Mathematics rate guide for Primary 6 learners
When two workers, taps or machines complete the same job together, their completion times are not added. Time tells how long one complete job takes; rate tells what fraction of the job is completed in one unit of time. Convert to rates, combine the rates, and only then convert back to time.
The one-line control: If A completes one job in a hours, A’s rate is 1/a job per hour. Add compatible rates for simultaneous work; take the reciprocal only after the combined rate is known.
This guide belongs to the permanent How to Perform in PSLE Learner’s Guide directory. The series began with the original PSLE Learning Guide. For broader teaching routes, use the Mathematics tuition hub. If a question depends on the current examination format, confirm it against the official SEAB PSLE formats examined in 2026; the method taught here is a reasoning routine, not a substitute for the official document.
The real problem: a familiar-looking question can still test a different decision
The trap is that every number is measured in time, so adding the times feels natural. Yet 4 hours and 6 hours describe separate completion durations, not simultaneous contributions. A worker who needs 4 hours completes 1/4 of the job per hour. A worker who needs 6 hours completes 1/6. Together they complete 1/4 + 1/6 = 5/12 of the job per hour, so one job takes 12/5 hours, or 2 hours 24 minutes.
Many Primary 6 errors in this area are not failures of memory. The learner may remember a definition, formula or classroom example, yet still apply it to the wrong cue. The important shift is from “What fact do I remember?” to “What decision does this exact wording require?” That shift protects marks because it makes the learner inspect the relationship among the words, numbers, labels or observations before answering.
A strong answer therefore has two layers. The first layer is the conclusion: the chosen word, rate, rule or datum. The second layer is the warrant: the short reason that shows why the conclusion fits this case. A conclusion without a warrant is fragile. A warrant forces the learner to check the relevant condition and makes careless switching easier to detect.
The mechanism
A work-rate model treats the whole job as 1. This “1” may represent one tank, one wall, one batch or one field. The unit must remain visible: job per hour, tank per minute, or section per day. Simultaneous positive contributions are added. A leak, drain or undoing process is subtracted only when it truly reverses the job. If a worker starts late or stops early, split the timeline and calculate the fraction completed in each interval.
The model assumes the stated rate remains constant over the relevant interval and that the contributions combine without interference. If the question says a machine slows, a pipe changes flow or two workers get in one another’s way, a constant-rate sum may not apply. Do not force the standard method onto changed conditions.
A four-step decision routine
- 1. Define one whole job. Write job = 1 and state the time unit.
- 2. Convert time to rate. A completion time of t units becomes 1/t job per unit time.
- 3. Combine only simultaneous rates. Add helpers, subtract a genuine opposing rate, and keep denominators and units correct.
- 4. Finish and verify. Use fraction done = rate × time, then confirm that the final time is reasonable and the completed fraction is exactly 1.
At first, write all four steps. After the routine becomes reliable, compress it mentally, but do not delete the checks. Speed should come from a familiar sequence, not from guessing earlier.
Worked cases: make the hidden decision visible
Case 1: 4-hour and 6-hour workers
Decision: Together they take 12/5 hours, or 2 hours 24 minutes.
Why it works: Rates are 1/4 and 1/6 job per hour; their sum is 5/12, so time is 1 ÷ 5/12. Control check: The answer must be less than 4 hours, the faster individual time.
Case 2: 3-hour and 6-hour taps
Decision: Together they fill the tank in 2 hours.
Why it works: Rates 1/3 and 1/6 sum to 1/2 tank per hour. Control check: Two hours at 1/2 per hour fills exactly one tank.
Case 3: 5-hour and 10-hour machines
Decision: Together they finish in 10/3 hours.
Why it works: Rates 1/5 + 1/10 = 3/10 job per hour; reciprocal gives 10/3 hours. Control check: 10/3 is below 5, so the direction is sensible.
Case 4: 12-minute and 18-minute printers
Decision: Together they complete a batch in 36/5 minutes.
Why it works: Rates 1/12 + 1/18 = 5/36 batch per minute. Control check: Keep the answer in minutes; 36/5 minutes is 7 minutes 12 seconds.
