12 Days of Christmas GMAT Competition 2025 - Day 2: Alex, Beth

Alex, Beth, Charles, and Dana, each working alone at their own constant rates, can complete a certain task in 8, 20, 40, and 80 hours, respectively. The project manager will schedule them to work on the task in a repeating four-person cycle in which each person works for exactly one hour before the next person begins.

Select for Minimum the minimum number of hours required for the task to be completed, and select for Maximum the maximum number of hours required for the task to be completed. Make only two selections, one in each column.

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16 Dec 2025, 11:27

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Assuming total work to be 800,
rate of A = 100/hr, B = 40/hr, C = 20/hr, D = 10/hr
4 person cycle => each cycle 100+40+20+10 = 170 work done in 4 hours
in 4 cycles 680 work done in 16 hours, remaining 120 work
for max time: B + C + D + A/2 = 40+20+10+100/2 = 120 work -> + 3.5 -> total 19.5 hrs
for min time: A + B/2 = 100 + 40/2 = 120 work -> +1.5 -> total 17.5 hrs

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16 Dec 2025, 11:28

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Bunuel

A=8; B=20; C=40; D=80
T.W. = 80 (let)
A=10; B= 4; C=2; D=1

Minimum Time Order ( A,B,C,D)
1st Cycle - 17 Units - 4 Hours
Now we multiply it by 4 to reach a multiple of 17, which is closest to 80, i.e., 68
4 Cycles - 68 Units - 16 Hours
A - 10 Units - 1 Hour
B - 2 Units - 0.5 hours
Total + 16+1+0.5 = 17.5 Hours.

Maximum Time Order (D,C,B,A)
1st Cycle - 17 Units - 4 Hours
Now we multiply it by 4 to reach a multiple of 17, which is closest to 80, i.e., 68
4 Cycles - 68 Units - 16 Hours
D - 1 Unit - 1 Hour
C - 2 Unit - 1 Hour
B - 4 Unit - 1 Hour
A - 5 Units - 0.5 Hours
Total = 19.5 Hours.

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16 Dec 2025, 11:37

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Let suppose each does 80 units of total work (took LCM of all rates), so
Alex - 8 hrs = 10 units/hr
Beth - 20 hrs = 4 units/hr
Charles - 40 hrs = 2 units/hr
Dana - 80 hrs = 1 unit/hr

In 4 hrs = 17 units work done if one person is only working for that hr as assigned by manages
So in 16 hrs = 68 units of work is done and remaining is 12 units, since total is 80 units considered

Now for min time, folks with higher work rates are considered so for 12 units another 1.5 hrs is needed (10 units by alex & 2 units by Beth)
For max time, folks with slower work rates should be consiered, so (1+2+4=7 by Dana,Charles & Beth & 5 units by Alex) adding to 3.5 hrs

So min = 17.5 and max = 19.5

Bunuel

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16 Dec 2025, 11:42

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Bunuel

Let's assume the total units of work that needs to be done = 80

A can do 10 units in an hour
B can do 4 units in an hour
C can do 2 units in an hour
D can do 1 unit in an hour

Total in 4 hours they can do 17 units of work

In 4 * 4 hours they can 17 * 4 units of works = 68 units of work

Remaining = 12

Fastest time :
A does 10 units then B does 2 units

16 + 1 + 0.5 = 17.5

Slowest time

D does 1 unit, C does 2 unit, B does 4 unit and A does 5 unit

16 + 1+ 1 + 1 +0.5 = 19.50

Minimum = 17.50
Maximum = 19.50

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16 Dec 2025, 11:46

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Bunuel

Rates:
$A=\frac{1}{8}=\frac{10}{80}$

$B=\frac{1}{20}=\frac{4}{80}$

$C=\frac{1}{40}=\frac{2}{80}$

$D=\frac{1}{80}$

For the fastest rate, we want A to go as often as possible, so we start the cycle from her. $\frac{17}{80}$ of the work gets done every cycle, and each cycle takes 4 hours. The closest multiple of 17 to 80, without exceeding 80, is $17*4 = 68$, at which point we've just finished D doing her work, and A is going again. A goes, it's $\frac{78}{80}$, then B goes, it's $\frac{82}{80}$, thus complete! That took a total of:

$(4 cycles * 4 hours each)+2=18$

Minimum time is 18 hours. There's only one option greater than this on the table for maximum hours, so instead of calculating it out, we'll select 19.5.

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16 Dec 2025, 11:57

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Taking LCM of 8, 20, 40 & 80 = 80 which is total work.
Alex work done in 1 hour = 10
Beth work done in 1 hour = 4
Charles work done in 1 hour = 2
Dana work done in 1 hour = 1

For minimum time, person with highest work done will do the work first, then second highest and so on..
So, 10,4,2 & 1 in this pattern all will work till the total work is completed i.e. 80. Total time taken will be 17.5 hours.

Similarly, for maximum time, pattern followed will be slowest first, then second slowest and so on..
It will be 1,2,4 & 10 in this pattern. Total time taken will be 19.5 hours.

Minimum = 17.5 hours
Maximum = 19.5 hours

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16 Dec 2025, 11:59

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Use the RTW table as all have independent rates.

R T W
A 10. 8. 80
B 4 20 80
C 2 40 80
D 1 80 80

here the rate is in units/ hr.
time is in hours and work is in units. Have taken 80 as it is the LCM of 8,20,40, 80 to ease out calculation of rate.

Total rate in a 4 hour cycle= 17 units per 4 hours.

Hence in 4 such cycles. 17 x 4: 68 units in 16 hours.

