Series route: This guide belongs to the complete PSLE Learner’s Guide volume directory. The original PSLE Learning Guide supplies the wider English, Mathematics and Science route, while PSLE Mathematics provide verified subject-specific continuation.
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This volume solves one precise learner problem: the learner performs long multiplication by 875 mechanically, forgets that 875 is 125 below 1000, loses the factor seven in the eighths method, or divides before checking which factor is convenient. The aim is not to memorise a slogan. It is to understand the mechanism well enough to recognise the idea when the wording, numbers, diagram or context changes. A Primary 6 learner should be able to explain the decision, perform it under time pressure, and detect a tempting answer that only looks familiar.
- Understand the mechanism before memorising a form.
- Study the worked examples and compare near-misses.
- Repair common misconceptions with targeted tests.
- Complete graded independent practice before reading the answers.
- Use delayed transfer and examination control to make the skill durable.
- Use the parent and tutor notes only after the learner has attempted the work.
The problem beneath the mistake
875 sits close to 1000: the gap is 125. It is also seven groups of 125, while eight groups of 125 make 1000. Therefore 875 is seven eighths of 1000.
For 24 × 875, the eighths route is efficient: 24 ÷ 8 × 7000 = 3 × 7000 = 21,000.
The complement route gives the same result: 24,000 − 24 × 125. Since 24 × 125 = 3000, the product is 21,000.
A shortcut is an equality, not a guess. The learner must be able to show that 7000 ÷ 8 = 875 and that 1000 − 125 = 875.
The most common missing factor is seven. Using n ÷ 8 × 1000 calculates one eighth of n thousand, which corresponds to 125n, not 875n.
Dividing n by 8 first is convenient when n is divisible by 8. If not, the complement route or another multiplication order may avoid awkward fractions.
For 14 × 875, use 14,000 − 14 × 125. Since 14 × 125 = 1750, the answer is 12,250.
Decimals can be controlled by scaling. For 2.4 × 875, compute 24 × 875 ÷ 10 = 21,000 ÷ 10 = 2100.
Estimation should place a positive product below n × 1000 and above n × 800. An answer above n thousand signals a sign, zero or complement error.
The gap check is exact: n × 1000 − n × 875 must equal n × 125. This is often faster than repeating the original method.
The method connects to known facts for 125, 250, 375, 625 and 875 without making them interchangeable. Each coefficient must retain its own numerator over eight.
The safe routine is expose the two identities, inspect divisibility, choose complement or eighths, keep seven and every zero visible, calculate, estimate, and check the exact gap from n thousand.
Core decision: Use 875 = 1000 − 125 = (7/8) × 1000 = 7000 ÷ 8. Thus n × 875 can be found as n × 1000 − n × 125, or as n ÷ 8 × 7000 when n is conveniently divisible by 8. Choose the cleaner exact route and verify the product is one eighth below n × 1000.
A useful rule has three parts: a trigger, an operation and a check. The trigger tells you when the rule applies. The operation tells you what to do. The check tells you whether the result still matches the meaning of the question. Learners who remember only the operation often apply it in the wrong place; learners who remember only an example often freeze when the surface details change.
Mechanism: why the rule works
1. Build the identity
Eight groups of 125 make 1000, so seven groups make 875.
Diagnostic question: Can you derive both forms? Reliable response: Write 875 = 7×125 = 7000÷8 = 1000−125.
2. Inspect divisibility
Early division by 8 is helpful only when it simplifies the chosen factor.
Diagnostic question: Which factor divides cleanly by 8? Reliable response: Reorder multiplication or use the complement route.
3. Protect the seven
The numerator seven distinguishes 875 from 125.
Diagnostic question: Where is the factor seven? Reliable response: Circle it until the product is complete.
4. Use the complement
Multiplying by 875 is multiplying by 1000 and removing one eighth-thousand, or 125 times the number.
Diagnostic question: What exact amount must be subtracted? Reliable response: Compute n×125 and subtract from n×1000.
5. Control magnitude
875 is close to 1000 but still smaller.
