A can complete a piece of work in 36 days, B in 54 days and C in 72 days. All the three began the work the work together but A left 8 days before the completion of the work and B 12 days before the completion of work. Only C worked up to the end. In how many days was the work completed?

36,54,72. 36/6=6
54/9=6
72/12=6
6+6+6=18
72*6=432/18=24
Ans:-24 days

Please find 3 methods to do this Q :
Method 1:

                R             S              T
Time        36           54            72
LCM = total work = 216 units
Eff            6      :     4          :     3
Let work completes in x days, R left 8 days before so he worked for (x-8) days , S left 12 days before  so he worked for (x-12) days & T did not leave in between so he worked for all x days.
So ATQ,  R*(x-8) +S*(x-12) + T*x = 216
6*(x-8) +4*(x-12) + 3*x = 216
6x - 48+4x-48+3x = 216
13x - 96 = 216
13x = 216+96 = 312
x = 24
Hence T worked for 24 days.

Method 2:
                R             S              T
Time        36           54            72
LCM = total work = 216 units
Eff            6      :     4          :     3
Work-done in last 12 days = T*12+R*(12-8) =3*12 + 6*4 = 36+24 = 60 units
Hence remaining work is done by all initially in = (216 - 60)/(6+4+3) = 156/13= 12 days
So total days = 12 + 12 = 24 days  
Hence T worked for 24 days.


Method 3:
                R             S              T
Time        36           54            72
Let work completes in x days, R left 8 days before so he worked for (x-8) days , S left 12 days before  so he worked for (x-12) days & T did not leave in between so he worked for all x days.
Hence , 1/36 * (x-8)  + 1/54 * (x-12) + 1/72 * (x) = 1
(x-8)/36  +  (x-12)/54 +  x/72 = 1
{6(x-8)  +  4(x-12) +  3x}/216 = 1
6x - 48 + 4x - 48 + 3x = 216
13x  - 96 = 216
13x = 216 + 96 = 312x = 24
Hence T worked for 24 days.

Siddu hugar we assume a complete of work os equal to 1

In above equation why the RHS is equal to one.

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