Find the exact learning problem
Two group averages look like two ordinary numbers, so it is tempting to add them and divide by two. That shortcut gives each group equal influence even when one group contains many more items. A combined average must give every original item one equal share of the total. For each group, total = average × number of items. Add the group totals, add the group counts, then divide combined total by combined count. The simple average of two group averages is valid only when the two group sizes are equal, or in special cases where both averages are already equal.
This guide belongs to the complete PSLE Learner’s Guide volume directory. The series foundation remains at the original PSLE Learning Guide, while current routes continue through the Mathematics hub. For current paper information, consult the official SEAB PSLE formats examined in 2026.
Understand the governing mechanism
For each group, total = average × number of items. Add the group totals, add the group counts, then divide combined total by combined count. The simple average of two group averages is valid only when the two group sizes are equal, or in special cases where both averages are already equal. The goal is not to memorise one answer. It is to recognise the relationship that makes the answer necessary, then use a different check to confirm it.
Use this four-step routine
- Step 1. Record each group’s count and average in aligned columns.
- Step 2. Recover each hidden total by multiplying average by count.
- Step 3. Add totals and counts separately; never mix a count with a value total.
- Step 4. Divide combined total by combined count, attach the correct unit and check that the result lies between the group averages when counts are positive.
Practise the full routine until each step has a clear purpose. Under examination conditions, compress it into a short note that preserves the decisive evidence.
Worked examples and targeted repairs
Worked example 1
Situation. 10 pupils average 60 and 30 pupils average 80. Reasoning. Totals are 600 and 2400. The combined average is 3000÷40=75, not 70, because the larger group has greater weight. Tempting error. Averaging 60 and 80 without using group sizes. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 2
Situation. 20 pupils average 60 and 20 pupils average 80. Reasoning. Equal counts give equal weights, so (60+80)÷2=70. Rebuilding totals also gives 2800÷40=70. Tempting error. Treating this equal-size shortcut as a universal rule. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 3
Situation. 8 books average 120 pages and 12 books average 180 pages. Reasoning. The totals are 960 and 2160 pages. Combined average is 3120÷20=156 pages. Tempting error. Adding page averages and dividing by the total book count. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 4
Situation. A group of 5 has average 14; a group of 15 has average 18. Reasoning. Totals are 70 and 270. Combined mean is 340÷20=17, closer to 18 because more values come from that group. Tempting error. Expecting the result to be exactly halfway between 14 and 18. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 5
Situation. Two groups both average 72 but have different sizes. Reasoning. Every group total is 72 times its count, so the combined total is 72 times the combined count. The combined average remains 72. Tempting error. Thinking unequal sizes always prevent averaging group means. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 6
Situation. A class average rises after joining a group with a higher average. Reasoning. The new average moves toward the joining group’s mean. The amount of movement depends on its size. Tempting error. Assuming any higher group makes the result equal to its average. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 7
Situation. A group average is 50 and another is 70; combined average is 65. Reasoning. The result is closer to 70, so the 70-average group must have more members, assuming both counts are positive. Tempting error. Inferring equal group sizes from two displayed averages. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 8
Situation. 6 scores average 12 and 4 scores average 7. Reasoning. Totals are 72 and 28; combined average is 100÷10=10. Tempting error. Computing (12+7)÷2=9.5. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 9
Situation. A 12-item group averages 25; three more values each equal 25 are added. Reasoning. The added values match the current average, so total and count grow proportionally and the average stays 25. Tempting error. Recomputing from incomplete data or assuming every addition changes the mean. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 10
Situation. A 12-item group averages 25; three values averaging 31 are added. Reasoning. Original total is 300; added total is 93. Combined average is 393÷15=26.2. Tempting error. Adding 31 directly to 25. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 11
Situation. Group A has 9 items averaging 40; Group B has 1 item averaging 100. Reasoning. Combined total is 360+100=460 over 10 items, giving 46. The single high value influences but does not dominate. Tempting error. Reporting 70 from the average of averages. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 12
Situation. Group A has 1 item averaging 40; Group B has 9 items averaging 100. Reasoning. Combined average is (40+900)÷10=94, showing how reversing counts reverses the weighting. Tempting error. Keeping the same answer because the two group averages are unchanged. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 13
