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 285 as an unfriendly multiplier, subtracts 15 units instead of fifteen copies of the multiplicand, or loses the direction between the ×300 benchmark and the exact product. 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
Multiplication by 285 becomes structured when 285 is read as 300 minus 15. The exact product is fifteen groups below a friendly ×300 benchmark.
The distributive property gives n×285=n×300−n×15. The multiplicand appears in both partial products because complete groups, not bare units, are being removed.
Multiplication by 15 can be built as ×10+×5. For 24×285, calculate 7200−360=6840.
A second route is 285=250+35. For 24, calculate 6000+840=6840. Agreement between independent routes strengthens the result.
Direction matters. With a positive multiplicand, ×285 must be less than ×300 and greater than ×250. A result outside that interval needs repair.
The method works with decimals: 0.8×285=240−12=228. Decimal scale stays attached to both partial products.
Money needs monetary notation. If 285 items cost $2.40 each, the total is $684.00.
Measurement keeps its unit: 285 strips of 1.6 m have total length 456 m.
Do not subtract 15 from the ×300 answer. The gap represents fifteen full copies of the multiplicand, so it equals 15n.
When ×15 is awkward, calculate ×10 and half of ×10, or calculate ×30 then halve. Choose a route that keeps place value visible.
Estimation can bracket the answer. Since 285 is close to 300, the exact result should be slightly below n×300 rather than far below it.
A complete solution writes the identity, both partial products, one subtraction, the unit, an interval check and a different-route verification.
Core decision: Rewrite 285 as 300 − 15, so n × 285 = n × 300 − n × 15; calculate ×15 as ×10 + ×5 and verify with another exact decomposition such as 250 + 35.
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. Name the multiplicand
Identify the amount in one group and the fact that 285 groups are required.
Diagnostic question: What quantity is repeated? Reliable response: Write the contextual equation first.
2. Decompose 285
Use 300−15 and confirm that the parts reconstruct 285.
Diagnostic question: Is the gap exactly fifteen groups? Reliable response: Record n(300−15).
3. Calculate ×300
Multiply by 3 and apply place value for hundreds.
Diagnostic question: What is three hundred copies? Reliable response: Keep decimal and unit scale visible.
4. Calculate ×15
Find ×10 and ×5, then add them.
Diagnostic question: Have fifteen copies, not fifteen units, been found? Reliable response: Label the partial product.
5. Subtract once
Remove the ×15 product from the ×300 product.
Diagnostic question: Is the direction lower than the benchmark? Reliable response: Attach the required unit.
6. Check differently
Use ×250+×35, reverse division or the exact benchmark gap.
Diagnostic question: Does ×300 minus the answer equal 15n? Reliable response: Investigate any mismatch.
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 24×285.
Reasoning. 24×300=7200 and 24×15=360. The controlling idea is whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 6840.
Independent check. Check 6000+840. 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: Larger factor
Task. Calculate 48×285.
Reasoning. 48×300=14,400 and 48×15=720. The controlling idea is whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 13,680.
Independent check. It lies between 12,000 and 14,400. 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: Multiple of ten
Task. Calculate 120×285.
Reasoning. 36,000−1800. The controlling idea is whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 34,200.
Independent check. Check 30,000+4200. 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 0.8×285.
Reasoning. 240−12. The controlling idea is whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 228.
Independent check. It lies between 200 and 240. 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 fraction
Task. Calculate 1.25×285.
Reasoning. 375−18.75. The controlling idea is whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 356.25.
Independent check. Check 312.5+43.75. 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. 285 notebooks cost $2.40 each.
Reasoning. $720−$36. The controlling idea is whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. $684.00.
Independent check. Work in cents as another route. 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: Length
Task. 285 strips are 1.6 m each.
Reasoning. 480−24. The controlling idea is whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 456 m.
Independent check. The unit remains metres. 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 division
Task. 19,950 is divided into 285 equal groups.
Reasoning. 70×285=21,000−1050. The controlling idea is whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 70 per group.
Independent check. Multiply the quotient back. 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: ×300
Case. Three hundred groups form a friendly upper benchmark.
Analysis. Place value controls the zeros and decimal scale. Notice that the answer depends on whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand; changing a decorative detail would not be enough to change the reasoning.
Challenge. Use n=36. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 2: ×15
Case. Fifteen groups equal ten groups plus five groups.
Analysis. The same multiplicand appears in both parts. Notice that the answer depends on whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand; changing a decorative detail would not be enough to change the reasoning.
Challenge. Use n=64. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 3: ×285
Case. The exact product is 15n below ×300.
