Work Units Combined Rates and Efficiency

What you will learn

A completion time tells you how long a whole job takes. A work rate tells you how much gets done in one unit of active time. The difference matters: when two people work together, their contributions add, but their solo completion times do not.

By the end you will be able to name the whole job and rate units, calculate a rate from partial work, combine constant independent rates, recover a missing contributor, and compare efficiency with the time needed for the same job. You will also check whether the supplied information permits the claimed answer.

This lesson retrieves exact fractions, equivalent ratios and inverse proportion from Numbers and Ratios. “Efficiency” here means completed work per unit active time. It does not mean work quality, physical effort, personal worth or energy efficiency. Changing teams and tank flow come in the next two lessons.

1 Retrieve the tools

Try these before the feedback. They are unscored revision prompts, not an entrance test.

1. Find 1/12 + 1/18.
2. Calculate (3/4) ÷ (5/36).
3. Change 0.4 hour into minutes.
4. For the same fixed job, six equal-rate workers need 10 active hours. Under unchanged independent rates, why do twelve such workers need 5 hours?

Feedback:

1. Use equal-sized parts: 1/12 + 1/18 = 3/36 + 2/36 = 5/36.
2. Divide by a fraction by multiplying by its reciprocal: (3/4) × (36/5) = 27/5.
3. One hour is 60 minutes, so 0.4 × 60 = 24 minutes. Decimal digits after the point are fractions of an hour, not a minute reading.
4. The required work represents 6 × 10 = 60 worker-hours at the stated equal hourly productivity. Twelve workers deliver 12 worker-hours each hour, so 60/12 = 5 hours. Headcount alone would not justify this if the workers' rates differed or they obstructed one another.

For fraction errors, revisit Fractions Decimals and Exact Calculation. For the fourth explanation, revisit Direct and Inverse Proportion. We will keep units beside our arithmetic here.

2 Fix the whole before finding a rate

Choose one complete, unchanged job as 1 job. Half that job is 1/2 job. This is a measurement choice: one job could mean packing a stated batch, painting a stated wall or completing a stated task. We can add fractions only when they refer to the same whole.

A constant rate means equal amounts are completed in equal active time intervals. If 2 hours produce twice the work of 1 hour, then dividing completed work by active time gives the amount completed in one hour:

work rate = completed amount ÷ active time

Multiplying that per-hour amount by the number of active hours gives the work done. Thus work = rate × active time; dividing the required work by a positive rate gives the required time. These relationships describe the same model in different directions.

The model needs conditions. In this lesson, a fixed job can be divided between workers. Each keeps a constant rate. Contributions are independent and are counted once. There is no setup delay, pause, coordination loss or obstruction unless a question explicitly states one. These are arithmetic assumptions, not promises about every real workplace. When a question deliberately leaves a condition unknown, do not silently supply it.

Worked case 1 From partial work to a rate

A worker completes 3/8 of one fixed job in 3 active hours at a constant rate. Find the rate, the whole-job time from the start, and the additional time still needed.

The 3 hours correspond to 3/8 job, not to a whole job. Rate = (3/8) ÷ 3 = 1/8 job per hour. At this rate, a whole job takes 1 ÷ (1/8) = 8 hours.

Panel 1 One job divided into eight equal work parts

Equal work part	1	2	3	4	5	6	7	8
Status after 3 hours	Done	Done	Done	Left	Left	Left	Left	Left
Size of each part	1/8	1/8	1/8	1/8	1/8	1/8	1/8	1/8

The panel shows 3/8 completed and 5/8 left. It is a work bar, not a picture of eight unequal physical jobs. At 1/8 job per hour, the remaining 5/8 needs (5/8) ÷ (1/8) = 5 more hours. Total elapsed time is 3 + 5 = 8 hours.

Check by rebuilding the amounts: 3 × 1/8 = 3/8 and 8 × 1/8 = 1. Using 1/3 job per hour would wrongly treat the first three hours as a whole-job time.

The same job could be renamed 80 equal work units. The worker would then complete 10 units per hour. The rate's number changes, but the job still takes 80/10 = 8 hours. Choosing units cannot change the answer.

