12 Days of Christmas GMAT Competition 2025 - Day 2: Alex, Beth : Two-part Analysis (TPA)

Deconstructing the Question
We have 4 workers with individual rates to complete a task.
Task completion times:
Alex (A): 8 hours
Beth (B): 20 hours
Charles (C): 40 hours
Dana (D): 80 hours

The work is done in a repeating 4-person cycle (1 hour each).
Goal: Find the Minimum and Maximum total hours to complete the task.

Step 1: Calculate Hourly Work Units
Let the Total Work be the LCM of (8, 20, 40, 80) = 80 units.
Efficiencies (units per hour):

$R_A = 80/8 = 10$ units/hr

$R_B = 80/20 = 4$ units/hr

$R_C = 80/40 = 2$ units/hr

$R_D = 80/80 = 1$ unit/hr

Step 2: Analyze Full Cycles

One full cycle consists of 4 hours (one hour per person).
Work done in 1 cycle = $10 + 4 + 2 + 1 = 17$ units.
Total work needed: 80 units. Number of full cycles = $80 / 17 \approx 4.7$.

So, we have 4 full cycles.

Time elapsed: $4 \times 4 = 16$ hours.
Work completed: $4 \times 17 = 68$ units.
Remaining Work: $80 - 68 = 12$ units.

Step 3: Calculate Minimum Time
To minimize time, the fastest workers must perform the remaining work first.
Sequence: A $\rightarrow$ B $\rightarrow$ C $\rightarrow$ D.
Remaining: 12 units.
Hour 17 (A works): Completes 10 units. (Remaining: $12 - 10 = 2$).
Hour 18 (B works): Rate is 4 units/hr. Needs to complete 2 units.

Time taken = $2 / 4 = 0.5$ hours.

Total Min Time = $16 + 1 + 0.5 = 17.5$ hours.

Step 4: Calculate Maximum Time
To maximize time, the slowest workers must perform the remaining work first.
Sequence: D $\rightarrow$ C $\rightarrow$ B $\rightarrow$ A.
Remaining: 12 units.
Hour 17 (D works): Completes 1 unit. (Remaining: 11).
Hour 18 (C works): Completes 2 units. (Remaining: 9).
Hour 19 (B works): Completes 4 units. (Remaining: 5).
Hour 20 (A works): Rate is 10 units/hr. Needs to complete 5 units.

Time taken = $5 / 10 = 0.5$ hours.

Total Max Time = $16 + 1 + 1 + 1 + 0.5 = 19.5$ hours.

Answer Selection:
Minimum: 17.5
Maximum: 19.5

laborumplaceat

Joined: 06 Dec 2025

Last visit: 11 May 2026

Posts: 124

Own Kudos:

[2]

Given Kudos: 3

Products:

Posts: 124

Kudos: 63

[2]

16 Dec 2025, 09:55

Kudos

Bookmarks

Let rate of:
A = 10
B = 4
C = 2
D = 1

Total work = 80
In 4 hours: 68 units of work done.

Additional time for maximum time:
D works -> 1 unit in 1 hour
then C works -> 2 units in 1 hour
then B works -> 4 units in 1 hour
then A works -> 5 unit in 0.5 hour - So total max time = 19.5 hours

Additional time for minimum time:
A works -> 10 units in 1 hour
then B works -> 2 units in 0.5 hour
So minimum total time = 17.5 hours

sitrem

Joined: 19 Nov 2025

Last visit: 24 Feb 2026

Posts: 91

Own Kudos:

[1]

Given Kudos: 238

Posts: 91

Kudos: 84

[1]

16 Dec 2025, 09:58

Kudos

Bookmarks

Rate of each worker (fraction of the total work done in 1h)
To make calculations easier assume that the work consists of 80 units (lcm of 8,20,40,80)
A 10 units each hour = 80 / 8 h
B 4
C 2
D 1

To finish in the minimum time, put the workers in order of speed (from fastest to slowest -> A, B, C, D).
In each cycle they make 10+4+2+1=17 units of work.
They will need 4 full cycles to get to 68, each cycle is 4 hours.
To get to 80, A works 1h and does 10 units and B works for half an hour to do the remaining 2.
68 units in 16 h (4 cycles) + 10 in 1 h + 2 in 0.5 h = 80 in 17.5 h

To finish in the maximum time use the opposite order (slowest to fastest).
They will still do 4 full cycles to get to 68 units in 16h.
Then D does 1 unit in 1h, C does 2 in 1h, B does 4 in 1h, A does the remaining 5 in half an hour.
68 units in 16h + 1 in 1h + 2 in 1h + 4 in 1h + 5 in 0.5h = 80 in 19.5h

