Question : $\mathrm{N}_1$ alone can do 20% of a piece of work in 6 days, $\mathrm{N}_2$ alone can do 20% of the same work in 4 days, and $\mathrm{N}_3$ 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?
Option 1: 10 days
Option 2: $\frac{120}{13}$ days
Option 3: 8 days
Option 4: $\frac{120}{17}$ days
Correct Answer: 8 days
Solution :
$N_1$ alone can do 20% of a piece of work in 6 days.
= 100% = 30 days
$N_2$ alone can do 20% of the same work in 4 days.
= 100% = 20 days
$N_3$ alone can do 50% of the same work in 12 days.
= 100% = 24 days
Total work = Time × Efficiency
Total work = LCM(30, 20, 24) = 120 units
Efficiency of $N_1$= $\frac{120}{30}$ = 4
Efficiency of $N_2$ = $\frac{120}{20}$ = 6
Efficiency of $N_3$ = $\frac{120}{24}$ = 5
Total of efficiency = 4 + 6 + 5 = 15
All of them working together finish the work = $\frac{120}{15}$ = 8 days
Hence, the correct answer is 8 days.
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