A can do a piece of work in 14 days while B can do i 21 days .they begin together ,but 3 days before the completion of tbe work, A leaves off. find the total number of days taken to complete the work
Answers
Solution :
A can do a piece of work in 14 days
So, Work done by A in 1 day = 1/14
B can do it in 21 days
So, work done by B in 1 day = 1/21
Work done by A and B in 1 day = Work done by A in 1 day + Work done by B in 1 day
= 1/14 + 1/21
= (3 + 2)/42
= 5/42
i.e Work done by A and B in 1 day = 5/42
A leaves of 3 days before the completion of work
A leaves 3 days before the completion of work
So, Work done by B alone in last 3 days = 3 * (1/21) = 1/7
Remaing work by A and B = 1 - 1/7 = (7 - 1)/7 = 6/7
Work done by A and B in = (6/7) / (5/42) = 6/7 * 42/5 = 36/5 days
Total no. of days = 36/5 + 3 = (36 + 15)/5 = 10.2 days
Hence, total number of days taken to complete the work is 10.2 days.
Given :----
- A can do a piece of work in = 14 days .
- B can do it in = 21 days .
- A left before 3 days of completion of the work .
To Find :----
- Total number of days to complete the work ?
Solution :------
LCM of 14 and 21 = 42 units = Let Total work ,
Efficiency of A = 42/14 = 3 unit / day
Efficiency of B = 42/21 = 2 unit/day .
now it has been said that, A left the work , before 3 days it was going to complete .
so, clearly in last three day B alone do the rest work.
in 3 days B did = 3×2 = 6 unit of work.
so, Rest work is done by both A + B now ,
Rest work = 42-6 = 36 units of work .
Time taken by (A+B) now , to do this work ,,
as efficiency of both = 3+2 = 5 unit /day .
so, time they will take = 36/5 = 7(1/5) days .
so, Total number of days to complete whole work = 7(1/5) + 3 days (B alone ) = 10(1/5) days..
Hence, total work was completed in 10(1/5) days..
(Hope it Helps you)