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 sees 255 as an awkward three-digit multiplier and loses the extra five groups, decimal scale or unit while combining partial products. 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 255 becomes compact when 255 is recognised as 250 plus 5. The product is two hundred and fifty groups of the number plus five more groups.
The distributive property gives n × 255 = n × 250 + n × 5. Because ×250 is one quarter of ×1000, many numbers produce simple partial products.
For 36 × 255, compute 36 × 250 = 9000 and 36 × 5 = 180. Adding gives 9180. The answer must be slightly above 9000, not below it.
A structurally different check uses 255 = 300 − 45. Then 36 × 300 − 36 × 45 = 10,800 − 1620 = 9180.
The sign is the main contrast with multiplication by 245. Since 255 is five more than 250, add five groups. For 245, subtract five groups. A number-line view makes the direction visible.
The method works for decimals. For 0.8 × 255, compute 0.8 × 250 = 200 and 0.8 × 5 = 4, giving 204.
Money keeps its unit and format. If 255 items cost $2.40 each, the total is $612.00. The multiplication finds amount; the answer still needs dollars and cents.
Measurement contexts require interpretation. If each rope is 1.6 m and there are 255 equal ropes, the total length is 408 m, not 408 ropes or 408 centimetres.
Estimation is immediate: replace 255 with 250. For a positive multiplicand, the exact result must be five groups larger than the ×250 benchmark.
Numbers divisible by four suit ×250 especially well. If a number is not, ×200 + ×50 + ×5 may be easier and represents the same 255 groups.
Do not add 5 to the ×250 product. Add five times the multiplicand. The difference between 250 groups and 255 groups is five complete groups, not five units.
A complete solution records the identity, computes accurate partial products, adds once, restores units and checks direction and scale against ×250.
Core decision: Rewrite 255 as 250 + 5, so a × 255 = a × 250 + a × 5; use a × 300 − a × 45 or a × 200 + a × 55 when an independent check is useful.
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. Identify the repeated quantity
Name what one group contains and what 255 counts.
Diagnostic question: What is multiplied 255 times? Reliable response: Write the contextual equation.
2. Choose the decomposition
Use 250 + 5 or, when easier, 200 + 50 + 5.
Diagnostic question: Which partial products are simplest? Reliable response: Record the equality.
3. Calculate ×250
Use ×1000 ÷4 or another valid route.
Diagnostic question: Is division by four exact or manageable? Reliable response: Keep decimal value visible.
4. Calculate ×5
Use half of ×10 when helpful.
Diagnostic question: What are five complete groups? Reliable response: Do not confuse with adding 5.
5. Add and label
Combine the two partial products and attach the unit.
Diagnostic question: Did the decomposition require addition? Reliable response: Write one clear total.
6. Check scale
Compare with ×250 and a different decomposition.
Diagnostic question: Is the result exactly 5n above the benchmark? 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 36 × 255.
Reasoning. 36 × 250 = 9000; 36 × 5 = 180. The controlling idea is whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 9180.
Independent check. The result is 180 above ×250. 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 number
Task. Calculate 64 × 255.
Reasoning. 64 × 250 = 16,000; 64 × 5 = 320. The controlling idea is whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 16,320.
Independent check. Check 64 × 300 − 64 × 45. 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
Task. Calculate 120 × 255.
Reasoning. 120 × 250 = 30,000; 120 × 5 = 600. The controlling idea is whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 30,600.
Independent check. Estimate near 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 0.8 × 255.
Reasoning. 200 + 4. The controlling idea is whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 204.
Independent check. Slightly above 0.8 × 250. 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 hundredths
Task. Calculate 1.25 × 255.
Reasoning. 312.5 + 6.25. The controlling idea is whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 318.75.
Independent check. Check 1.25 × 200 + ×50 + ×5. 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. 255 books cost $2.40 each.
Reasoning. 2.4 × 250 = 600; 2.4 × 5 = 12. The controlling idea is whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. $612.00.
Independent check. Use two decimal places. 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. 255 strips are 1.6 m each.
Reasoning. 1.6 × 250 = 400; 1.6 × 5 = 8. The controlling idea is whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 408 m.
Independent check. Keep 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
Task. A total of 17,850 is shared among 255 equal groups.