Case 5: One helper joins later
Decision: Compute the first worker’s fraction before the joining time, then apply the combined rate to the remainder.
Why it works: Rates can be added only during the overlap. Control check: Draw a timeline so the late worker is not credited for earlier work.
Case 6: One worker leaves early
Decision: Use the combined rate for the shared interval and the remaining worker’s rate afterwards.
Why it works: The active team changes, so the rate changes. Control check: Fractions across the intervals must total 1.
Case 7: A 4-hour inlet and 12-hour leak
Decision: Net filling time is 6 hours if both operate together.
Why it works: The inlet adds 1/4 tank per hour; the leak removes 1/12; net = 1/6. Control check: Subtract only because the leak genuinely removes filled water.
Case 8: An 8-hour inlet and 6-hour drain
Decision: The tank will not fill while both run from empty.
Why it works: Net rate 1/8 – 1/6 is negative. Control check: A negative net rate means the stated goal cannot be reached under those conditions.
Case 9: Three equal 9-hour machines
Decision: Together they take 3 hours.
Why it works: Combined rate is 3 × 1/9 = 1/3 job per hour. Control check: Three identical independent rates triple output.
Case 10: Two 8-hour workers and one 4-hour worker
Decision: Together they take 2 hours.
Why it works: Rate = 1/8 + 1/8 + 1/4 = 1/2 job per hour. Control check: The 4-hour worker contributes as much as the two slower workers combined.
Case 11: Half a job remains
Decision: At a combined rate of 3/10 job per hour, half a job takes 5/3 hours.
Why it works: Time = fraction remaining ÷ rate = 1/2 ÷ 3/10. Control check: Do not take the reciprocal as if a full job remained.
Case 12: Rate given directly
Decision: At 2/15 job per minute, a full job takes 15/2 minutes.
Why it works: Time is 1 ÷ 2/15. Control check: Keep the numerator 1 only for one whole job.
Case 13: Different job sizes
Decision: Do not add rates until both refer to the same whole job.
Why it works: One worker’s “one wall” may not equal another’s larger wall. Control check: Normalise the job or use absolute units such as square metres per hour.
Case 14: Hours and minutes mixed
Decision: Convert all times to one unit before adding rates.
Why it works: A rate per hour cannot be added directly to a rate per minute. Control check: Use 30 minutes = 1/2 hour or convert everything to minutes.
Case 15: Two workers each finish half
Decision: If A takes 4 hours for a whole job, half takes 2 hours at constant rate; if B then takes over at a 6-hour whole-job rate, the other half takes 3 hours.
Why it works: The workers act sequentially, not simultaneously. Control check: Add interval times because the work happens one after another.
Case 16: A works 1 hour before B joins
Decision: If A’s time is 4 hours and B’s is 6, A first completes 1/4. The remaining 3/4 at 5/12 per hour takes 9/5 hours.
Why it works: Separate solo and combined intervals. Control check: Total time is 1 + 9/5 = 14/5 hours.
Case 17: B leaves after 1 hour together
Decision: With 4-hour A and 6-hour B, one shared hour completes 5/12. A alone completes the remaining 7/12 in 7/3 hours.
Why it works: Use the correct active rate after B leaves. Control check: Total elapsed time is 10/3 hours.
Case 18: A leak opens halfway
Decision: Calculate the time to half-full using the inlet alone, then use the net rate for the second half.
Why it works: The opposing rate begins only at the stated event. Control check: Do not subtract the leak during the earlier interval.
Case 19: Work already completed
Decision: If 2/5 is done, the remaining job is 3/5.
Why it works: Use time = 3/5 ÷ combined rate. Control check: The starting fraction changes the numerator, not the workers’ rates.
Case 20: Compare two teams
Decision: Find each team’s combined rate before comparing completion times.
Why it works: The faster team has the larger rate but the smaller time. Control check: Do not compare sums of individual completion times.
Case 21: Unknown worker rate
Decision: If together they finish in 3 hours and A alone in 5, B’s rate is 1/3 – 1/5 = 2/15 job per hour.
