Changing Teams and Finishing the Remaining Work

What you will learn

A team can change before a job is finished. The amount of work is still fixed, but the rate need not stay the same throughout. You will record each active interval, carry completed work forward, and use the new rate only on what remains.

By the end you will handle a known joining or leaving time, a stated change in daily working hours, and one unknown joining time. You will distinguish additional time from total elapsed time, detect an impossible schedule, explain missing information, and stop when the job is complete even if another event was planned later.

Use the previous lesson's meaning of one fixed job and constant independent rates. Unless a question deliberately withholds a condition, work is divisible, contributions do not overlap or obstruct one another, and there are no unstated breaks or setup delays. Rates stay unchanged within each stated interval. Worker-hours require equal, unchanged hourly productivity; headcount alone does not establish it.

1 Retrieve what a rate measures

Try before the feedback:

1. Find 1/8 + 1/24.
2. If 3/8 of a job is done, what remains?
3. Change 27/2 hours into hours and minutes.
4. How many worker-hours do six equal-rate workers contribute over four days at five active hours per day?

Feedback: 1/8 + 1/24 = 4/24 = 1/6 job per hour; the remaining amount is 5/8 job; 27/2 hours is 13 hours 30 minutes; the effort is 6 × 4 × 5 = 120 worker-hours. A worker-hour counts one worker for one active hour. It becomes a measure of output only under the specified productivity.

Revisit Work Units Combined Rates and Efficiency for the rate conversion or contribution sum. Revisit exact fractions if subtraction or division is uncertain. We introduce the small unknown-duration equation locally below; you do not need to finish the later Algebra module first.

2 A new interval begins when the active rate changes

Mark the start, each joining/leaving event, and the finish. For each interval, record:

• Who is active and the rate they jointly produce
• How long that interval lasts
• The work added during it
• The cumulative completed amount and the remaining amount

The interval calculation is still work = rate × active time. The change is that we must not apply one interval's rate to the whole schedule. A worker who has not joined cannot be credited with earlier work; one who has left cannot contribute afterward.

Before applying a whole planned interval, ask whether the remaining job will finish during it. A schedule describes potential events; the first-completion question ends as soon as the work reaches one whole job.

Worked case 1 A helper joins after three hours

A alone needs 12 hours and B alone needs 18 hours for the same divisible job. A starts alone. After 3 active hours, B joins and both continue at unchanged independent rates until completion. Find the additional time after B joins and the total time from A's start.

Panel 1 A joining event and its work ledger

Event order: start at 0 h → B joins at 3 h → job finishes. The arrow lengths do not represent a time scale.

Interval	Active workers	Rate in jobs per hour	Duration	Work completed in the interval
Before joining	A	1/12	3 h	3/12 = 1/4
After joining	A and B	1/12 + 1/18 = 5/36	To be found	Remaining 3/4

A has completed 1/4, leaving 3/4. At the new combined rate, additional time = (3/4)/(5/36) = 27/5 hours = 5 hours 24 minutes. Total time = 3 + 27/5 = 42/5 hours = 8 hours 24 minutes.

Check each person's active time rather than just repeating the final division. A works for 42/5 hours and contributes (42/5)/12 = 7/10 job. B works only for 27/5 hours and contributes (27/5)/18 = 3/10. The sum is one. The total is longer than the all-together 36/5 hours, because B missed the beginning, but shorter than A's solo 12 hours.

Do not report 27/5 hours if the question asks for time from the original start. Do not give B three hours of work before joining. Do not divide one whole job by the new rate after part of the job is already complete.

Pause and complete a fresh ledger A takes 20 hours alone and B takes 30 hours alone. A completes one-quarter of the same job before B joins. Both then work continuously at unchanged independent rates. Fill in the first duration, joint rate, additional duration and total duration.

Feedback: A's first duration is (1/4)/(1/20) = 5 hours. Joint rate is 1/20 + 1/30 = 1/12 job per hour. The remaining 3/4 needs (3/4)/(1/12) = 9 hours. Total = 5 + 9 = 14 hours. A works 14 hours and B 9 hours: 14/20 + 9/30 = 7/10 + 3/10 = 1.

Worked case 2 A worker leaves

A takes 12 hours and B 18 hours alone for the same job. Both work together for 3 hours. A then stops; B continues alone at the same constant rate until completion. Find further and total time.

