'A' takes 7 times as much time as 'B' takes or 3 times

Solution

Let the time taken by 'B' be '3x' days Therefore, time taken by 'A' = 7 X 3x = '21x' days Time taken by 'C' = (21x/3) = '7x' days Let total work = 21x units {LCM of '3x, '21x' and '7x'} Efficiency of 'A' = (21x/21x) = 1 unit/day Efficiency of 'B' = (21x/3x) = 7 units/day Efficiency of 'C' = (21x/7x) = 3 units/day Combined efficiency of 'A', 'B' and 'C' = (21x/21) = 'x' units/day Or, x = 1 + 7 + 3 = 11 Therefore, time taken by 'B' = 3x = 33 days