Pause and explain

1. At 1/6 job per hour, how much is done in 2 hours? How long for the remaining part?
2. If the worker's rate changes after those 2 hours, can the old rate alone determine the additional time?

Feedback: 2 × 1/6 = 1/3 is done; 2/3 remains. At the unchanged rate, (2/3)/(1/6) = 4 more hours. If the future rate changes without being specified, that four-hour prediction is not justified. An observed average for one interval does not determine a different interval.

3 Add the work done during the same interval

Imagine asking each worker, “How much of this same job will you contribute during the next hour?” Those amounts can be added if both are active and independent. A solo completion time is not an amount of work, so adding or averaging solo times does not answer that question.

Worked case 2 Two workers and two equivalent representations

A alone needs 12 hours and B alone needs 18 hours for the same divisible job. They start together and retain their own constant independent rates until the job is complete.

Panel 2 One simultaneous hour

Contributor	Rate in jobs per hour	Rate if one job is 36 equal units
A	1/12 = 3/36	3 units per hour
B	1/18 = 2/36	2 units per hour
A and B together	5/36	5 units per hour

In one hour they jointly complete 5/36 job. Required time = 1 ÷ (5/36) = 36/5 hours. The right-hand column gives the same result: 36 units ÷ 5 units per hour = 36/5 hours. We chose 36 because it is a convenient common multiple of 12 and 18. Another common multiple would also work; it would simply relabel the equal work units.

Convert exactly: 36/5 = 7 + 1/5 hours. The remaining fifth of an hour is (1/5) × 60 = 12 minutes. So the answer is 7 hours 12 minutes. It is not 7 hours 20 minutes.

Check each contribution over that same elapsed time:

• A does (1/12) × (36/5) = 3/5 of the job
• B does (1/18) × (36/5) = 2/5 of the job
• The contributions add to 1

With both positive independent rates active throughout, the joint time must be less than either solo time. Here 36/5 is below both 12 and 18. The average, 15 hours, fails that check. This bound must not be carried into a later problem where someone stops early.

Pause before using a shortcut Suppose the same two workers start together with only half of this job left. Fill in:

• Required amount = □ job
• Combined rate = □ job per hour
• Required time = □ hours

Feedback: the amount is 1/2, the rate remains 5/36, and time = (1/2)/(5/36) = 18/5 hours = 3 hours 36 minutes. Half the amount at the same rate needs half the time. A full-job shortcut applied without adjustment would overcount the work.

For positive solo times a and b on the same whole job, the above reasoning gives t = 1 ÷ (1/a + 1/b) = ab/(a + b). This is a derived compact form, not a replacement for reading the conditions. It applies when both work throughout at unchanged additive rates; it must be adjusted for a partial job. The rate method also makes the units and contributions visible.

Mathematical basis: NCERT Ganita Prakash Grade 8 Part II, Chapter 3 §3.6, printed pp.66–67 develops whole-job units and combined unequal-worker rates. Our tables, cases and explanations are original.

4 A missing worker is a missing contribution

If the joint contribution and one person's contribution are known for the same hour, subtract to find the other. Subtracting their completion times has no such meaning.

Worked case 3 Recover B and test the assumptions

A and B together complete one fixed job in 6 hours. A alone takes 10 hours. Both contribute at positive, constant, independent rates. Find B's solo time.

Joint rate = 1/6 job per hour. A's rate = 1/10. Therefore B's rate = 1/6 − 1/10 = 5/30 − 3/30 = 2/30 = 1/15 job per hour. B alone needs 1 ÷ (1/15) = 15 hours.

Check: 1/10 + 1/15 = 3/30 + 2/30 = 1/6. Both together are faster than either alone: 6 is below 10 and 15. The subtraction 10 − 6 = 4 does not find B's rate or solo time.

Now test a separate claim: A still takes 10 hours alone, but the pair supposedly takes 12 hours under the same positive independent-rate assumptions. The inferred B rate would be 1/12 − 1/10 = −1/60 job per hour. That contradicts the stated positive contribution. It is inconsistent information under this model, not a worker who completes the job in negative time.

