Let the total work to be done = L.C.M. (2,10) = 10 units Let efficiencies of A, B and C be and units/day respectively. A, B and C together can finish of a work (i.e. 5 units) in 2 days, ⇒ (a+b+c)=
5
2
=2.5 units/day ------(i) Similarly, (b+c)=
3
2
=1.5 units/day ------(ii) Subtracting equation (ii) from (i), we get : ⇒ a=2.5−1.5=1 unit/day ∴Time taken by A alone to complete the work =