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.
Find your route through this guide
This volume solves one precise learner problem: the learner treats multiplication by 245 as a long opaque algorithm and loses place value, partial products or units under time pressure. 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
Multiplying by 245 becomes easier when 245 is treated as a structured quantity rather than a string of digits. Two hundred and forty-five groups can be split into two hundred groups and forty-five groups.
The distributive property gives n × 245 = n × 200 + n × 45. The second part can be split again as n × 40 + n × 5. Every step preserves the number of groups.
A complementary route uses 245 = 250 − 5. Then n × 245 = n × 250 − n × 5. This is often an efficient mental check because multiplying by 250 can be seen as multiplying by 1000 and dividing by 4.
The two routes should agree. Agreement is useful evidence because they organise the multiplier differently. If the answers differ, locate the first partial product or recombination that changed value.
Place value matters. Multiplying by 200 is not multiplying by 2 and casually adding two zeros to every kind of number. The safe meaning is two hundred equal groups, with decimal placement handled by value.
For 36 × 245, the first route gives 36 × 200 = 7200 and 36 × 45 = 1620, so the total is 8820. The checking route gives 36 × 250 − 36 × 5 = 9000 − 180 = 8820.
The method transfers to decimals because the multiplier structure is unchanged. For 0.6 × 245, compute 0.6 × 200 + 0.6 × 45 = 120 + 27 = 147.
Money requires a unit check. If 245 items cost $3.20 each, the total is 245 × $3.20 = $784.00. The numerical product and the money format must both be correct.
Measurement contexts require interpretation before calculation. Multiplying 2.4 kg by 245 means 245 equal masses of 2.4 kg, producing 588 kg. It does not change kilograms into another unit.
Estimation controls reasonableness. Since 245 is close to 250, a positive product should be slightly less than n × 250. If 48 × 245 is reported as 1,176, the missing place value becomes obvious.
Choose the route that reduces working for the number in front. Numbers divisible by 4 often suit ×250 − ×5. Numbers easy to multiply by 40 and 5 suit ×200 + ×40 + ×5.
A complete solution names the decomposition, records aligned partial products, recombines once, attaches the correct unit and checks scale against ×250.
Core decision: Rewrite 245 as 200 + 45, so a × 245 = a × 200 + a × 45; verify efficiently with the equivalent identity a × 245 = a × 250 − a × 5.
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. Read the multiplication
Identify the multiplicand, multiplier and any unit.
Diagnostic question: What quantity is repeated 245 times? Reliable response: Write the operation in context.
2. Decompose 245
Use 200 + 45 or 250 − 5.
Diagnostic question: Which form makes the given number easier to handle? Reliable response: Record the equality before calculating.
3. Calculate partial products
Find ×200 and ×45, or ×250 and ×5.
Diagnostic question: Does each line represent the stated groups? Reliable response: Keep place values aligned.
4. Recombine once
Add or subtract according to the chosen identity.
Diagnostic question: Did the sign match the decomposition? Reliable response: Write one clear total line.
5. Restore meaning
Attach dollars, kilograms, metres or item meaning.
Diagnostic question: What does the product count or measure? Reliable response: Include the correct unit and format.
6. Check independently
Estimate near ×250 and, where useful, use the alternative decomposition.
Diagnostic question: Is the result slightly below ×250 for positive values? Reliable response: Investigate any disagreement.
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: Whole number
Task. Calculate 36 × 245.
Reasoning. 36 × 200 = 7200; 36 × 40 = 1440; 36 × 5 = 180. The controlling idea is whether every partial product represents the correct number of equal groups and recombines to the original multiplier. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 7200 + 1440 + 180 = 8820.
Independent check. Check 36 × 250 − 36 × 5 = 9000 − 180. 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: Alternative route
Task. Calculate 48 × 245 using 250 − 5.
Reasoning. 48 × 250 = 12,000 and 48 × 5 = 240. The controlling idea is whether every partial product represents the correct number of equal groups and recombines to the original multiplier. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 12,000 − 240 = 11,760.
Independent check. The result is just below 12,000. 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: Factor-friendly number
Task. Calculate 125 × 245.
Reasoning. 125 × 250 = 31,250; 125 × 5 = 625. The controlling idea is whether every partial product represents the correct number of equal groups and recombines to the original multiplier. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 31,250 − 625 = 30,625.