If pair time and A's time were both exactly 10 hours, the inferred B rate would be zero. A zero rate makes no progress, so it has no finite solo completion time for a nonzero job. It also contradicts the statement that B's rate is positive. Never divide by zero to invent a time.

Pause and classify A takes 9 hours alone. A and B together are said to take 5 hours. Does the inferred B rate have the required sign? What if the claimed joint time were 11 hours?

Feedback: 1/5 − 1/9 = 4/45, positive; B's solo time would be 45/4 hours. But 1/11 − 1/9 = −2/99 is negative, so the second claim fails the positive-contributor model. Here all needed numbers are present; the problem is contradiction, not missing information.

5 Efficiency compares rates for the same work

For the same fixed work W, rate × solo time = W. If a rate is multiplied by a positive factor, the time is divided by that factor. This is the inverse relationship already studied in Ratios.

Worked case 4 Compare rates and solo times

For the same job, A's work rate is to B's as 3:2. B alone needs 15 hours. Find A's solo time and their joint time under unchanged independent rates.

B's rate is 1/15 job per hour. A's is (3/2) × (1/15) = 1/10 job per hour, so A alone needs 10 hours. Joint rate = 1/10 + 1/15 = 1/6, giving 6 hours together.

Panel 3 Equal work connects opposite ratios

Comparison, always A to B	A	B	A:B
Rate	1/10 job per hour	1/15 job per hour	3:2
Solo time for one job	10 hours	15 hours	2:3
Work done in the same 2 hours	1/5 job	2/15 job	3:2

The rate and equal-time output ratios agree. The same-job time ratio is reversed. The products (1/10) × 10 and (1/15) × 15 both equal one job.

A percentage comparison with an explicit base In a separate job comparison, B takes 12 hours and A's rate is 25% greater than B's. “Greater” is measured against B's rate. A's rate factor is 1 + 25/100 = 5/4, so A's time factor is its reciprocal, 4/5. A's time = 12 × 4/5 = 48/5 hours = 9 hours 36 minutes.

The time has fallen by 1 − 4/5 = 1/5, or 20%. A 25% higher rate does not mean 25% less time. Check: rate factor × time factor = (5/4) × (4/5) = 1, preserving the work. If “25% more efficient” is used, its intended meaning as work per unit time must be stated.

Pause and explain the reference

1. A has twice B's rate. B takes 14 hours for the same job. What is A's solo time?
2. Can the ratio A:B = 3:2 in rates, by itself, determine A's absolute solo time?
3. A processes 60 labels in 12 minutes; B processes 90 labels in 15 minutes. At constant rates, which processes more labels per minute?

Feedback:

1. Seven hours: double rate means half the time for the same work.
2. No. Rates 1/10 and 1/15 give solo times 10 and 15 hours; rates 1/20 and 1/30 still have ratio 3:2 but give 20 and 30 hours. A scale, such as one actual rate or solo time, is needed.
3. A's rate is 60/12 = 5 labels per minute; B's is 90/15 = 6. B has the higher rate despite the larger listed duration, because the amounts differ. Compare raw completion times only for the same amount of work under the relevant conditions.

6 A solving routine that keeps the meaning

1. Name the unchanged whole and the required amount
2. Put every time in one unit and identify the active interval
3. Find each rate as amount per unit active time
4. Add simultaneous independent rates, or subtract a known contribution from a known total
5. Divide the required amount by the valid positive rate
6. Check units, sign, size and the sum of contributions

The next lesson changes who is active over time. Do not assume today's all-start, all-finish conditions when the schedule says otherwise.

Independent practice

Try every item before reading the key. These are untimed learning checks: one mark for a correct answer, zero otherwise, no negative marking and no pass cutoff. These local settings do not describe the RRB exam’s marking rules.

1. At a constant rate, a worker completes 5/12 of one fixed job in 2.5 active hours. Which pair correctly gives the rate and the additional time for the remaining work at that same rate?

A. 1/6 job per hour; 3 h 30 min

B. 2/5 job per hour; 3 h 30 min

C. 1/6 job per hour; 6 h

D. 1/6 job per hour; 3 h 5 min

2. A alone finishes a divisible job in 8 h and B alone in 12 h. Both start together and continue at unchanged independent constant rates with no delay until completion. How long do they take?

