A alone can do a piece of work in 8 days and B alone in 6 days. A start the work, works for 2 days and leaves the job. In how many days will B complete the remaining work?
Answers
A alone can do a piece of work in 8 days.
So, work done by A in one day = 1/8
Similarly, B alone do a piece of work in 6 days.
So, work done by B in one day = 1/6
Now,
Work done by both A and B is 1/8 + 1/6
=> 1/8 + 1/6
=> (3 + 4)/24
=> 7/24
A starts the work for 2 days and then leave.
We know that..
Work done by A alone is 8 days and in one day = 1/8
So, work done by A in 2 days = 1/8 × 2
=> 2/8
=> 1/4 part of the work
Remaining work = 1 - 1/4
=> (4 - 1)/4
=> 3/4
We have to find the number of days taken by B to complete the remaining work.
B alone do a piece of work in 6 days and remaining work is 3/4
So,
Work done by B to complete the remaining work = 3/4 × 6
=> 9/2 days
∴ In 9/2 days B complete the remaining work.
Given :-
A can do work in 8 days
B can do work in 6 days
To Find :-
How many days B complete the remaining work
Solution :-
A can do work in 8 days
Therefore,
Work done by A in one day will be 1/8.
Work done by B in one day will be 1/6.
Total work done by A & B will be
1/8 + 1/6
-> 7/24
According to the question A start the work for 2 days and left the job
Therefore,
Work done by A in 2 days will be
= 1/8 × 2
= 2/8
= 1/4
The work left to do = 1 - 1/4
= 3/4
Therefore, B can do work in 6 day and the work is remained to do is 3/4
3/4 × 6
= 9/2
Therefore, B takes 9/2 days to complete the remaining work.