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17 Dec 2025, 06:40
Kudos
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Let the total work be 80 units (LCM of the time taken by each)
Efficiency = total work / total time taken
Efficiency of A = 80 / 8 = 10 units per hour
Similarly, B = 4 units / hr
C = 2 units / hr
D = 1 units / hr
Each works in a cycle one after another. To complete the task at the earliest, fastest worker should start first. Therefore, as per the efficiency work done in one cycle would be A,B,C,D which would be,
10 + 4 + 2 + 1 = 17 units work in 4 hours
68 units of work would be completed in 16 hours
In the 17th hour A would finish 10 units of work leaving 2 units of work which would then be finished by B in 1/2 hours after the 17th hour.
Hence minimum time taken would be 17.5 hours.
For maximum time cycle would start from slowest to the fastest worker, which would be D,C,B,A
4 cycles in 16 hours would similarly be 16 hours leaving 12 units of work.
After this D,C,B will complete 1+2+4, i.e. 7 units in 3 hours and remaining 5 units will be completed by A in 1/2 hours.
Hence, maximum time would be 16 + 3 + 1/2 = 19.5 hours.
Efficiency = total work / total time taken
Efficiency of A = 80 / 8 = 10 units per hour
Similarly, B = 4 units / hr
C = 2 units / hr
D = 1 units / hr
Each works in a cycle one after another. To complete the task at the earliest, fastest worker should start first. Therefore, as per the efficiency work done in one cycle would be A,B,C,D which would be,
10 + 4 + 2 + 1 = 17 units work in 4 hours
68 units of work would be completed in 16 hours
In the 17th hour A would finish 10 units of work leaving 2 units of work which would then be finished by B in 1/2 hours after the 17th hour.
Hence minimum time taken would be 17.5 hours.
For maximum time cycle would start from slowest to the fastest worker, which would be D,C,B,A
4 cycles in 16 hours would similarly be 16 hours leaving 12 units of work.
After this D,C,B will complete 1+2+4, i.e. 7 units in 3 hours and remaining 5 units will be completed by A in 1/2 hours.
Hence, maximum time would be 16 + 3 + 1/2 = 19.5 hours.
17 Dec 2025, 07:20
Kudos
Bookmarks
Work done by A in 1 hours = 1/8 = 10/80
B = 1/20 = 4/80
C = 1/40 = 2/80
D = 1/80
For simplicity sake, lets assume that 80 units of work to be accomplished (it can be easily taken as 1 unit, but for simplification sake , lets take it as 80. For 1 unit calculation, divide everything by 80)
then A does 10 units in 1 hours, B 4, C 2 ad D 1.
For minimum time, fastest has to work first, so cycle becomes: ABCDABCDA...each working for 1 hour
so work completed in one cycle(of 4 hours) is 10(A)+4(B)+2(C)+1(D)= 17
work completed in 4 such cycle of total 16 hours = 17*4 = 68
then comes A's turn for 1 hour, total work done = 68+10 = 78, total hrs = 17
then 2 units left and B's turn(of 4 unit in 1 hour) means only 0.5 hrs req. , total hrs = 17.5
for max, cycle should be DCBADCBA..
first 16 hours same as above,(4 cycles of DCBA)
for hour 17, D's turn , work done = 1, total work = 69
for hour 18, C's turn, work done = 2, total work = 71
for hour 19, B's turn, work done = 4, total = 75
for hour 20, A's turn , work done = 10, needed 5, so only 0.5 hour needed
therefore total time = 19.5
B = 1/20 = 4/80
C = 1/40 = 2/80
D = 1/80
For simplicity sake, lets assume that 80 units of work to be accomplished (it can be easily taken as 1 unit, but for simplification sake , lets take it as 80. For 1 unit calculation, divide everything by 80)
then A does 10 units in 1 hours, B 4, C 2 ad D 1.
