BTT Mathematics / Primary Mathematics Learning Hub / Guide 12
Checking is part of doing Mathematics, not an activity reserved for after a mistake. Before calculating, a learner can estimate the size or range of a sensible answer. During working, units and relationships can be monitored. After calculating, inverse operations, alternative methods and the original conditions can challenge the result.
An exact answer without a reason to trust it remains fragile. An estimate without an exact answer may be insufficient when precision is required. Strong work uses both at the right stage.
This guide develops rounding estimates, compatible numbers, magnitude, mental compensation, inverse checks, condition checks and error diagnosis. It complements the existing BTT article I Want the Student to Know Roughly What the Answer Should Be Before the Calculator Does by providing a worked Primary practice route.
Estimate first · Mental restructuring · Inverse checks · Condition checks · Find the first wrong move · 24 questions · Worked answers
1. Estimation gives an answer a neighbourhood
Estimate 398 + 207. Round to convenient nearby numbers: 400 + 200 = 600. The exact answer should be close to six hundred.
The exact sum is 605. The estimate did not predict the final five, but it would reject an answer such as 6,050.
Choose a precision suitable for the question
For 4,872 + 3,149, rounding both to the nearest thousand gives about 5,000 + 3,000 = 8,000. Rounding to the nearest hundred gives 4,900 + 3,100 = 8,000 as well.
Sometimes a finer estimate is useful. If two possible answers are 8,021 and 8,201, an estimate to the nearest thousand will not distinguish them. The checking precision should match the size of the possible error.
Estimate subtraction through nearby landmarks
For 803 − 397, think 800 − 400 ≈ 400. The exact difference, 406, fits that neighbourhood.
An answer of 1,200 would be impossible because subtracting a positive number from 803 cannot produce a result larger than 803.
Estimate multiplication through place value
For 49 × 21, use 50 × 20 = 1,000 as a rough estimate. The exact product 1,029 is close.
A learner who calculates 49 × 21 = 129 has lost a place-value contribution. The estimate makes that error visible before the full algorithm is repeated.
Estimate division with compatible numbers
For 598 ÷ 6, use 600 ÷ 6 = 100. The exact quotient should be very close to one hundred. In fact, 598 ÷ 6 = 99 remainder 4, or about 99.67 if a decimal quotient is appropriate.
Choosing a nearby number divisible by the divisor can be more useful than ordinary rounding.
2. Mental restructuring reduces unnecessary calculation
Compensation in addition
47 + 38 can become 50 + 35 = 85. Three was transferred from thirty-eight to forty-seven. The total is preserved.
The same relationship can be written 47 + 38 = 47 + 40 − 2. Both routes use a nearby friendly number while accounting for the adjustment.
Compensation in subtraction
503 − 198 can be rewritten as 503 − 200 + 2 = 305. Subtracting two hundred removes two too many, so add two back.
Another route increases both numbers by two: 505 − 200 = 305. Adding the same amount to both numbers preserves their difference.
Distributive reasoning
18 × 7 = (20 − 2) × 7 = 140 − 14 = 126. The product is split into easier known products.
This is not a shortcut detached from the multiplication. Eighteen groups of seven can be viewed as twenty groups minus two groups.
Halving and doubling
25 × 16 can become 50 × 8, then 100 × 4 = 400. Doubling one factor while halving the other preserves the product.
Use this only when the halving remains manageable. The point is to preserve the product while choosing easier factors.
Use known facts to build nearby facts
If 8 × 7 = 56, then 9 × 7 = 56 + 7 = 63. If 12 × 6 = 72, then 11 × 6 = 72 − 6 = 66.
This reduces dependence on isolated recall and develops relational number sense.
3. Inverse operations can reconstruct the original relationship
Check addition with subtraction
If 276 + 358 = 634, subtract one addend from the total: 634 − 358 = 276. This asks whether the proposed total contains the other addend correctly.
Check subtraction with addition
If 503 − 278 = 225, add removed and remaining amounts: 278 + 225 = 503.
This is stronger than repeating the same subtraction because a different operation challenges the result.
Check multiplication with division
If 24 × 7 = 168, divide the product by one factor: 168 ÷ 7 = 24.
When a multiplication has factors and units, restore those units too. Seven boxes of twenty-four items should account for all 168 items.
Check division with multiplication and remainder
If 157 ÷ 6 = 26 remainder 1, reconstruct: 6 × 26 + 1 = 157. Also check that the remainder is smaller than six.
