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 2375 as an awkward multiplier, copies seventeen eighths from 2125, or adds only one 375 after multiplying the 2000 term by the multiplicand. 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
The multiplier 2375 is nineteen groups of 125. Eight groups make 1000, so it is nineteen eighths of 1000.
It is also 2000+375, and 375 is three groups of 125.
A second round-number form is 2500−125. The best route depends on the multiplicand.
For 24×2375, divide 24 by 8 to get 3, then calculate 3×19×1000=57,000.
The addition route confirms it: 48,000+24×375=48,000+9,000=57,000.
A shortcut remains valid only while every line is an equality. The factor 1000 cannot disappear from nineteen eighths.
The eighths family now advances through 17 groups for 2125 and 19 groups for 2375. Derive the numerator.
Numbers divisible by eight favour the fraction route. Multiples that work easily with 4 or 8 may favour 375 or 125 decompositions.
Because 2375 is below 2500 and above 2250, scaled bounds catch misplaced zeros and wrong signs.
Last digits offer another check because 2375 ends in 375.
Units and the requested noun still matter after efficient arithmetic.
The reliable routine is derive two or three forms, inspect n, choose an exact route, retain every factor and verify independently.
Core decision: Because 2375=19×125=(19/8)×1000=2000+375=2500−125, a product n×2375 can be reorganised through the exact form best suited to n. Multiples of eight favour n÷8×19×1000. Other values may favour 2000n+375n, 2500n−125n or a nearby multiple. For positive n, the answer lies between 2250n and 2500n.
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. Build from 125
Nineteen lots of 125 make 2375.
Diagnostic question: Can n be divided by eight exactly? Reliable response: Use n÷8×19×1000 when convenient.
2. Add 375 to 2000
Distributivity gives 2000n+375n.
Diagnostic question: Is n×375 easy? Reliable response: Multiply both terms by n.
3. Subtract from 2500
The form 2500−125 can simplify some products.
Diagnostic question: Is n×2500 easier? Reliable response: Compute 2500n−125n.
4. Protect scale
Nineteen eighths must multiply 1000.
Diagnostic question: Where is the thousand factor? Reliable response: Keep it visible.
5. Select rather than force
Different multiplicands favour different forms.
Diagnostic question: Which setup uses the simplest exact operations? Reliable response: Compare before calculating.
6. Verify
A second representation detects route-specific errors.
Diagnostic question: Can addition check subtraction? Reliable response: Use another form, bounds or reverse division.
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: Eight groups
Task. Find 8×2375.
Reasoning. Eight cancels the denominator. The controlling idea is the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 19,000.
Independent check. Check 16,000+3,000. A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Worked example 2: Twenty-four groups
Task. Find 24×2375.
Reasoning. 24÷8=3; 3×19×1000. The controlling idea is the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 57,000.
Independent check. Check 60,000−3,000. A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Worked example 3: Thirty-two groups
Task. Find 32×2375.
Reasoning. 32÷8=4. The controlling idea is the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 76,000.
Independent check. Bound between 72,000 and 80,000. 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: Twelve groups
Task. Find 12×2375.
Reasoning. Use 30,000−1,500. The controlling idea is the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 28,500.
Independent check. Check 24,000+4,500. 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: Seven groups
Task. Find 7×2375.
Reasoning. Use eight groups minus one. The controlling idea is the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 16,625.
Independent check. Check 14,000+2,625. 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: Nine groups
Task. Find 9×2375.
Reasoning. Ten groups minus one. The controlling idea is the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 21,375.
Independent check. Check 18,000+3,375. 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: Forty groups
Task. Find 40×2375.
Reasoning. 40÷8=5; 5×19×1000. The controlling idea is the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 95,000.
Independent check. Four groups make 9,500; multiply by ten. 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: Decimal
Task. Find 2.4×2375.
Reasoning. Find 24 groups then divide by ten. The controlling idea is the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 5,700.
Independent check. Check 2×2375+0.4×2375. 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: Sixteen
Case. Two groups of eight.
Analysis. 38,000. Notice that the answer depends on the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound; changing a decorative detail would not be enough to change the reasoning.
Challenge. Verify with 40,000−2,000. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 2: Forty-eight
Case. Six groups of eight.
Analysis. 114,000. Notice that the answer depends on the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound; changing a decorative detail would not be enough to change the reasoning.
Challenge. Use a bound. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 3: Fifty-six
Case. Seven groups of eight.
Analysis. 133,000. Notice that the answer depends on the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound; changing a decorative detail would not be enough to change the reasoning.
Challenge. Check 140,000−7,000. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 4: Sixty-four
Case. Eight groups of eight.
Analysis. 152,000. Notice that the answer depends on the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound; changing a decorative detail would not be enough to change the reasoning.
Challenge. Reverse divide. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 5: Seventy-two
Case. Nine groups of eight.
