When an addition or subtraction answer is wrong, the next useful question is not always another sum. First find out whether the learner misunderstood the story, chose the wrong relationship, lost place value, miscalculated a fact or recorded a valid idea incorrectly. The same wrong answer can come from different decisions, and those decisions need different teaching responses.
This repair laboratory supplies an original sequence of worked examples and practice tasks for the Primary 1 Mathematics Learning Hub. It complements the existing word-problem structures guide and diagnostic handbook. Those pages explain the concepts and error families; this one gives a teaching sequence that can be used at the table.
The goal is a repair that can be explained and used on a changed question. A corrected worksheet is not enough evidence by itself. Nor is a single error enough to establish a persistent misconception. Observe the method, offer a specific teaching move, and see what happens on the next comparable task.
Scope: meaningful calculation within 100
Primary 1 includes addition and subtraction within 100, written algorithms and specified mental-calculation work. The examples here stay within that number range. See the MOE mathematics syllabus, Primary One section. The sequence below is an original practice route, not an official assessment or a prescribed school teaching order.
Use the sections selectively. A learner who already understands place value does not need every bundle demonstration repeated. A learner who can calculate accurately but misreads comparisons should begin with the story contrasts rather than spend the session on column arithmetic.
Prepare counters, bundles of ten, paper for quick drawings and a pencil. A ten frame and number line are optional supports. None of the tasks requires a calculator. Record the child’s unaided first attempt before offering a hint; otherwise it becomes difficult to tell which part of the work the learner selected independently.
Begin with four stories using small numbers
Combine: There are nine red counters and four blue counters. How many counters are there altogether? The two parts make one whole, so 9 + 4 = 13. A number bond or two groups of counters represents the relationship directly.
Take away: There are nine counters. Four are removed. How many remain? The starting quantity decreases, so 9 − 4 = 5. A before-and-after drawing shows why the answer cannot be greater than nine.
Compare: One tray has nine counters and another has four. How many more counters are on the first tray? Nothing is removed from either tray. The difference is still 9 − 4 = 5. Aligning the groups shows five unmatched counters.
Unknown start: A tray had some counters. Four were added and now there are thirteen. How many were there at first? The change increased the quantity, but the unknown is the earlier amount. Reverse the change: 13 − 4 = 9.
Ask the learner to tell each story without immediately saying an operation. This separates understanding the event or comparison from choosing a sign. If reading is difficult, read the original wording aloud without substituting “add” or “subtract”. Record that reading support was provided.
The same numbers do not guarantee the same operation. Equally, the same operation can describe different situations. These two observations are the foundation of the repair lab: numbers need roles, and signs need relationships.
Repair route A: the learner uses every visible number
Suppose the learner adds nine and four in all four stories. Do not conclude immediately that addition is their only known operation. Give the take-away story with actual counters. If they remove four correctly and state five, the operation meaning exists in action but was not selected from the written story.
The next teaching move is to connect the action with a record. Draw nine marks, cross out four and write 9 − 4 = 5. Ask what the nine, four and five represent. Then give a different take-away story with twelve and three. The purpose is to test the connection, not to rehearse the same answer.
For comparison, resist making a fake removal event when none exists. Align a group of nine with a group of four. Four pairs match, leaving five unmatched objects. The number five describes the gap between the quantities. It does not describe a new combined group.
Return to a mixed pair: “gets four more” and “has four more than”. Ask which quantity is known and which is requested. Do not replace one keyword rule with another. The phrase “more than” can require addition or subtraction depending on whether the larger quantity, smaller quantity or difference is unknown.
Repair route B: the learner knows the story but loses a fact
Suppose the learner correctly says that eight counters and seven counters should be joined but produces sixteen. Keep the story classification as a success. Inspect the calculation separately. Ask the learner to show eight and seven on ten frames, or to explain the fact they used.
A make-ten route splits seven into two and five. Eight and two make ten; five remain, so the total is fifteen. Record 8 + 7 = 8 + 2 + 5 = 15. Every expression in that chain has the same value, so the equal signs remain meaningful.
A second route uses a near double. Seven and seven make fourteen, so eight and seven make fifteen. Do not require both methods from a child who is still establishing one. Show the alternative after the first route is clear, and ask which known relationship made the calculation easier.
