Re: A contractor undertakes to do a job in 20 days so he employs 36 labore
[#permalink]
06 Dec 2021, 04:31
If the workers do \(\frac{3}{11}\) of the job in 5 days, after 20 days they will do\(\frac{12}{11}\) of the work. So to complete the job in exactly 20 days there has to be a slight decrease in the amount of workers. There are only two answers which decrease the work force:
C - fire 4
E - fire 16
Losing 16 would be almost half of the work force which is far too great of a number to lose. Answer will logically have to be C.
A way to solve it:
To make numbers a little more palatable: convert \(\frac{3}{11}\) into \(\frac{3*36*5}{11*36*5}\) as we are working with 36 workers of 5 days.
\(\frac{540}{1980}\).
Each day the workers collectively do \(\frac {\frac{540}{5}}{1980}= \frac{108}{1980}\) of the job.
Which means that each worker does \(\frac {\frac{108}{36}}{1980} = \frac{3}{1980}\) of the job each day.
We know that \(\frac{540}{1980}\) has already been done, which means that there is still \(\frac{1440}{1980}\) work to be done in 15 days.
Which means each day, collectively, the workers need to do \(\frac {\frac{1440}{15}}{1980} = \frac{96}{1980}\) of the job.
Currently they do \(108\) and to drop that to \(96\), there needs to be a daily reduction of \(\frac{12}{1980}\). We know that each worker does: \(\frac{3}{1980}\) daily, so \(\frac{12}{3} = 4\). 4 people need to be fired for the job to be completed in precisely 20 days.
Answer C