Panel 2 The rate changes after leaving

Interval	Active workers	Work added	Work remaining afterward
First 3 hours	A and B	3 × (5/36) = 5/12	7/12
Remaining interval	B alone	The remaining 7/12	0 at completion

B's rate in the second interval is 1/18 job per hour. Further time = (7/12)/(1/18) = 21/2 hours = 10 hours 30 minutes. Total time = 3 + 21/2 = 27/2 hours = 13 hours 30 minutes.

Check: A contributes 3/12 = 1/4. B works for the whole 27/2 hours and contributes (27/2)/18 = 3/4. The sum is one. It would be wrong to write 18 − 3 as B's remaining time: B did not have to do all the work completed in the first interval, because A helped.

The total 13.5 hours exceeds A's solo 12 hours. That is possible here: A did not remain for the whole job. The earlier “joint time is below both solo times” check applies only when both positive independent contributions remain active throughout.

Pause and choose the applicable rate After A leaves, should the remaining work be divided by 5/36 or by 1/18? Explain without calculating.

Feedback: divide by 1/18, because only B is active. The joint rate describes a different interval. The amount already completed stays completed, but A's future contribution is zero.

3 Worker-hours connect people and active hours

For equal workers with the same constant hourly productivity, define one worker's one-hour output as one work unit. Then workers × active hours measures completed output in that scale. If hours per day change, multiplying workers by days alone loses essential information.

Worked case 3 A larger team with a longer working day

Eight equal-rate workers would complete a divisible job in 10 working days at 6 active hours per day. After 4 such days, four more identical workers join. All twelve now work 8 active hours per day, with unchanged hourly productivity. Find further working days.

Panel 3 Count the work before changing the schedule

Quantity	Calculation in worker-hours	Result
Whole job	8 × 6 × 10	480
Work already done	8 × 6 × 4	192
Remaining work	480 − 192	288
New contribution each working day	12 × 8	96 per working day

Further days = 288/96 = 3 working days. Total = 4 + 3 = 7 working days.

Check: 8 × 6 × 4 + 12 × 8 × 3 = 480. The old team and old daily hours apply to the work already done. The new team and new daily hours apply only afterward.

Seven working days does not mean 168 active hours or specify an exact clock timestamp. Days include the stated active hours, and the question does not describe the rest of the daily timetable. If workers had unequal hourly rates, their headcounts could not simply be combined in this way; their actual rates would be needed.

Pause and check the unit Suppose 168 worker-hours remain. Seven identical workers now work 4 active hours each day at the same hourly productivity. How many further working days are required?

Feedback: daily contribution = 7 × 4 = 28 worker-hours per day; 168/28 = 6 further working days. Dividing 168 by 7 alone gives 24 active hours per worker, not the number of four-hour working days.

4 One unknown joining time can be recovered from work

An unknown duration is a number whose meaning we have not yet found. Name what it measures. If A works from the start to a known finish time T, and B joins x hours after that start, B's active time is T − x. It is not x.

A work-balance equation states that the contributions add to the required amount. We can solve it by subtracting known work and dividing by a known rate. The equality is preserved because we perform the same operation on both sides. The result must also fit the time interval in the story.

Worked case 4 Find when B joined

A needs 12 hours and B 18 hours alone for the same job. A starts at time zero and keeps working. B joins later and then stays. They complete the job 9 hours after A began, with unchanged independent rates. When did B join?

A's contribution is known: 9/12 = 3/4 job. B must contribute the remaining 1/4. B needs (1/4)/(1/18) = 9/2 hours of active work, so B joined 9 − 9/2 = 9/2 hours after the start, or 4 hours 30 minutes after A.

Panel 4 The unknown marks a start, not a duration worked

Worker	Starts at	Finishes at	Active duration	Contribution
A	0	9 h	9 h	9/12
B	x h	9 h	(9 − x) h	(9 − x)/18

The same reasoning as an equation is:

• 9/12 + (9 − x)/18 = 1
• Subtract 9/12 from both sides: (9 − x)/18 = 1/4
• Multiply both sides by 18: 9 − x = 9/2
• Therefore x = 9 − 9/2 = 9/2

The active duration 9 − x must be nonnegative, so 0 ≤ x ≤ 9 is necessary. “Joins later” also excludes x = 0. The answer 9/2 meets those conditions. Substituting into the original work balance gives 9/12 + (9 − 9/2)/18 = 3/4 + 1/4 = 1.

