Question

A water pump takes 6 hours to fill an overhead tank. A standby pump takes 10 hours to fill the same tank. If the first pipe fails after 2 hours, then how long will the standby pump take to fill the rest?

Answer 1

Hint: This problem is similar to those problems in which the working persons, their working hours and number of days they work would be given. This is also similar to it. So we will first find the total work done and then the work done by the two pipes in one hour. According to that we will calculate the work done by the first pipe in 2 hours and then the remaining work to be completed by the standby pipe.

Complete step-by-step answer:
Let the first pump be A and the standby pump be S.
Now pipe A takes 6 hours to fill an overhead tank.
Standby pump S takes 10 hours to fill the same tank.
Now the LCM of these hours will be the total capacity of the tank we can say.
So the capacity is 60.
Now pump A completes \[\dfrac{{60}}{6} = 10work/hr\]
And pump S completes \[\dfrac{{60}}{{10}} = 6work/hr\]
That is pump S is slower. Anyways !
Now we started with pump A for the first two hours. So it will fill \[2 \times 10 = 20work\] of the total capacity.
Now pump S is in work. So it needs to fill \[60 - 20 = 40work\].
But it can do \[6work/hr\].
So the time taken by it to complete the rest of the work is,
\[\dfrac{{40}}{6} = 6.6\bar 6hrs\]
This is the time the standby pump took.
So, the correct answer is “\[ 6.6\bar 6hrs\]”.

Note: Here note that the pump A takes lesser times than pump S. so we can conclude here that the work done and the time are inversely proportional. Such that in less time pump A completes more work.
Sometimes the proportion is not inverse in some cases. Like the wages paid for hours of working. The more a person works, the more he gets paid.