Diagnostic question: What range should the product occupy? Reliable response: Bound it between 800n and 1000n for positive n.
6. Attach units
The transformation changes arithmetic only.
Diagnostic question: What quantity does the product represent? Reliable response: Carry units and referents into the answer.
Worked examples: make the decision visible
Do not race through these examples. Before reading each explanation, name the evidence that matters and the information that does not. Then compare the correct route with the tempting route. That comparison is where the transferable learning sits.
Worked example 1: Divisible factor
Task. Find 16 × 875.
Reasoning. 16 ÷ 8 × 7000 = 2 × 7000. The controlling idea is the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 14,000.
Independent check. Check 16,000 − 16 × 125 = 16,000 − 2000. A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Worked example 2: Complement route
Task. Find 14 × 875.
Reasoning. 14,000 − 14 × 125; 14 × 125 = 1750. The controlling idea is the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 12,250.
Independent check. The result lies between 11,200 and 14,000. A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Worked example 3: Decimal
Task. Find 2.4 × 875.
Reasoning. Use 24 × 875 ÷ 10. The controlling idea is the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 2100.
Independent check. 2.4 × 1000 − 2.4 × 125 = 2400 − 300. A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Worked example 4: Fraction
Task. Find 4/7 of 875.
Reasoning. Since 875 = 7 × 125, four sevenths equal 4 × 125. The controlling idea is the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 500.
Independent check. Seven equal groups of 125 rebuild 875. A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Worked example 5: Rate
Task. A machine makes 875 parts per hour for 8 hours.
Reasoning. Eight groups of 875 equal 7000. The controlling idea is the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 7000 parts.
Independent check. The unit is parts, not hours. A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Worked example 6: Large factor
Task. Find 128 × 875.
Reasoning. 128 ÷ 8 × 7000 = 16 × 7000. The controlling idea is the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 112,000.
Independent check. Check 128,000 − 16,000. A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Worked example 7: Missing seven
Task. A learner writes 32 × 875 = 32 ÷ 8 × 1000.
Reasoning. That calculates 32 × 125 and loses seven. The controlling idea is the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. Correct: 32 ÷ 8 × 7000 = 28,000.
Independent check. Check the gap from 32,000 is 4000. A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Worked example 8: Reverse groups
Task. How many groups of 875 make 35,000?
Reasoning. Since 875 × 8 = 7000, five sets of eight groups make 35,000. The controlling idea is the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 40 groups.
Independent check. Verify 40 × 875 = 35,000. A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Deepening lab: compare cases, conditions and limits
A method becomes independent when you can say where it applies, where it stops and which change would alter the conclusion. The cases below deliberately vary one structural feature at a time. For each case, pause before the analysis and predict whether the core rule remains available. Then name the exact condition that controls the decision. This is harder than repeating a worked solution, but it is also closer to what an unfamiliar examination item demands.
Deepening case 1: Eight times
Case. The factor is 8.
Analysis. 8 × 875 = 7000 is the anchor fact. Notice that the answer depends on the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000; changing a decorative detail would not be enough to change the reasoning.
Challenge. Explain the seven-eighths structure. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 2: Forty times
Case. Forty divides by 8.
Analysis. 40÷8×7000 = 35,000. Notice that the answer depends on the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000; changing a decorative detail would not be enough to change the reasoning.
Challenge. Check by subtracting 5000 from 40,000. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 3: Twenty-four times
Case. Twenty-four contains three groups of eight.
Analysis. The product is 3×7000. Notice that the answer depends on the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000; changing a decorative detail would not be enough to change the reasoning.
Challenge. Verify the exact gap. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 4: Twelve times
Case. Twelve does not divide evenly by 8 as a whole number.
Analysis. Use 12,000−12×125 = 10,500. Notice that the answer depends on the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000; changing a decorative detail would not be enough to change the reasoning.
Challenge. Compare method lengths. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 5: Hundred times
Case. Place value makes direct multiplication simple.
