a can completes a work in 3 days b can do the same work in 4 days and c in 5 days they work alternately first day a second day and third c and then a smilarly what day will the work end plzzzzz solve
Answers
Answer:
starts the work and does 1/16 of the work on the first day
B works on the 2nd day and does 1/12 of the work
Thus in two days both of them would complete 1/16 + 1/12 of the work i.e. 4+3/48 (48 being the L.C.M of 16 and 12). In two days they finish 7/48 of the work.
In 6 cycles (A on 1st day + B the next day) they would complete 6 x 7 = 42/48 of the work. Now only 6 units of work remains to be done.
The remaining work is 6/48. On the 13th day, A comes and does 1/16 of the whole work which is 1/16 of 48, i.e., 3 units of the work and only 3 units of work is still remaining. B comes the next day i.e. on the 14th day and 1/12 of 48 i.e. 1/12 *48 which is 4 units. But only 3 units of work is required to be completed on the 14th day and B takes 3/4 of the day to complete the 3 remaining units of work
Thus the total time taken by both of them is 13 and 3/4 days.