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A and B can complete a work in 12 days, B and C in 15 days, and C and A in 20 days. In how many days can A, B, and C together complete the work?
8
10
12
15
10
Calculate the total LCM of 12, 15, and 20 (which is 60). The sum of efficiencies is (60/12)+(60/15)+(60/20)=5+4+3=12. Since this represents 2├Ч(A+B+C), divide 2├Ч60 by 12 to get 10.
Time taken by (A+B) = 12 days, (B+C) = 15 days, (C+A) = 20 days.
TtotalтАЛ=xLCMтАЛ+yLCMтАЛ+zLCMтАЛ2├ЧLCM(x,y,z)тАЛ
Calculate the total LCM of 12, 15, and 20 (which is 60). The sum of efficiencies is (60/12)+(60/15)+(60/20)=5+4+3=12. Since this represents 2├Ч(A+B+C), divide 2├Ч60 by 12 to get 10.
Students often forget that the sum of efficiencies represents two times the work of (A+B+C) and fail to divide the final sum by 2.
Calculate Total Work
Assuming total work is the LCM of 12, 15, and 20.
LCM(12,15,20)=60
Determine Individual Pair Efficiencies
Efficiencies for pairs are: EA+BтАЛ=60/12=5, EB+CтАЛ=60/15=4, EC+AтАЛ=60/20=3.
EtotalтАЛ=(A+B)+(B+C)+(C+A)=5+4+3=12
Calculate Combined Efficiency
Since 2(A+B+C)=12, then (A+B+C)=12/2=6.
EA+B+CтАЛ=212тАЛ=6
Final Calculation
Days taken by A, B, and C together is Total Work divided by their combined efficiency.
Time=660тАЛ=10
B is correct because the combined efficiency of A, B, and C is 6 units per day, resulting in 10 days to complete 60 units of work.
This method is directly applicable to Pipe and Cistern problems where multiple inlets/outlets work in pairs at different rates.