A number of men undertook to dig a ditch and could have finished the job in 14 hours, if they had all begun working simultaneously. But only one of them starts working. A 2nd person starts working after x hours. 3rd person starts working x hours after 2nd person starts working. Similarly each successive person starts working x hours after the previous worker starts. After the last worker had begun working the job was finished in x hours, each one of the participants working till the completion of the job. How long did they work, if the first worker to begin worked 6 times as long as the last one to begin?
Explanation:
Let the number of men be ‘n’ and equal time durations be ‘x’ and the efficiency of each person be 1 unit/hour.
Had all of them worked together the work would’ve been completed in 14 hours.
∴ Total work done = n × 1 × 14 = 14n units.
Now, each of them joins x hours after the previous man and the last man works for x hours. ∴ The first person works for a total of nx hours.
Since the 1st person worked 6 times as long as the last one i.e., he works for 6x hours. ⇒ nx = 6x ⇒ n = 6
∴ The works lasts for a total of 6x hours. The 1st person works for 6x hours, 2nd for 5x hours and so on till the last person who works for x hours.
∴ Total work done = 6x × 1 + 5x × 1 + 4x × 1 + 3x × 1 + 2x × 1 + x × 1 = 21x
Now, 21x = 14n = 14 × 6 = 84
⇒ x = 4 hours
So, total time taken = 6x = 6 × 4 = 24 hrs
Hence, option (c).
» Your doubt will be displayed only after approval.
Couldn't understand the question
Question updated with better clarity.
Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing us your valuable feedback about Apti4All and how it can be improved.
Couldn't understand the question
Question updated with better clarity.