2 men and 5 boys can finish a piece of work in 4 days, while 3 men and 6 boys can finish it in 3 days. Find the time taken by one man alone to finish the work and that taken by one boy alone to finish the work.
1st Method
Let the time taken by one man alone to finish the work and that taken by one boy alone to finish the work be u and v days respectively.
Time taken by 1 man to finish one part of the work = 1/u days
Time taken by 1 boy to finish one part of the same work = 1/v days
According to question -
2 men and 5 boys can finish a piece of work in 4 days. Therefore, to finish one part of work they will take 1/4 days
Similarly, 3 men and 6 boys can finish the same work in 3 days. Therefore, to finish one part of work they will take 1/3 days
.....(1)
and,
.....(2)
Subtracting [2 × equation (2)] from [3 × equation (1)], we get -
⇒ v = 36
Substituting the value of v in equation (1), we get -
u = 18
Thus, time taken by one man to finish the work alone = 18 days
and, time taken by one boy to finish the work alone = 36 days
2nd Method
According to question -
2 men and 5 boys can finish a piece of work in 4 days. But to Finish this work in 1 day, we need 8 men and 20 boys. [apply Unitary Method]
Similarly,
3 men and 6 boys can finish the same work in 3 days. But to Finish this work in 1 day, we need 9 men and 18 boys. [apply Unitary Method]
Now,
8 men + 20 boys = 9 men + 18 boys
because their work rate is same as they both can finish the same work in 1 day.
∴ 1 man = 2 boys i.e. as much as work 2 boys can do , same amount of work 1 man can do it alone.
∴ 2 men + 5 boys = 9 boys
9 boys can finish a work in 4 days. Therefore, 1 boy will take 36 days to finish the work alone.
also,
∴ 3 men + 6 boys = 6 men
6 men can finish a work in 3 days. Therefore, 1 man will take 18 days to finish the work alone.