Math, asked by snehamohan28, 1 year ago

A can do a work in 16 days and B can do it in 24 days. They started work together but A left 2 days before completion of the work. In how much time was the work completed? ​

Answers

Answered by Anonymous
23

Answer :-

Work will be completed in 12.6 days

Solution :-

A can do a work in 16 days

So A's one day work = 1/16

B can do the same work in 24 days

So B's one day work = 1/24

A and B one day work = A's one day work + B's one day work

= 1/16 + 1/24

= (3 + 2)/48

= 5/48

So A and B can do the work in 48/5 days

Number of days A and B work together = 48/5 - 2

Since A leaves 2 days before completion of work

= 48/5 - 2/1

= (48 - 10)/5

= 38/5 days

Work done by both in 38/5 days = A and B one day work * Number of days A and B work together

= 5/48(38/5)

= 1/48 * 38

= 38/48

Remaining work to be done by B = 1 - 38/48

= 1/1 - 38/48

= (48 - 38)/48

= 10/48

1 work B can do in = 24 days

10/48 work B can do in = 24 * 10/48

= 1 * 10/2

= 10/2

= 5 days

Total number of days required for completion of work = Number of days A and B work together + Number of days B required to do remaining work

= 38/5 + 5

= 38/5 + 5/1

= (38 + 25)/5

= 63/5

= 12.6 days

Therefore work will be completed in 12.6 days

Similar questions