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A, B and C can do a piece of work in 15, 10 and 30 days, respectively. In how many days can A do the work if he is assisted by B and C on every 3rd day?
12 days
9 days
6 days
15 days
9 days
Calculate LCM of 15, 10, and 30 to get total work units as 30. Calculate individual efficiencies: A=2, B=3, C=1. In a 3-day cycle, total work = (A+A+A+B+C) = 2+2+(2+3+1) = 10 units.
Time taken by A = 15 days, B = 10 days, C = 30 days. Work schedule: A works every day, assisted by B and C on every 3rd day.
Work=Efficiency├ЧTime
Calculate LCM of 15, 10, and 30 to get total work units as 30. Calculate individual efficiencies: A=2, B=3, C=1. In a 3-day cycle, total work = (A+A+A+B+C) = 2+2+(2+3+1) = 10 units.
Students often add the efficiencies of A, B, and C for all three days instead of realizing B and C only assist on the 3rd day.
Calculate efficiencies
Assuming total work is the LCM of 15, 10, 30, which is 30 units. The efficiencies are: Efficiency of A=30/15=2 units/day, B=30/10=3 units/day, C=30/30=1 unit/day.
EfficiencyAтАЛ=2,EfficiencyBтАЛ=3,EfficiencyCтАЛ=1
Determine work in a 3-day cycle
A works alone on day 1 and day 2. On day 3, A is assisted by B and C. Work on day 1 = 2, day 2 = 2, day 3 = 2+3+1=6. Total work for 3 days = 2+2+6=10 units.
Work3daysтАЛ=10┬аunits
Calculate total time
Since 10 units of work are done every 3 days, to complete 30 units, it will take 30/10=3 cycles of 3 days each. Total days = 3├Ч3=9 days.
Total┬аDays=1030тАЛ├Ч3=9
B is correct because the work completed in each 3-day cycle is 10 units, meaning 3 cycles (9 days) are required to finish the total work of 30 units.
This type of alternating work problem is common in competitive exams; always reduce it to a repeating cycle based on the frequency of the assistance or interruption.
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