For subtraction, 15 − 8 can be treated as the missing part in 8 + □ = 15. The fact network supports the calculation: knowing the whole and one part helps recover the other. A child may still count on from eight; that is a valid route to observe and gradually make more efficient.
Worked repair: counting on without including the start
To calculate 8 + 3, start at eight and make three forward moves: nine, ten, eleven. Eight is the starting quantity, not the first added unit. A learner who says “eight, nine, ten” and reports ten may be counting the starting position as an added step.
Use a small number line and mark eight before moving. Put a separate mark above each jump, not above each number touched. The three intervals end at eleven. Then ask the learner to solve 7 + 2 with the same method. This smaller comparison makes the counting convention inspectable.
The repair is not “always add one to your answer”. That would preserve the underlying error and fail on other calculations. Repair the interpretation of the starting quantity and the count of changes.
Repair route C: tens and ones have become digit tricks
Compare 43 + 6 and 43 + 20. In the first calculation, six ones join three ones, giving nine ones while the four tens remain unchanged. The result is forty-nine. In the second, two tens join four tens, giving six tens while the three ones remain unchanged. The result is sixty-three.
A learner who changes the tens when adding six may be treating the written digits as adjacent marks rather than as counts of different units. Build forty-three with four tens and three ones. Add six loose ones beside the existing ones. Ask what changed and what did not.
Written alignment records this unit structure. Ones belong beneath ones and tens beneath tens because quantities of the same unit are being combined. Neat columns support the reasoning, but neatness alone does not prove that the child understands the units.
After the model is clear, ask for one calculation with the same structure and different numbers: 52 + 4 or 52 + 30. A useful explanation identifies the place that changes rather than simply repeating an instruction to put digits in the right column.
Worked example: 27 + 8 crosses a ten
Twenty-seven contains two tens and seven ones. Adding eight ones gives fifteen ones. Ten of those ones can be regrouped into one ten, leaving five ones. The two original tens plus the new ten give three tens and five ones: thirty-five.
Write the calculation after the exchange is visible. In a column method, seven ones plus eight ones gives fifteen ones; the extra ten is recorded in the tens column. That small written one represents one ten, not one extra loose object.
A number-line route reaches the same result differently: 27 + 3 = 30, then 30 + 5 = 35. Both routes split or reorganise quantity around a ten. Use one route to teach and, when useful, the other to check.
A child who produces 215 may have written two tens beside fifteen ones without completing the regrouping. Return to the physical exchange. Do not describe the written 215 as “almost right”; in ordinary decimal notation it represents a very different number.
Worked example: 42 − 7 opens a ten
Forty-two contains four tens and two ones. To remove seven ones from this representation, rename one ten as ten ones. The same forty-two is now three tens and twelve ones. Remove seven ones to leave five; the three tens remain. The answer is thirty-five.
The equality 42 = 30 + 12 explains the exchange. No extra quantity has appeared. A ten has changed form. This is why crossing out and rewriting digits in a written method must be connected to a conserved total rather than taught as an unexplained borrowing ritual.
A second route subtracts in two parts: 42 − 2 = 40, then 40 − 5 = 35. This is useful for a learner who understands counting back through a landmark. It is not compulsory to use every available method on every question.
Check by adding the removed quantity back: 35 + 7 = 42. This check tests the whole-and-parts relationship. It can reveal an incorrect result even when the original column procedure looked tidy.
Worked example: 46 + 27
Begin with four tens and six ones, then add two tens and seven ones. The ones total thirteen, which is one ten and three ones. The tens total seven after including that new ten. The answer is seventy-three.
A decomposed calculation gives the same result: 46 + 20 = 66, then 66 + 7 = 73. Keep the intermediate statements separate or use a valid equality chain such as 46 + 27 = 66 + 7 = 73. Avoid writing 46 + 20 = 66 + 7 = 73 because the first expression is sixty-six while the later expression is seventy-three.
This distinction is a communication repair as well as an equality repair. A learner may perform the two arithmetic steps correctly and still connect unequal values with equal signs. Preserve the correct calculation while fixing the written relationship.
Worked example: 70 − 36
Seventy contains seven tens and zero ones. Rename one ten to obtain six tens and ten ones. Remove three tens and six ones. Three tens and four ones remain, giving thirty-four.
