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 275 as an awkward multiplier, subtracts 25 units instead of twenty-five groups, 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 275 becomes compact when 275 is seen as 300 minus 25. The exact product contains twenty-five fewer groups than a friendly ×300 product.
The distributive property gives n ×275=n ×300−n ×25. Both partial products contain the multiplicand; the subtraction removes twenty-five complete groups.
Multiplication by25 is often one quarter of multiplication by100. For24 ×275, calculate7200−600=6600.
A different decomposition is275=250+25. For24, that gives6000+600=6600. Agreement between the two routes strengthens the check.
Direction matters. Since275 is below300, the exact positive product must be below ×300. Since275 is above250, it must be above ×250.
The method works for decimals. For0.8 ×275, calculate240−20=220.
Money retains its notation. If275 items cost $2.40 each, the total is $660.00.
Measurement keeps its referent. Two hundred and seventy-five strips of1.6 m have total length440 m.
Do not subtract25 from the ×300 product. The gap between300 groups and275 groups is25 copies of the multiplicand.
If the multiplicand quarters easily, ×25 is efficient. If it does not, ×250+×25 or place-value decomposition may make the working clearer.
Estimation can bracket the answer between ×250 and ×300. This range catches place-value and sign errors before a full recomputation.
A complete solution states the identity, calculates both partial products, subtracts once, restores the unit and checks the product against both nearby benchmarks.
Core decision: Rewrite 275 as 300 − 25, so n × 275 = n × 300 − n × 25; check independently with 250 + 25 or 200 + 75.
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 one group
Identify the quantity repeated275 times.
Diagnostic question: What belongs to one item or interval? Reliable response: Write the contextual equation.
2. Choose300−25
Split275 into a higher friendly benchmark and exact gap.
Diagnostic question: Do the parts reconstruct275? Reliable response: Record n(300−25).
3. Calculate ×300
Multiply by3 and append or place the two zeros accurately.
Diagnostic question: What is three hundred groups? Reliable response: Keep decimal scale visible.
4. Calculate ×25
Use ×100÷4 or ×50÷2.
Diagnostic question: Have twenty-five complete groups been found? Reliable response: Do not merely subtract25.
5. Subtract and label
Remove the ×25 partial product and attach the unit.
Diagnostic question: Which product is larger? Reliable response: Write one clear result.
6. Check independently
Use ×250+×25 and benchmark inequalities.
Diagnostic question: Is the answer25n below ×300? 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. Calculate24×275.
Reasoning. 24×300=7200;24×25=600. The controlling idea is whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 6600.
Independent check. Check6000+600. 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: Multiple of four
Task. Calculate48×275.
Reasoning. 14,400−1200. The controlling idea is whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 13,200.
Independent check. It lies between12,000 and14,400. 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: Large factor
Task. Calculate120×275.
Reasoning. 36,000−3000. The controlling idea is whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 33,000.
Independent check. Check30,000+3000. 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. Calculate0.8×275.
Reasoning. 240−20. The controlling idea is whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 220.
Independent check. The result lies between200 and240. 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. Calculate1.25×275.
Reasoning. 375−31.25. The controlling idea is whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 343.75.
Independent check. Check312.5+31.25. 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. 275 notebooks cost $2.40 each.
Reasoning. 720−60. The controlling idea is whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. $660.00.
Independent check. Work in cents as another check. 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. 275 strips are1.6 m each.
Reasoning. 480−40. The controlling idea is whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 440 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. 19,250 is divided among275 equal groups.
Reasoning. 70×275=21,000−1750. The controlling idea is whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250. 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 back to verify. 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 provide a friendly high benchmark.
Analysis. Place value must remain accurate. Notice that the answer depends on whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250; 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: ×25
Case. One quarter of ×100.
Analysis. This is efficient with many decimals and multiples of four. Notice that the answer depends on whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250; 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 3: ×275
Case. Twenty-five groups below ×300.
Analysis. The difference is25n. Notice that the answer depends on whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250; 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: ×250+×25
Case. A lower benchmark gives an independent route.