Now test a separate claim of completion in 6 hours with the same start-and-join rules. A would contribute 6/12 = 1/2. B would need 9 active hours to contribute the other half, but at most 6 hours is available. The equation gives x = 6 − 9 = −3 hours. That is not an acceptable “join later” time. The claim is impossible under the stated schedule. Even both starting immediately need 36/5 hours, more than 6.

Pause and solve a new balance A alone needs 8 hours and B 12 hours. A starts first; B joins later and stays. A keeps working and the job finishes at 6 hours after A's start. At what time did B join?

Feedback: A contributes 6/8 = 3/4; B must do 1/4, requiring (1/4)/(1/12) = 3 active hours. B joins at 6 − 3 = 3 hours. Check: 6/8 + (6 − 3)/12 = 1 and 0 < 3 < 6.

The small equation bridge is supported by NIOS Secondary Mathematics 211, Chapter 5 §§5.2–5.3, printed pp.142–146. We use named variables, balance-preserving steps and substitution; two-variable systems and graphing are not needed here.

5 Decide whether the target event is determined

Missing information and inconsistent information are different. Missing information permits more than one answer. Inconsistent information permits no answer within the stated model. A missing detail can also be irrelevant if its event occurs after the requested completion.

Worked case 5 Two different schedule questions

Case A: an unspecified joining time. A alone needs 12 hours and B 18 hours. A starts alone, B joins later and stays, and A continues throughout. No joining time, amount completed before joining or final completion time is given. Can total completion time be uniquely found?

Panel 5 Two schedules that both obey the statement

B joins after	A has already done	Further time together	Total time
3 h	1/4 job	(3/4)/(5/36) = 27/5 h	42/5 h
6 h	1/2 job	(1/2)/(5/36) = 18/5 h	48/5 h

Both schedules obey every supplied condition, but the totals differ. Therefore the information is insufficient for a unique total. A joining time or the work amount at joining would determine this version of the question. Do not assume “later” means halfway through the job.

Case B: a leaving event after the job would finish. This is a separate schedule using the same solo rates. Both start together; B is scheduled to leave after 8 hours only if the job is still unfinished. When is the job first completed?

While both are active, completion needs 1/(5/36) = 36/5 hours = 7 hours 12 minutes. That is before the 8-hour event. Stop at 36/5 hours. Do not first record 8 hours of work, obtain more than one job, and then attempt to solve a negative remainder. The planned departure never affects this first-completion calculation.

Pause and choose a safe next step

1. In a planned interval of length d with positive rate r and remaining work R, what should you compare before applying all d hours?
2. A could finish alone in 5 hours. A starts alone; B is scheduled to arrive after 6 hours if work remains. B's rate is not given. Can the first completion time still be found?

Feedback:

1. Compare R/r with d. If R/r ≤ d, completion occurs within or exactly at the end of that interval; stop there. If R/r > d, calculate rd work, subtract it from R, and move to the next interval. Equality does not require an extra interval.
2. Yes, provided A keeps the stated constant rate without interruption: first completion is at 5 hours, before B's possible arrival. B's unknown rate is irrelevant to that event.

6 A schedule-solving routine

1. Name the whole job and the time origin
2. Mark changes and identify who is active in each interval
3. Check whether completion happens before the next change
4. Record interval work and carry the remaining amount forward
5. Report additional or total time exactly as requested
6. Substitute all actual active durations into the contribution sum
7. Reject negative durations; prove insufficiency with two admissible schedules when necessary

The next lesson carries this same ledger into a tank. Water can leave as well as enter, so signs and the empty/full boundaries must be included.

Independent practice

Try before reading the key. These are untimed formative checks: one mark for a correct answer, zero otherwise, no negative marking and no pass cutoff. These local learning settings are not the official RRB marking scheme.

1. A needs 10 h and B 15 h alone for the same divisible job. A works alone for 2 h, then both continue at unchanged independent constant rates without delay. What is the total time from A’s start?

A. 4 h 48 min

B. 6 h 48 min

C. 6 h

D. 8 h

2. A needs 8 h and B 12 h alone for one divisible job. They work together for 2 h at constant independent rates; A then leaves and B continues unchanged. Find total time from their start.

