A contractor undertook a contract to complete a part of a stadium in 9 months with a team of 560 persons. Later on, it was required to complete the job in 5 months. How many extra persons should he employ to complete the work?

Hint: We see that the amount of work before $\left( {{W}_{1}} \right)$ and after$\left( {{W}_{2}} \right)$ the change in schedule to complete the part is same. If we assume the number of persons and amount of time before the change in schedule as ${{x}_{1}},{{t}_{1}}$ and after the schedule as ${{x}_{2}},{{t}_{2}}$. We have ${{W}_{1}}={{x}_{1}}{{t}_{1}}={{x}_{2}}{{t}_{2}}={{W}_{2}}$. We put the gien values, solve the equation and find ${{t}_{2}}$.

Complete step-by-step solution:
We see that the amount of time for the work or productivity of the building part of a stadium was initially 9 months and then later the contractor has to complete the same amount of the work in 5 months.
Let us denote the number of persons required to complete the part as $x$ and the amount of time as ${{t}_{1}}$ initially before the change of schedule. We are given that ${{x}_{1}}=560$ and ${{t}_{1}}=9$ months. So the total productivity or work initially before the change in schedule set up was ${{W}_{1}}={{x}_{1}}\times {{t}_{1}}=560\times 9=4940$\[\]
We see that later on schedule changed and the work was ordered to complete in 5 months. Let us assume the number of persons required to complete the work in 5 months as ${{x}_{2}}$ and the time period as ${{t}_{2}}$. We are given ${{t}_{2}}=5$ months. The total productivity or work after the change in schedule is ${{W}_{2}}={{x}_{2}}\times {{t}_{2}}={{x}_{2}}\times 5=5{{x}_{2}}$.\[\]
The amount of work is same before and after the change in schedule . So we have
\[\begin{align}
  & {{W}_{1}}={{W}_{2}} \\
 & \Rightarrow 4940=5{{x}_{2}} \\
 & \Rightarrow {{x}_{2}}=\dfrac{4940}{5}=988 \\
\end{align}\]
So the number of persons he should employ to finish the working 5 months is 988.

Note: We note that it is assumed that when the questions say “LATER ON” the work has not started. Here we used equal product in work-time problems because of productivity but in case of rate of work we have to use equal ratio which is ${{r}_{1}}={{r}_{2}}\Rightarrow \dfrac{{{W}_{1}}}{{{t}_{1}}}=\dfrac{{{W}_{2}}}{{{t}_{2}}}$. We also use equal ratio in case of speed $s=\dfrac{d}{t}$ and product in case of distance $ d=st.$