Independent check. Estimate 125 × 240 = 30,000. A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Worked example 4: Decimal
Task. Calculate 0.6 × 245.
Reasoning. 0.6 × 200 = 120 and 0.6 × 45 = 27. The controlling idea is whether every partial product represents the correct number of equal groups and recombines to the original multiplier. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 147.
Independent check. It is slightly less than 0.6 × 250 = 150. 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: Decimal with hundredths
Task. Calculate 1.75 × 245.
Reasoning. 1.75 × 200 = 350; ×40 = 70; ×5 = 8.75. The controlling idea is whether every partial product represents the correct number of equal groups and recombines to the original multiplier. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 428.75.
Independent check. Check 437.5 − 8.75 = 428.75. 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: Money
Task. 245 notebooks cost $3.20 each.
Reasoning. 3.20 × 200 = 640; ×40 = 128; ×5 = 16. The controlling idea is whether every partial product represents the correct number of equal groups and recombines to the original multiplier. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. $784.00.
Independent check. Use two decimal places for money. 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: Mass
Task. Each container has mass 2.4 kg. Find 245 containers.
Reasoning. 2.4 × 200 = 480; ×40 = 96; ×5 = 12. The controlling idea is whether every partial product represents the correct number of equal groups and recombines to the original multiplier. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 588 kg.
Independent check. Compare with 2.4 × 250 = 600 kg. 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 problem
Task. A total of 17,150 is made from 245 equal groups. Find one group.
Reasoning. Compute 17,150 ÷ 245 and verify by multiplication. The controlling idea is whether every partial product represents the correct number of equal groups and recombines to the original multiplier. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 70, because 70 × 245 = 14,000 + 3150.
Independent check. Check 70 × 250 − 70 × 5 = 17,500 − 350. 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: ×200
Case. Multiply by two hundred equal groups.
Analysis. For whole numbers, ×2 then ×100 is a value-based shortcut. Notice that the answer depends on whether every partial product represents the correct number of equal groups and recombines to the original multiplier; changing a decorative detail would not be enough to change the reasoning.
Challenge. Explain why it works. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 2: ×40
Case. Four tens of the number.
Analysis. Compute ×4 then ×10. Notice that the answer depends on whether every partial product represents the correct number of equal groups and recombines to the original multiplier; changing a decorative detail would not be enough to change the reasoning.
Challenge. Track a decimal. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 3: ×5
Case. Half of ×10 when useful.
Analysis. This supports both routes. Notice that the answer depends on whether every partial product represents the correct number of equal groups and recombines to the original multiplier; changing a decorative detail would not be enough to change the reasoning.
Challenge. Use an odd decimal. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 4: ×45
Case. Combine ×40 and ×5.
Analysis. Keep the two partial products distinct. Notice that the answer depends on whether every partial product represents the correct number of equal groups and recombines to the original multiplier; changing a decorative detail would not be enough to change the reasoning.
Challenge. Calculate 64 × 45. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 5: ×250
Case. One quarter of ×1000.
Analysis. Useful when the multiplicand divides cleanly by 4. Notice that the answer depends on whether every partial product represents the correct number of equal groups and recombines to the original multiplier; changing a decorative detail would not be enough to change the reasoning.
Challenge. Calculate 72 × 250. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 6: ×245
Case. Five less than 250 groups.
Analysis. Subtract only five groups, not five. Notice that the answer depends on whether every partial product represents the correct number of equal groups and recombines to the original multiplier; changing a decorative detail would not be enough to change the reasoning.
Challenge. Explain the distinction. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 7: Commutativity
Case. 245 × n equals n × 245.
Analysis. Choose the orientation that clarifies context. Notice that the answer depends on whether every partial product represents the correct number of equal groups and recombines to the original multiplier; changing a decorative detail would not be enough to change the reasoning.
Challenge. Rewrite a word problem. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 8: Distributive property
Case. n(200 + 45) = 200n + 45n.
Analysis. The equality preserves all groups. Notice that the answer depends on whether every partial product represents the correct number of equal groups and recombines to the original multiplier; changing a decorative detail would not be enough to change the reasoning.
Challenge. Draw an area model. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 9: Subtractive decomposition
Case. n(250 − 5) = 250n − 5n.
Analysis. The bracket controls the subtraction. Notice that the answer depends on whether every partial product represents the correct number of equal groups and recombines to the original multiplier; changing a decorative detail would not be enough to change the reasoning.