Minimum 17.5
Maximum 19.5

JiriNovacek

Joined: 14 Dec 2025

Last visit: 07 Feb 2026

Posts: 17

Own Kudos:

[1]

Posts: 17

Kudos: 9

[1]

16 Dec 2025, 09:58

Kudos

Bookmarks

Minimum is 17,5 hours, the fastest worker begins (Alex, 8 hours), he does in the first cycle in the hour 1/8 of the work, then works Beth (1/20 of work during one hour), etc. In the sum, they can do 17/80 per cycle of the work. It means 4 cycles (16hours) and 0.15 of the work still remains. Alex makes pro hour as the firs worker of the 5.cycle 0.125 of the work (1/8) and Beth 0,025 of the remaining work do for 0.5 hour (so overall minimum 17.5 hours)

Maximum:19.5 hours
remaining rest of the work after 4 cycles (16 hours) is again 0.15 (15%). But this time the Dana begins (as the slowest worker). Then the same logic as above.

Kinshook

Major Poster

Joined: 03 Jun 2019

Last visit: 16 May 2026

Posts: 6,003

Own Kudos:

5,877

[1]

Given Kudos: 163

Location: India

Schools: HBS '26 CBS '26 Wharton '26

GMAT 1: 690 Q50 V34 Verified score

WE:Engineering (Transportation)

Products:

Schools: HBS '26 CBS '26 Wharton '26

GMAT 1: 690 Q50 V34 Verified score

Posts: 6,003

Kudos: 5,877

[1]

16 Dec 2025, 09:58

Kudos

Bookmarks

The fraction of the work done by Alex, Beth, Charles & Dana in an hour is 1/8, 1/20, 1/40 & 1/80 respectively.

For minimum number of hours, we assign the work to most productive workers first.

Hour 1: Alex : Work done = 1/8 = 12.5%
Hour 2: Beth : Work done = 1/20 = 5% : Cumulative work done = 1/8 + 1/20 = 17.5%
Hour 3: Charles: Work done = 1/40 = 2.5%: Cumulative work done = 17.5% + 2.5% = 20%
Hour 4: Dana: Work done = 1/80 = 1.25%: Cumulative work done = 21.25%
Hours 5-8: Cumulative work done = 21.25% + 21.25% = 42.5%
Hours 9-12: Cumulative work done = 42.5% + 21.25% = 63.75%
Hours 13-16: Cumulative work done = 63.75% + 21.25% = 85%
Hour 17: Alex : Work done = 1/8 = 12.5% : Cumulative work done = 85% + 12.5% = 97.5%
Remaining work = 100% - 97.5% = 2.5%
Beth can do the remaining work in 2.5%/5% = .5 hours

Minimum hours required for the task to be completed = 17.5 hours

For maximum number of hours, we assign the work to the least productive workers first.

Hours 1-4: Cumulative work done = 21.25%
Hours 5-8: Cumulative work done = 42.5%
Hours 9-12: Cumulative work done = 63.75%
Hours 13-16: Cumulative work done = 85%
Hour 17: Dana: Work done = 1/80 = 1.25%: Cumulative work done = 85% + 1.25% = 86.25%
Hour 18: Charles: Work done = 1/40 = 2.5%: Cumulative work done = 86.25% + 2.5% = 88.75%
Hour 19: Beth : Work done = 1/20 = 5%: Cumulative work done = 88.75% +5% = 93.75%
Remaining work = 100% - 93.75% = 6.25%
Time taken by Alex to complete the task = 6.25%/12.5% = .5 Hours

Maximum number of hours to complete the task = 19.5 hours

Minimum	17.5
Maximum	19.5

Archit3110

Major Poster

Joined: 18 Aug 2017

Last visit: 16 May 2026

Posts: 8,648

Own Kudos:

5,198

[1]

Given Kudos: 243

Status:You learn more from failure than from success.

Location: India

Concentration: Sustainability, Marketing

GMAT Focus 1: 545 Q79 V79 DI73 Verified score

GMAT Focus 2: 645 Q83 V82 DI81 Verified score

GPA: 4

WE:Marketing (Energy)

Products:

GMAT Focus 2: 645 Q83 V82 DI81 Verified score

Posts: 8,648

Kudos: 5,198

[1]

16 Dec 2025, 10:01

Kudos

Bookmarks

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.