A. 10 h

B. 4 h 8 min

C. 4 h 48 min

D. 20 h

3. A needs 15 h and B 12 h alone for the same fixed job. They work together throughout at unchanged independent constant rates. Rename that whole job as 60 equal work units. Which row gives A’s units per hour, B’s units per hour, and joint time?

A. 15; 12; 20/9 h

B. 4; 5; 20/3 h

C. 1/15; 1/12; 20/3 h

D. 4; 5; 9 h

4. A and B together complete a fixed job in 4 h. A alone needs 6 h. Their constant positive rates are independent and add. What is B’s solo completion time?

A. 2 h

B. 10 h

C. 1/12 h

D. 12 h

5. A alone takes 6 h for a job. A report claims A and B together need 7 h while both work throughout at positive, unchanged independent rates. Which conclusion follows?

A. B alone must take 42 h

B. The report contradicts the stated rate model

C. B alone must take 13 h

D. B alone must take 1 h

6. For the same job, A:B work rates are 5:3. B alone takes 20 h at its constant rate. How long does A take alone?

A. 100/3 h

B. 18 h

C. 12 h

D. 8 h

7. A completes 20% more work per active hour than B. B needs 15 h for the same job. Both solo rates stay constant. Find A’s solo time.

A. 12 h 30 min

B. 12 h

C. 18 h

D. 15 h 20 min

8. A:B constant work rates for the same fixed job are 7:4. No actual rate or solo time is supplied. What can be concluded about A’s solo time?

A. It must be 4 h

B. It must be 7 h

C. It must be 11/2 h

D. It cannot be uniquely determined

Answer key and explanations

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

1. A is correct. Convert 2.5 h to 5/2 h. Rate = (5/12)/(5/2) = 1/6 job per hour. Remaining work = 7/12, so additional time = (7/12)/(1/6) = 7/2 h = 3 h 30 min. Check: 5/12 + (1/6)(7/2) = 1. B uses 1/2.5 = 2/5 as though 2.5 h completed the whole job. C gives the full-job time 6 h rather than the additional time. D misreads the half hour as 5 minutes; it is 30 minutes.

2. C is correct. Rate = 1/8 + 1/12 = 5/24 job per hour. Time = 24/5 h = 4.8 h = 4 h 48 min. Contributions are (24/5)/8 = 3/5 and (24/5)/12 = 2/5, summing to one. A averages 8 and 12; D adds them. Neither is a sum of work contributions and both fail the faster-than-either-solo check. B treats 0.8 h as 8 minutes, but 0.8 × 60 = 48 minutes.

3. B is correct. A completes 60/15 = 4 chosen units per hour and B completes 60/12 = 5. Together they complete 9 units per hour, so 60/9 = 20/3 h, or 6 h 40 min. Check: 9 × 20/3 = 60. A mistakes solo times for unit rates and divides by their sum. C gives rates in whole jobs per hour where the question asks for the chosen units per hour; its time alone is correct, so the whole row is not. D treats the numerical joint rate 9 as a time instead of dividing required work by it.

4. D is correct. B rate = 1/4 − 1/6 = 1/12 job per hour, so B time = 12 h. Check: 1/6 + 1/12 = 1/4. A subtracts completion times 6 − 4; B adds them. Neither operation finds a missing hourly contribution. C labels the numerical rate 1/12 as hours instead of taking its reciprocal for a whole job.

5. B is correct. The inferred B rate is 1/7 − 1/6 = −1/42 job per hour, not a positive rate. The pair also cannot be slower than A alone when both add positive work throughout. A drops the negative sign and reports 42 h, but that would give joint time 21/4 h, not 7 h. C adds 6 + 7 without a rate interpretation. D subtracts 7 − 6; a one-hour B would make the pair much faster, with joint time 6/7 h. These are inconsistent data under the stated model, not evidence of a negative real completion time.

