07 Time & Work

This is the topic which gives the relation between the time required to fill or empty the tank with the taps opened or closed.

Till the time we have only defined the concept of positive work but in the problems related to pipes and cisterns we have to define the concept of negative work also. Concept of negative work is defined as when the work is done against the requirement. 

Example 20: Pipe \(A\) fills a tank in 24 minutes. Pipe \(B\) fills the same tank 7 times faster than pipe \(A\). If both the pipes are kept open when the tank is empty, how long will it take for the tank to overflow?

Solution: Pipe \(B\) will fill the tank in \(\frac{{24}}{7}\) minutes as it is 7 times faster than pipe \(A\). 

Together, the two pipes will fill \(\frac{1}{{24}} + \frac{7}{{24}} = \frac{8}{{24}} = \frac{1}{3}rd\) of the tank in a minute. So, it will take 3 minutes for the tank to overflow. 

Example 21: A tap can fill a tank in 16 minutes and another can empty it in 8 minutes. If the tank is already 1/2 full and both the taps are opened together, will the tank be filled or emptied? How long will it take before the tank is either filled or emptied completely?

Solution: If both the pumps are opened together, then the tank will be emptied because the working efficiency of pump empting is more than that of the pump filling it. Thus in 1 min net proportion of the volume of tank filled 

= \(\frac{1}{{16}} - \frac{1}{8} =  - \frac{1}{{16}}\)

(Which means that tank will be emptied 1/16th in one minute.) Hence in 8 minutes half of the tank will be emptied. 

Example 22: A pump can be operated both for filling a tank and for emptying it. The capacity of tank is \(2400{\rm{ }}{m^3}\). The emptying capacity of the pump is \(10{\rm{ }}{m^3}\) per minute higher than its filling capacity. Consequently, the pump needs 8 minutes less to empty the tank to fill it. Find the filling-capacity of pump.

Solution: Let the filling capacity of the pump is \(x{\rm{ }} {m^3}/min\), then the emptying capacity of the pump will be \(\left( {x + {\rm{ }}10} \right){m^3}\)/min.

Hence \(\frac{{2400}}{x} = \frac{{2400}}{{x + 10}} + 8\)

Solving this we get \(x = {\rm{ }}50{\rm{ }}{m^3}\)/min

Example 23: \(A\) and \(B\) are two pipes which can fill a tank in 8 and 12 hr respectively. \(C\) alone can empty the same tank in 24 hrs. if all three pipes are opened for alternate hours starting with \(A\), in how much time an empty tank can be filled?

Solution: In one cycle fraction of tank filled 

\(\frac{1}{8} + \frac{1}{{12}} - \frac{1}{{24}} = \frac{4}{{24}} = \frac{1}{6}\)

But we can’t conclude that it will take 6 complete cycle to fill the tank completely. Because at the last hour only pipe \(C\) is working and that is an emptying pipe hence, after 5 cycle, 5/6th of the tank is full. Next hour \(A\) alone can fill 1/8th part and remaining \((1/6 - 1/8) = 1/24th\) part, \(B\) alone can fill in \(\frac{1}{2}\) hr 

Total time required = \(5 \times 3{\rm{ }} + {\rm{ }}1{\rm{ }} + {\rm{ }}1/2\) =  \(16\frac{1}{2}\)hr 

Example 24: In how many minutes can a tank be filled by three pipes whose diameters are 20 cm, 30 cm and 60 cm? Given that the largest pipe alone can fill it in 49 min.  

Solution:  We know that rate of flow   is directly proportional to the cross section area of the pipe and cross sectional area of the pipe is directly proportional to the square of the diameter.

Ratio of filling rates = \(4 : 9 : 36\)

Total time taken  \( = \frac{{36 \times 49}}{{4 + 9 + 36}} = 36\) min

Example 25: Two pipes \(A\) and \(B\) can fill a cistern in 15 hours and 30 hours respectively. There is also an outlet \(c\). If all the three pipes are opened together the tank is full in 45 hours. How much time will be taken by \(C\) to empty the full tank? 

Solution:  Suppose \(C\) can empty the tank in \(x\) hours working alone

\( \Rightarrow \)\(\frac{1}{{45}} = \frac{1}{{15}} + \frac{1}{{30}} - \frac{1}{x}\) 

\( \Rightarrow \) = \(\frac{1}{x} = \left( {\frac{1}{{15}} + \frac{1}{{30}} - \frac{1}{{45}}} \right)\) = \(\frac{{6 + 3 - 2}}{{90}} = \frac{7}{{90}}\)

\(\therefore \) C can empty the full tank in \(\frac{{90}}{7}\) hours.