Situation. A team’s five-day average is 18 units and its next five-day average is 22 units. Reasoning. The intervals contain equal numbers of days, so the ten-day average is 20 units. Tempting error. Using the shortcut without first confirming equal counts. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 14
Situation. A team averages 18 over five days and 22 over ten days. Reasoning. Totals are 90 and 220; combined average is 310÷15=20⅔ units. Tempting error. Calling the two time blocks equal because both have one stated average. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 15
Situation. Three groups have averages 10, 20 and 30 with equal counts. Reasoning. Equal counts allow the simple mean (10+20+30)÷3=20. Tempting error. Assuming the same shortcut for unequal counts. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 16
Situation. Three groups have counts 2, 3 and 5 with averages 10, 20 and 30. Reasoning. Totals are 20, 60 and 150; combined average is 230÷10=23. Tempting error. Dividing the sum of averages by three. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 17
Situation. A class of 24 averages 68; after 6 pupils join, the average is 70. Reasoning. Original total is 1632; new total is 2100. The joining pupils total 468 and average 78. Tempting error. Subtracting averages to claim newcomers averaged 2. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 18
Situation. A group of 30 averages 70; 5 pupils averaging 64 leave. Reasoning. Original total is 2100; departing total 320; remaining total 1780 over 25, so average 71.2. Tempting error. Subtracting 64 from 70. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 19
Situation. A combined average equals the lower group average although the other group average is higher. Reasoning. With positive group sizes, this cannot happen unless the higher group has zero members or data are inconsistent. Tempting error. Accepting an impossible result without a range check. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 20
Situation. A combined average is 84 from groups averaging 78 and 90. Reasoning. The result can be valid, but counts determine the position. If counts are equal, 84 follows immediately. Tempting error. Assuming any number between the means is automatically correct. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 21
Situation. Average mass uses grams and counts objects. Reasoning. Multiply grams per object by number of objects to recover grams. The combined result remains grams per object. Tempting error. Treating number of objects as another mass. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 22
Situation. Average speed over two journeys is considered. Reasoning. A simple mean of speeds is not generally valid; weighting depends on time or distance structure. Rebuild the appropriate total relationship rather than transferring the data-average rule blindly. Tempting error. Averaging any two quantities labelled average. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 23
Situation. Averages are rounded before combining. Reasoning. Reconstructed totals may be approximate. State or preserve enough precision and avoid claiming an exact combined mean from rounded inputs. Tempting error. Multiplying rounded means and reporting excessive precision. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 24
Situation. One group count is missing but combined average is given. Reasoning. Let the missing count be a variable, form total equations and solve while preserving whole-number constraints. Tempting error. Guessing the count from how close the combined mean looks. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 25
Situation. Group A: 12 pupils average 75; Group B: unknown count average 85; combined 80. Reasoning. Equal distance between 75 and 85 implies equal total weights here, so Group B also has 12 pupils. Algebra confirms (900+85n)/(12+n)=80. Tempting error. Saying midpoint always means equal counts without checking two-group conditions. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 26
Situation. Group A: 10 items average 30; combined with a 50-average group gives 45. Reasoning. The combined mean lies one quarter of the way below 50 but three quarters above 30, so the higher group has three times the count: 30 items. Tempting error. Reversing the weight ratio. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 27
Situation. The combined mean is requested to one decimal place. Reasoning. Carry exact totals through the division and round only the final combined result. Tempting error. Rounding each group total or intermediate quotient early. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 28
Situation. A table lists average and count for four classes. Reasoning. Treat each row as a total contribution, sum average×count across rows, then divide by total count. Tempting error. Averaging the four averages because there are four rows. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Worked example 29
Situation. A school average is compared across years with different cohort sizes. Reasoning. Each year’s overall mean is valid for that cohort, but combining years requires cohort counts. Trend comparison and pooled average are different questions. Tempting error. Pooling annual means without enrolment numbers. Repair. Identify the controlling relationship, state it in one sentence and verify it with the routine before moving on.
