Aptitude Discussion

Q. |
Three pipes $A$, $B$ and $C$ are connected to a tank. These pipes can fill the tank separately in 5 hr, 10 hr and 15hr respectively. When all the three pipes were opened simultaneously, it was observed that pipes $A$ and $B$ were supplying water at \(\left(\dfrac{3}{4}\right)^{th}\) of their normal rates for the 1st hour after which they supplied water at normal rate. Pipe $C$ supplied water at \(\left(\dfrac{2}{3}\right)^{th}\)of its normal rate for 1st 2 hours, after which it supplied at its normal rate. In how much time, tank would be filled? |

✖ A. |
1.05 hr |

✖ B. |
2.05 hr |

✔ C. |
3.05 hr |

✖ D. |
None of these |

**Solution:**

Option(**C**) is correct

The part of the tank filled by $A$ and $B$ in first two hour

\(\Rightarrow \dfrac{3}{4}\times \left(\dfrac{1}{5}+\dfrac{1}{10}\right)+\left(\dfrac{1}{5}+\dfrac{1}{10}\right)\)

The part of tank filled by $C$ in first two hours = \(2\times \dfrac{2}{3}\times \dfrac{1}{15}\)

Remaining part = \(\dfrac{139}{360}\)

In 1 hour, all the three pipes together will fill = \(\dfrac{11}{30}\)

Hence, the time taken to fill the remaining tank = \(\dfrac{139}{360}\times \dfrac{30}{11}\)= 1.0530 hour

Thus, the total time taken to fill the remaining tank = **3.05 hour.**

**Vineet**

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45 hours....................?

**Dan**

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Why can't we use the following way to solve this problem?

In 1st hour the performance of the 3 pipes will be following:

Pipe A = $5*0.75=3.75 hrs $

Pipe B = $ 10*0.75=7.5 hrs $

Pipe C = $ 15* 2/3 = 10 hrs $

So for 1st hour we get the following:

$ 1*(1/3.75+1/7.5+1/10)=x$

$x=0.5$

So in 1 hour half of tank was filled

For the following another hour we have the following:

Pipe A = $5 hrs $

Pipe B = $10 hrs$

Pipe C = $ 10 hrs$

Amount of tank filled for another hour is :

$1*(1/5+1/10+1/10)=x$

$x=0.4$

For 1st 2 hours amount of tank filled is 0.5+0.4=0.9

The 0.1 amount of tank will be filled in:

$ x*(1/5+1/10+1/15)=0.1$

$ x= 16 mins $

So the tank will be filled in 1+1+0.25= 2.25 hours

its wrong. the answer has to be above 2.5hrs

**Shaibal**

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Is it correct that all the three pipes together will fill $= \dfrac{11}{30}$?

Since according to my calculation all the three pipes together will fill in one hour will be $\dfrac{221}{360 \times 2}$

In one hour the three pipes will fill,

$\left(\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{1}{15}\right)=\dfrac{11}{30}.$

I guess you are making some mistake somewhere.

If you show me the steps of your calculation, maybe I can be of help.

Pipe P1 can fill a cistern in 40 hours. Pipe P2 can empty the same completely filled cistern in 60 hours. If in every 3 hours P1 runs in the first two hours and P2 runs in the last hour, then how long will it take to fill the same half filled cistern?

Options are:

1 43 hours 20 minutes

2 45 hours 40 minutes

3 47 hours 20 minutes

4 45 hours

5 43 hours