Challenge. Show a common sign error. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 10: Zero
Case. 0 × 245 = 0.
Analysis. No groups contain no quantity. Notice that the answer depends on whether every partial product represents the correct number of equal groups and recombines to the original multiplier; changing a decorative detail would not be enough to change the reasoning.
Challenge. Check both routes. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 11: One
Case. 1 × 245 = 245.
Analysis. One group preserves the multiplier. Notice that the answer depends on whether every partial product represents the correct number of equal groups and recombines to the original multiplier; changing a decorative detail would not be enough to change the reasoning.
Challenge. Use as a boundary check. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 12: Fraction
Case. One half of 245 is 122.5.
Analysis. Products need not be whole numbers. Notice that the answer depends on whether every partial product represents the correct number of equal groups and recombines to the original multiplier; changing a decorative detail would not be enough to change the reasoning.
Challenge. Use three quarters. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 13: Decimal scale
Case. 0.04 × 245 = 9.8.
Analysis. Estimate near 0.04 × 250 = 10. Notice that the answer depends on whether every partial product represents the correct number of equal groups and recombines to the original multiplier; changing a decorative detail would not be enough to change the reasoning.
Challenge. Explain decimal placement. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 14: Money scale
Case. $0.80 × 245 = $196.00.
Analysis. Eighty cents repeated 245 times. Notice that the answer depends on whether every partial product represents the correct number of equal groups and recombines to the original multiplier; changing a decorative detail would not be enough to change the reasoning.
Challenge. Check in cents. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 15: Length
Case. 1.2 m × 245 = 294 m.
Analysis. The unit stays metres. Notice that the answer depends on whether every partial product represents the correct number of equal groups and recombines to the original multiplier; changing a decorative detail would not be enough to change the reasoning.
Challenge. Convert after multiplying. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 16: Estimation
Case. Replace 245 with 250 for an upper nearby benchmark when n is positive.
Analysis. The exact result should be 5n lower. Notice that the answer depends on whether every partial product represents the correct number of equal groups and recombines to the original multiplier; changing a decorative detail would not be enough to change the reasoning.
Challenge. Use n = 84. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 17: Error location
Case. If two methods disagree, compare ×5 first.
Analysis. The difference between 250 and 245 is exactly five groups. Notice that the answer depends on whether every partial product represents the correct number of equal groups and recombines to the original multiplier; changing a decorative detail would not be enough to change the reasoning.
Challenge. Diagnose a sample mismatch. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 18: Calculator use
Case. Enter the direct product only after predicting scale.
Analysis. A display does not supply interpretation. Notice that the answer depends on whether every partial product represents the correct number of equal groups and recombines to the original multiplier; changing a decorative detail would not be enough to change the reasoning.
Challenge. Record the estimate. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 19: Word problem
Case. Identify whether 245 is number of groups, amount per group or a distractor.
Analysis. Structure precedes arithmetic. Notice that the answer depends on whether every partial product represents the correct number of equal groups and recombines to the original multiplier; changing a decorative detail would not be enough to change the reasoning.
Challenge. Write a matching equation. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 20: Final audit
Case. Decomposition, partial products, recombination, unit and estimate form the full method.
Analysis. Each line protects a different failure point. Notice that the answer depends on whether every partial product represents the correct number of equal groups and recombines to the original multiplier; changing a decorative detail would not be enough to change the reasoning.
Challenge. Audit one solution. 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: Subtract five only
What the learner may think. 245 is five below 250, so subtract 5 after multiplying by 250.
Why it fails. The missing amount is five groups of the multiplicand. 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 × 5. Use this short test immediately: What does one group contain?
Misconception 2: Treat 45 as 4 + 5
What the learner may think. Compute n × 4 + n × 5.
Why it fails. The digit 4 represents forty, not four. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Use ×40 + ×5. Use this short test immediately: What is the place value?
Misconception 3: Append zeros blindly
What the learner may think. ×200 always means write two zeros.
Why it fails. That shortcut can obscure decimal value and units. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Use ×2 ×100 meaningfully. Use this short test immediately: Does the estimate agree?
Misconception 4: Add both routes
What the learner may think. Use ×200 + ×45 + ×250 − ×5.
Why it fails. The two decompositions are alternatives, not extra pieces. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Choose one route; use the other only as a check. Use this short test immediately: How many groups are represented?