Misconceptions that survive easy practice
A weak shortcut can appear successful when an example does not challenge it. Build contrast pairs instead: keep most details the same, change the controlling relationship, and explain why the answer changes. This makes the rule available when the surface wording is unfamiliar.
- Separate what is directly stated or observed from what is inferred.
- Use the smallest decisive cue rather than every nearby detail.
- Check scope: identify exactly which clause, group, path, row, axis or diagram region a cue governs.
- Preserve quantities and units when rebuilding hidden totals or tracing a physical mechanism.
- Reject a tempting alternative with a specific reason instead of saying it merely looks wrong.
- Finish with a complete sentence that answers the requested question, not just a calculation or label.
Independent practice with explained answers
Practice 1
Task. 10 values average 8; 20 values average 11. Combined average? Answer. 10 Why. Totals 80 and 220 give 300÷30. Before reading the answer, write the decisive cue. Then cover the explanation and reproduce it in a shorter form.
Practice 2
Task. 15 values average 12; 15 values average 18. Combined average? Answer. 15 Why. Equal group sizes permit the mean of the two averages. Before reading the answer, write the decisive cue. Then cover the explanation and reproduce it in a shorter form.
Practice 3
Task. 4 items average 30; 6 items average 50. Combined average? Answer. 42 Why. Totals 120 and 300 give 420÷10. Before reading the answer, write the decisive cue. Then cover the explanation and reproduce it in a shorter form.
Practice 4
Task. Why is 40 wrong in the previous task? Answer. It gives both groups equal weight. Why. The 50-average group has more items. Before reading the answer, write the decisive cue. Then cover the explanation and reproduce it in a shorter form.
Practice 5
Task. Two unequal groups both average 25. Combined average? Answer. 25 Why. The common mean remains unchanged regardless of positive counts. Before reading the answer, write the decisive cue. Then cover the explanation and reproduce it in a shorter form.
Practice 6
Task. A group of 8 averages 14. Find its total. Answer. 112 Why. Average × count = total. Before reading the answer, write the decisive cue. Then cover the explanation and reproduce it in a shorter form.
Practice 7
Task. A total of 360 belongs to 12 items. Find average. Answer. 30 Why. Total ÷ count = average. Before reading the answer, write the decisive cue. Then cover the explanation and reproduce it in a shorter form.
Practice 8
Task. Group means are 40 and 70; combined mean 60. Which group is larger? Answer. The 70-mean group. Why. The result lies closer to 70. Before reading the answer, write the decisive cue. Then cover the explanation and reproduce it in a shorter form.
Practice 9
Task. 20 pupils average 65; 5 pupils averaging 85 join. New average? Answer. 69 Why. Totals 1300 and 425 give 1725÷25. Before reading the answer, write the decisive cue. Then cover the explanation and reproduce it in a shorter form.
Practice 10
Task. 30 pupils average 72; 6 pupils averaging 62 leave. New average? Answer. 74.5 Why. Original total 2160 minus 372 leaves 1788, and 1788÷24=74.5. Before reading the answer, write the decisive cue. Then cover the explanation and reproduce it in a shorter form.
Practice 11
Task. Why should the new average be above 72 in the previous situation? Answer. The departing pupils averaged below the original mean. Why. Removing below-average values raises the mean of the values that remain. Before reading the answer, write the decisive cue. Then cover the explanation and reproduce it in a shorter form.
Practice 12
Task. Three equal-size groups average 18, 21 and 27. Combined? Answer. 22 Why. Their means may be averaged because counts are equal. Before reading the answer, write the decisive cue. Then cover the explanation and reproduce it in a shorter form.
