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 uses long multiplication for ×45 even when a nearby friendly multiplier makes the product faster and easier to check. 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 current official SEAB page for PSLE formats examined in 2026 is the primary route for format-specific information. This article teaches a reasoning habit rather than predicting a particular question. If a school instruction or an official paper direction differs from a practice routine here, follow the instruction printed for that task.
The problem beneath the mistake
Forty-five is close to fifty. That closeness is useful only when the difference is handled exactly. Multiplying by 50 gives five groups too many, so five times the original number must be removed.
The method is the distributive property: n × (50 − 5) = n × 50 − n × 5. It is not a trick tied to one example and works with whole numbers, decimals, money and measures.
A fast route is ×100 then halve for ×50, and ×10 then halve for ×5. Both branches use secure place-value facts before recombining by subtraction.
The common failure is referent loss. Learners subtract 5 instead of five times the number, or halve the wrong product. Labels such as 50n and 5n keep both branches attached to n.
For positive numbers, the answer must be slightly below ×50 and above ×40. This range catches reversed subtraction, missing zeroes and misplaced decimals.
Core decision: Because 45 = 50 − 5, multiply any suitable number by 45 by finding fifty times the number and subtracting five times the same original number: n × 45 = n × 50 − n × 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. Decompose 45 exactly
Write 45 as 50−5. The rewrite must be equal, not merely close.
Diagnostic question: What correction turns fifty groups into forty-five? Reliable response: Remove exactly five groups.
2. Distribute the original factor
n multiplies both 50 and 5: n(50−5)=50n−5n.
Diagnostic question: Did n multiply both terms? Reliable response: Label both partial products.
3. Build ×50 efficiently
Find ×100 then halve, or ×5 then ×10, choosing the easier exact route.
Diagnostic question: Which route keeps the arithmetic simple? Reliable response: Avoid creating needless fractions.
4. Build ×5 efficiently
Five times a number is half of ten times it.
Diagnostic question: Can ×10 be halved accurately? Reliable response: Keep place value and units visible.
5. Subtract in the correct direction
For positive n, 50n is larger than 5n, so compute 50n−5n.
Diagnostic question: Should the result be just below ×50? Reliable response: Use magnitude to order subtraction.
6. Check independently
Split 45 as 40+5 or use standard multiplication.
Diagnostic question: Does the check repeat the same decomposition? Reliable response: Use a genuinely different route.
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: Twenty-four times forty-five
Task. Calculate 24×45.
Reasoning. 24×50=1200 and 24×5=120; subtract. The controlling idea is preserving the same original multiplicand in both partial products and subtracting the five-group product from the fifty-group product. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 1080
Independent check. Check 24×40+24×5=960+120. 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: Sixty-eight times forty-five
Task. Calculate 68×45.
Reasoning. 68×50=3400 and 68×5=340. The controlling idea is preserving the same original multiplicand in both partial products and subtracting the five-group product from the fifty-group product. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 3060
Independent check. It lies between 2720 and 3400. 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 factor
Task. Calculate 3.6×45.
Reasoning. 3.6×50=180 and 3.6×5=18. The controlling idea is preserving the same original multiplicand in both partial products and subtracting the five-group product from the fifty-group product. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 162
Independent check. Check 144+18. 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: Money quantity
Task. Forty-five tickets cost $12.40 each.
Reasoning. $12.40×50=$620 and ×5=$62. The controlling idea is preserving the same original multiplicand in both partial products and subtracting the five-group product from the fifty-group product. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. $558
Independent check. The answer lies between $496 and $620. 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: Measurement
Task. A machine makes 2.75 m per cycle for 45 cycles.
Reasoning. 2.75×50=137.5 and ×5=13.75. The controlling idea is preserving the same original multiplicand in both partial products and subtracting the five-group product from the fifty-group product. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 123.75 m
Independent check. Check 110+13.75. 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: Large number
Task. Calculate 1,204×45.