Reasoning. 17,850 ÷255 = 70 because 70 ×255 = 17,500 +350. The controlling idea is whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values. 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. Verify with direct multiplication. 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: ×250
Case. Quarter of ×1000.
Analysis. Useful with numbers divisible by four. Notice that the answer depends on whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values; changing a decorative detail would not be enough to change the reasoning.
Challenge. Calculate 84 ×250. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 2: ×5
Case. Half of ×10.
Analysis. The partial product is five groups. Notice that the answer depends on whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values; changing a decorative detail would not be enough to change the reasoning.
Challenge. Use 3.6. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 3: ×255
Case. Five groups above ×250.
Analysis. Direction gives an error check. Notice that the answer depends on whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values; changing a decorative detail would not be enough to change the reasoning.
Challenge. Draw a number line. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 4: ×200 + ×50 + ×5
Case. A place-value decomposition.
Analysis. It may suit decimals or mental methods. Notice that the answer depends on whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values; changing a decorative detail would not be enough to change the reasoning.
Challenge. Compare routes. 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 − ×45
Case. A different check.
Analysis. It avoids repeating the same structure. Notice that the answer depends on whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values; changing a decorative detail would not be enough to change the reasoning.
Challenge. Use n=48. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 6: Commutativity
Case. 255n = n255.
Analysis. Orient the expression to fit context. Notice that the answer depends on whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values; 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 7: Distributive property
Case. n(250+5)=250n+5n.
Analysis. All groups are preserved. Notice that the answer depends on whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values; changing a decorative detail would not be enough to change the reasoning.
Challenge. Explain with area. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 8: Zero
Case. 0 ×255 =0.
Analysis. No quantity repeated remains zero. Notice that the answer depends on whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values; changing a decorative detail would not be enough to change the reasoning.
Challenge. Check both parts. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 9: One
Case. 1 ×255 =255.
Analysis. A boundary check. Notice that the answer depends on whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values; 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 10: Fraction
Case. Half of 255 is127.5.
Analysis. Products need not be whole. Notice that the answer depends on whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values; 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 11: Small decimal
Case. 0.04 ×255 =10.2.
Analysis. Benchmark is10 at ×250. Notice that the answer depends on whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values; changing a decorative detail would not be enough to change the reasoning.
Challenge. Explain the extra0.2. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 12: Money
Case. $0.60 ×255 =$153.00.
Analysis. Cents can prevent decimal confusion. Notice that the answer depends on whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values; changing a decorative detail would not be enough to change the reasoning.
Challenge. Verify 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 13: Mass
Case. 2.5 kg ×255 =637.5 kg.
Analysis. The unit stays kilograms. Notice that the answer depends on whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values; changing a decorative detail would not be enough to change the reasoning.
Challenge. Convert after. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 14: Estimate
Case. ×250 supplies a nearby lower benchmark for positive n.
Analysis. Exact result is5n higher. Notice that the answer depends on whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values; changing a decorative detail would not be enough to change the reasoning.
Challenge. Use n=92. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 15: Direction
Case. Adding is essential because255>250.
Analysis. A subtraction answer belongs to245 logic. Notice that the answer depends on whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values; changing a decorative detail would not be enough to change the reasoning.
Challenge. Compare the pair. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 16: Error difference
Case. If the answer equals ×250, the ×5 part is missing.
Analysis. The gap identifies the omission. Notice that the answer depends on whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values; changing a decorative detail would not be enough to change the reasoning.
Challenge. Diagnose 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: Calculator
Case. Predict scale before entry.
Analysis. A display does not label units. Notice that the answer depends on whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values; changing a decorative detail would not be enough to change the reasoning.
Challenge. Record benchmark. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 18: Word problem
Case. Decide whether255 is group count or amount per group.
Analysis. Equation structure precedes arithmetic. Notice that the answer depends on whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values; changing a decorative detail would not be enough to change the reasoning.
Challenge. Write a model. 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 operation
Case. Division can recover one group.
Analysis. Always verify by multiplying back. Notice that the answer depends on whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values; changing a decorative detail would not be enough to change the reasoning.
Challenge. Create 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, partial products, direction, sum, unit and estimate form the method.
Analysis. Each protects a failure point. Notice that the answer depends on whether the chosen decomposition represents exactly 255 equal groups and the product stays slightly above the corresponding ×250 benchmark for positive values; changing a decorative detail would not be enough to change the reasoning.
