we know that the part of the reservoir filled per hour when both the pipes are functioning will be equal to the sum of the parts of the reservoir filled per hour when pipe A is functioning alone and when pipe B is functioning alone.

\[

\dfrac{1}{{12}} = \dfrac{1}{x} + \dfrac{1}{{\left( {x - 10} \right)}} \Rightarrow \dfrac{1}{{12}} = \dfrac{{\left( {x - 10} \right) + x}}{{x\left( {x - 10} \right)}} \Rightarrow \dfrac{1}{{12}} = \dfrac{{2x - 10}}{{x\left( {x - 10} \right)}} \Rightarrow x\left( {x - 10} \right) = 12\left( {2x - 10} \right) \\

\Rightarrow {x^2} - 10x = 24x - 120 \Rightarrow {x^2} - 34x + 120 = 0 \Rightarrow {x^2} - 4x - 30x + 120 = 0 \Rightarrow x\left( {x - 4} \right) - 30\left( {x - 4} \right) = 0 \\

\Rightarrow \left( {x - 4} \right)\left( {x - 30} \right) = 0 \\

\]

Either $x = 4$ or $x = 30$

When $x = 4$ then $\left( {x - 10} \right) = 4 - 10 = - 6$ which is not possible because time is always positive. So, $x = 4$ is neglected.

When $x = 30$ then $\left( {x - 10} \right) = 30 - 10 = 20$.

Therefore, time taken by pipe A alone to fill the reservoir is 30 hours and the time taken by pipe B alone to fill the reservoir is 20 hours.

Hence, 20 hours are taken by the second pipe to fill the reservoir.

Note- In these types of problems, we simply take the reciprocal of the respective time taken by the pipes to find the part of the reservoir filled by each pipe. Then, we finally equal the part of the reservoir filled per hour by both pipes acting together with the sum of the parts of the reservoir filled per hour by each pipe when acting individually.

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