6. C is correct. A is 5/3 as fast, so its same-job time is 3/5 of B’s: 20 × 3/5 = 12 h. Equivalently, A rate = (5/3)(1/20) = 1/12. The time ratio 12:20 is 3:5, the reverse of 5:3. A uses the rate factor directly on time and makes the faster worker take longer. B subtracts the ratio-term difference 5 − 3 from 20, which has no unit or proportional basis. D uses the difference 20 − 12 = 8 as the requested time; 8 h would imply rate ratio 20:8 = 5:2, not 5:3.

7. A is correct. Use B’s rate as the percentage base: A rate factor = 1.20 = 6/5. Time factor = 5/6, so A time = 15 × 5/6 = 25/2 h = 12 h 30 min. Check: (6/5)(5/6) = 1. B reduces time directly by 20%, incorrectly using the same percentage change for an inverse relationship. C increases time by 20%, reversing the direction. D treats the rate percentage as 20 minutes to add, mixing a percentage with a duration.

8. D is correct. Rates 1/4 and 1/7 job per hour have ratio 7:4 and give A time 4 h. Halving both rates to 1/8 and 1/14 preserves 7:4 but gives A time 8 h. Both satisfy the supplied information, so a scale is missing. A names one possible time as necessary. B treats the first rate-ratio term as a number of hours. C averages the two ratio terms, which neither gives an actual rate scale nor a justified duration. The valid relative conclusion is solo-time ratio A:B = 4:7.

Let the error choose the revision

• Item 1: revisit the whole, partial amount and additional time in case 1
• Items 2–3: revisit the rate/unit columns and contribution check in case 2
• Items 4–5: revisit missing-rate subtraction and the positive-rate conditions in case 3
• Items 6–8: revisit inverse time factors, the percentage base and missing scale in case 4

After revisiting, solve the relevant item without its key and explain why all three alternatives fail.

Sources and lesson boundary

RRB CEN 09/2025, §14.1, printed p.28 names Time and Work in Mathematics. This supplies topic scope, not a prediction of question frequency. The official list is illustrative and not necessarily exhaustive; later amendments have not all been audited here.

NCERT Ganita Prakash Grade 8 Part II, Chapter 3 §3.6, printed pp.63–68 supports fixed-work proportional reasoning and combined rates. Its official preliminary pages, PDF p.2 identify First Edition December 2025. Missing-rate subtraction, feasibility checks and the percentage-efficiency comparison here are our explained extensions, not named official exam variants. All prose, panels, worked cases and practice are original.

This lesson covers constant independent work rates. Joining/leaving schedules, workday changes and signed tank flows are taught next. Wage division, alternating attendance cycles, staffing-rounding rules and physical variable-flow laws are outside this lesson.

Analogy

Imagine a fixed batch of plain labels moving into a finished tray. One worker finishes three equal bundles each hour; another finishes two. If they work independently during the same hour, five bundles enter the tray. The tray’s target amount is unchanged, so a larger hourly contribution reaches it sooner.

This analogy shows why we add output per hour, not solo completion times. Renaming each bundle as two smaller equal bundles doubles the numerical rates and target together, leaving the completion time unchanged.

Its limit matters: real labels may be indivisible, workers may share a tool, and setup or checking may consume time. Our continuous divisible-work model does not automatically cover those details. State constant rates and independent contributions rather than treating a real process as perfectly additive without evidence.

Quick reference

• Fix the same whole job and a common time unit before comparing or adding rates
• Rate = completed work / active time; work = rate × active time on a constant-rate interval
• Time for a required amount = required work / positive rate
• If a whole job takes t time units, rate = 1/t jobs per time unit; a partial-job time is not a whole-job time
• For simultaneous independent workers throughout: combined rate = sum of individual rates. For two positive solo times a,b on one whole job, t = ab/(a+b) follows from that sum
• Any common work-unit scale is valid if both the total and every rate use it consistently
• Missing contribution = joint rate − known rate. A negative or zero result is incompatible with a stated positive contributor; zero rate has no finite completion time for a nonzero job
• Same-job rates a:b imply solo-time ratio b:a. An increase by rate factor f gives time factor 1/f
• A ratio alone gives relative sizes, not an absolute time without a rate/time scale
• Fractional hours convert by multiplying the fractional part by 60; keep exact fractions until that step
• Check units, sign and contributions. “Together is faster than either alone” needs both positive independent workers active throughout