Practice 13
Task. Counts 1, 2 and 3 have means 18, 21 and 27. Combined? Answer. 23.5 Why. Weighted total is 18+42+81=141 over 6. Before reading the answer, write the decisive cue. Then cover the explanation and reproduce it in a shorter form.
Practice 14
Task. Can a combined mean of positive groups exceed both group means? Answer. No. Why. A weighted mean lies between the contributing means. Before reading the answer, write the decisive cue. Then cover the explanation and reproduce it in a shorter form.
Practice 15
Task. A 12-pupil class average rises from 70 to 72 after 3 join. What total did newcomers score? Answer. 240 Why. The new total is 15×72=1080 and the old total is 12×70=840, so the newcomers contributed 1080−840=240. Before reading the answer, write the decisive cue. Then cover the explanation and reproduce it in a shorter form.
Practice 16
Task. Check the newcomer total by recombining it with the old class total. Answer. 840+240=1080. Why. Dividing 1080 by 15 returns the stated new average of 72. Before reading the answer, write the decisive cue. Then cover the explanation and reproduce it in a shorter form.
Practice 17
Task. The 3 newcomers’ average? Answer. 80 Why. Their total 240 divided by 3. Before reading the answer, write the decisive cue. Then cover the explanation and reproduce it in a shorter form.
Practice 18
Task. Write the combined-average formula in words. Answer. Add all group totals, then divide by all group counts. Why. Averages must first be converted back to totals. Before reading the answer, write the decisive cue. Then cover the explanation and reproduce it in a shorter form.
Delayed transfer
Wait at least one day before the next attempt. Retrieve the four-step routine from memory, then solve one familiar item, one item with the same mechanism hidden inside different wording, and one item containing irrelevant details. Mark whether an error came from reading, concept choice, execution or checking.
Next, design a near-miss example that would trap the shortcut described in this guide. Change only one controlling condition. Explain both versions to another person without using the answer key. If the listener cannot see why the answers differ, improve the contrast.
Examination control
- First pass: answer direct items and mark those where two possibilities remain.
- Second pass: apply the topic routine only to marked items; do not restart the whole paper.
- Use a different check from the method that produced the answer.
- Keep exact values, clause boundaries, path traces and label scopes visible until the final step.
- Before submission, verify that the final statement includes the requested form, unit, relationship or evidence.
A correction is trustworthy only when you can name the reason for changing the answer. Do not switch because the alternative looks more formal, more complicated or more familiar.
A graded two-week route
- Days 1–2: study the mechanism and annotate worked examples.
- Days 3–4: complete contrast pairs with the guide open.
- Days 5–6: close the guide and explain each decision in one sentence.
- Day 7: complete a short timed set and classify every error by cause.
- Days 8–10: repair the weakest error class with new examples.
- Days 11–12: mix this skill with unrelated topics so the cue is not announced.
- Days 13–14: complete delayed transfer, teach one example aloud and schedule the next retrieval.
For parents and tutors
Ask diagnostic questions before supplying an answer: “Which word, number, component or visual cue controls this?”, “What assumption are you using?”, “What would change the answer?”, and “How could you check it differently?” These questions show whether the learner owns the mechanism.
Keep an error log organised by cause. A page labelled only ‘grammar’, ‘average’, ‘circuit’ or ‘diagram’ is too broad. Record the exact failed decision and the repair cue. Short, precise entries create reusable examination control.
Frequently asked questions
Why does the larger group have more influence?
The combined mean gives every original item equal weight. A larger group contributes more items and therefore more of the total.
When may I average two averages directly?
When the group counts are equal, or when both means are identical. Otherwise rebuild totals.
Why must the answer lie between the group averages?
A positive weighted average is a balance of the contributing means; no group can pull it beyond both extremes.
What if an average was rounded?
The recovered total is approximate. Retain sensible precision and describe the pooled result accordingly.
Does this method work for average speed?
Only after identifying the correct underlying totals. Average speed has distance and time structure, so do not blindly weight by number of journeys.