Rate of A 1/8
B 1/20
c 1/40
d 1/80
net rate
1/8 +1/20 + 1/40+ 1/80 = 10+4+2+1 /80 = 17/80

4 person cycle repeating , 1 hour each
4 * 17/ 80 = 17/20
work left 1-17/20 = 3/20
this way work gets done in 17.5 hours
maximum time will be when work is done in order of DCBA ; work done in full hours by DCB and A completes half work
19.5hours
17.5 ; 19.5

ManifestDreamMBA

Joined: 17 Sep 2024

Last visit: 21 Feb 2026

Posts: 1,387

Own Kudos:

901

[1]

Given Kudos: 243

Posts: 1,387

Kudos: 901

[1]

16 Dec 2025, 10:03

Kudos

Bookmarks

Since they work in 4 person cycle, every 4 hours, they complete
1/8 +1/20+1/40+1/80 = 17/80

To reach 100%, they will need to run 80/17 = 4 complete cycles and 1 partial cycle = So time required = 4*4 = 16 + time for partial cycle

Work remaining after 4 cycles = 1 - 4*17/80 = 1 - 68/80 = 1 - 17/20 = 3/20

To minimise time, we start with fastest person
Speed A>B>C>D
Contribution
A: Rate * Time = 1/8 * 1
Remaining = 3/20 - 1/8 = 1/40

Note 1/40 <1/20, so whole hour is not required
B: Rate * Time = 1/20 * t
1/40 = 1/20 * t => t = 0.5

Min time = 16 + 1 + 0.5 = 17.5

To maximise time, we start with slowest person
Contriution
D: rate * Time = 1/80 * 1
Remaining = 3/20 - 1/80 = 11/80

C = rate * time = 1/40 * 1
Remaining = 11/80 - 1/40 = 9/80

B = rate * Time = 1/20 * 1
Remaining = 9/80 - 1/20 = 5/80

Note 5/80 < 1/8, so whole hour is not required
A = rate * time = 1/8 * t = 5/80
t=0.5
Max time = 16 + 1 + 1 + 1 + 0.5 = 19.5

Bunuel

Ayeka

Joined: 26 May 2024

Last visit: 15 May 2026

Posts: 528

Own Kudos:

402

[1]

Given Kudos: 158

Location: India

Schools: ISB

GMAT Focus 1: 645 Q82 V83 DI80 Verified score

GPA: 4.2

Products:

Schools: ISB

GMAT Focus 1: 645 Q82 V83 DI80 Verified score

Posts: 528

Kudos: 402

[1]

16 Dec 2025, 10:18

Kudos

Bookmarks

A= 8 hrs
B= 20 hrs
C= 40 hrs
D= 80 hrs
Work done in first 4 hours regardless of order = (1/8)+(1/20)+(1/40)+(1/80)=(10+4+2+1)/80=17/80;
Work don in 8 hours = 2*(17/80)=34/80
Work done in 12 hours = 3*(17/80) =51/80
Work done in 16 hours =4*(17/80)=68/80
Remaining work to be done =1-(68/80)=12/80

In order to minimize the time taken to complete the task, the fastest person (Alex and Beth) will do the work ahead in the 4 hour cycle.
Hour 17, Alex works at (1/8) rate.
Work remaining after Alex shift = (12/80)-(10/80)=2/80
Hour 18, Beth works at (1/20) rate. Beth rate is (1/20) or (4/80), it means she will finish the job before the hour ends.
Time taken by Beth = (2/80)/(4/80)=0.5 hours
Minimum time taken to complete the task = 16+1+0.5=17.5 hours.

In order to maximize the time, slowest persons (Dana and Charles) will work ahead in the cycle.
Hour 17, Dana works (1/80) rate
Work remainjng after Dana shift = (12/80)-(1/80) =11/80
Hour 18, Charles works at (2/80) rate and remaining work after his shift = (11/80)-(2/80)=9/80
Hour 19, Beth works at (4/80) rate and remaining work after his shift = (9/80)-(4/80)=(5/80)
Hour 20, Alex works at (1/8) rate and time he needs to complete the remaining task =(5/80)/(1/8)=0.5 hours
Maximum time taken to complete the task =16+1+1+1+0.5=19.5 hours

Minimum =17.5 and Maximum =19.5

Chaithanya20

Joined: 15 Dec 2025

Last visit: 26 Dec 2025

Posts: 14

Own Kudos:

[1]

Given Kudos: 1

Posts: 14

Kudos: 12

[1]

