C1, C2 and C3 can do a work alone in 10, 12 and 15 days respectively. All three of them began the work together but C2 left 2 days before the completion of the work. In how many days was the work completed?
Answer & explanation
Correct answer: option 3
Total work = LCM of 10, 12, 15
total efficiency of C1, C2, C3 = 6 + 5 + 4 = 15
Work = Time x Efficiency
Let the time taken for the total work was 'x',
then C1, C3 worked for 'x' days and C2 worked for 'x-2' days,
Using the formula,
6x + 5(x-2) + 4x = 60
⇒ 6x + 5x + 4x - 10 = 60
⇒ 14x = 70
⇒ x = 5
Thus, it takes 5 days to complete the work.