The Dog Of The Commander

A long and straight row of soldiers is marching with constant speed. The dog of the commander runs along the row from the end till the beginning, from the beginning till the end, from the end till the beginning etc. This takes always 2 minutes, 1 minute, 2 minutes, etc. 

At some point the soldiers have to pass a very small bridge. However, the commander does not trust the strength bridge completely. He wonders how much time it will take them to pass the bridge when they stay marching with the same speed. 

How much time is that?  

You can skip to the answer! 

Passing A Very Small Bridge

What is the question?

For march of soldiers to pass any point, there must be first man crossing that point first and then later the last man should cross the same point. 

Now the dog runs from the first to last man in 1 minute, then runs back to first man in 2 minutes and again back to last man in 1 minutes. That means the dog must be at the same point where he started journey from first man. In other words, the dog has run for 2 minutes in both the directions. Hence, he must be at the same point after 4 minutes.

Interestingly, the dog was at first man when started & after 4 minutes ended at the last man of the march. Since dog is at the same point after 4 minutes, the entire march of soldiers is crossed that point in 4 minutes as first and last man have crossed that point.

The same when applied to a very small bridge which can be assumed as a point, we can say that it requires 4 minutes for soldiers to cross it.
 

Draining Reservoir

A large fresh water reservoir has two types of drainage system: small pipes and large pipes.

6 large pipes, on their own, can drain the reservoir in 12 hours.

3 large pipes and 9 small pipes, at the same time, can drain the reservoir in 8 hours.

How long will 5 small pipes, on their own, take to drain the reservoir?

Find here the exact time required! 

Time To Drain The Reservoir

Here is the question! 

Let T be the capacity of the tank. Let L be the large pipe and S be the small pipe.

T/6L = 12

L = T/72

T/(3L + 9S) = 8

Putting L = T/72,

S = T/108

Hence T/5S = 108/5 gives S = 21.6 hours.

To conclude, small pipes require 21.6 hours i.e. 21 hours and 36 minutes to drain the reservoir.