Analysis. The gap can check the answer. Notice that the answer depends on whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand; changing a decorative detail would not be enough to change the reasoning.
Challenge. Draw a bar model. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 4: ×250+×35
Case. A lower benchmark offers an independent exact route.
Analysis. The result should match the subtraction method. Notice that the answer depends on whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand; changing a decorative detail would not be enough to change the reasoning.
Challenge. Use n=44. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 5: ×300−×10−×5
Case. The ×15 subtraction can be split safely.
Analysis. Both subtractions remove complete groups. Notice that the answer depends on whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand; changing a decorative detail would not be enough to change the reasoning.
Challenge. Use n=28. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 6: ×57×5
Case. Since 285=57×5, factorisation can help selected numbers.
Analysis. Method choice follows number structure. Notice that the answer depends on whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand; changing a decorative detail would not be enough to change the reasoning.
Challenge. Test n=20. 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. 285n=n285.
Analysis. Context still identifies group size and count. Notice that the answer depends on whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand; changing a decorative detail would not be enough to change the reasoning.
Challenge. Rewrite a ticket 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: Distributivity
Case. n(300−15)=300n−15n.
Analysis. Each term contains n. Notice that the answer depends on whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand; changing a decorative detail would not be enough to change the reasoning.
Challenge. Explain with 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: Zero
Case. 0×285=0.
Analysis. Both partial products are zero. Notice that the answer depends on whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand; changing a decorative detail would not be enough to change the reasoning.
Challenge. Check the identity. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 10: One
Case. 1×285=285.
Analysis. The benchmark gap is exactly 15. Notice that the answer depends on whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand; changing a decorative detail would not be enough to change the reasoning.
Challenge. Use 0.1. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 11: Half
Case. 0.5×285=142.5.
Analysis. A whole-number multiplier can produce a decimal product. Notice that the answer depends on whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand; changing a decorative detail would not be enough to change the reasoning.
Challenge. Use one quarter. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 12: Small decimal
Case. 0.04×285=11.4.
Analysis. 12−0.6. Notice that the answer depends on whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand; changing a decorative detail would not be enough to change the reasoning.
Challenge. Explain the decimal scale. 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 cents
Case. $0.60×285=$171.00.
Analysis. Whole cents can simplify the calculation. Notice that the answer depends on whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand; changing a decorative detail would not be enough to change the reasoning.
Challenge. Convert back accurately. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 14: Mass
Case. 2.5 kg×285=712.5 kg.
Analysis. 750−37.5. Notice that the answer depends on whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand; changing a decorative detail would not be enough to change the reasoning.
Challenge. Convert the answer to grams. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 15: Upper benchmark
Case. For positive n, 285n is below 300n.
Analysis. This detects wrong addition. Notice that the answer depends on whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand; changing a decorative detail would not be enough to change the reasoning.
Challenge. State 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 16: Lower benchmark
Case. For positive n, 285n is above 250n.
Analysis. The extra amount is 35n. Notice that the answer depends on whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand; changing a decorative detail would not be enough to change the reasoning.
Challenge. Bracket a result. 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 difference
Case. 300n−285n must equal15n.
Analysis. A wrong gap reveals a partial-product error. Notice that the answer depends on whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand; changing a decorative detail would not be enough to change the reasoning.
Challenge. Use n=72. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 18: Unit check
Case. The multiplicand’s unit becomes the product’s unit.
Analysis. The multiplier counts equal groups. Notice that the answer depends on whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand; changing a decorative detail would not be enough to change the reasoning.
Challenge. Label a distance problem. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 19: Reverse division
Case. A quotient can be verified by multiplying by285.
Analysis. Inverse structure tests both arithmetic and referent. Notice that the answer depends on whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand; changing a decorative detail would not be enough to change the reasoning.
Challenge. Construct a divisible total. 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. Identity, ×300, ×15, subtraction, range, unit and alternative route complete the check.
Analysis. Each catches a different error. Notice that the answer depends on whether the decomposition preserves exactly 285 equal groups and the product lies fifteen multiplicand-groups below ×300 for a positive multiplicand; changing a decorative detail would not be enough to change the reasoning.
Challenge. Audit five calculations. 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 bare fifteen
What the learner may think. Compute n×300−15.
Why it fails. The difference between 300 groups and285 groups is fifteen copies of n. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Write n beside every group count. Use this short test immediately: Is the gap 15n?
Misconception 2: Add because 285 is positive
What the learner may think. Use 300+15.
Why it fails. 285 is below300, so the adjustment must lower the benchmark. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Place285 on a number line. Use this short test immediately: Which direction?