What is the best examination check?
Confirm the unit, verify the answer lies in range, and decide whether it is closer to the mean of the larger group.
Next routes
Use the complete volume directory to choose the next precise learning job, revisit the original PSLE Learning Guide for the foundation, or continue through the Mathematics hub. Check the official SEAB examination-format page whenever a current assessment detail matters.
Retrieval extension 1
Return to this case without viewing its earlier solution: 10 pupils average 60 and 30 pupils average 80. State the controlling relationship, then justify the correct interpretation: Totals are 600 and 2400. The combined average is 3000÷40=75, not 70, because the larger group has greater weight. Reject the trap explicitly: Averaging 60 and 80 without using group sizes. Finally, alter one condition so that the answer or conclusion changes. Name that changed condition and explain why it matters. This variation is not extra decoration; it tests whether the mechanism can transfer beyond memorised wording.
Retrieval extension 2
Return to this case without viewing its earlier solution: 20 pupils average 60 and 20 pupils average 80. State the controlling relationship, then justify the correct interpretation: Equal counts give equal weights, so (60+80)÷2=70. Rebuilding totals also gives 2800÷40=70. Reject the trap explicitly: Treating this equal-size shortcut as a universal rule. Finally, alter one condition so that the answer or conclusion changes. Name that changed condition and explain why it matters. This variation is not extra decoration; it tests whether the mechanism can transfer beyond memorised wording.
Retrieval extension 3
Return to this case without viewing its earlier solution: 8 books average 120 pages and 12 books average 180 pages. State the controlling relationship, then justify the correct interpretation: The totals are 960 and 2160 pages. Combined average is 3120÷20=156 pages. Reject the trap explicitly: Adding page averages and dividing by the total book count. Finally, alter one condition so that the answer or conclusion changes. Name that changed condition and explain why it matters. This variation is not extra decoration; it tests whether the mechanism can transfer beyond memorised wording.
Retrieval extension 4
Return to this case without viewing its earlier solution: A group of 5 has average 14; a group of 15 has average 18. State the controlling relationship, then justify the correct interpretation: Totals are 70 and 270. Combined mean is 340÷20=17, closer to 18 because more values come from that group. Reject the trap explicitly: Expecting the result to be exactly halfway between 14 and 18. Finally, alter one condition so that the answer or conclusion changes. Name that changed condition and explain why it matters. This variation is not extra decoration; it tests whether the mechanism can transfer beyond memorised wording.
Retrieval extension 5
Return to this case without viewing its earlier solution: Two groups both average 72 but have different sizes. State the controlling relationship, then justify the correct interpretation: Every group total is 72 times its count, so the combined total is 72 times the combined count. The combined average remains 72. Reject the trap explicitly: Thinking unequal sizes always prevent averaging group means. Finally, alter one condition so that the answer or conclusion changes. Name that changed condition and explain why it matters. This variation is not extra decoration; it tests whether the mechanism can transfer beyond memorised wording.
Retrieval extension 6
Return to this case without viewing its earlier solution: A class average rises after joining a group with a higher average. State the controlling relationship, then justify the correct interpretation: The new average moves toward the joining group’s mean. The amount of movement depends on its size. Reject the trap explicitly: Assuming any higher group makes the result equal to its average. Finally, alter one condition so that the answer or conclusion changes. Name that changed condition and explain why it matters. This variation is not extra decoration; it tests whether the mechanism can transfer beyond memorised wording.
Retrieval extension 7
Return to this case without viewing its earlier solution: A group average is 50 and another is 70; combined average is 65. State the controlling relationship, then justify the correct interpretation: The result is closer to 70, so the 70-average group must have more members, assuming both counts are positive. Reject the trap explicitly: Inferring equal group sizes from two displayed averages. Finally, alter one condition so that the answer or conclusion changes. Name that changed condition and explain why it matters. This variation is not extra decoration; it tests whether the mechanism can transfer beyond memorised wording.