Both conditions are needed. A statement such as 157 ÷ 6 = 25 remainder 7 reconstructs the total but is unfinished because seven can form another full group of six.
Check fraction operations by recombination
If 5/6 − 1/3 = 1/2, add the removed third back: 1/2 + 1/3 = 3/6 + 2/6 = 5/6.
Inverse checking extends beyond whole-number arithmetic.
4. A correct calculation can still answer the wrong question
Check units
A rectangle 12 m by 8 m has area 96 m² and perimeter 40 m. If a solution gives 96 m for fencing, the multiplication is correct but the quantity is wrong.
Ask what the final number measures. Units can expose a mismatch between operation and question.
Check size conditions
Fifty-three pupils need vans with eight seats each. Division gives six remainder five. Six vans are not enough because five pupils remain without seats. The final answer must be seven vans.
Arithmetic alone does not decide how a remainder should be interpreted.
Check every stated relationship
Two numbers total eighty-four and one is three times the other. The pair fifty-six and twenty-eight passes the total check but fails the three-times relationship.
The correct pair sixty-three and twenty-one satisfies both. Each sentence in the question creates a condition the answer must survive.
Check the base in percentage work
A price rises from eighty dollars to ninety-six dollars. The increase is sixteen. Sixteen is twenty per cent of the original eighty.
Using ninety-six as the denominator would answer “the increase as a percentage of the new price,” a different question.
Check whether the information is sufficient
A box contains twelve red and eighteen blue counters. Nothing is said about green counters or the total. The number of green counters cannot be determined.
Forcing every stated number into an operation can create a numerical answer where the conditions do not support one.
5. Find the first wrong move, not only the last wrong answer
A final answer may be wrong because the problem was copied incorrectly, the relationship was misread, the right method was chosen but executed badly, or a sensible answer was later reported in the wrong unit.
Error type A: wrong starting quantity
Question: 4,508 − 279. Learner copies 4,580 − 279. Every later step might be arithmetically correct and still answer the wrong problem.
The repair is copying and comparison, not subtraction technique.
Error type B: right numbers, wrong relationship
Question: Jo has twelve more than Kim. Jo has thirty-one. Learner calculates 31 + 12. The arithmetic is correct but moves away from the smaller unknown amount.
The repair is the comparison model: Jo = Kim + 12, so Kim = 31 − 12.
Error type C: right method, arithmetic slip
Question: 276 + 358. Learner sets up addition correctly but writes 624. The structure is right; one place-value calculation needs repair.
Use an inverse check or recalculate the affected column rather than reteaching the entire problem type.
Error type D: correct number, wrong answer form
Question: 53 pupils, vans hold eight. Learner writes “6 remainder 5 vans.” The division result is correct, but the context requires enough whole vans for all pupils.
The repair is interpretation, not long division.
Error type E: plausible result, incomplete justification
A learner says a number is prime because it is odd. This may happen to be correct for a particular number, but oddness alone does not prove primality; nine and fifteen are odd composite numbers.
The repair is evidence: test factors that could divide the number.
6. Practice: 24 questions
For questions 1–8, estimate before finding the exact answer. For later questions, identify the most useful check.
Questions 1–8: Estimate and calculate
1. Estimate 398 + 207, then calculate exactly.
2. Estimate 803 − 397, then calculate exactly.
3. Estimate 49 × 21, then calculate exactly.
4. Estimate 598 ÷ 6 using a compatible number.
5. Estimate 4,872 + 3,149 to the nearest hundred, then calculate exactly.
6. Estimate 2,994 × 3, then calculate exactly.
7. Estimate 10,012 − 4,989, then calculate exactly.
8. Estimate 1,998 ÷ 2, then calculate exactly.
Questions 9–16: Mental restructuring and inverse checks
9. Calculate 47 + 38 using compensation.
10. Calculate 503 − 198 using a nearby number.
11. Calculate 18 × 7 using 20 × 7.
12. Calculate 25 × 16 using halving and doubling.
13. Check 276 + 358 = 634 using an inverse operation.
14. Check 503 − 278 = 225 using an inverse operation.
15. Is 157 ÷ 6 = 25 remainder 7 a completed division? Explain.
16. Check 24 × 7 = 168 using division.
Questions 17–24: Reason about conditions
17. A learner gives 96 m as the fencing needed for a 12 m by 8 m rectangle. Identify the mistake and give the correct fencing length.