Reasoning. 60,200−6,020. The controlling idea is preserving the same original multiplicand in both partial products and subtracting the five-group product from the fifty-group product. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 54,180
Independent check. It is nine-tenths of the ×50 result. 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: Reverse problem
Task. A number times 45 is 2,025.
Reasoning. Test 45: 45×50−45×5=2250−225. The controlling idea is preserving the same original multiplicand in both partial products and subtracting the five-group product from the fifty-group product. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 45
Independent check. Substitute into the equation. 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: Error diagnosis
Task. A learner writes 32×45=32×50−5.
Reasoning. The correction must remove five groups of 32. The controlling idea is preserving the same original multiplicand in both partial products and subtracting the five-group product from the fifty-group product. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 1600−160=1440
Independent check. Check 1280+160. 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: Connection to ninety percent
Case. 45 is 90% of 50.
Analysis. After ×50, take nine-tenths for another exact route. Notice that the answer depends on preserving the same original multiplicand in both partial products and subtracting the five-group product from the fifty-group product; changing a decorative detail would not be enough to change the reasoning.
Challenge. Use it for 84×45. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 2: Decimal below one
Case. Calculate 0.48×45.
Analysis. ×50 gives 24 and ×5 gives 2.4. Notice that the answer depends on preserving the same original multiplicand in both partial products and subtracting the five-group product from the fifty-group product; changing a decorative detail would not be enough to change the reasoning.
Challenge. Subtract and estimate. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 3: Alternative decomposition
Case. Use 40+5.
Analysis. It is equally valid and sometimes easier. Notice that the answer depends on preserving the same original multiplicand in both partial products and subtracting the five-group product from the fifty-group product; changing a decorative detail would not be enough to change the reasoning.
Challenge. Compare both routes for 36. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 4: Multiplying by 4.5
Case. Calculate 28×4.5.
Analysis. Find 28×45 then divide by 10. Notice that the answer depends on preserving the same original multiplicand in both partial products and subtracting the five-group product from the fifty-group product; changing a decorative detail would not be enough to change the reasoning.
Challenge. Check with 28×(5−0.5). State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 5: Multiplying by 450
Case. Calculate 16×450.
Analysis. Find 16×45 then multiply by 10. Notice that the answer depends on preserving the same original multiplicand in both partial products and subtracting the five-group product from the fifty-group product; changing a decorative detail would not be enough to change the reasoning.
Challenge. Explain the place-value shift. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 6: Grouped context
Case. 45 boxes contain 128 items each.
Analysis. Subtracting five groups from fifty preserves the group meaning. Notice that the answer depends on preserving the same original multiplicand in both partial products and subtracting the five-group product from the fifty-group product; changing a decorative detail would not be enough to change the reasoning.
Challenge. Label every partial product with items. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 7: Fractional multiplicand
Case. Calculate three-quarters ×45.
Analysis. The identity remains exact with fractions. Notice that the answer depends on preserving the same original multiplicand in both partial products and subtracting the five-group product from the fifty-group product; changing a decorative detail would not be enough to change the reasoning.
Challenge. Give the mixed-number result. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 8: Rounding trap
Case. A learner rounds 45 to 50 and stops.
Analysis. Fifty is only an intermediate value until five groups are removed. Notice that the answer depends on preserving the same original multiplicand in both partial products and subtracting the five-group product from the fifty-group product; changing a decorative detail would not be enough to change the reasoning.
Challenge. Separate estimate and exact answer. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 9: Calculator verification
Case. A calculator is allowed.
Analysis. The decomposition still predicts magnitude and catches entry errors. Notice that the answer depends on preserving the same original multiplicand in both partial products and subtracting the five-group product from the fifty-group product; changing a decorative detail would not be enough to change the reasoning.
Challenge. Estimate 327×45. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 10: Algebraic generalisation
Case. Simplify 45a+45b.
Analysis. Factor 45(a+b) before applying a mental method. Notice that the answer depends on preserving the same original multiplicand in both partial products and subtracting the five-group product from the fifty-group product; changing a decorative detail would not be enough to change the reasoning.
Challenge. Evaluate when a+b=62. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 11: Selection in a story
Case. A question contains 45 and 50 but asks for their difference.
