A can do a piece of work in 16 days and B in 20 days. They worked together for 5 days and then B leaves. In how many days will A finish the remaining work?
Answers
Answered by
1
It's 12 days
Explanation:
(A + B)'s 1 day's work = ( 1 + 1 ) = 1 .
15 10 6
Work done by A and B in 2 days = ( 1 x 2 ) = 1 .
6 3
Remaining work = ( 1 - 1 ) = 2 .
3 3
Now, 1 work is done by A in 1 day.
15
Therefore 2 work will be done by a in ( 15 x 2 ) = 10 days.
3 3
Hence, the total time taken = (10 + 2) = 12 days.
Answered by
6
A can do a piece of work in = 16 days
A's 1 day work = 1/16
B can do a piece of work in = 20 days
B's 1 day work = 1/20
(A+B)'s 1 day work = (1/16+1/20) = 9/80
So, (A+B)'s 5 day work = 9/16
Remaining work = (1-9/16) = 7/16
A's 1 day work = 1/16
7/16 work is done by B in = (7/16×16) = 7 days
A's 1 day work = 1/16
B can do a piece of work in = 20 days
B's 1 day work = 1/20
(A+B)'s 1 day work = (1/16+1/20) = 9/80
So, (A+B)'s 5 day work = 9/16
Remaining work = (1-9/16) = 7/16
A's 1 day work = 1/16
7/16 work is done by B in = (7/16×16) = 7 days
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