Why it works: Subtract rates to isolate the contribution. Control check: B alone would take 15/2 hours.
Case 22: Unknown completion time
Decision: If B’s rate is 1/8 job per hour, B’s time is 8 hours.
Why it works: Completion time is the reciprocal of a full-job rate. Control check: Write units to prevent treating 1/8 as 8 hours without explanation.
Case 23: Output rather than job fraction
Decision: If machines make 24 and 36 parts per hour, together they make 60 parts per hour.
Why it works: Absolute rates with the same unit can be added directly. Control check: For 180 parts, time is 180 ÷ 60 = 3 hours.
Case 24: Nonconstant rate warning
Decision: If a battery weakens and output drops, a single constant rate may be invalid.
Why it works: The standard fraction model depends on stable rates. Control check: Use the stage information supplied instead of averaging without justification.
Common misconceptions and targeted repairs
Misconception 1: Add completion times
Why it fails: This combines durations for separate solo jobs and does not model simultaneous work.
Repair: Convert each time to a per-unit-time rate first.
Misconception 2: Average the two times
Why it fails: The arithmetic mean ignores unequal rates and rarely gives the joint time.
Repair: Add rates and reciprocate.
Misconception 3: Add rates with different units
Why it fails: Job per hour and job per minute are not directly compatible.
Repair: Convert to one time unit.
Misconception 4: Always subtract a drain
Why it fails: A second process may not oppose the task; it may fill another section or act at a different time.
Repair: Read the direction and timeline.
Misconception 5: Use reciprocal too early
Why it fails: Reciprocating each rate is already what produced the solo time.
Repair: Combine rates first; reciprocate the final full-job rate.
Misconception 6: Assume the answer must be a whole number
Why it fails: Work-rate results are often fractions of an hour.
Repair: Convert the fractional hour accurately when needed.
Misconception 7: Ignore the fraction already completed
Why it fails: Using 1 as the remaining job overcounts work.
Repair: Set the actual remainder before calculating time.
Misconception 8: Assume every worker keeps the same rate
Why it fails: A question may state a change, pause or slowdown.
Repair: Split the work into constant-rate stages.
Independent practice
Do these without looking back. For each item, record both the answer and one sentence naming the deciding condition. A bare answer can be lucky; the deciding condition reveals whether the method is secure.
Practice 1
A paints a fence in 6 hours and B in 3 hours. How long together?
Explained answer: Rates are 1/6 and 1/3 = 2/6, so combined rate is 1/2 fence per hour. Time is 2 hours.
Practice 2
A tap fills a tank in 10 minutes and another in 15 minutes.
Explained answer: Combined rate = 1/10 + 1/15 = 1/6 tank per minute, so time is 6 minutes.
Practice 3
Two machines each finish a batch in 14 minutes.
Explained answer: Combined rate = 2/14 = 1/7 batch per minute, so they take 7 minutes.
Practice 4
A 5-hour worker works alone for 2 hours. What fraction remains?
Explained answer: The worker completes 2 × 1/5 = 2/5, so 3/5 remains.
Practice 5
A 5-hour worker does 2 hours, then a 10-hour worker joins. How much longer?
Explained answer: After 2 hours, 3/5 remains. Combined rate is 1/5 + 1/10 = 3/10. Extra time = 3/5 ÷ 3/10 = 2 hours.
Practice 6
An inlet fills in 6 hours; a leak empties a full tank in 12 hours. Net time?
Explained answer: Net rate = 1/6 – 1/12 = 1/12 tank per hour, so 12 hours.
Practice 7
A team completes 3/8 of a job per day. Time for the whole job?
Explained answer: 1 ÷ 3/8 = 8/3 days, assuming the rate stays constant.
Practice 8
Together A and B take 4 hours. A alone takes 6. Find B’s solo time.
Explained answer: B’s rate = 1/4 – 1/6 = 1/12 job per hour, so B alone takes 12 hours.
Practice 9
A produces 18 items per hour; B produces 27. Time for 180 items together?
Explained answer: Combined rate = 45 items per hour. Time = 180 ÷ 45 = 4 hours.