For minimum time, fastest has to work first, so cycle becomes: ABCDABCDA...each working for 1 hour
so work completed in one cycle(of 4 hours) is 10(A)+4(B)+2(C)+1(D)= 17
work completed in 4 such cycle of total 16 hours = 17*4 = 68
then comes A's turn for 1 hour, total work done = 68+10 = 78, total hrs = 17
then 2 units left and B's turn(of 4 unit in 1 hour) means only 0.5 hrs req. , total hrs = 17.5
for max, cycle should be DCBADCBA..
first 16 hours same as above,(4 cycles of DCBA)
for hour 17, D's turn , work done = 1, total work = 69
for hour 18, C's turn, work done = 2, total work = 71
for hour 19, B's turn, work done = 4, total = 75
for hour 20, A's turn , work done = 10, needed 5, so only 0.5 hour needed
therefore total time = 19.5
17 Dec 2025, 07:44
Kudos
Bookmarks
Okay so according to the problem, the Work per hour => Alex = W/8, Beth = W/20, Charles = W/ 40 and Dana = W/80, where W denotes the total work required for completing the task.
Now work done for one cycle = W/8+W/20+W/40+W/80 = 17W/80 in 4 hours (1 hour each)
Number of full cycles which can be completed (integer multiple of work done per cycle which is just less than the total work, W) = 4 (17W/80 * 4 = 68/80W)
Time taken for 4 cycles = 4*4 =16 hours
Work left to do = 12W/80
Clearly, in any case, the common work done in cycles will be the 4 full cycles and the determination of the minimum and maximum time taken will only be based on who gets to work on the remaining work , 12W/80.
Minimum Case:
=> Remaining work is completed in the chronology of the fastest first (Alex, Beth, Charles, Dana)
Remaining work after Alex works for an hour => 12W/80 - 10W/80 = 2W/ 80 = W/40 - 1 hour added
Next, time taken for Beth to finish = W/40/W/20 = 1/2 = 0.5 hours added
Mininum time = 16+1+0.5 = 17.5 hours
Maximum Case:
=> Remaining work is completed in the chronology of slowest first (Dana, Charles, Beth, Alex)
Remaining work after Dana works for an hour = 12W/80 - W/80 = 11 W/80 - 1 hour added
Remaining work after Charles works hor an hour = 11W/80 - 2W/80 = 9W/80 - 1 hour added
Remaining work after Beth works for anhour = 9W/80 - 4W/80 = 5W/80 = W/16 - 1 hour added
Next, time taken for Alex to finish = W/16/W/8 = 1/2 = 0.5 hours added
Maximum time = 16+1+1+1+0.5 = 19.5 hours
Now work done for one cycle = W/8+W/20+W/40+W/80 = 17W/80 in 4 hours (1 hour each)
Number of full cycles which can be completed (integer multiple of work done per cycle which is just less than the total work, W) = 4 (17W/80 * 4 = 68/80W)
Time taken for 4 cycles = 4*4 =16 hours
Work left to do = 12W/80
Clearly, in any case, the common work done in cycles will be the 4 full cycles and the determination of the minimum and maximum time taken will only be based on who gets to work on the remaining work , 12W/80.
Minimum Case:
=> Remaining work is completed in the chronology of the fastest first (Alex, Beth, Charles, Dana)
Remaining work after Alex works for an hour => 12W/80 - 10W/80 = 2W/ 80 = W/40 - 1 hour added
Next, time taken for Beth to finish = W/40/W/20 = 1/2 = 0.5 hours added
Mininum time = 16+1+0.5 = 17.5 hours
Maximum Case:
=> Remaining work is completed in the chronology of slowest first (Dana, Charles, Beth, Alex)
Remaining work after Dana works for an hour = 12W/80 - W/80 = 11 W/80 - 1 hour added
Remaining work after Charles works hor an hour = 11W/80 - 2W/80 = 9W/80 - 1 hour added
Remaining work after Beth works for anhour = 9W/80 - 4W/80 = 5W/80 = W/16 - 1 hour added
Next, time taken for Alex to finish = W/16/W/8 = 1/2 = 0.5 hours added
Maximum time = 16+1+1+1+0.5 = 19.5 hours
