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Two pipes C and D can fill a tank in 6 hours and 9 hours, respectively. They are opened alternately for 1 hour each, starting with C first. In what duration, will the tank be filled?
5 hours
6 hours
8 hours
7 hours
7 hours
Calculate the work done in a 2-hour cycle (C+D). 3+2=5 units per 2 hours. The total capacity is 18 units. Calculate cycles to reach close to 18 units.
Pipe C fills a tank in 6 hours, pipe D fills it in 9 hours. They are opened alternately for 1 hour each, starting with C.
Work=Efficiency×Time
Calculate the work done in a 2-hour cycle (C+D). 3+2=5 units per 2 hours. The total capacity is 18 units. Calculate cycles to reach close to 18 units.
Students often stop after finding the work done in 3 cycles (6 hours) and forget to account for the exact remaining work and the partial hour required for completion.
Calculate LCM and Efficiencies
Assuming total tank capacity is the LCM of 6 and 9, which is 18 units. Efficiency of C = 18/6=3 units/hr. Efficiency of D = 18/9=2 units/hr.
EfficiencyC=3,EfficiencyD=2
Calculate Work in 2-hour cycles
In every 2-hour cycle (C then D), the total work done is 3+2=5 units. After 3 full cycles (6 hours), the work done is 5×3=15 units.
Work6hr=15 units
Calculate remaining work and time
Remaining work = 18−15=3 units. In the 7th hour, C starts and works at an efficiency of 3 units/hr. Since 3 units remain, C will fill the remaining tank in exactly 1 hour.
Time=6+33=7 hours
D is correct because after 6 hours, 15 units are filled, and the remaining 3 units are completed by Pipe C in exactly 1 additional hour, totaling 7 hours.
This logic is identical to 'man-days' problems where one person rests or works alternately; always calculate the work per complete cycle first.
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