Math, asked by MukeshRabari, 1 year ago

A and B can complete a certain job in 20 and 30 days respectively. They start the job together and work for 3 days .Then B Falls sick and unable to do work for 5 days during which A works alone.Then again B joins him and they complete the work together. In how many days was the work completed?

Answers

Answered by itsmeaysha
6

Answer: we can use LCM METHOD....

As A can complete the work in 30 day. Surely... B could complete the work in 20 days.

Step-by-step explanation:

Here, we are going to calculate LCM of 30 and 20.

The total unit of work= LCM(20,30) = 60 units

So, the number of unit work done by A = 60/30=2 units

And, the number of unit work done by B = 60/20=3 units

As per the given information, B leaves the work 5 days before the work is finished.

Therefore, A would be working alone for the last 5 days and the amount of work done by A during those days= 2*5= 10 units

 Therefore, Remaining amount of work = 60-10 = 50 units

These 50 units of work is done by both A and B together.

Therefore, working together work done per day by A and B = 2+3 = 5 units

Therefore, to complete 50 units of work A and B together requires = 50/5 = 10 days

So the total number of days to finish the work = 10+5= 15 days.


Therefore ur answer is 15 dayss.....

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MukeshRabari: no this is wrong ans is 6
Answered by renupatel898
15

6 ddays is correct answet

Step-by-step explanation:

A's 10 days work = 10×3 units/ days = 30 units

So,(60−30)=30 units of work would have been done by A

and B both and it took time:

30(3+2)=6days

So, B worked for 6 days

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