The zero does not prevent subtraction. It tells us that the standard starting representation contains no loose ones. An equivalent representation supplies the ones needed by opening a ten. Show this with bundles when the written zero case feels mysterious.
Verify the result with addition: 34 + 36 = 70. Thirty and thirty make sixty; four and six make ten; sixty and ten make seventy. The check uses place value and a bond to ten rather than simply repeating the subtraction procedure.
This example belongs later in the sequence. A learner who cannot yet explain forty-two as three tens and twelve ones should not be rushed into zero-ones examples merely because the numbers remain within one hundred.
Unknowns: identify the role of the missing number
Compare 8 + □ = 15, □ − 5 = 9 and 15 − □ = 9. The missing values are seven, fourteen and six respectively. The blank is not always the final result. It can represent a missing part, an unknown starting whole or the amount removed.
Act out each equation with a short story. “Eight are visible; fifteen altogether” asks for the hidden part. “Some were here; five left; nine remain” asks for the start. “Fifteen were here; some left; nine remain” asks for the change.
A number bond can help, but its labels must remain meaningful. If the learner places nine as the whole in □ − 5 = 9 simply because it follows the equal sign, return to the story. Nine is the remaining part; the starting whole must also include the five removed.
Use true and false equations as a separate check on equality. In 7 + 5 = 8 + □, both sides must equal twelve, so the blank is four. The child should not add every visible number and put the total in the blank.
Twenty independent questions
Offer one short group at a time, using the rest as later checks. The questions are not normed, and no total score here is a diagnosis or an official readiness threshold. Ask for enough working to make the method inspectable without demanding elaborate drawings for facts the child already knows.
1. Calculate 6 + 7. Show a known fact that could help.
2. Calculate 14 − 6 and check with addition.
3. Complete 8 + □ = 15.
4. Complete □ − 5 = 9.
5. Complete 15 − □ = 9.
6. Complete 7 + 5 = 8 + □.
7. Calculate 43 + 6. State which place changes.
8. Calculate 43 + 20. State which place changes.
9. Calculate 28 + 7.
10. Calculate 52 − 8.
11. Calculate 36 + 27.
12. Calculate 71 − 26.
13. A shelf holds eight blue cups and six yellow cups. How many cups are there altogether?
14. Fourteen birds are on a fence. Five fly away. How many remain?
15. Mei has fourteen stickers. Kai has nine stickers. How many more stickers does Mei have?
16. A box contains fourteen counters. Nine are red and the rest are blue. How many are blue?
17. A tray had some counters. Five were added and now there are fourteen. How many were there at first?
18. A tray had some counters. Five were removed and fourteen remain. How many were there at first?
19. Mei has fourteen beads. She has four more beads than Kai. How many beads does Kai have?
20. A learner has eight counters, receives seven more and then gives away three. They write 8 + 7 = 15 − 3 = 12. Find the final quantity and repair the written working.
Explained answers and what each item checks
1. Thirteen. One route uses the double 6 + 6 = 12, then adds one more. Another uses 6 + 4 + 3 = 13. Accept a valid route rather than requiring one named method.
2. Eight. Fourteen minus six is eight. Check that 8 + 6 = 14. The calculation and the inverse check should refer to the same whole and parts.
3. Seven. Eight and seven make fifteen. The blank is the missing part of a known whole. Subtracting eight from fifteen is one valid method.
4. Fourteen. The unknown starting amount contains both the five removed and the nine remaining. Nine plus five rebuilds fourteen. Check 14 − 5 = 9.
5. Six. The whole is fifteen and the remaining amount is nine. The amount removed is six. Check that 15 − 6 = 9; do not confuse the removed part with the remaining part.
6. Four. Seven plus five equals twelve, so eight plus the missing number must also equal twelve. Four makes the two sides equal.
7. Forty-nine. Six ones join three ones to make nine ones. Four tens remain unchanged. The ones place changes without making a new ten.
8. Sixty-three. Twenty represents two tens. Adding two tens to four tens gives six tens, while the three ones remain. This contrasts with adding six ones in Question 7.
9. Thirty-five. Eight ones plus seven ones makes fifteen ones. Regroup ten of them into one ten; two original tens become three tens, with five ones left.