Analysis. The appearance of 45 does not automatically signal multiplication. Notice that the answer depends on preserving the same original multiplicand in both partial products and subtracting the five-group product from the fifty-group product; changing a decorative detail would not be enough to change the reasoning.
Challenge. Write a non-multiplication example. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 12: Negative extension
Case. Evaluate −8×45.
Analysis. The algebraic identity still holds: −400−(−40)=−360. Notice that the answer depends on preserving the same original multiplicand in both partial products and subtracting the five-group product from the fifty-group product; changing a decorative detail would not be enough to change the reasoning.
Challenge. Check with −8×(40+5). State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 13: Ninety-nine times forty-five
Case. Calculate 99×45.
Analysis. The nearby-number method can be nested carefully: 100×45−1×45=4500−45. This may be simpler than building ×50 and ×5 from 99. Notice that the answer depends on preserving the same original multiplicand in both partial products and subtracting the five-group product from the fifty-group product; changing a decorative detail would not be enough to change the reasoning.
Challenge. Compare the two exact routes. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 14: Associative regrouping
Case. Calculate 16×45 as (16×5)×9.
Analysis. Because 45=5×9, the factors may be regrouped. This alternative is efficient when ×5 creates a friendly number and shows that decomposition need not always use addition or subtraction. Notice that the answer depends on preserving the same original multiplicand in both partial products and subtracting the five-group product from the fifty-group product; changing a decorative detail would not be enough to change the reasoning.
Challenge. Verify 80×9. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 15: Area context
Case. A rectangle is 45 cm long and 28 cm wide.
Analysis. Area requires 45×28, so 28×50−28×5 gives 1400−140. The method preserves square centimetres because length is multiplied by width. Notice that the answer depends on preserving the same original multiplicand in both partial products and subtracting the five-group product from the fifty-group product; changing a decorative detail would not be enough to change the reasoning.
Challenge. State the area with the correct unit. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 16: Rate context
Case. A tap delivers 1.8 litres per minute for 45 minutes.
Analysis. The product is 1.8×45=90−9=81 litres. The same identity applies, but the final unit comes from rate multiplied by time. Notice that the answer depends on preserving the same original multiplicand in both partial products and subtracting the five-group product from the fifty-group product; changing a decorative detail would not be enough to change the reasoning.
Challenge. Check whether the answer is reasonable for 50 minutes. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 17: Two-step problem
Case. A school buys 45 packs of 24 pencils and shares them equally among 30 pupils.
Analysis. First find the total 24×45=1080, then divide by 30 to obtain 36. The mental product is only one stage of the full problem. Notice that the answer depends on preserving the same original multiplicand in both partial products and subtracting the five-group product from the fifty-group product; changing a decorative detail would not be enough to change the reasoning.
Challenge. Explain why stopping at 1080 does not answer the sharing question. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 18: Choosing standard multiplication
Case. Calculate 7,843×45 under written-working conditions.
Analysis. The decomposition is still valid, but standard multiplication may be clearer and less error-prone for a learner. Strategy choice should consider accuracy, auditability and time rather than novelty. Notice that the answer depends on preserving the same original multiplicand in both partial products and subtracting the five-group product from the fifty-group product; changing a decorative detail would not be enough to change the reasoning.
Challenge. Estimate the result before selecting a method. 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. n×45=n×50−5.
Why it fails. The correction is five groups 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 50n−5n. Use this short test immediately: How many groups were added?
Misconception 2: Halve ×50
What the learner may think. Half of ×50 gives ×45.
Why it fails. Half of 50 is 25. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Subtract ×5 or take nine-tenths. Use this short test immediately: What is 45/50?
Misconception 3: Reverse subtraction
What the learner may think. Use 5n−50n.
Why it fails. That makes a positive product negative. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Picture removing five groups from fifty. Use this short test immediately: Should positive×positive be positive?
Misconception 4: Drop units
What the learner may think. Report 123.75 without metres or dollars.