Practice 10
Why must a joint time be less than the faster solo time when both contributions are positive?
Explained answer: Adding a positive rate makes the combined rate larger than either individual rate. For the same one job, a larger rate gives a smaller time.
Practice 11
A completes a job in 8 hours and B in 12 hours. Find the fraction done in 2 hours together.
Explained answer: Combined rate = 1/8 + 1/12 = 5/24. In 2 hours they complete 10/24 = 5/12.
Practice 12
The same pair has worked 2 hours. What fraction remains?
Explained answer: 1 – 5/12 = 7/12 remains.
Practice 13
A drain empties in 4 hours while an inlet fills in 8. Can the tank fill from empty with both open?
Explained answer: No. Net rate 1/8 – 1/4 is negative, so the level does not rise.
Practice 14
A 3-hour machine and 6-hour machine work for one hour, then the faster machine stops. Total time?
Explained answer: They complete 1/3 + 1/6 = 1/2 in the first hour. The 6-hour machine completes the remaining half in 3 hours. Total 4 hours.
Practice 15
Explain why 4 hours + 6 hours = 10 hours is not the joint completion time.
Explained answer: The numbers are solo durations, not amounts of work completed per hour. Simultaneous contributions must be combined as rates 1/4 and 1/6.
Practice 16
Create a reasonableness bound for workers taking 7 and 9 hours alone.
Explained answer: With both helping positively and constantly, the joint time must be below 7 hours and above 0. A result above 7 signals an error.
Delayed transfer: prove that the method survives a changed surface
Transfer the model beyond workers. A printer completing a batch, a tap filling a tank, a pump removing water and a machine producing parts all use rate × time = amount. First decide what the “whole” is. Then decide whether rates add, subtract or operate in separate stages. The story words change, but the unit analysis reveals the structure.
- Return after twenty-four hours and solve four cases in a different order. Cover the headings so the topic name cannot cue the answer.
- Return after three days and create one near-miss example: change only one condition so that the correct decision changes. Explain which condition did the work.
- Return after seven days and mix this skill with two unrelated questions. The aim is to recognise when the method is relevant, not merely to perform it when told.
- Return after fourteen days and complete one timed mini-set. Mark the answer, then mark the quality of the warrant separately. Repair any item whose conclusion was right for a weak reason.
This spacing matters because immediate repetition can feel fluent even when the learner is following the order of examples. Delayed, mixed practice removes that support. It tests whether the learner can rebuild the decision from the question itself.
Examination control
- Whole-job control: Write job = 1 before forming fractions.
- Unit control: Choose hours or minutes and convert before adding.
- Timeline control: Mark when each worker, tap or machine is active.
- Direction control: Add helpers and subtract only genuine opposing processes.
- Remainder control: Use the unfinished fraction, not always 1.
- Reasonableness control: Positive helpers must produce a time below the fastest solo time.
If time is tight, keep the shortest reliable chain: identify the target, test the decisive condition, state the answer, and run one contradiction check. Do not replace this chain with a vague feeling that an option “sounds right”.
A two-week route to independence
- Days 1–2: Copy the four-step routine and annotate six worked cases. Circle the exact word, number or observation that controls each decision.
- Days 3–4: Solve eight fresh cases untimed. Write a full warrant. Sort errors into cue, method, calculation or expression errors.
- Day 5: Build four near-miss pairs. In each pair, alter one condition and explain why the answer changes.
- Days 6–7: Mix this skill with other English, Mathematics and Science questions. Decide first whether the routine applies.
- Days 8–9: Complete a short timed set. Keep accuracy records, but also record whether every answer has a valid warrant.
- Day 10: Re-teach the method aloud without notes. Any point that cannot be explained clearly becomes the next revision target.
- Days 11–12: Attempt delayed transfer questions with unfamiliar names, values, diagrams or contexts.
- Days 13–14: Complete a final mixed set, compare it with Day 3, and write one personal examination rule beginning, “Before I answer, I will…”
Guidance for parents and tutors
Parents and tutors should resist giving the formula first. Ask the learner to explain what 1/6 job per hour means and to shade the fraction completed after one or two hours. Move from equal rates to unequal rates, then to delayed starts and leaks. If errors recur, inspect whether the learner confused time, rate or amount; each needs a different repair.
