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Three pipes A, B, and C can fill a tank in 10, 15, and 30 hours respectively. How long will they take if all are open?
5┬аhours
6┬аhours
8┬аhours
10┬аhours
5┬аhours
Use the LCM method: find the LCM of 10, 15, and 30 as total units, then divide by the sum of individual efficiencies.
Pipe A = 10 hours, Pipe B = 15 hours, Pipe C = 30 hours
Time=Combined┬аRateTotal┬аWorkтАЛ
Use the LCM method: find the LCM of 10, 15, and 30 as total units, then divide by the sum of individual efficiencies.
Failing to add the efficiencies of all three pipes together and mistakenly treating them as separate or subtracting them.
Find Total Work via LCM
Assume the total capacity of the tank is the LCM of 10, 15, and 30, which is 30 units.
LCM(10,15,30)=30
Determine Individual Efficiencies
Calculate the work done per hour by each pipe: Pipe A = 30/10=3 units/hr, Pipe B = 30/15=2 units/hr, Pipe C = 30/30=1 unit/hr.
EAтАЛ=3,EBтАЛ=2,ECтАЛ=1
Calculate Combined Efficiency
Add the individual rates of pipes A, B, and C working together: 3+2+1=6 units/hr.
Total┬аEfficiency=3+2+1=6
Calculate Total Time
Divide the total units of work by the combined efficiency to get the total time required: 30/6=5 hours.
Time=630тАЛ=5┬аhours
A is correct because the combined work rate of all three pipes yields a total time of 5 hours.
This concept is identical to problems involving multiple workers completing a piece of work together.
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