Why it fails. The product needs its measurement meaning. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Carry units through partial products. Use this short test immediately: What does one group measure?
Misconception 5: Use for every 45
What the learner may think. Every appearance of 45 triggers this method.
Why it fails. The story may require another operation. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Identify equal groups first. Use this short test immediately: What relationship is asked?
Misconception 6: Estimate is exact
What the learner may think. Round 45 to 50 and report it.
Why it fails. Rounding changes the value. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Label ×50 as intermediate. Use this short test immediately: Where is the five-group correction?
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: Whole number
Question. 18×45
Explained answer. 810
Why this answer earns trust. 900−90. Transfer variation: Check 720+90.
Practice 2: Whole number
Question. 72×45
Explained answer. 3240
Why this answer earns trust. 3600−360. Transfer variation: Use nine-tenths of ×50.
Practice 3: Decimal
Question. 4.8×45
Explained answer. 216
Why this answer earns trust. 240−24. Transfer variation: Estimate between ×40 and ×50.
Practice 4: Money
Question. 45 items at $7.60
Explained answer. $342
Why this answer earns trust. $380−$38. Transfer variation: Check 40 and 5 groups.
Practice 5: Measure
Question. 1.25 kg for each of 45 packs
Explained answer. 56.25 kg
Why this answer earns trust. 62.5−6.25. Transfer variation: Use 1.25×(40+5).
Practice 6: Repair
Question. Fix 27×45=1350−5.
Explained answer. 1350−135=1215
Why this answer earns trust. Subtract five groups of 27. Transfer variation: Name the missing referent.
Practice 7: Reverse
Question. x×45=3600
Explained answer. x=80
Why this answer earns trust. 80×50−80×5=4000−400. Transfer variation: Verify by division.
Practice 8: Generalise
Question. Express n×45 using 50 and 5.
Explained answer. 50n−5n
Why this answer earns trust. The distributive property preserves n. Transfer variation: Express ×95 similarly.
Practice 9: Select
Question. When is ×50−×5 useful?
Explained answer. When both partial products and their subtraction are easy and exact.
Why this answer earns trust. A strategy should reduce cognitive load. Transfer variation: Compare for 999×45.
Practice 10: Independent rule
Question. State and check the method.
Explained answer. Find ×50, subtract ×5 of the same number, then verify with ×40+×5.
Why this answer earns trust. It includes operation and check. Transfer variation: Apply after two days.
Delayed transfer: prove that the learning lasts
Transfer test 1: Mixed multipliers
Solve ×35, ×45, ×55 and ×95 using nearby friendly multiples. Do this after a delay, without rereading the worked examples. The purpose is to test retrieval and decision-making, not immediate imitation.
Success criterion. Write exact decompositions before calculating. If the criterion is not met, return to the smallest missing decision rather than repeating the entire chapter.
Transfer test 2: Context conversion
Create one money and one measurement problem using ×45. Do this after a delay, without rereading the worked examples. The purpose is to test retrieval and decision-making, not immediate imitation.
Success criterion. Carry units and interpret each product. If the criterion is not met, return to the smallest missing decision rather than repeating the entire chapter.
Transfer test 3: Delayed test
After two days solve 84×45 and 2.4×45. Do this after a delay, without rereading the worked examples. The purpose is to test retrieval and decision-making, not immediate imitation.
Success criterion. Obtain 3780 and 108 with checks. 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 fifty groups minus five groups before writing numbers. This prevents subtracting only 5.
Allow either 50−5 or 40+5 after comparing effort. The goal is exact, explainable calculation.
Mix ×45 items with problems where 45 is not a multiplier so the learner must select the operation.
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 does ×50−×5 equal ×45?
Because 50−5=45 and the same original number multiplies both parts.
How do I find ×50?
Multiply by 100 and halve.
How do I find ×5?
Multiply by 10 and halve.
Can I use ×40+×5?
Yes; it is exact and a useful check.
Does it work with decimals and money?
Yes, if place value and units are preserved.
What is the fastest check?
The positive answer must lie below ×50, above ×40 and agree with another route.
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.