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A group of workers can complete a job in 9 days. But it so happens that every alternate day starting form the second day 2 workers are withdrawn from the job and every alternate day starting from the third day one worker added to the group. In such a way that the job is finished by the time and there is no worker left. If it takes double the time to finish the job now, find the number of workers who started the job?

A group of workers can complete a job in 9 days. But it so happens that every alternate day starting form the second day 2 workers are withdrawn from the job and every alternate day starting from the third day one worker added to the group. In such a way that the job is finished by the time and there is no worker left. If it takes double the time to finish the job now, find the number of workers who started the job? Correct Answer 10

Let,

The number of workers who started the job = y

Efficiency of each worker = a/day

⇒ Total work = 9ay

According to the question,

Work will be finished in = 9 × 2 =18 days

Number of workers on 1st day = y

Number of workers on 2nd day = y - 2

Number of workers on 3rd day = y - 2 + 1 = y - 1

Number of workers on 4th day = y - 3

Number of workers on 5th day = y - 3 + 1 = y - 2

Number of workers on 6th day = y - 4

Number of workers on 7th day = y - 4 + 1 = y - 3

Number of workers on 8th day = y - 5

Number of workers on 9th day = y - 5 + 1 = y - 4

Number of workers on 10th day = y - 6

Number of workers on 11th day = y - 6 + 1 = y - 5

Number of workers on 12th day = y - 7

Number of workers on 13th day = y - 7 + 1 = y - 6

Number of workers on 14th day = y - 8

Number of workers on 15th day = y - 8 + 1 = y - 7

Number of workers on 16th day = y - 9

Number of workers on 17th day = y - 9 + 1 = y - 8

Number of workers on 18thday = y - 10

⇒ Total work done = × a = (18y - 90) × a

⇒ 9ay = (18y - 90)a

⇒ 9y = 90

⇒ y = 10

∴ The number of workers who started the job = 10

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