To find min. Start with lowest rate. 68+1+2+4+10/2=80 units of work
time is 16 + 1+1+1+1/2=19.5

to find max. Start from highest rate. 68+10+4/2=80 80 units of work
time is 16+1+1/2=17.5.

Hence Max time is 19.5hrs and min time is 17.5 hrs.

Bunuel

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16 Dec 2025, 12:00

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To solve such problems, consider total work as LCM of hours each person.
Here it will be LCM of (8,20,40,80) will be 80.
So, consider total work = 80 units and solve as explained in image.

Attachments

SolutionPS3.jpg [ 124.96 KiB | Viewed 227 times ]

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16 Dec 2025, 12:02

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Quote:

Alex, Beth, Charles, and Dana, each working alone at their own constant rates, can complete a certain task in 8, 20, 40, and 80 hours, respectively

Their work rate per hours equals to 1/8, 1/20, 1/40 & 1/80 of the task per hour, respectively.
Each cycle of each person working on this task per hour therefore would complete 1/8+1/20+1/40+1/80 = 17/80 of the task.
4 cycles will see 68/80 of the work being done. 5 cycles equals to 85/80 of the work, which means the work will be completed within 4 to 5 cycles (16 to 20 hours). This leaves us with 12/80 of the work to be done.
The fastest way to complete the task is to let the individuals with the highest work rate to do it first (Alex -> Beth ->...). After 1 hours of Alex doing the work, there will be 12/80-1/8=1/40 of the task left for Beth to do, which takes her (1/40) / (1/20) = 0.5 hours to complete => The minimum amount of time is 16 + 1 +0.5 = 17.5 hours.
Conversely, to find the maximum amount of time, let those with the lowest work rate do it first. Applying the reversed procedure:
After 1 hours of Dana, there'll be 12/80 - 1/80 = 11/80 of the task left
After 1 hours of Charles, there'll be 11/80 - 1/40 = 9/80 of the task left
After 1 hours of Beth, there'll be 9/80 - 1/20 = 5/80 of the task left
Alex to complete it in (5/80) / (1/8) = 0.5 hours.
The Maximum amount of time is 16+ 1 + 1 + 1 + 0.5 = 19.5 hours

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16 Dec 2025, 12:06

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To find the amount of work to be done by all find the LCM and add 1/8+1/20+1/40+1/80= 17/80 of work per hour meaning in 4 hours they will have covered 68/80 remaining with 12/80 =0.15
To minimize total work we can redistribute the work starting with Alex who takes the shortest towards Dana until the work ends
After 4x4 hours= 16 hours next we go to Alex (1 hour) so 0.15-1/8 = 0.025. Beth rate is is 0.05 so she will finish the remaining in 0.5 hours
Total minimal time = 16+1+0.5= 17.5 hours
To maximize we do the opposite and start with the slowest person who is Dana and work towards Alex
After 4x4 hours next we go to Dana (1 hour) 0.15-1/80 = 0.1375 then Charles (1 hour) 0.1375-1/40= 0.1125, Beth (1 hour ) 0.1125-1/20= 0.0625, Alex (0.0625/0.125) = O.5 hr
Max time= 16+1+1+1+0.5= 19.5 hours

Bunuel

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16 Dec 2025, 12:14

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Bunuel

calculate task done by person in a hour
A = 1/8 = 0.125
B = 0.05
C = 0.025
D = 0.0125

task done in 4 hours of their work = A + B + C + D = 0.2125
since the task are done in round robin fashion
min time would be when quickest person does the job first
so order will be
A B C D x 4times = 0.85 work done in 16 hours
in next hour A does 0.125 ( total 0.975 in 17 hours)
remaining is 0.025 which B can do in 30 mins

so all work is done in 17.5 hours

max time will be take in opposite order of above
D C B A
same wil continue 4 times
D C B A x 4 times = 0.85 work done in 16 hours
in next 1 hour D does 0.0125 , next 1 hour C does 0.025, next 1 hour B does 0.05
remaining is 0.0625 work, which A can do in 1/2 hour
so total is 16 + 1 + 1 + 1 + 0.5 = 19.5 hours

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16 Dec 2025, 12:17

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First, let's denote the respective speeds of workers bringing them to the common denominator of 80:
$a = 1/8=10/80$
$b=1/20=4/80$
$c=1/40=2/80$
$d=1/80$
The units of speeds are share of the project per hour.

Now, in one cycle of 4 people we're getting $\frac{10+4+2+1}{80}=17/80$ of the project done, and this cycle lasts 4 hours.
Once we finish 4 such cycles, we will come as close as possible to the whole project, completing $17*4/80=68/80$ of the work, and it will take 16 hours.

How there's the slowest and quickest scenario for the remaining $12/80$:

(1) To cover 12 parts slowly, we need to go in the order of $d -> c -> b -> 0.5a$, which will take 3.5 hours (and this will be added to the required 16).

(2) To cover 12 parts quickly, we need to start from the fastest: $a -> 0.5b,$ which will take 1.5 hours.

Therefore, the minimum necessary time is 17.5 hours, and the maximum is 19.5 hours.

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16 Dec 2025, 12:32

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LCM (8,20,40,80)=80
So in one set work done = 17
17*4=68 units ..so 4*6=16 hrs.

remaining 12 units.
minimum time if A works full 1 hr & B works half = 1+0.5 = 1.5 hrs
max time if D, C, B works full 1 hr each & A half = 1+1+1+0.5=3.5

Min time = 17.5
Max time = 19.5

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16 Dec 2025, 12:41

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Bunuel