Analysis. Addition should match subtraction route. Notice that the answer depends on whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250; 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: ×200+×75
Case. Place value can suit written work.
Analysis. Method choice depends on number structure. Notice that the answer depends on whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250; changing a decorative detail would not be enough to change the reasoning.
Challenge. Compare totals. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 6: ×11 then ×25
Case. Since275=11×25, factorisation is possible.
Analysis. It may help for selected multiplicands. Notice that the answer depends on whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250; changing a decorative detail would not be enough to change the reasoning.
Challenge. Test n=16. 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. 275n=n275.
Analysis. Context still identifies group size and count. Notice that the answer depends on whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250; changing a decorative detail would not be enough to change the reasoning.
Challenge. Rewrite a word problem. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 8: Distributivity
Case. n(300−25)=300n−25n.
Analysis. No group is counted twice after subtraction. Notice that the answer depends on whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250; 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 9: Zero
Case. 0×275=0.
Analysis. Both partial products are zero. Notice that the answer depends on whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250; 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. 1×275=275.
Analysis. A boundary case confirms the gap. Notice that the answer depends on whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250; changing a decorative detail would not be enough to change the reasoning.
Challenge. Use0.1. 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. 0.5×275=137.5.
Analysis. Products need not be whole. Notice that the answer depends on whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250; 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. 0.04×275=11.
Analysis. 12−1. Notice that the answer depends on whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250; changing a decorative detail would not be enough to change the reasoning.
Challenge. Explain 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. $0.60×275=$165.00.
Analysis. Cents can make the arithmetic whole-number based. Notice that the answer depends on whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250; changing a decorative detail would not be enough to change the reasoning.
Challenge. Convert back. 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×275=687.5 kg.
Analysis. The unit stays kilograms. Notice that the answer depends on whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250; changing a decorative detail would not be enough to change the reasoning.
Challenge. Convert 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. ×300 exceeds the exact positive product.
Analysis. It checks subtraction direction. Notice that the answer depends on whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250; changing a decorative detail would not be enough to change the reasoning.
Challenge. State the difference. 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. ×250 is below the exact positive product.
Analysis. The exact result is25n higher. Notice that the answer depends on whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250; changing a decorative detail would not be enough to change the reasoning.
Challenge. Bracket an answer. 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. ×300−exact should equal25n.
Analysis. This detects a missing or wrong partial product. Notice that the answer depends on whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250; 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 18: Unit check
Case. The multiplicand unit becomes the total unit.
Analysis. The multiplier counts groups. Notice that the answer depends on whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250; changing a decorative detail would not be enough to change the reasoning.
Challenge. Label 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 19: Reverse division
Case. A quotient can be checked by multiplying by275.
Analysis. Inverse structure tests the result. Notice that the answer depends on whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250; 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, partial products, subtraction, range, unit and independent route complete the check.
Analysis. Each catches a different error. Notice that the answer depends on whether the decomposition preserves exactly 275 equal groups and the positive product sits exactly twenty-five multiplicand-groups below ×300 and above ×250; changing a decorative detail would not be enough to change the reasoning.
Challenge. Audit one solution. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Finish the lab with a blank-page recall. Write the rule, its conditions, one confirming example, one near-miss and one independent check from memory. Compare that page with the article only after finishing. If the rule is correct but the condition is missing, the learning is not yet safe: an overgeneralised rule can produce confident errors. If the condition is correct but no example can be generated, practise construction rather than rereading.
Common misconceptions and targeted repairs
Misconception 1: Subtract25 only
What the learner may think. Find ×300 and subtract25.
Why it fails. The gap is25 full groups of the multiplicand. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Subtract25n. Use this short test immediately: What is one group?
Misconception 2: Add because275 is positive
What the learner may think. All positive multipliers require addition in a decomposition.
Why it fails. The chosen benchmark300 is larger than275. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Use the sign from the identity. Use this short test immediately: Is275 above or below300?
Misconception 3: Quarter the wrong value
What the learner may think. Divide the multiplicand by4 and forget ×100.
Why it fails. ×25 is one quarter of ×100, not one quarter of n alone. 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×100÷4. Use this short test immediately: How many groups result?
