BTT Mathematics / Primary Mathematics Learning Hub / Excess and Shortage
An excess-and-shortage problem compares two complete plans for the same fixed collection. One plan may leave items over. Another may require more items than are available. The gap between the two plans is created repeatedly across every group, so the difference can reveal the number of groups.
Suppose some counters are shared among an unknown number of children. If each child receives four counters, six counters are left. If each child should receive five counters, eight more counters are needed. Moving from four per child to five per child requires one extra counter for every child.
The first plan has six spare counters and the second is eight counters short. The two plans differ by fourteen counters altogether. Since the change is one extra counter per child, there must be 14 children. The original number of counters is fourteen times four plus six, or 62 counters. Check the other plan: fourteen times five is seventy, which is eight more than sixty-two.
The method is related to the Unitary Method: each group’s requirement changes by a fixed amount. It is also related to the Assumption Method: two complete imagined arrangements are compared through a constant per-item difference.
Use the MOE Primary curriculum and syllabus page with the learner’s current school scope. The examples and 24 questions below are original teaching material and do not reproduce school or examination questions.
Record the two plans · Explain the gap · Different excess/shortage cases · Worked examples · Feasibility checks · 24 questions · Worked solutions
1. Treat each statement as a complete sharing plan
If every child gets four counters and six remain, the total collection can be written as:
4 counters × number of children + 6 spare counters.
If every child gets five counters and eight more are needed, the same collection can be written as:
5 counters × number of children − 8 counters.
Both expressions describe the same original total. The second uses minus eight because the planned requirement exceeds the available collection by eight.
Excess means available exceeds required
If twelve chairs are left after seating everyone three to a table, the available chair count exceeds the chairs used in that plan by twelve. The remainder belongs to the available total.
Shortage means required exceeds available
If nine more chairs are needed to seat everyone four to a table, the four-per-group plan requires nine more than are currently available. Do not add the shortage when reconstructing the available total from the requirement; subtract it.
Use signs only after meaning is clear
For learners comfortable with algebra, an excess of six can be written +6 and a shortage of eight as −8 relative to a plan. For others, the sentences “six available beyond the plan” and “eight needed beyond the collection” are sufficient.
The important step is keeping the fixed collection and the imagined requirements separate. The collection does not gain eight counters merely because one plan is short by eight.
2. The plan difference repeats once for each group
Suppose Plan A gives four items per child and Plan B gives six. Plan B requires two more items per child. If there are n children, the requirement increases by 2n items.
That increase must also equal the difference between the two plan totals. If Plan A leaves ten and Plan B is short fourteen, the plans are twenty-four items apart: ten spare must be used first, then another fourteen items would still be required.
Therefore 2n=24, giving twelve children.
Why excess plus shortage is sometimes used
When one plan is above the available total and the other below it, the distance between the plans crosses the actual collection. The two distances add.
For example, one plan uses six fewer than available while another requires eight more than available. The total plan-to-plan difference is fourteen.
When both plans have excess, subtract the excesses
If giving three per child leaves twenty and giving five per child leaves four, the second plan uses sixteen more items from the same collection. Its per-child requirement is two greater, so 16 ÷ 2 = eight children.
Adding twenty and four would be wrong because both plan requirements lie on the same side of the fixed total.
When both plans have shortage, subtract the shortages
If giving six per child is short by eighteen and giving eight per child is short by forty, the eight-per-child plan needs twenty-two more items than the six-per-child plan. Each child accounts for two of that increase, so there are eleven children.
Draw a number-line picture of requirements
Place the available total in the middle. A plan with excess lies below it because it requires fewer items; a plan with shortage lies above it because it requires more. The plan-to-plan distance is then visible as either a sum across the available total or a difference on the same side.
3. Four common cases share one underlying relationship
Case A: Excess then shortage
Four each leaves six; five each is short eight. The gap between plans is 14 and the per-group increase is one, so there are fourteen groups.
Case B: Shortage then excess
Giving seven each is short twelve, but giving five each leaves ten. The plan difference is twenty-two. The per-group difference is two, so there are eleven groups.
The order in which the statements are printed does not matter. Compare the smaller and larger per-group requirements and identify where each sits relative to the available total.
Case C: Two excesses
Giving three each leaves twenty-three; giving five each leaves seven. The second plan uses sixteen more items. Two more are used per group, so eight groups are present.
