A, B and C can do one-third of a work in 15 days, 30 days and 10 days, respectively. A started the work. C joined him after 1 day and B joined them after 3 days of the beginning. For how many days did C work?

If they can do one-third of the work in the given days, then the total time taken by them to complete the full work individually is:
Time taken by A = 15×3=45 days

Time taken by B = 30×3=90 days

Time taken by C = 10×3=30 days

Let Total Work = LCM(45,90,30)=90 units.

Efficiency of A =
45
90
​
=2 units/day

Efficiency of B =
90
90
​
=1 unit/day

Efficiency of C =
30
90
​
=3 units/day

Timeline of work:

Day 1: Only A works.

Work done=2×1=2 units
Day 2 and Day 3 (2 days): C joins A, so A and C work together.

Work done=(Efficiency of A+Efficiency of C)×2=(2+3)×2=10 units
Total work done in first 3 days = 2+10=12 units.
Remaining work = 90−12=78 units.

From Day 4 onwards: B also joins, so A, B, and C work together.

Combined efficiency=2+1+3=6 units/day
Time taken to finish remaining work=
6
78
​
=13 days
Total days C worked = Days on (Day 2 + Day 3) + Remaining days = 2+13=15 days.