16 Dec 2025, 10:19

Kudos

Bookmarks

In one 4 hour cycle,
1/8+1/20+1/40+1/80 = (10+4+2+1)/80 = 17/80
After 4 cycles , time = 16 hrs
work = 68/80
Remaining = 12/80
For minimum, Alex starts to work, => 10/80 done in one hr and the rest by Beth at 2/80 in 0.5 hr => Minimum time = 17.5 hours
For maximum, Beth starts at 4/80 in one hr and Charles at 2/ 80 in one hr and Dave at 1/80 in one hr with Alex finishing the remaining 5/80 at 0.5 hr => Maximum time =19.5 hours

Rahilgaur

Joined: 24 Jun 2024

Last visit: 26 Jan 2026

Posts: 162

Own Kudos:

125

[1]

Given Kudos: 47

GMAT Focus 1: 575 Q81 V82 DI72 Verified score

Products:

GMAT Focus 1: 575 Q81 V82 DI72 Verified score

Posts: 162

Kudos: 125

[1]

16 Dec 2025, 10:30

Kudos

Bookmarks

Bunuel

Let Total work be = 80x

Efficiency of A, B, C & D respectively will be = 10x, 4x, 2x, and x hours

To get minimum time --> A will start the work followed by b, c & D is respective orders

--> in 16 hours, 4 cycle will be completed and 4*17x = 68 x work will be completed and A will do additional 10x work in 1 hour and B will complete the remaining 2x works in 0.5 hrs

---> Total time taken is 17.5 hours

To get maximum time ---> D will start the work followed by C, B and A respectively

68x works will be completed in 4 cycles and 16 hours --> D, C and B will complete 7x works in next 3 hours ---> Remaining work = 5x will be completed by A in 0.5 hours --> Total time taken = 19.5 Hours

gchandana

Joined: 16 May 2024

Last visit: 14 May 2026

Posts: 201

Own Kudos:

146

[1]

Given Kudos: 170

Location: India

Products:

Posts: 201

Kudos: 146

[1]

16 Dec 2025, 10:33

Kudos

Bookmarks

In one 4-person cycle, 17/80 th work is completed. (LCM of 1/8, 1/20, 1/40, 1/80)
So in 4 such cycles, 62/80 th work gets completed.
For sure, we need 16 hours.

We focus on the 12/80 th work remaining.
Next, for the minimum time required, we will make higher-rate people work.
Let's start with Alex. Alex completes 1/8 th work in 1 hour, so the work left 2/80 th.
Then Beth takes 30 minutes to complete the remaining ((2/80) * 20).
A total of 1.5 hrs, which gives 16+1.5 = 17.5 hrs.

For the maximum, we will make lower-rate people work.
Let's start with Dana. Dana completes 1/80 th. Remaining 11/80 th.
Charles completes 1/40 th of it. Remaining 9/80 th.
Beth completes 1/20 th of it. Remaining 5/80 th.
Now, Alex takes 30 minutes to complete the remaining. ((5/80) * 8)
A total of 3.5 hrs, which gives 16+3.5 = 19.5 hrs.

Bunuel

prepapr

Joined: 06 Jan 2025

Last visit: 16 May 2026

Posts: 93

Own Kudos:

[1]

Given Kudos: 6

GMAT Focus 1: 615 Q85 V80 DI77 Verified score

Products:

GMAT Focus 1: 615 Q85 V80 DI77 Verified score

Posts: 93

Kudos: 82

[1]

16 Dec 2025, 10:37

Kudos

Bookmarks

Let the work completed by Alex, Beth, Charles and Dana in one hour be denoted as A,B,C,D.
So, this gives
A=1/8
B=1/20
C=1/40
D=1/80
Bringing to common denominator,
A=10/80
B=4/80
C=2/80
D=1/80
Consider one cycle giving one hour each to all 4 of them. At the end of this cycle, (10+4+2+1)/80 work would be done
Simplifying and analysing
17/80 in one cycle - 4 hours completed
34/80 by second cycle - 8 hours completed
51/80 by third cycle - 12 hours completed
68/80 by fourth cycle - 16 hours completed

Now we have 12/80 work pending. Assigning the order of A,B,C,D would decide the minimum and maximum number of hours taken
Since the efficiency is in the order A>B>C>D

For minimum, we take them in the order ABCD
Calculating minimum, A does 10/80 in another hour, B does the remaining 2/80 in half an hour totaling to 80/80 of work.
Total time taken= 16+1+0.5=17.5 hours