Case D: Two shortages
Giving five each is short nine; giving eight each is short thirty. The larger plan needs twenty-one more items. Three more are needed per group, so seven groups are present.
One equation behind all four cases
If the same total T is represented by p items per group plus an adjustment a, and q items per group plus adjustment b, then:
pn+a = qn+b.
Rearranging gives (q−p)n = a−b. The signs of a and b encode excess or shortage. Formal algebra is optional; the verbal plan-gap method expresses the same relationship.
The quotient must represent a whole group count where the context requires it
If a calculation produces 7.5 children, something is inconsistent with a whole-child sharing problem. Do not round to eight. Recheck the plan gap, per-group difference and printed conditions.
4. Worked examples across sharing, seating, storage and money
Example A: Counters
If each child gets three counters, eleven remain. If each gets five, thirteen more are needed. The plan gap is eleven plus thirteen = twenty-four. The per-child increase is two, so there are twelve children.
Total counters=12×3+11=47. Check the second plan: 12×5=60, exactly thirteen more than forty-seven.
Example B: Chairs per table
A hall has a fixed number of chairs. If four chairs are placed at each table, sixteen chairs remain. If six are placed at each table, no chairs remain. The plan gap is sixteen and the per-table increase is two, giving 8 tables.
Total chairs=8×6=48. The four-per-table plan uses thirty-two and leaves sixteen.
Example C: Two excesses
A collection of books can be placed on shelves. At eight books per shelf, twenty-six books remain. At ten books per shelf, eight remain. The second plan uses eighteen more books, two more per shelf. There are 9 shelves.
Total books=9×8+26=98. Check: 9×10=90, leaving eight.
Example D: Two shortages
A teacher wants to place exercise books into bundles. At six books per bundle, fourteen more books are needed. At nine per bundle, thirty-eight more are needed. The second plan requires twenty-four more books, three per bundle, so there are 8 bundles.
The available collection is 8×6−14=34 books. Check the other requirement: 8×9=72, which is thirty-eight more than thirty-four.
Example E: Rows and seats
If pupils sit five in each row, four seats are empty. If they sit four in each row, seven pupils have no seat under a plan using the same number of rows and one seat place per intended occupant. Compare the total seat requirements: the five-per-row plan has four excess seat places while the four-per-row plan is short seven.
The per-row difference is one, and the plan gap is eleven, so there are 11 rows. The number of pupils is 11×5−4=51. Check: 11×4=44, leaving seven pupils without seats.
Example F: Money distributed equally
An illustrative fund is divided among a fixed number of groups. Giving each group $12 leaves $18. Giving each $15 would require $9 more. The plan gap is $27. Each group accounts for a $3 increase, so there are 9 groups.
The fund is 9×$12+$18=$126. The $15 plan requires $135, which is nine dollars more.
Example G: Find an excess rather than the group count
There are ten teams and eighty-six badges. If each team receives seven badges, the excess is 86−70=16 badges. If each team is intended to receive nine, the shortage is 90−86=4 badges.
The excess and shortage differ because the per-team change of two acts across ten teams, creating a twenty-badge plan gap. Sixteen plus four equals twenty.
Example H: Reconstruct from one plan once the group count is known
A problem’s first stage reveals fifteen groups. A second statement says giving eight items per group leaves five. The total is therefore 15×8+5=125. There is no need to use the other plan again except as a check.
Choose the simpler reconstruction. Two plan descriptions should agree; they are not both required for the final arithmetic once the group count has been found.
5. Check whether the conditions can describe one fixed collection
The larger per-group plan must require more items
If a problem says five per child leaves ten, while seven per child leaves twenty, the statements cannot both describe the same positive number of children. The larger per-child plan cannot leave an even larger excess when the total collection is fixed.
Algebra confirms the contradiction: 5n+10=7n+20 would give n=−5. A negative group count exposes the inconsistent data rather than producing a meaningful answer.
Plan differences must be divisible by the per-group change
If four each leaves five and six each is short eight, the plan gap is thirteen. Each group accounts for two. Thirteen divided by two gives 6.5 groups, impossible in a whole-group context.
Do not round. State that the conditions are inconsistent under the stated whole-group model.
The reconstructed total must satisfy both plans
Even after obtaining a whole group count, substitute it into both descriptions. A copied excess or shortage may have been attached to the wrong plan.