10. Forty-four. Rename fifty-two as four tens and twelve ones. Twelve ones minus eight ones leaves four ones. Four tens remain, giving forty-four. Check 44 + 8 = 52.
11. Sixty-three. Six ones plus seven ones gives thirteen ones, or one ten and three ones. Three tens plus two tens plus the regrouped ten gives six tens. The total is sixty-three.
12. Forty-five. Rename seventy-one as six tens and eleven ones. Remove two tens and six ones. Four tens and five ones remain. Check 45 + 26 = 71.
13. Fourteen cups. Eight blue cups and six yellow cups are two parts of a whole. The colour distinguishes the parts; it does not change the addition relationship.
14. Nine birds. The starting amount decreases by five. Fourteen minus five equals nine. A result larger than fourteen would contradict the event described.
15. Five more stickers. The question compares fourteen with nine. Their difference is five. Nobody needs to remove or exchange stickers for the comparison to be valid.
16. Five blue counters. Fourteen is the whole and nine is the known red part. The missing blue part is five. The arithmetic matches Question 15, but the interpretation is part–whole rather than comparison.
17. Nine counters at first. Reverse the increase of five from the final fourteen. Fourteen minus five gives nine. The word added describes the event, not the operation automatically required to find the unknown start.
18. Nineteen counters at first. The starting amount must include the fourteen remaining and the five removed. Fourteen plus five gives nineteen. Check that removing five returns to fourteen.
19. Ten beads. Mei’s fourteen is the larger quantity. Kai has four fewer, so 14 − 4 = 10. The phrase four more than does not automatically mean adding four to the known fourteen.
20. Twelve counters, with corrected working. Write 8 + 7 = 15, followed by 15 − 3 = 12. The original chain wrongly says that 8 + 7, which equals fifteen, has the same value as 15 − 3, which equals twelve. The arithmetic steps are right; the connecting equality is not.
Choose the next task from the pattern, not just the score
If Questions 1 and 2 are difficult but the learner interprets the stories correctly, work on a small fact network. Use one bond, double or inverse relationship across several examples. The next lesson need not re-teach every word-problem structure.
If Questions 3 to 6 are difficult, inspect equality and the role of the unknown. Ask the learner to explain a completed equation in ordinary words before introducing another blank. An unknown-start story can expose a misunderstanding that a familiar result-unknown sum hides.
If Questions 7 and 8 are confused, use tens-and-ones materials before increasing the calculation range. If those are stable but Questions 9 to 12 fail, inspect regrouping separately. This narrows the teaching decision rather than treating all two-digit work as one undifferentiated difficulty.
If the numerical calculations succeed but Questions 15, 17, 18 or 19 fail, compare story roles. Read the wording aloud once without providing the operation. A correct response after that support tells a different story from a correct response only after the adult draws and labels the entire model.
Use Question 20 to inspect mathematical writing independently of arithmetic. Do not erase the learner’s valid steps when repairing the equal signs. Show how separate number sentences communicate the sequence accurately, then ask for a new two-step example with the same recording issue.
A practical repair conversation
A learner answers a comparison question with addition. The adult says, “Tell me what the two amounts stand for.” The learner identifies both people correctly. The adult asks, “Are we joining their collections, or comparing how many each has?” The learner says comparing and draws aligned groups.
That is a useful supported response, but it is not yet an independent success. Record that a classification prompt was needed. Later, use a different comparison question without the prompt. If the learner selects the difference relationship then, the new response provides stronger evidence that the repair is available.
This is an illustrative teaching conversation, not a report of an observed pupil. It shows why support should be recorded honestly. A correct answer obtained after the adult supplies the central decision should not be described as the child having made that decision unaided.
Evidence boundary and continuing route
The What Works Clearinghouse elementary mathematics intervention guide supports systematic instruction, clear mathematical language and purposeful representations. The questions, repair routes and answer explanations here are original applications. They do not carry published reliability scores, diagnostic cut-offs or guaranteed learning outcomes.
For a later check, choose fewer questions with changed quantities and mixed structures. Preserve the concept being tested while removing the familiar surface. Ask for an explanation or a check where it genuinely reveals a decision; do not turn every fluent fact into a long writing exercise.
Continue through the addition and subtraction word-problem structures guide, or return to the Primary 1 Mathematics Learning Hub for the wider sequence.