Analysis. 100×875 = 87,500. Notice that the answer depends on the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000; changing a decorative detail would not be enough to change the reasoning.
Challenge. Do not force a longer shortcut. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 6: 0.8 times
Case. Eight tenths of 875 equals 700.
Analysis. Use 8×875÷10. Notice that the answer depends on the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000; changing a decorative detail would not be enough to change the reasoning.
Challenge. Check below 875. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 7: 0.08 times
Case. Eight hundredths gives 70.
Analysis. Use the anchor 7000÷100. Notice that the answer depends on the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000; changing a decorative detail would not be enough to change the reasoning.
Challenge. Track two decimal places. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 8: 1.6 times
Case. Sixteen tenths gives 1400.
Analysis. 16×875÷10 = 14,000÷10. Notice that the answer depends on the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000; changing a decorative detail would not be enough to change the reasoning.
Challenge. Estimate below 1600. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 9: Seven eighths of 1000
Case. The expression directly equals 875.
Analysis. Divide 1000 by 8 and multiply by 7. Notice that the answer depends on the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000; changing a decorative detail would not be enough to change the reasoning.
Challenge. Explain order flexibility. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 10: Seven eighths of 24,000
Case. Divide 24,000 by 8 and multiply by 7.
Analysis. The result is 21,000. Notice that the answer depends on the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000; changing a decorative detail would not be enough to change the reasoning.
Challenge. Connect with 24×875. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 11: Complement of 125
Case. 875 lacks one 125-unit block from 1000.
Analysis. This makes n×125 the exact subtraction. Notice that the answer depends on the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000; changing a decorative detail would not be enough to change the reasoning.
Challenge. Draw eight blocks. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 12: Compare 875 and 625
Case. Both are eighth-based but use different numerators.
Analysis. 875 is 7/8 thousand; 625 is 5/8 thousand. Notice that the answer depends on the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000; changing a decorative detail would not be enough to change the reasoning.
Challenge. Do not copy the numerator. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 13: Money
Case. Thirty-two items cost $8.75 each.
Analysis. Scale cents or use 875 cents, then convert. Notice that the answer depends on the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000; changing a decorative detail would not be enough to change the reasoning.
Challenge. Control currency units. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 14: Division
Case. A total is divided into 875-unit groups.
Analysis. Use anchor multiples to estimate the quotient. Notice that the answer depends on the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000; changing a decorative detail would not be enough to change the reasoning.
Challenge. Check multiplication afterward. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 15: Negative extension
Case. −8×875 has magnitude 7000 and negative sign.
Analysis. Sign control is separate from the identity. Notice that the answer depends on the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000; changing a decorative detail would not be enough to change the reasoning.
Challenge. Explain the sign rule. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 16: Estimate
Case. 875 is nearer 900 than 800.
Analysis. A 900n estimate provides a tight check. Notice that the answer depends on the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000; changing a decorative detail would not be enough to change the reasoning.
Challenge. State whether exact is lower. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 17: Calculator display
Case. A display shows 262,500 for 300×875.
Analysis. 300×1000−300×125 = 300,000−37,500. Notice that the answer depends on the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000; changing a decorative detail would not be enough to change the reasoning.
Challenge. Confirm the display. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 18: Factor pairing
Case. The expression is 18×16×875.
Analysis. Pair 16 with 875 first to get 14,000. Notice that the answer depends on the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000; changing a decorative detail would not be enough to change the reasoning.
Challenge. Explain commutativity. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 19: Unit rate
Case. 875 mL per bottle for 24 bottles.
Analysis. The product is 21,000 mL, then convert if requested. Notice that the answer depends on the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000; changing a decorative detail would not be enough to change the reasoning.
Challenge. Keep millilitres before litres. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 20: Exact gap
Case. For positive n, n×875 is n×125 below n×1000.
Analysis. The gap identity is an independent check. Notice that the answer depends on the exact equivalence 875 = 1000 − 125 = 7000 ÷ 8 and preservation of the factor seven, place value and complement to 1000; changing a decorative detail would not be enough to change the reasoning.