Check whether zero excess or zero shortage is allowed
A plan can fit exactly, giving neither excess nor shortage. Treat this as an adjustment of zero. Comparing it with another plan often makes the structure especially clear.
Watch for different numbers of groups
The method assumes the same group count in both plans. If one statement uses twelve tables and another uses fifteen tables, the repeated difference is not acting across one fixed number of groups. Model the two situations separately.
Watch for changing group membership
If pupils join or leave between the two seating plans, the number of people is no longer fixed. An excess/shortage shortcut based on one unchanged collection does not apply without accounting for that change.
6. Practice: 24 questions
For each question, write the two complete plans, the plan gap and the per-group difference. Keep the solutions covered until the first attempt is complete.
Questions 1–8: One excess and one shortage
1. Giving 4 counters to each child leaves 7 counters. Giving 5 to each child is short 6 counters. Find the number of children and total counters.
2. Giving 3 stickers to each pupil leaves 12. Giving 5 each is short 10. Find the number of pupils and stickers.
3. Placing 6 books on each shelf leaves 5. Placing 8 on each shelf is short 9. Find the number of shelves and books.
4. Giving each group $9 leaves $20. Giving each group $13 is short $12. Find the number of groups and the fund.
5. Seating 5 pupils in each row leaves 8 seats empty. Seating 4 per row leaves 6 pupils without seats. Find the number of rows and pupils.
6. Packing 7 items per box leaves 4 items. Packing 9 per box is short 10 items. Find the number of boxes and total items.
7. Giving 12 cards per team leaves 15 cards. Giving 15 per team is short 6 cards. Find teams and cards.
8. Putting 10 balls in each basket leaves 11 balls. Putting 12 in each basket is short 5 balls. Find baskets and balls.
Questions 9–16: Two excesses, two shortages and exact fits
9. Giving 3 items per group leaves 24. Giving 5 per group leaves 8. Find the number of groups and total items.
10. Giving 7 books per shelf leaves 31. Giving 10 per shelf leaves 7. Find shelves and books.
11. Giving 5 counters per child is short 9. Giving 8 per child is short 30. Find children and available counters.
12. Giving 4 cards per team is short 10. Giving 6 per team is short 24. Find teams and available cards.
13. Seating 4 people at each table leaves 18 seats unused. Seating 6 at each table fits exactly. Find tables and people.
14. Packing 8 items per box leaves 20. Packing 10 per box fits exactly. Find boxes and items.
15. Giving $14 per group fits exactly. Giving $17 per group would be short $27. Find the number of groups and fund.
16. Giving 9 counters per child is short 28. Giving 7 counters per child fits exactly. Find children and counters.
Questions 17–24: Reverse and feasibility reasoning
17. There are 12 groups and 93 items. Find the excess if each gets 7, and the shortage if each is to get 8.
18. There are 15 rows and 97 pupils. Find the number of empty seats at 7 seats per row, and the shortage at 6 seats per row.
19. Giving 4 per group leaves 5; giving 6 per group is short 8. Can both statements describe a whole number of groups? Explain.
20. Giving 5 per group leaves 10; giving 7 per group leaves 20. Can both statements describe one fixed collection? Explain.
21. Giving 8 per group leaves 19; giving 11 per group leaves 1. Find groups and total items.
22. Giving 6 per group is short 17; giving 9 per group is short 41. Find groups and available items.
23. A solution finds 14 groups. One plan gives 9 per group and leaves 3. Find the total and determine the shortage under a plan of 11 per group.
24. Explain why an excess of twelve and a shortage of eight produce a twenty-item gap between two plans, not a four-item gap, when the plans lie on opposite sides of the available total.
7. Worked solutions
Solutions 1–8
1. 13 children; 59 counters. Plan gap=7+6=13. Per-child difference=1. Total=13×4+7=59. Check: 13×5=65, short by six.
2. 11 pupils; 45 stickers. Gap=12+10=22, divided by two extra per pupil gives eleven. Total=11×3+12=45.
3. 7 shelves; 47 books. Gap=5+9=14, per-shelf difference two, so seven shelves. Total=7×6+5=47.
4. 8 groups; $92. Gap=$20+$12=$32. Each plan differs by $4 per group. Eight groups. Fund=8×9+20=$92.
5. 14 rows; 62 pupils. The five-seat plan has eight empty seats, while the four-seat plan lacks six seats. Gap=14 places, one per row, so fourteen rows. Five-seat capacity=70; subtract eight to get sixty-two pupils.