A. 7 h

B. 10 h

C. 24/5 h

D. 9 h

3. Six identical workers need 12 working days at 5 active hours per day for one divisible job. After 4 such days, the team becomes 10 identical workers at 6 active hours per day. Hourly productivity stays unchanged, with independent contributions and no extra delay. How many further working days are required?

A. 4

B. 8

C. 24/5

D. 6

4. A needs 10 h and B 15 h alone for the same divisible job. A starts and works continuously; B joins later and then stays. At unchanged independent constant rates, the job finishes 8 h after A started. When did B join?

A. 3 h after A

B. 4 h after A

C. 5 h after A

D. 8 h after A

5. A needs 10 h and B 15 h alone for one divisible job. A starts first; B may join only after A starts, and both keep their positive constant independent rates. A keeps working until completion. A report says the job finishes just 5 h after A starts. Which conclusion is valid?

A. B joined 2.5 h after A

B. The reported completion is impossible under these conditions

C. B joined exactly when A started

D. There are many valid joining times giving 5 h

6. For the same divisible job, A alone needs 10 h and B 15 h at constant independent rates. A starts alone, B joins later and stays, and A continues. No joining time, work fraction at joining or final time is supplied. What is the unique total completion time?

A. 6 h

B. 6 h 48 min

C. 7 h 36 min

D. No unique total is determined

7. A and B need 10 h and 15 h alone for one divisible job. They start together at unchanged positive independent constant rates. B will leave after 7 h only if work remains. At what time is the job first completed?

A. 6 h

B. 7 h

C. 8 h

D. 10 h

8. Four workers finish a divisible job in 9 h at positive constant independent rates. The same four, with unchanged rates, are joined from the start by two more workers; all six contribute independently with no extra delay. The new workers’ rates are not given. Can the new completion time be fixed?

A. It must be 6 h

B. It must be 36/5 h

C. No unique time; the extra rates are needed

D. It must remain 9 h

Answer key and explanations

1. B; 2. D; 3. A; 4. C; 5. B; 6. D; 7. A; 8. C

1. B is correct. A first completes 2/10 = 1/5 job. The remaining 4/5 is completed at rate 1/10 + 1/15 = 1/6, requiring 24/5 h more. Total = 2 + 24/5 = 34/5 h = 6 h 48 min. Check: (34/5)/10 + (24/5)/15 = 17/25 + 8/25 = 1. A reports only additional time. C uses the all-start-together time and credits B before joining. D subtracts 2 from A’s solo 10 h, ignoring B’s actual contribution and the requested time origin.

2. D is correct. First-stage work = 2(1/8 + 1/12) = 5/12. Remaining 7/12 at B’s rate 1/12 requires 7 further hours. Total = 9 h. Check: A contributes 2/8 = 1/4 and B contributes 9/12 = 3/4. A gives further time instead of total. B uses 12 − 2, forgetting A’s earlier help and not rebuilding the completed amount. C is the all-together completion time 1/(5/24); A does not remain active that long in this schedule.

3. A is correct. Whole effort = 6×5×12 = 360 worker-hours; completed = 6×5×4 = 120; remaining = 240. New daily effort = 10×6 = 60, so further days = 240/60 = 4. B gives total days 4+4 = 8. C uses 240/(10×5) = 24/5 and keeps the old five-hour day. D divides the whole 360 by the new daily effort, counting already completed work again. Check: 120 + 4×60 = 360.

4. C is correct. A does 8/10 = 4/5; B must do 1/5 and needs (1/5)/(1/15) = 3 active hours. B joins at 8−3 = 5 h. Check: 8/10 + (8−5)/15 = 1 and 0<5<8. A confuses B’s active duration with its start time. B assumes halfway without a work balance; it would produce 8/10 + 4/15 = 16/15 jobs. D leaves B no active time, giving only A’s 4/5 job.

5. B is correct. A does 5/10 = 1/2 job. B would need 7.5 active hours for the other half, so the required joining time is 5−7.5 = −2.5 h, outside the allowed schedule. Even simultaneous start needs 1/(1/10+1/15) = 6 h. A discards the negative sign. C still gives 6 h, not 5. D calls contradiction missing information: no permitted joining time reaches the stated result, so there cannot be many such times.