Retrieval extension 8
Return to this case without viewing its earlier solution: 6 scores average 12 and 4 scores average 7. State the controlling relationship, then justify the correct interpretation: Totals are 72 and 28; combined average is 100÷10=10. Reject the trap explicitly: Computing (12+7)÷2=9.5. Finally, alter one condition so that the answer or conclusion changes. Name that changed condition and explain why it matters. This variation is not extra decoration; it tests whether the mechanism can transfer beyond memorised wording.
Retrieval extension 9
Return to this case without viewing its earlier solution: A 12-item group averages 25; three more values each equal 25 are added. State the controlling relationship, then justify the correct interpretation: The added values match the current average, so total and count grow proportionally and the average stays 25. Reject the trap explicitly: Recomputing from incomplete data or assuming every addition changes the mean. Finally, alter one condition so that the answer or conclusion changes. Name that changed condition and explain why it matters. This variation is not extra decoration; it tests whether the mechanism can transfer beyond memorised wording.
Retrieval extension 10
Return to this case without viewing its earlier solution: A 12-item group averages 25; three values averaging 31 are added. State the controlling relationship, then justify the correct interpretation: Original total is 300; added total is 93. Combined average is 393÷15=26.2. Reject the trap explicitly: Adding 31 directly to 25. Finally, alter one condition so that the answer or conclusion changes. Name that changed condition and explain why it matters. This variation is not extra decoration; it tests whether the mechanism can transfer beyond memorised wording.
Retrieval extension 11
Return to this case without viewing its earlier solution: Group A has 9 items averaging 40; Group B has 1 item averaging 100. State the controlling relationship, then justify the correct interpretation: Combined total is 360+100=460 over 10 items, giving 46. The single high value influences but does not dominate. Reject the trap explicitly: Reporting 70 from the average of averages. Finally, alter one condition so that the answer or conclusion changes. Name that changed condition and explain why it matters. This variation is not extra decoration; it tests whether the mechanism can transfer beyond memorised wording.
Retrieval extension 12
Return to this case without viewing its earlier solution: Group A has 1 item averaging 40; Group B has 9 items averaging 100. State the controlling relationship, then justify the correct interpretation: Combined average is (40+900)÷10=94, showing how reversing counts reverses the weighting. Reject the trap explicitly: Keeping the same answer because the two group averages are unchanged. Finally, alter one condition so that the answer or conclusion changes. Name that changed condition and explain why it matters. This variation is not extra decoration; it tests whether the mechanism can transfer beyond memorised wording.
Retrieval extension 13
Return to this case without viewing its earlier solution: A team’s five-day average is 18 units and its next five-day average is 22 units. State the controlling relationship, then justify the correct interpretation: The intervals contain equal numbers of days, so the ten-day average is 20 units. Reject the trap explicitly: Using the shortcut without first confirming equal counts. Finally, alter one condition so that the answer or conclusion changes. Name that changed condition and explain why it matters. This variation is not extra decoration; it tests whether the mechanism can transfer beyond memorised wording.
Retrieval extension 14
Return to this case without viewing its earlier solution: A team averages 18 over five days and 22 over ten days. State the controlling relationship, then justify the correct interpretation: Totals are 90 and 220; combined average is 310÷15=20⅔ units. Reject the trap explicitly: Calling the two time blocks equal because both have one stated average. Finally, alter one condition so that the answer or conclusion changes. Name that changed condition and explain why it matters. This variation is not extra decoration; it tests whether the mechanism can transfer beyond memorised wording.
Retrieval extension 15
Return to this case without viewing its earlier solution: Three groups have averages 10, 20 and 30 with equal counts. State the controlling relationship, then justify the correct interpretation: Equal counts allow the simple mean (10+20+30)÷3=20. Reject the trap explicitly: Assuming the same shortcut for unequal counts. Finally, alter one condition so that the answer or conclusion changes. Name that changed condition and explain why it matters. This variation is not extra decoration; it tests whether the mechanism can transfer beyond memorised wording.