6. 7 boxes; 53 items. Gap=4+10=14; difference two per box gives seven. Total=49+4=53.
7. 7 teams; 99 cards. Gap=15+6=21; difference three per team gives seven. Total=84+15=99.
8. 8 baskets; 91 balls. Gap=11+5=16; difference two per basket gives eight. Total=80+11=91.
Solutions 9–16
9. 8 groups; 48 items. Both plans have excess. Difference in excess is 24−8=16. Per-group change is two, giving eight groups. Total=8×5+8=48.
10. 8 shelves; 87 books. Excess difference=31−7=24. Per-shelf change=3, so eight shelves. Total=8×10+7=87.
11. 7 children; 26 counters. Both plans are shortages. Shortage difference=30−9=21. Per-child change=3, giving seven. Available total=7×5−9=26.
12. 7 teams; 18 cards. Shortage difference=24−10=14. Per-team difference=2, so seven. Available=7×4−10=18.
13. 9 tables; 54 people. Exact six-per-table plan uses all seats. Compared with four per table, two extra seats are used at each table. Eighteen divided by two gives nine tables.
14. 10 boxes; 100 items. The exact ten-per-box plan uses twenty more items than the eight-per-box plan. Two more per box gives ten boxes.
15. 9 groups; $126. Going from fourteen to seventeen dollars adds three dollars per group. The exact plan becomes short by twenty-seven, so 27÷3=9. Fund=9×14=$126.
16. 14 children; 98 counters. The nine-per-child requirement exceeds the exact seven-per-child plan by twenty-eight. The per-child difference is two, so there are fourteen children. Exact total=14×7=98.
Solutions 17–24
17. Excess 9; shortage 3. Seven each uses 84, leaving nine. Eight each requires 96, three more than ninety-three.
18. Eight empty seats at seven per row; shortage 7 at six per row. Seven-seat capacity=105, eight above ninety-seven. Six-seat capacity=90, seven below the pupil count.
19. No. The plan gap is 5+8=13, but each group changes requirement by two. Thirteen is not divisible by two, so the implied group count is 6.5, impossible for whole groups.
20. No. Increasing the allocation from five to seven per group must use more items and therefore reduce an excess, not increase it from ten to twenty. The data are inconsistent for a positive fixed group count.
21. 6 groups; 67 items. Excess difference=19−1=18. Per-group difference=3, giving six groups. Total=6×11+1=67.
22. 8 groups; 31 items. Shortage difference=41−17=24. Per-group change=3 gives eight. Available total=8×6−17=31.
23. Total 129; shortage 25. Fourteen groups at nine each require 126; with three left, total=129. Eleven each requires 154, so shortage=154−129=25.
24. The lower plan is twelve below the available total and the higher plan is eight above it. Travelling from the lower requirement to the higher crosses both distances: twelve to reach the available total, then eight more to reach the higher requirement. Total gap=20.
8. Teaching and diagnosis
If a learner always adds excess and shortage, give a two-excess example. Ask where both plan requirements lie relative to the actual total. A number-line sketch makes it clear that same-side distances subtract while opposite-side distances add.
When the total is reconstructed with the wrong sign
Write “required by plan” and “available” as separate lines. Excess means available is greater; shortage means required is greater. Reconstruct the actual total only after that comparison is verbalised.
When the per-group difference is ignored
Compare two plans for one group first. If each child receives four instead of six, the requirement changes by two. Then ask how that two-item change repeats across every child.
When a non-whole group count appears
Do not round. Use the result as evidence that the stated conditions are inconsistent under the whole-group model. Check whether the plan gap was formed correctly and whether both statements use the same number of groups.
Use a forward return check
After finding the group count and total, rebuild both plans independently. The first should produce exactly its stated excess or shortage, and so should the second. A match to only one plan is not enough.
Continue through the structural-reasoning collection
For finding one equal unit and scaling it, use Unitary Method and Equal-Unit Reasoning. For several conditions that must all hold, continue to Simultaneous Conditions and Multiple Constraints. For minimum, maximum and guaranteed conclusions, use Bounds, Extremes and Guaranteed Conclusions.
Return to the BTT Primary Mathematics Learning Hub.
Original explanations and 24 original practice questions with separate worked solutions. Money, seating and storage examples are hypothetical mathematical models. No examination prediction or performance guarantee is implied.