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Rahul can do a piece of work in 20 days whereas Rohan can do the same work in 30 days. Both started the work together and worked for 6 days with full efficiency, and in the next 6 days with half of their efficiencies. In how many days will the remaining work be finished if both work with their full efficiencies for the remaining period?
7 days
5 days
10 days
3 days
3 days
Use the LCM of 20 and 30 (60) as total work. Efficiency of Rahul = 3, Rohan = 2. Total efficiency = 5. In the first 6 days (5 ├Ч 6 = 30) and next 6 days (2.5 ├Ч 6 = 15), total work done is 45. Remaining work = 15, which takes 15 / 5 = 3 days.
Rahul takes 20 days and Rohan takes 30 days to finish a work. They work for 6 days at full efficiency, then 6 days at half efficiency.
W=E├ЧT
Use the LCM of 20 and 30 (60) as total work. Efficiency of Rahul = 3, Rohan = 2. Total efficiency = 5. In the first 6 days (5 ├Ч 6 = 30) and next 6 days (2.5 ├Ч 6 = 15), total work done is 45. Remaining work = 15, which takes 15 / 5 = 3 days.
Students often forget that the efficiency is halved for BOTH workers during the middle 6-day period, not just one.
Calculate Efficiencies
Let the total work be the LCM of 20 and 30, which is 60 units. Rahul's efficiency is 60/20=3 units/day, and Rohan's is 60/30=2 units/day. Combined full efficiency is 3+2=5 units/day.
EtotalтАЛ=5┬аunits/day
Calculate Work Done
Work done in the first 6 days at full efficiency is 6├Ч5=30 units. Work done in the next 6 days at half efficiency (5/2=2.5) is 6├Ч2.5=15 units. Total work done is 30+15=45 units.
WdoneтАЛ=45┬аunits
Calculate Remaining Work
The remaining work is 60тИТ45=15 units. They return to full efficiency to complete this, so the time taken is 15/5=3 days.
T=515тАЛ=3┬аdays
D is correct because the total work of 60 units minus the 45 units completed in the first 12 days leaves 15 units, which both take 3 days to finish at full efficiency.
This logic is essential for pipe and cistern problems, where the rate of flow (efficiency) is adjusted periodically by closing or opening taps.
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