Math, asked by saryuramferro, 1 month ago

A can do a piece of work in 30 days , B in 50 days and C in 40 days .How soon can the work be done if A is assisted by B on one day and by C on the next the day alternately?

Ans: 17 whole 32/35 days.

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Answers

Answered by manishadhiman31
0

Answer:

First day A+B works=2+3=5A+B works=2+3=5 units ,

2nd day A+C works=2+6=8A+C works=2+6=8 units

3rd dayA+B works=5A+B works=5 units

In one cycle work is completed = 13 units

It will continue at last, work will finish in

6013(8+5)6013(8+5)= 4 cycle and 8 units of work is left.

Now, A will do with 'B' and they will complete 5 units.

Remaining 3 units will be done by A and C in

Total time = 4 cycle×2=84 cycle×2=8 days

(A+B)=1(A+B)=1 days

A+C=38A+C=38days

Hence, total time

= 8days+(A+B)′s days+(A+C)′s days8days+(A+B)′s days+(A+C)′s days

=8+1+38=9388+1+38=938 days

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