18. Fifty-three pupils need vans with eight pupil seats each. A learner writes 6 remainder 5. What final answer should be reported?
19. Two numbers total 84 and one is three times the other. Does 56 and 28 satisfy both conditions? Find the correct pair.
20. A price rises from $80 to $96. What percentage increase is this, using the correct base?
21. A box contains 12 red and 18 blue counters. No total and no green count are given. Can the green count be determined?
22. A learner says 29 is prime because it is odd. Is the conclusion correct, and is the reason sufficient?
23. An answer to 4,508 − 279 is produced from working that began with 4,580 − 279. What should be repaired first?
24. A calculator returns 6,348 for 276 + 358. Give two independent reasons to reject the result without simply repeating the same calculator entry.
7. Worked answers
Answers 1–8
1. Estimate about 600; exact 605. Use 400+200 for the estimate, then 398+207=605.
2. Estimate about 400; exact 406. Use 800−400, then calculate 803−397=406.
3. Estimate about 1,000; exact 1,029. Use 50×20, then calculate 49×21=1,029.
4. About 100. Use 600÷6=100. Exact whole-number division is 99 remainder 4.
5. Estimate 8,000; exact 8,021. Round to 4,900+3,100. Exact addition gives 8,021.
6. Estimate about 9,000; exact 8,982. Use 3,000×3.
7. Estimate about 5,000; exact 5,023. Use 10,000−5,000.
8. Estimate about 1,000; exact 999. Use 2,000÷2.
Answers 9–16
9. 85. Transfer three from thirty-eight to forty-seven: 50+35=85.
10. 305. Calculate 503−200+2.
11. 126. 20×7−2×7=140−14=126.
12. 400. 25×16=50×8=100×4=400.
13. Valid. 634−358=276, so the proposed total reconstructs the other addend.
14. Valid. 278+225=503.
15. No. Although 6×25+7=157, remainder seven is not smaller than divisor six. One more group can be formed. Correct result is 26 remainder 1.
16. Valid. 168÷7=24.
Answers 17–24
17. Area was calculated instead of perimeter; correct fencing length is 40 m. Add all four sides: 12+8+12+8.
18. 7 vans. Six vans seat only forty-eight pupils; five still need seats, so another whole van is required.
19. No; correct pair is 63 and 21. Fifty-six is only twice twenty-eight. Four equal units total eighty-four, so one unit is twenty-one.
20. 20%. Increase is sixteen. Divide by the original base eighty: 16/80=0.20.
21. No. The stated information does not determine one green count.
22. The conclusion is correct, but the reason is insufficient. Twenty-nine is prime because it has no factors other than one and itself. Many odd numbers are composite.
23. Repair the copied starting number first. The working answers a different subtraction even if every later arithmetic step is correct.
24. First, estimate 276+358 as about 300+400=700, so 6,348 is far too large. Second, subtraction check fails: 6,348−358 is not 276. The exact sum is 634.
8. Build a checking ladder
Before calculation: estimate the size, sign or range. During calculation: protect units, place value and stage labels. After calculation: use an inverse operation, a different method or the original conditions.
Not every problem needs every check. Choose a check that is capable of detecting the likely failure.
Do not confuse checking with anxiety
Repeatedly recalculating the same line without a new strategy can consume time without increasing confidence. A good check is targeted: estimate, reverse, substitute, compare units or test a boundary.
Use errors as evidence
A wrong answer that is the right size but off by ten suggests something different from an answer hundreds of times too large. A correct arithmetic result with the wrong unit suggests interpretation rather than calculation.
Classifying the first wrong move helps choose the smallest useful repair.
Keep exact and approximate statements separate
398+207 is approximately six hundred and exactly six hundred five. Write an approximation sign or use words such as “about” when precision has deliberately been reduced.
Do not silently replace an exact fraction or measurement with a rounded decimal when the task requires exactness.
Continue through the Primary Mathematics series
For place-value checking, use Place Value and Regrouping. For relationship checks, use Word Problems, Bar Models and Checking. For equations and substitution, use Patterns, Early Algebra and Equations. For rate plausibility, use Speed, Distance and Time.
Return to the BTT Primary Mathematics Learning Hub. For broader examination-focused reasoning, see How No-Calculator Reasoning Works in PSLE Mathematics.
Original learning guide. Curriculum reference checked 6 September 2026. Practice questions are teaching illustrations, not official examination items or performance predictions.