Question

A can do a work in 8 days, b can do a work in 7 days, c can do a work in 6 days. a works on the first day, b works on the second day and c on the third day resly.that is they work on alternate days. when will they finish the work. which day will they finish the work.

Answer 1

A can do 1 work in 8 days
A in 1 day does: 1/8 work
B can do 1 work in 7 days
B in 1 day does: 1/7 work
C can do 1 work in 6 days
C in one day does: 1/6 work

Since, A,B,C work alternatively:
In 3 days they do:
1/8 + 1/7 + 1/6 = 21+24+28/168 = 73/168 work
3 days=== 73/168 work
Since 1 cylce(A,B,C) is completed in 3 days,find the multiple of 73 which is close to 168(coz the tot work = 168/168 = 1 work)
3days* 2 = 2*73/168 = 146/168
Hence in 6 days== 146/168 work is completed
Work remaining = 1 - 146/168 = 22/168
Next is A's turn:
1 day== 1/8 work or 21/168 work
Hence by 7th day 21/168 work of remaining 22/168 work is completed.
Work remaining = 1/168

Next is B's turn:
1 day==1/7 work
x day== 1/168
x/7 = 1/168
x = 1/24
So the total work is completed in 7 1/24 days
Hope it helps