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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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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