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Move the counters into equal shares
Task reminder
Move counters to make every plate equal. Keep all 36.
Four plates hold 6, 8, 10 and 12 counters. Make the shares equal while keeping all 36 counters. Return a counter to the centre, then give it to another plate.
Still to share
Plate 1
Plate 2
Plate 3
Plate 4
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Each plate gets 9 counters. Four equal shares use the same total: 4 × 9 = 36.
The arithmetic mean, or average, is the equal share of the total:
Here, 36 ÷ 4 = 9. None of the original plates held nine counters. The mean describes the whole set; it need not be one of its values.
Count every recorded value
Check your understanding 1
Count every day
Check your understanding 1
Count every day
Four daily counts are shown. Find their total, number of observations and mean.
One observation per day
Daily counts: 0 objects; 4 objects; 4 objects; 12 objects.
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Figure data
This table is included in print while it is open.
| Set | Values |
|---|---|
| Daily counts | 0 objects; 4 objects; 4 objects; 12 objects |
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Total: 0 + 4 + 4 + 12 = 20.
Number of observations: 4.
Mean: 20 ÷ 4 = 5 objects per day.
Leaving out the zero would divide by three. Merging the repeated 4s would also change the set. Count observations, not just different values.
Find a mean between whole numbers
Two pupils read 1 book and 2 books. Find the total, number of pupils and mean number of books. Give the exact mean.
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Total: 1 + 2 = 3 books.
Mean: 3 ÷ 2 = 1½ books per pupil.
The pupils’ recorded counts stay 1 and 2. The mean is a summary, not a claim that someone read half of a book.
Task reminder
Do you need the total, the count or the mean?
Decide which quantity is missing
Six measurements have a mean of 9 cm. Each measurement contributes one share to the equal-share model. Find their total length, in centimetres.
Check your understanding 2
Which quantity is missing?
Check your understanding 2
Which quantity is missing?
Show the total checkHide answer
Total = mean × count.
6 × 9 = 54 cm. Check: 54 ÷ 6 = 9 cm.
In a different set, the total is 45 cm and the mean is 5 cm. Find the number of measurements.
Show the count checkHide answer
Count = total ÷ mean.
45 ÷ 5 = 9 measurements. Check: 9 × 5 = 45 cm.
A count of observations must be a positive whole number. If a calculation gives 2½ observations, recheck the givens and calculation; rounding changes the data.
Use the total to find a missing score
Five quiz scores have a mean of 14 marks. Four scores are 10, 12, 15 and 16 marks. Find the total of all five scores, then the missing score.
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Whole total: 5 × 14 = 70 marks.
Known part: 10 + 12 + 15 + 16 = 53 marks.
Missing score: 70 − 53 = 17 marks.
Check: (53 + 17) ÷ 5 = 14 marks. The mean is the equal share of all five scores, not the value to subtract from their known total.
Task reminder
Add a value. What happens to the total and the count?
Compare the totals before and after a change
Check your understanding 3
Find the new reading
Check your understanding 3
Find the new reading
Four length readings have mean 18 cm. A fifth reading is added and the new mean is 20 cm. Find both totals and the fifth reading, in centimetres.
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Figure data
This table is included in print while it is open.
| Set | Values |
|---|---|
| Before: four readings | 18 cm share; 18 cm share; 18 cm share; 18 cm share |
| After: five readings | 20 cm share; 20 cm share; 20 cm share; 20 cm share; 20 cm share |
Show workingHide answer
Before: 4 × 18 = 72 cm.
After: 5 × 20 = 100 cm.
Added reading: 100 − 72 = 28 cm.
Check: (72 + 28) ÷ 5 = 20 cm. Using four readings with the new mean would omit the extra observation.
Find each group’s total before combining
Check your understanding 4
Each bag must count once
Check your understanding 4
Each bag must count once
Two bags have a mean mass of 5 kg. Four other bags have a mean mass of 8 kg. Find both group totals, then the total mass, count and mean for all six bags.
Two different group sizes
Group 1: 2 bags; 5 kg mean. Group 2: 4 bags; 8 kg mean.
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Figure data
This table is included in print while it is open.
| Set | Values |
|---|---|
| Group 1 | 2 bags; 5 kg mean |
| Group 2 | 4 bags; 8 kg mean |
Show workingHide answer
Group 1: 2 × 5 = 10 kg.
Group 2: 4 × 8 = 32 kg.
Combined total: 10 + 32 = 42 kg.
Combined count: 2 + 4 = 6 bags.
Combined mean: 42 ÷ 6 = 7 kg.
Averaging just 5 and 8 gives 6.5 kg. That treats each group as one observation, rather than treating each bag equally. The combined mean should be closer to 8 kg because the larger group has that mean. Directly averaging the group means works when the groups have equal numbers of observations.
Check which numbers are means
Three bags have a mean mass of 8 kg. Two other bags have actual masses 4 kg and 10 kg. Find each group’s total, then the total, count and mean for all the bags. The two individual masses form the second group.
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Summary group: 3 × 8 = 24 kg.
Individual bags: 4 + 10 = 14 kg.
All bags: (24 + 14) ÷ (3 + 2) = 38 ÷ 5 = 7.6 kg.
The 8 kg is a mean for three bags; 4 kg and 10 kg are two actual masses. They cannot all be treated as the same kind of observation.
Task reminder
Does each number count one day or the days so far?
Running totals are not daily counts
A counter records every visitor, without being reset. These readings are taken at the same time each day. Find the number of visitors during the four days, the number of daily observations and the mean daily number of visitors.
| When | Running total |
|---|---|
| Start | 20 |
| After day 1 | 23 |
| After day 2 | 23 |
| After day 3 | 30 |
| After day 4 | 36 |
Show workingHide answer
Daily increases: 3, 0, 7 and 6 visitors.
Total during the four days: 36 − 20 = 16 visitors.
Mean: 16 ÷ 4 = 4 visitors per day.
The unchanged reading still describes one day. Averaging all five running totals would describe counter readings, not daily visitors.
The original count can be missing too
Some length readings total 18 cm, with a mean of 6 cm. One reading is added and the new mean is 7 cm. First find the original number of readings.
Now record the original total, calculate the new total and find the added reading, in centimetres.
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Original count: 18 ÷ 6 = 3.
New count: 3 + 1 = 4.
New total: 4 × 7 = 28 cm.
Added reading: 28 − 18 = 10 cm.
Check: (18 + 10) ÷ 4 = 7 cm. Subtracting the means would compare equal shares, without accounting for the extra reading.
Is the mean a complete description?
Compare these two sets
Set A: 0; 0; 12. Set B: 4; 4; 4.
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Figure data
This table is included in print while it is open.
| Set | Values |
|---|---|
| Set A | 0; 0; 12 |
| Set B | 4; 4; 4 |
Find each mean. What difference between the sets does the mean hide? The values are all zero or more. If you know only total 0, mean 0, can you find their count?
Show answerHide answer
Both totals are 12 across three values, so both means are 4. Set A has two zeros and a much larger value; Set B is constant. The mean does not show that variation or recover the individual values.
A total of zero and mean of zero could describe one zero, three zeros or any positive number of zeros. The count is not determined: 0 ÷ 0 cannot tell you it.
For a grown-up
Pip’s synthetic voice uses the University of Edinburgh’s Alba speech dataset, via Piper’s Alba voice, under CC BY 4.0. Audio is optional.
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Syllabus and review details
- Primary 6 Mathematics 2026 · 2021-updated-2025-10
- Primary 6 Foundation Mathematics 2026 · 2021-updated-2025-10
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