Ask, “Which condition decided that?” more often than “What is the answer?” If a learner is wrong, locate the first broken step. Re-teach only that step, then give a near-miss pair. If the learner is right, occasionally ask for a counterexample. This keeps success attached to reasoning rather than praise for speed.
Keep the record small: date, item type, first wrong step, repair attempted, and result after a delay. Avoid turning the log into a punishment. Its purpose is to reveal patterns. A learner who repeatedly misses the cue needs different support from a learner who identifies the cue but calculates or expresses the conclusion poorly.
Frequently asked questions
Why is the job set to 1?
It lets every contribution be expressed as a fraction of the same whole.
Do I always take a reciprocal?
Only when converting a full-job rate into the time for one whole job.
Can I add completion times if workers act one after another?
You may add the durations of separate sequential intervals, after finding the work assigned to each interval.
What if the rates are given in items per hour?
Add them directly if the item and time units match, then divide the required number of items by the combined rate.
What if one worker is twice as fast?
Its rate is twice as large, so its solo completion time is half as long for the same job.
Why can a leak make the answer impossible?
If the opposing rate is at least as large as the filling rate, the required amount never increases to the target.
Should I convert a fraction of an hour?
Only if the question or answer format requires it; convert accurately using 60 minutes per hour.
How do I check a result?
Multiply each active rate by its active time and confirm that all signed contributions total exactly one job.
Deep reasoning laboratory
The following investigations move beyond recognising the answer. They train you to justify, challenge and transfer the method. Work through them slowly once, then revisit selected tasks after a delay.
Investigation 1: Derive rather than memorise
If a worker completes one job in 8 hours, equal work in each hour means 1/8 per hour. This statement, not a mysterious formula, creates the rate. Re-derive it whenever memory becomes uncertain. Understanding the reciprocal prevents swapping numerator and denominator.
Investigation 2: Use a common-work diagram
Draw one bar for the entire job and partition it into a convenient common number of equal units. For 4-hour and 6-hour workers, 12 units work well: they complete 3 and 2 units per hour. Together they complete 5 of 12 units, matching 5/12 job per hour.
Investigation 3: Distinguish simultaneous from sequential
Rates add only during shared time. If A works for two hours and then B takes over, compute each interval separately and add elapsed times. If they overlap for one hour, add rates only for that hour. A timeline is often more valuable than another equation.
Investigation 4: Interpret the reciprocal
A combined rate of 7/20 job per hour means seven twentieths are completed each hour. The time for one job is 1 divided by 7/20, or 20/7 hours. The reciprocal answers “how many hours per job”; it does not erase the unit logic.
Investigation 5: Check with contribution totals
After finding a time, multiply each active rate by the time for which it operated. Add helpers and subtract opposing contributions. The signed total must equal the required fraction. This check catches late-start, early-stop and wrong-remainder errors.
Investigation 6: Choose a unit strategically
If all times are multiples of minutes, minutes may keep the arithmetic whole. If most values are in hours, convert one minute value into a fraction of an hour. The correct unit is the one that keeps every rate compatible and the work legible.
Investigation 7: Model an opposing process carefully
A leak rate is subtracted because it removes filled volume. A second worker who makes a different part is not necessarily subtracted or added to the same job fraction; the jobs may need a new definition. Decide the physical relationship before choosing a sign.
Investigation 8: Recognise impossible or steady states
When an opposing rate equals the positive rate, the net rate is zero: the amount stays constant. When it is larger, the target moves farther away. Reporting a negative completion time would hide the real conclusion that the job cannot be completed under those conditions.
Investigation 9: Use bounds before calculating
With two positive workers, the joint time is less than the faster solo time. With a leak, the fill time is longer than the inlet’s solo time if filling remains possible. Write this bound first; it becomes an immediate error detector.
Investigation 10: Solve backwards from a team rate
If together A and B finish in 4 hours, their rate is 1/4. If A’s rate is 1/10, then B’s rate is 1/4 – 1/10 = 3/20, so B takes 20/3 hours alone. Subtract contributions, not completion times.
