Mathsblog Question (By Umesh)

Four people are digging a ditch of some pre-specified size, one after another, and finished a ditch. These four might have different speed in their work. Each of them might have worked for a different time and finished some portion of the work. It is observed that each of them dug for such time that, during that time the other three, working together, could have finished half the ditch. This is true for each of the workers.

Question: If they worked together, how faster they would have finished the ditch, when compared to the total time they took (i.e., sum of the individual time each worker spent)?

(For example, if, working one after another, they took 12 hours to finish the work, how much time it would have taken if they worked simultaneously?)

Let the ith person dig a length of xi meters in 1 hour and let him work for ti hours. Let d be the depth of the ditch. Then we have
\[{t_1} \times 1 + {t_2} \times 2 + {t_3} \times 3 + {t_4} \times 4 = d\]\[{t_1} + {t_2} + {t_3} + {t_4} = 12\]

It is observed that each of them dug for such time that, during that time the other three, working together, could have finished half the ditch. This is true for each of the workers.

Hence we must have 
\[{t_1}{\rm{ }}\left( {{x_2}{\rm{ }} + {\rm{ }}{x_3}{\rm{ }} + {\rm{ }}{x_4}} \right){\rm{ }} = {\rm{ }}\frac{d}{2}\]\[{t_2}({x_1} + {x_3} + {x_4}) = \frac{d}{2}\]\[{t_3}{\rm{ }}\left( {{x_1}{\rm{ }} + {\rm{ }}{x_2}{\rm{ }} + {\rm{ }}{x_4}} \right){\rm{ }} = {\rm{ }}\frac{d}{2}\]\[{t_4}{\rm{ }}\left( {{x_1}{\rm{ }} + {\rm{ }}{x_2}{\rm{ }} + {\rm{ }}{x_3}} \right){\rm{ }} = {\rm{ }}\frac{d}{2}\]
Adding the above equations
\[({t_2} + {t_3} + {t_4}){x_1} + ({t_1} + {t_3} + {t_4}){x_2} + ({t_1} + {t_2} + {t_4}){x_3} + ({t_1} + {t_2} + {t_3}){x_4} = 2d\]This Implies\[\left( {12 - {t_1}} \right){x_1} + \left( {12 - {t_2}} \right){x_2} + \left( {12 - {t_3}} \right){x_3} + \left( {12 - {t_4}} \right){x_4} = 2d\] That is\[12\left( {{x_1} + {x_2} + {x_3} + {x_4}} \right) - \left( {{t_1}{x_1} + {t_2}{x_2} + {t_3}{x_3} + {t_4}{x_4}} \right) = 2d\]This means\[12\left( {{x_1} + {x_2} + {x_3} + {x_4}} \right) = 3d\] and so finally we have\[\left( {{x_1} + {x_2} + {x_3} + {x_4}} \right) = \frac{d}{4}\] Now let us suppose that they can finish the job if they work together for t hours.This means \[t\left( {{x_1} + {x_2} + {x_3} + {x_4}} \right) = d\,\,\, \Rightarrow \,\,\,t\frac{d}{4} = d\,\,\, \Rightarrow \,\,\,t = 4\]

Post Script: Many readers of the mathsblog viewed my answer unwisely complicated. Some even went to the extent that those who study higher mathematics lack commonsense and they first make simple things complicate and then solve! The present problem was an Olympiad problem for small children of age six and my solution was stated to be something outside their reach. Subsequently alternate solutions were given by other people which were graded as the ideal ones and were indeed so. Before posting this here, I once again went through my earlier solution and tried to modify it to look natural. But on finishing the job I felt my second solution very artificial and threw it away. Poor me!