Challenge. Apply it to n=56. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Finish the lab with a blank-page recall. Write the rule, its conditions, one confirming example, one near-miss and one independent check from memory. Compare that page with the article only after finishing. If the rule is correct but the condition is missing, the learning is not yet safe: an overgeneralised rule can produce confident errors. If the condition is correct but no example can be generated, practise construction rather than rereading.
Common misconceptions and targeted repairs
Misconception 1: 875 is one eighth of 1000
What the learner may think. The numerator seven is omitted.
Why it fails. One eighth is 125. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Write seven equal 125 blocks. Use this short test immediately: How many blocks make 875?
Misconception 2: Subtract 125 once
What the learner may think. n×875 = n×1000−125.
Why it fails. The 125 gap occurs for every group. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Subtract n×125. Use this short test immediately: What does distributivity require?
Misconception 3: Divide first always
What the learner may think. n must be divided by 8 even when awkward.
Why it fails. The order is a convenience. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Choose complement when cleaner. Use this short test immediately: Which route stays exact?
Misconception 4: 875 is larger than 1000 after multiplication
What the learner may think. A large product may exceed n×1000.
Why it fails. For positive n, multiplying preserves 875<1000. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Use a bound. Use this short test immediately: Is the answer below n thousand?
Misconception 5: Shortcut changes value
What the learner may think. A decomposition is approximate.
Why it fails. The identities are exact. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Prove them numerically. Use this short test immediately: Does 7000÷8 equal 875?
Misconception 6: Units disappear
What the learner may think. Mental arithmetic needs no unit.
Why it fails. The calculation still answers a quantity question. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Restore unit and referent. Use this short test immediately: What was multiplied?
Graded independent practice
Attempt each item without looking ahead. Write one line that identifies the controlling evidence before writing the final answer. Items 1–3 check recognition, 4–7 require explanation, and the later items change the surface form so that memory alone is insufficient.
Practice 1: Recognition
Question. Complete: 875 = ___ ÷ 8.
Explained answer. 7000.
Why this answer earns trust. Seven eighths of 1000 is 875. Transfer variation: Also write 1000−125.
Practice 2: Basic
Question. Find 32×875.
Explained answer. 28,000.
Why this answer earns trust. 32÷8×7000. Transfer variation: Check gap 4000 from 32,000.
Practice 3: Non-divisible
Question. Find 18×875.
Explained answer. 15,750.
Why this answer earns trust. 18,000−18×125 = 18,000−2250. Transfer variation: Estimate with 900.
Practice 4: Decimal
Question. Find 1.2×875.
Explained answer. 1050.
Why this answer earns trust. 12×875÷10 = 10,500÷10. Transfer variation: Check 1200−150.
Practice 5: Fraction
Question. Find 3/7 of 875.
Explained answer. 375.
Why this answer earns trust. 875 is seven groups of 125. Transfer variation: Relate to 3×125.
Practice 6: Error
Question. Correct n×875=n÷8×1000.
Explained answer. Use n÷8×7000 when valid.
Why this answer earns trust. The old form loses seven. Transfer variation: Show complement form.
Practice 7: Context
Question. Sixteen boxes hold 875 beads each.
Explained answer. 14,000 beads.
Why this answer earns trust. 16÷8×7000. Transfer variation: State beads.
Practice 8: Reverse
Question. How many 875s make 21,000?
Explained answer. 24.
Why this answer earns trust. Three anchor groups of eight make 24. Transfer variation: Verify multiplication.
Practice 9: Choice
Question. For 100×875, choose a method.
Explained answer. Direct place value gives 87,500.
Why this answer earns trust. The simplest exact method wins. Transfer variation: Give a factor where division by 8 wins.
Practice 10: Routine
Question. State the routine.
Explained answer. Write identities, inspect divisibility, choose route, protect seven, calculate, estimate and gap-check.
Why this answer earns trust. It controls structure and magnitude. Transfer variation: Apply to 56×875.