Investigation 11: Handle partial targets
For three quarters of a tank, use amount 3/4 rather than 1. At net rate 1/8 tank per minute, the time is 3/4 ÷ 1/8 = 6 minutes. The rate stays the same; the target amount changes.
Investigation 12: State the model assumption
Constant-rate work problems assume steady output and independent contributions unless the question says otherwise. Real workers may tire and real pipes may vary, but an exam model uses the stated conditions. Mentioning the assumption helps you notice when the wording introduces a stage change.
Mixed challenge workshop
These cases combine the core skill with a second demand. Read the whole situation, state the deciding condition and then complete the answer.
Challenge 1: Two painters and different wall sizes
If A’s stated 4 hours refers to a small wall and B’s 6 hours refers to a wall twice as large, 1/4 and 1/6 are not fractions of the same job. Convert to area painted per hour or redefine a common wall before combining. Shared denominators require a shared whole.
Challenge 2: A tap opens after a delay
Suppose a 6-hour tap fills alone for 2 hours, so 1/3 of the tank is filled. A 12-hour tap then joins. The remaining 2/3 is filled at 1/6 + 1/12 = 1/4 tank per hour, requiring 8/3 hours more. Total time is 14/3 hours.
Challenge 3: A leak begins at half-full
An 8-hour inlet reaches half-full in 4 hours. If a 16-hour leak then opens, the net rate becomes 1/8 – 1/16 = 1/16 tank per hour. Filling the remaining half takes 8 more hours, so the total is 12 hours.
Challenge 4: One worker is removed
A and B have rates 1/5 and 1/10. Together for 2 hours they complete 3/5. If A leaves, B needs 4 more hours for the remaining 2/5 because 2/5 ÷ 1/10 = 4. The total is 6 hours.
Challenge 5: A team target is not one job
Three machines together produce 5/12 of a batch per hour. For 2.5 batches, time is 5/2 ÷ 5/12 = 6 hours. Keep the target amount visible; taking the reciprocal would answer only the time for one batch.
Challenge 6: Compare adding a worker with doubling time
If a worker’s rate is 1/8, two identical independent workers have rate 1/4 and halve the time to 4 hours. Doubling the time for one worker doubles the amount completed instead. Rate, time and amount respond differently.
Challenge 7: A partial-day answer
A result of 7/4 hours is 1 hour 45 minutes, not 1 hour 75 minutes. Convert the fractional hour by multiplying 3/4 by 60. Unit conversion happens after the rate reasoning and must preserve value.
Challenge 8: A net zero case
An inlet and leak each have rate 1/6 tank per hour. With both open, net rate is zero, so a partly filled tank stays at the same level under the model. The reciprocal of zero is not a time; the target cannot be reached by net change.
Challenge 9: A reasonableness contradiction
If 4-hour and 10-hour workers are said to finish together in 5 hours, reject the result before detailed checking. The 4-hour worker alone is faster than that; adding positive help cannot make the job slower. Bounds detect structural errors immediately.
Challenge 10: A complete written solution
State the whole, individual rates, active intervals, combined or net rate, fraction remaining, calculated time and final unit. This sequence lets a marker follow the model and lets you locate an error without restarting the entire question.
Self-marking protocol
Award one point for identifying the correct target, one for selecting the correct evidence or relationship, one for executing the method accurately, and one for a clear final statement with units or wording where needed. A correct final answer with a broken method should not receive full self-credit. The purpose is diagnosis, not punishment.
For every lost point, name the first failure: target, cue, method, execution or communication. Then write a repair question that changes only one feature. Reattempt it after a delay. If the same failure returns, reduce the difficulty and rebuild the missing step before doing more full questions.
Next routes
Review the earlier guide on equal-distance average speed to compare two different rate structures, then use the Mathematics tuition hub for broader problem-solving routes.
Use the complete volume directory to locate the wider sequence and the original PSLE Learning Guide for the foundation of the series. Continue only after you can solve mixed cases and explain the deciding condition without prompts.