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A can do a piece of work in 5 days. B can do it in 20 days. With the assistance of C, they completed the work in 2 days. C alone can do it in ______________days.
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Assume total work as LCM of 5, 20, and 2, which is 20. Then, C's efficiency is simply the difference between the combined efficiency (20/2=10) and individual efficiencies (A: 20/5=4, B: 20/20=1).
Time taken by A = 5 days, Time taken by B = 20 days, Time taken by (A+B+C) = 2 days.
Work=Efficiency├ЧTime, EfficiencyCтАЛ=EfficiencyA+B+CтАЛтИТ(EfficiencyAтАЛ+EfficiencyBтАЛ)
Assume total work as LCM of 5, 20, and 2, which is 20. Then, C's efficiency is simply the difference between the combined efficiency (20/2=10) and individual efficiencies (A: 20/5=4, B: 20/20=1).
Students often add the efficiencies of A, B, and C as 1/5 + 1/20 + 1/C = 1/2 but fail to isolate 1/C correctly, or subtract the combined efficiency from individual ones accidentally.
Calculate individual efficiencies
Assuming total work is the LCM(5, 20, 2) = 20 units. Efficiency of A = 20/5 = 4 units/day. Efficiency of B = 20/20 = 1 unit/day.
EffAтАЛ=4,EffBтАЛ=1
Calculate combined efficiency
Combined efficiency of A, B, and C working together for 2 days is 20/2 = 10 units/day.
EffA+B+CтАЛ=10
Determine efficiency of C
Subtract the efficiencies of A and B from the combined efficiency: Efficiency of C = 10 - (4 + 1) = 5 units/day.
EffCтАЛ=10тИТ5=5
Calculate time for C
Time taken by C to complete 20 units of work is 20/5 = 4 days.
TimeCтАЛ=520тАЛ=4
B is correct because C requires 4 days to finish the work alone given an efficiency of 5 units/day.
This method is directly transferable to Pipe and Cistern problems where 'work' becomes 'volume of tank' and 'efficiency' becomes 'rate of flow'.
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