For maximum, we take them in the order DCBA
Calculating maximum, D does 1/80 in one hour, C does 2/80 in another hour, B does 4/80 in another hour, A does the remaining 5/80 in half an hour totalling to 80/80 of work
Total time taken = 16+1+1+1+0.5 = 19.5hours

Bunuel

namna

Joined: 31 Jul 2014

Last visit: 15 Apr 2026

Posts: 13

Own Kudos:

[1]

Concentration: Finance, Other

Schools: INSEAD Aug '26 HEC Sept '26 ESSEC FT MBA '26 LBS '26

GPA: 3.3

WE:Engineering (Non-Profit and Government)

Schools: INSEAD Aug '26 HEC Sept '26 ESSEC FT MBA '26 LBS '26

Posts: 13

Kudos: 16

[1]

16 Dec 2025, 10:39

Kudos

Bookmarks

Rates per hour:

A = 1/8
B = 1/20
C = 1/40
D = 1/80

Each works alone for exactly one hour in a four hour repeating cycle. Minimum Time/Max Time depends on who finishes the work last.

Min Time: finished by the fastest worker, so, the cycle goes ABCD | ABCD | ABCD ..
Max Time: finished by slowest worker DCBA | DCBA | DCBA ...

One full cycle = 4 hours and completes (1/8 + 1/20 + 1/40 + 1/80) of the task
So, 4 hours <=> 17/80 of the task

How many cycles are need to get the entire task done:

After 4 cycle (=16 hours), 4 * 17/80 = 68/80 of the task.
After 5 cycles (=20 hours), 5 * 17/80 = 85/80 >1

So the task is completed between 16 and 20 hours.

Minimum Time (Alex 1/8 finishes it):
After 16 hours, the remaining work = 1 - 68/80 = 12/80
Alex rate = 1/8 per hour

Time for Alex to finish 12/80 of the task = (12/80) / (1/8) = 1.2 hours

Total Minimum Time = 16 + 1.2 = 17.2 hours ~ 17.5 hours (closest to option)

Maximum Time (D finishes it):
The DCBA cycle also takes 4 cycles = 16 hours to finish 68/80 of the task

Since the cycles are strictly in one hour only, task completed by Dana in the first hour of the 5th cycle = 1/80 (+1 hour)
Remaining task for the next person C = 1 - (68/80 + 1/80) = 11/80 (+1 hour)
C complete 2/80 task in the next hour, leaving for B = 1- (69/80 + 2/80) = 9/80 (+1 hour)
B completes 4/80 task, remaining for A = 5/80
A complete 5/80 task at the rate of 1/8 (=10/80) in 0.5 minutes

Total Maximum Time = 16 +1 + 1 + 1 + 0.5 = 19.5 hours

Final answers:
Min = 17.5 hrs
Max = 19.5 hrs

harishg

Joined: 18 Dec 2018

Last visit: 09 Apr 2026

Posts: 176

Own Kudos:

174

[1]

Given Kudos: 31

GMAT Focus 1: 695 Q88 V84 DI81 Verified score

Products:

GMAT Focus 1: 695 Q88 V84 DI81 Verified score

Posts: 176

Kudos: 174

[1]

16 Dec 2025, 10:49

Kudos

Bookmarks

A, B, C and D work at the rate of 10, 4, 2, and 1 units per hour respectively if total work is considered as 80.

For minimum time, we take the highest efficiency performing individuals first.

10+4+2+1 =17 units of work is done every 4 hrs

17*4+ 10+ 2 units of work is to be done which will take 16+1+1/2 =17.5 hrs

For maximum time, we take the slowest efficiency performing individuals first.

1+2+4+10

17*4+1+4+2+5 gives a time of 16+1+1+1+1/2 =19.5

Therefore, Minimum = 17.5 and Maximum = 19.5

Swanoff

Joined: 04 Aug 2021

Last visit: 08 May 2026

Posts: 30

Own Kudos:

[1]

Given Kudos: 92

Location: Uzbekistan

Concentration: Strategy, General Management

Schools: McDonough '26 (S)

WE:Education (Non-Profit)

Products:

Schools: McDonough '26 (S)

Posts: 30

Kudos: 39

[1]

16 Dec 2025, 10:56

Kudos

Bookmarks

Easy question
First 16 hours clear because 1/80+1/40+1/20+1/8 = 17/80

4 times 4 hour cycle then 4*17/80= 68/80
then plus fastest rates 1/8+1/20
16+1+0.5=17.5

Second case is the same four cycles with the same time but
with smaller values at start like that

68/80+1/80+1/40+1/20 + (1/8 - > half of it)= 19.5

Bunuel