6. D is correct. Joining after 2 h leaves 4/5 job, requiring (4/5)/(1/6) = 24/5 h more; total 34/5 h = 6 h 48 min. Joining after 4 h leaves 3/5, requiring 18/5 h more; total 38/5 h = 7 h 36 min. Both satisfy the statement but differ. B and C are possible answers, not uniquely forced ones. A is the all-start-together time, which disregards the stated later joining. A joining time or equivalent work-state information is needed.

7. A is correct. Together rate = 1/10+1/15 = 1/6 job per hour, so first completion is at 6 h. Since 6<7, B has not left before completion. B uses the planned event time as the finish. C forces a later stage after the job is already done; no work remains at 6 h. D ignores B’s contribution and gives A’s solo time. Check: 6/10 + 6/15 = 1. Never apply seven hours of production to this first-completion question.

8. C is correct. A possible model gives each original worker rate 1/36 job per hour. If each extra worker also has 1/36, total rate is 6/36 and time 6 h. If each extra has 1/72, total rate is 4/36+2/72 = 5/36 and time 36/5 h. Both models preserve the original four-worker time of 9 h. A and B assert one possible model as necessary. D ignores two positive independent contributions; they make the time less than 9 h, although their unspecified sizes prevent an exact total. Equal headcount does not establish equal rates.

Let the error choose the revision

• Items 1–2: revisit interval ledgers and the difference between additional and total time
• Item 3: revisit the worker-hour table and changed hours per day
• Items 4–5: revisit the meaning and admissible range of the joining time
• Items 6 and 8: rebuild two admissible models before claiming a unique answer
• Item 7: compare first-completion time with the next planned event

Retry the relevant item without its key and explain each rejected option.

Sources and boundary

RRB CEN 09/2025 §14.1, printed/physical p.28 names Time and Work and Pipes & Cistern. This is topic-scope evidence, not a frequency prediction. The list is illustrative and not necessarily exhaustive. Not all later amendments have been audited here.

NCERT Ganita Prakash Grade 8 Part II, Chapter 3 §3.6, printed pp.63–68 supplies fixed-work proportional reasoning and combined-rate foundations. Official prelims, PDF p.2 verify First Edition December 2025. NIOS Secondary Mathematics 211, Chapter 5 §§5.2–5.3, printed pp.142–146 supports forming and checking a simple balance equation; the accessible chapter does not establish its publication year. Our schedules, signed-flow applications and feasibility/boundary checks are explained extensions, not claims of named official question variants. All exposition, panels, cases and practice are original.

Alternating attendance/valve cycles, wage sharing, staffing-rounding rules, multiple-unknown systems and pressure-dependent flow physics remain outside this module.

Analogy

Think of a fixed-length progress strip being coloured steadily. During the first interval one worker colours it; during the next interval a helper also colours different unfinished parts. At the joining mark, the coloured part stays coloured. Only the rate at which the uncoloured part shrinks changes.

The ledger is a record of those actual contributions. It prevents giving a late helper credit for earlier work or making a departed worker keep contributing. It also tells you to stop when the whole strip is coloured, even if a later change had been planned.

The analogy assumes equal pieces really represent equal amounts of work and workers do not colour the same piece twice. Real tasks may require setup, sequence constraints or checking. Those effects need stated data; a simple divisible-work strip does not automatically include them.

Quick reference

• Start with one fixed whole and an explicit time origin
• Split at each joining, leaving or active-hours change; record active workers, rate, duration and work
• Completed work stays completed. Remaining work = whole − completed work
• For each positive-rate interval, compare remaining work / rate with the time until the next event before applying the whole interval
• If completion falls within or exactly at that interval’s end, stop there; do not create a negative remainder or add an unnecessary later stage
• Additional time after a change is not total time from the original start
• Worker-hours = workers × active hours only as an output scale under equal unchanged hourly productivity. With days, include the stated active hours per day
• If B joins at x during a total duration T, B works T − x. Build a contribution balance and substitute the answer back
• All durations must be nonnegative and must fit the stated joining/leaving rules
• Inconsistent data admit no valid schedule. Insufficient data admit more than one; demonstrate two when claiming nonuniqueness
• A missing rate may be irrelevant if its participant is scheduled after first completion
• Do not apply an all-together fastest-solo bound to a schedule where a worker leaves early