Question

A, B and C can finish a job working alone in 72,24 and 36 days respectively. In how many days they can finish the job if they worked together?
(a) 12
(b) 9
(c) 15
(d) 18

Answer 1

Hint:In this problem we first try to calculate the work done by each worker in one day and then multiply with the total to get the number of days involved when all of them worked together.So, finally our answer for the total number of days involved is obtained.

Complete step-by-step answer:
We are given that a job is to be done. Other information is provided for different individual work.
For worker A, the total time consumed for completing the particular job is 72 days.
For worker B, the total time consumed for completing the particular job is 24 days.
For worker C, the total time consumed for completing the particular job is 36 days.
Now, we try to find out each individual one-day work.
For worker A, one day work will be $\dfrac{1}{72}$ days.
For worker B, one day work will be $\dfrac{1}{24}$ days.
For worker A, one day work will be $\dfrac{1}{36}$ days.
So, the total amount of work done in one day:
$\begin{align}
  & \dfrac{1}{72}+\dfrac{1}{24}+\dfrac{1}{36} \\
 & =\dfrac{1+3+2}{72} \\
 & =\dfrac{6}{72} \\
\end{align}$
So, the reciprocal of this will give us the total amount of time consumed if all of the workers worked together.
Therefore, total time consumed as per our analysis is:
$\dfrac{72}{6}=12$
Hence, the total time consumed to complete the work when all workers are involved is 12 days.
Therefore, option (a) is correct.

Note: The key step involved in solving this problem is the knowledge of topic time and work. Students must be careful while doing calculations.
The possible alternate solution of this problem can be expressed by assuming the total work to be 72 units. Now, A, B, C works 1, 3 and 2 units in one day. The L.C.M. of this one-day work would be 6. Therefore, total time for completing the task by all workers $=\dfrac{72}{6}=12$ days.