Misconception 3: ×15 equals ×10+5
What the learner may think. Find ten copies and add five units.
Why it fails. Both parts must be multiples of n. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Write 10n+5n. Use this short test immediately: Are there fifteen groups?
Misconception 4: Drop decimal scale
What the learner may think. Treat0.8×15 as120.
Why it fails. Partial products must keep the multiplicand’s scale. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Estimate before calculating. Use this short test immediately: Should the answer be near240?
Misconception 5: Ignore units
What the learner may think. A bare684 answers a money problem fully.
Why it fails. The number needs the quantity label and currency precision. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Restore dollars and cents. Use this short test immediately: What does684 measure?
Misconception 6: Same-route check
What the learner may think. Repeat300−15 identically.
Why it fails. The same hidden error may repeat. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Use250+35 or inverse division. Use this short test immediately: Is the route 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. Calculate 8×285.
Explained answer. 2280.
Why this answer earns trust. 2400−120=2280. Transfer variation: Check2000+280.
Practice 2: Whole number
Question. Calculate 32×285.
Explained answer. 9120.
Why this answer earns trust. 9600−480=9120. Transfer variation: State the benchmark gap.
Practice 3: Decimal
Question. Calculate1.2×285.
Explained answer. 342.
Why this answer earns trust. 360−18=342. Transfer variation: Check300+42.
Practice 4: Money
Question. 285 pens cost $1.20 each. Find the total.
Explained answer. $342.00.
Why this answer earns trust. The arithmetic matches1.2×285 and the answer needs currency notation. Transfer variation: Work in cents.
Practice 5: Measurement
Question. Each of285 pieces is0.4 m. Find total length.
Explained answer. 114 m.
Why this answer earns trust. 120−6=114. Transfer variation: Convert to centimetres.
Practice 6: Error analysis
Question. A learner writes24×285=7200−15. Explain the error.
Explained answer. The learner subtracted15 units instead of15 groups of24; subtract360.
Why this answer earns trust. The distributive factor24 must remain. Transfer variation: Repair with an area model.
Practice 7: Range
Question. Must50×285 lie between12,500 and15,000?
Explained answer. Yes; it equals14,250.
Why this answer earns trust. 285 lies between250 and300. Transfer variation: State distances from both bounds.
Practice 8: Reverse
Question. Check whether22,800÷285=80.
Explained answer. Yes, because80×285=24,000−1200=22,800.
Why this answer earns trust. Multiplication verifies the quotient. Transfer variation: Create another exact quotient.
Practice 9: Rule recall
Question. State the identity and one check.
Explained answer. n×285=n×300−n×15; verify with n×250+n×35.
Why this answer earns trust. Both routes preserve285 groups. Transfer variation: Recall after a delay.
Practice 10: Transfer
Question. Solve mixed ×275, ×285 and ×295 items without being told the decomposition.
Explained answer. Each benchmark adjustment matches the multiplier’s distance from300.
Why this answer earns trust. Method selection transfers beyond one multiplier. Transfer variation: Explain every sign.
Delayed transfer: prove that the learning lasts
Transfer test 1: Benchmark sort
Sort multipliers below300 by their exact gap and write a matching identity. 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 sign and group count reconstructs the multiplier. If the criterion is not met, return to the smallest missing decision rather than repeating the entire chapter.
Transfer test 2: Delayed mixed practice
After one week, solve whole-number, decimal, money and measure cases. Do this after a delay, without rereading the worked examples. The purpose is to test retrieval and decision-making, not immediate imitation.
Success criterion. Products pass range, unit and independent-route checks. If the criterion is not met, return to the smallest missing decision rather than repeating the entire chapter.
Transfer test 3: Question construction
Create a word problem whose answer uses ×285 and solve it two ways. 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 story identifies one group,285 groups and a labelled total. 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 first, “How far is285 from300?” Then ask, “What is one group?” The two answers prevent sign and bare-adjustment errors.
Require a second route that is genuinely different, such as250+35. Repeating300−15 with prettier working is not an independent check.
Mix decimals and units only after whole-number structure is secure. The method should remain visible when the surface context changes.
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 use300−15?
Because300 is easy to multiply by and the exact gap to285 is15 groups.
Why not subtract15 from the product?
The adjustment is fifteen copies of the multiplicand, so it is15n.
How do I calculate ×15?
Use ×10+×5, where ×5 is half of ×10.
What is a good second method?
Use ×250+×35 or another exact decomposition that does not repeat the same route.
Does the method work for decimals?
Yes, if every partial product preserves decimal scale and the final estimate is sensible.
What is the fastest error check?
For positive n, the answer must be15n below n×300.
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.