Misconception 4: Drop decimals
What the learner may think. Whole-number working determines decimal placement without estimation.
Why it fails. Place value can shift the answer by ten or hundred. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Bracket between ×250 and ×300. Use this short test immediately: Does the size fit?
Misconception 5: Ignore units
What the learner may think. A numeral alone completes a money or measure problem.
Why it fails. The unit identifies 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. Restore dollars, metres or kilograms. Use this short test immediately: What was repeated?
Misconception 6: Same-route check
What the learner may think. Repeat300−25 and call it independent.
Why it fails. The same sign error can recur. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Use250+25 or inverse division. Use this short test immediately: Is the check structurally different?
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. Calculate16×275.
Explained answer. 4400.
Why this answer earns trust. 4800−400. Transfer variation: Check4000+400.
Practice 2: Whole number
Question. Calculate72×275.
Explained answer. 19,800.
Why this answer earns trust. 21,600−1800. Transfer variation: Bracket between18,000 and21,600.
Practice 3: Decimal
Question. Calculate0.6×275.
Explained answer. 165.
Why this answer earns trust. 180−15. Transfer variation: Check150+15.
Practice 4: Decimal
Question. Calculate1.2×275.
Explained answer. 330.
Why this answer earns trust. 360−30. Transfer variation: Compare with ×250.
Practice 5: Money
Question. 275 pens cost $0.80 each.
Explained answer. $220.00.
Why this answer earns trust. 240−20. Transfer variation: Work in cents.
Practice 6: Measure
Question. 275 ribbons each measure2.4 m.
Explained answer. 660 m.
Why this answer earns trust. 720−60. Transfer variation: Convert afterward if asked.
Practice 7: Error diagnosis
Question. A learner writes40×275=12,000−25. Repair it.
Explained answer. 40×25=1000, so the total is11,000.
Why this answer earns trust. The error removed25 units, not25 groups. Transfer variation: Check10,000+1000.
Practice 8: Method choice
Question. Use a second route for32×275.
Explained answer. 32×250+32×25=8000+800=8800.
Why this answer earns trust. The route differs from300−25. Transfer variation: Verify the gap from9600.
Practice 9: Rule recall
Question. State the identity and range.
Explained answer. n×275=n×300−n×25, between ×250 and ×300 for positive n.
Why this answer earns trust. The structure preserves group count. Transfer variation: Recall later.
Practice 10: Transfer
Question. Create a context whose calculation is48×275.
Explained answer. Any valid275 equal groups of48 units total13,200 units.
Why this answer earns trust. Context and arithmetic align. Transfer variation: Reverse to division.
Delayed transfer: prove that the learning lasts
Transfer test 1: Mixed benchmarks
Solve shuffled ×255, ×265 and ×275 items. 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 chooses addition or subtraction from the exact benchmark relation. If the criterion is not met, return to the smallest missing decision rather than repeating the entire chapter.
Transfer test 2: Delayed set
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. Every product has correct scale, unit and different-route check. If the criterion is not met, return to the smallest missing decision rather than repeating the entire chapter.
Transfer test 3: Context construction
Write three word problems using275, including one reverse problem. Do this after a delay, without rereading the worked examples. The purpose is to test retrieval and decision-making, not immediate imitation.
Success criterion. Each identifies group size, count and requested unit precisely. 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 “How far is275 from300?” before arithmetic. Twenty-five below determines both the partial product and subtraction sign.
Require a bracket between ×250 and ×300 for positive values. It catches many sign and scale errors quickly.
Use decimals and units after the whole-number structure is secure so the method transfers beyond a mental-arithmetic trick.
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 use300−25?
Multiplication by300 and25 is often simpler, and their difference preserves275 groups.
How do I multiply by25?
Multiply by100 and divide by4.
What is the fastest direction check?
For a positive number, the result must be below ×300 and above ×250.
Can I use250+25?
Yes. It is an excellent independent check.
Does the method work for decimals?
Yes, provided place value and scale are checked.
What is the commonest error?
Subtracting25 units instead of25 complete groups of the multiplicand.
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.