Challenge. Audit one answer. 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: Add five only
What the learner may think. Compute n×250+5.
Why it fails. The extra 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. Add n×5. Use this short test immediately: What is one group?
Misconception 2: Subtract because five is small
What the learner may think. Use ×250−×5.
Why it fails. That represents245 groups, not255. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Compare multiplier direction. Use this short test immediately: Is255 above or below250?
Misconception 3: Divide by four after ×250
What the learner may think. Compute ×250 then divide by four again.
Why it fails. ×250 is already ×1000÷4. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Write the identity once. Use this short test immediately: How many groups result?
Misconception 4: Lose decimal scale
What the learner may think. Treat0.8×250 as2000.
Why it fails. Place value must follow the quantity. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Estimate around200. Use this short test immediately: What is0.8×1000?
Misconception 5: No unit needed
What the learner may think. The numeral completes a context problem.
Why it fails. Units distinguish money, mass and length. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Restore quantity and format. Use this short test immediately: What does612 measure?
Misconception 6: Same route is independent check
What the learner may think. Repeat ×250+×5 twice.
Why it fails. The same 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. Use ×300−×45 or direct multiplication. Use this short test immediately: Did the structure change?
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 ×255
Explained answer. 6120
Why this answer earns trust. 24×250=6000; 24×5=120. Transfer variation: Check with place value.
Practice 2: Whole
Question. 48 ×255
Explained answer. 12,240
Why this answer earns trust. 12,000+240. Transfer variation: Use ×300−×45.
Practice 3: Factor-friendly
Question. 96 ×255
Explained answer. 24,480
Why this answer earns trust. 24,000+480. Transfer variation: Compare ×250.
Practice 4: Decimal
Question. 1.2 ×255
Explained answer. 306
Why this answer earns trust. 300+6. Transfer variation: Estimate just above300.
Practice 5: Small decimal
Question. 0.08 ×255
Explained answer. 20.4
Why this answer earns trust. 20+0.4. Transfer variation: Explain scale.
Practice 6: Money
Question. 255 pens at $1.80
Explained answer. $459.00
Why this answer earns trust. 450+9. Transfer variation: Check in cents.
Practice 7: Measure
Question. 255 lengths of0.4 m
Explained answer. 102 m
Why this answer earns trust. 100+2. Transfer variation: Convert to centimetres later.
Practice 8: Reverse
Question. 25,500 ÷255
Explained answer. 100
Why this answer earns trust. 100 groups form25,500. Transfer variation: Multiply back.
Practice 9: Rule recall
Question. State two identities.
Explained answer. n×255=n×250+n×5=n×200+n×50+n×5.
Why this answer earns trust. Both preserve255 groups. Transfer variation: Recall later.
Practice 10: Transfer
Question. Choose a route for124×255.
Explained answer. ×250+×5:31,000+620=31,620.
Why this answer earns trust. 124 divides by4 cleanly after ×1000. Transfer variation: Check differently.
Delayed transfer: prove that the learning lasts
Transfer test 1: Direction pairs
Mix ×245 and ×255 questions around the same ×250 benchmark. 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 adds or subtracts five groups correctly. If the criterion is not met, return to the smallest missing decision rather than repeating the entire chapter.
Transfer test 2: Context set
Use money, mass, length and equal groups with whole and decimal amounts. 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 product has correct scale, unit 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 method
After one week, solve unfamiliar ×255 products and check selected answers with ×300−×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. Method and independent verification remain available. 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 “five more groups than what?” so the learner sees why ×5 is added to ×250.
Pair ×245 and ×255 examples to make direction visible, but require the learner to state the identity before calculating.
Diagnose decomposition, partial product, direction, recombination and unit separately instead of restarting blindly.
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 use250+5?
Both partial products are simple and together represent exactly255 groups.
How do I multiply by250?
Multiply by1000 and divide by4, or use another equivalent method.
Why add rather than subtract?
Because255 is five greater than250.
Do I add5 or five times the number?
Add five times the multiplicand because the difference is five complete groups.
Does it work for decimals?
Yes. The distributive property applies to decimal quantities.
How can I check?
Use a different decomposition such as300−45 and compare scale with ×250.
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.