Delayed transfer: prove that the learning lasts
Transfer test 1: Anchor table
Build multiples of 875 from 1 to 20. Do this after a delay, without rereading the worked examples. The purpose is to test retrieval and decision-making, not immediate imitation.
Success criterion. Every result has a complement check. If the criterion is not met, return to the smallest missing decision rather than repeating the entire chapter.
Transfer test 2: Mixed coefficients
Mix products by 125, 375, 625 and 875. Do this after a delay, without rereading the worked examples. The purpose is to test retrieval and decision-making, not immediate imitation.
Success criterion. The numerator over eight is selected correctly. If the criterion is not met, return to the smallest missing decision rather than repeating the entire chapter.
Transfer test 3: Delayed contexts
Solve money, volume, rate and reverse problems after two days. Do this after a delay, without rereading the worked examples. The purpose is to test retrieval and decision-making, not immediate imitation.
Success criterion. Units and factor seven remain correct. If the criterion is not met, return to the smallest missing decision rather than repeating the entire chapter.
Examination control
1. Read the job before the story. Underline the command and the noun it governs. “Explain the change,” “state the word,” “find the fraction” and “identify the evidence” are different jobs even when the surrounding context is similar.
2. Separate given information from remembered information. Use only the facts supplied for the current task plus secure subject knowledge. Do not let a previous question, a familiar worksheet or an expected pattern silently become new evidence.
3. Write enough working to expose the decision. A marker or teacher should be able to see the relationship that justifies the answer. Extra sentences are not automatically stronger; the strongest working is concise, relevant and checkable.
4. Reserve a different-route check. For language, read the sentence for meaning and grammar. For mathematics, estimate or reverse the operation. For science, check the claim against the stated observation and a conservation or fair-test principle.
5. Stop when the requested job is complete. Unrequested speculation can contradict a correct answer. Add detail only when it explains the mechanism or satisfies another explicit part of the question.
Under time pressure, use a compact four-part margin routine: trigger → evidence → operation → check. Trigger names the feature that activates the method. Evidence identifies the exact words, values or observations allowed. Operation performs the language, mathematical or scientific step. Check asks whether the final statement answers the requested job, preserves units or referents, and avoids a stronger claim than the evidence supports.
A compact error log
After practice, do not record only “careless mistake.” Name the broken decision. Useful labels include trigger missed, evidence imported, relationship reversed, referent lost, unit dropped, operation correct but job wrong, and check repeated the same route. Add one prevention cue beside the label. A short accurate log is more useful than a long emotional account because it tells the learner what to do on the next item.
Review the log after two days and again after a week. Reattempt only one or two representative items first. If the method transfers, move to a mixed set. If it fails, repair the exact link between trigger and operation. This spacing matters because a solution can feel obvious while it is still visible and disappear when the page is closed.
Notes for parents and tutors
Ask the learner to draw eight blocks of 125 and shade seven before memorising speed.
Mix divisible and non-divisible factors so method choice remains active.
Require the exact gap from n thousand as a fast independent check.
The adult’s most useful question is not “Why were you careless?” but “Which decision did the evidence require?” Ask the learner to point to the controlling word, value, relationship or observation. If the learner can explain that link, reduce support. If the learner cannot, model one example and then change the surface details for the next attempt.
Frequently asked questions
Why is 875 seven eighths of 1000?
Because one eighth of 1000 is 125 and seven groups of 125 total 875.
What is the complement method?
Multiply by 1000, then subtract the number multiplied by 125.
Must I divide by 8 first?
No. Do so only when it simplifies the arithmetic.
Where does seven appear?
In 875 = 7×125 = 7000÷8.
How do I estimate?
For positive n, the result is below 1000n and above 800n.
What is the fastest exact check?
Confirm that 1000n−875n equals 125n.
Next routes
Return to the complete volume directory to locate the series, use the original PSLE Learning Guide for the foundation route, and continue through PSLE Mathematics for verified subject guidance. The next useful task is a mixed set in which this idea is sometimes relevant and sometimes irrelevant; that is the point at which a learner must choose, not merely copy.