A can complete 25% of the work in 4 days, B can complete 50% of the same work in 12 days and C can complete the same work in 32 days. They started working together but C left after 4 days of the start and A left 6 days before the completion of the work. In how many days was the work completed?

Find the total number of days required by each individual to complete the entire work:
A completes 25% (
4
1
​
) of the work in 4 days → Total time for A =4×4=16 days.

B completes 50% (
2
1
​
) of the work in 12 days → Total time for B =12×2=24 days.

C completes the full work in 32 days → Total time for C =32 days.

Let the total work be the LCM of 16,24,32, which is 96 units.

Efficiencies (work done per day):

Efficiency of A =
16
96
​
=6 units/day

Efficiency of B =
24
96
​
=4 units/day

Efficiency of C =
32
96
​
=3 units/day

Let the total work be completed in x days.

C worked for 4 days.

A left 6 days before completion, so A worked for (x−6) days.

B worked for the entire duration of x days.

Formulating the work equation:

Work done by A+Work done by B+Work done by C=Total Work
6(x−6)+4x+3(4)=96
6x−36+4x+12=96
10x−24=96
10x=120
x=12 days