Misconception 5: Ignore the unit
What the learner may think. A correct numeral is a complete answer.
Why it fails. Context may require dollars, kilograms or metres. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Restore the quantity and format. Use this short test immediately: What does 588 measure?
Misconception 6: Trust a repeated error
What the learner may think. Two identical calculations prove the result.
Why it fails. Repeating the same decomposition may repeat the same mistake. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Check with a structurally different route or estimate. Use this short test immediately: Is the evidence independent?
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: Foundation
Question. 24 × 245
Explained answer. 5880
Why this answer earns trust. 24 × 200 = 4800; ×40 = 960; ×5 = 120. Transfer variation: Check with ×250 − ×5.
Practice 2: Alternative
Question. 64 × 245
Explained answer. 15,680
Why this answer earns trust. 64 × 250 = 16,000; subtract 320. Transfer variation: Use ×200 + ×45.
Practice 3: Factor-friendly
Question. 80 × 245
Explained answer. 19,600
Why this answer earns trust. 80 × 250 = 20,000; subtract 400. Transfer variation: Estimate near 20,000.
Practice 4: Decimal
Question. 1.2 × 245
Explained answer. 294
Why this answer earns trust. 240 + 48 + 6 = 294. Transfer variation: Check against 300.
Practice 5: Small decimal
Question. 0.08 × 245
Explained answer. 19.6
Why this answer earns trust. 0.08 × 250 = 20; subtract 0.4. Transfer variation: Explain scale.
Practice 6: Money
Question. 245 pens at $1.60 each
Explained answer. $392.00
Why this answer earns trust. 1.6 × 200 + ×40 + ×5 = 320 + 64 + 8. Transfer variation: Check in cents.
Practice 7: Measure
Question. 245 strips each 0.4 m
Explained answer. 98 m
Why this answer earns trust. 0.4 × 250 − 0.4 × 5 = 100 − 2. Transfer variation: Convert to centimetres afterward.
Practice 8: Reverse
Question. 24,500 ÷ 245
Explained answer. 100
Why this answer earns trust. 100 groups of 245 give 24,500. Transfer variation: Verify by multiplication.
Practice 9: Rule recall
Question. State both identities.
Explained answer. n × 245 = n × 200 + n × 45 = n × 250 − n × 5.
Why this answer earns trust. They represent the same 245 groups. Transfer variation: Recall after a delay.
Practice 10: Transfer
Question. Choose the better route for 124 × 245.
Explained answer. ×250 − ×5 is efficient: 31,000 − 620 = 30,380.
Why this answer earns trust. The multiplicand makes ×250 convenient. Transfer variation: Solve with the other route.
Delayed transfer: prove that the learning lasts
Transfer test 1: Route-choice set
Solve twenty mixed whole-number and decimal products, choosing and naming the shorter decomposition. 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 learner selects a route from number structure rather than habit. If the criterion is not met, return to the smallest missing decision rather than repeating the entire chapter.
Transfer test 2: Context restoration
Solve products embedded in money, mass, length and equal-group problems. 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 the correct unit, scale and interpretation. If the criterion is not met, return to the smallest missing decision rather than repeating the entire chapter.
Transfer test 3: Delayed dual check
After one week, solve unfamiliar ×245 questions and verify selected items with ×250 − ×5. 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 two identities are recalled and used independently. 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 say “two hundred groups plus forty-five groups” before writing partial products. This prevents the digits from becoming an unexplained trick.
When an error occurs, identify whether it began in decomposition, partial product, recombination, unit or scale. Avoid restarting the whole question without diagnosis.
Mix whole numbers, decimals, money and measures so the learner practises preserving meaning, not only reproducing a fixed column algorithm.
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 split 245 into 200 and 45?
The distributive property lets us calculate easier partial products that still represent all 245 groups.
Can I use 250 minus 5 instead?
Yes. Multiply the number by 250, then subtract five groups of that number.
Which method is better?
Choose the decomposition that produces simpler partial products for the multiplicand; use the other as a check.
Does the method work for decimals?
Yes. The distributive property applies to decimal quantities as well as whole numbers.
What is the quickest estimate?
For a positive multiplicand, compare with ×250; the exact ×245 result is five groups lower.
What is the main error to avoid?
When using 250 − 5, subtract five times the multiplicand, not merely 5.
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.