N1 alone can do 20% of a piece of work in 6 days, N2 alone can do 20% of the same work in 4 days, and N3 alone can do 50% of the same work in 12 days. In how many days can they complete the work, if all of them work together?
Show answer & explanation
Correct answer: C
Explanation
The correct answer is option 3 i.e. 8 days.
Concept Used :
Total work = efficiency × time
Given that :
N1 can do 20% of a piece of work in 6 days
N2 can do 20% of the work in 4 days
N3 can do 50% of the work in 12 days
Explanation :
N1 can do the 100% work in 6 × 5 = 30 days
N2 can do 100% work in 4 × 5 = 20 days
N3 can do 100% work in 12 × 2 = 24 days
Now, Total work = LCM of (30, 20, 24) = 120 units
Efficiency of N1 = 120/30 = 4 units/day
Efficiency of N2 = 120/20 = 6 units/day
Efficiency of N3 = 120/24 = 5 units/day
The efficiency of (N1 + N2 + N3) together = 4 + 6 + 5 = 15 units/day
Time taken by (N1 + N2 + N3) together = (Total work)/Efficiency
= 120/15 = 8 days
Hence they can complete the work in 8 days.