Analysis. 171,000. Notice that the answer depends on the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound; changing a decorative detail would not be enough to change the reasoning.
Challenge. Compare with 72×2500. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 6: Eighty
Case. Ten groups of eight.
Analysis. 190,000. Notice that the answer depends on the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound; changing a decorative detail would not be enough to change the reasoning.
Challenge. Protect zeros. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 7: Ninety-six
Case. Twelve groups of eight.
Analysis. 228,000. Notice that the answer depends on the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound; changing a decorative detail would not be enough to change the reasoning.
Challenge. Use 100 groups minus four. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 8: One hundred
Case. Place value is simplest.
Analysis. 237,500. Notice that the answer depends on the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound; changing a decorative detail would not be enough to change the reasoning.
Challenge. Do not force eighths. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 9: Five
Case. Half of ten groups.
Analysis. 11,875. Notice that the answer depends on the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound; changing a decorative detail would not be enough to change the reasoning.
Challenge. Double it. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 10: Three
Case. Use 6000+1125.
Analysis. 7,125. Notice that the answer depends on the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound; changing a decorative detail would not be enough to change the reasoning.
Challenge. Check 3×19×125. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 11: Fifteen
Case. Ten plus five.
Analysis. 35,625. Notice that the answer depends on the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound; changing a decorative detail would not be enough to change the reasoning.
Challenge. Check last digits. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 12: One half
Case. Half the multiplier.
Analysis. 1,187.5. Notice that the answer depends on the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound; changing a decorative detail would not be enough to change the reasoning.
Challenge. Double it. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 13: One and one half
Case. One plus half.
Analysis. 3,562.5. Notice that the answer depends on the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound; changing a decorative detail would not be enough to change the reasoning.
Challenge. Add parts. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 14: Reverse
Case. 76,000 contains groups of 2375.
Analysis. There are 32 groups. Notice that the answer depends on the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound; changing a decorative detail would not be enough to change the reasoning.
Challenge. Multiply back. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 15: Algebra
Case. Keep n symbolic.
Analysis. 2375n=2000n+375n=2500n−125n. Notice that the answer depends on the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound; changing a decorative detail would not be enough to change the reasoning.
Challenge. State when eighths simplify. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 16: Money
Case. 2375 cents is $23.75.
Analysis. Money offers decimal checking. Notice that the answer depends on the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound; changing a decorative detail would not be enough to change the reasoning.
Challenge. Keep units. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 17: Area
Case. A 2375 mm by n mm rectangle has area 2375n mm².
Analysis. Units remain squared. Notice that the answer depends on the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound; changing a decorative detail would not be enough to change the reasoning.
Challenge. State them. 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 a route for 48 groups
Case. The task is 48×2375, and several exact forms are available. Begin with a decision rather than immediate calculation. Because 48 is divisible by eight, the fractional structure is efficient: 48÷8=6, so the product is 6×19×1000=114,000. The same answer follows from 48×(2500−125)=120,000−6,000.
Analysis. The first route reduces the number of operations because the division is exact. The second route is valuable as an independent verification because it reorganises 2375 differently. A learner who writes 48×2000+375 has not distributed 48 to both addends; the correct expanded form is 48×2000+48×375. The answer must also fall between 48×2250=108,000 and 48×2500=120,000. Notice that the answer depends on the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound; changing a decorative detail would not be enough to change the reasoning.
Challenge. Explain why the nineteen-eighths route is efficient, calculate the product, verify with 2500−125, and state how the bound detects a misplaced zero or an addition error. Then decide which route you would choose for 49×2375 and justify the change. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 19: Selecting for 49 groups
Case. The multiplicand 49 is not divisible by eight, but it is one more than 48 and one less than 50. Use a nearby product: 49×2375=48×2375+2375=116,375, or 50×2375−2375=118,750−2,375=116,375.
Analysis. Both routes preserve exact equality. The useful choice comes from the relationship between the multiplicand and a friendly nearby value, not from forcing the same shortcut into every problem. The product lies between 110,250 and 122,500, and its last three digits agree with repeated addition of 375 modulo 1000. Notice that the answer depends on the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound; changing a decorative detail would not be enough to change the reasoning.
Challenge. Compare the operation count of the two nearby-multiple routes, show a subtraction with place values aligned, and use the bound to reject 1,163,750. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 20: Choosing the shortest exact route
Case. Compare 48×2375 by three methods before calculating. The eighths route gives 48÷8×19×1000, the subtraction route gives 48×2500−48×125, and the addition route gives 48×2000+48×375.
Analysis. All three are equal, but the eighths route is shortest here because 48 divides cleanly by eight: 6×19×1000=114,000. The subtraction route confirms 120,000−6,000=114,000. Notice that the answer depends on the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound; changing a decorative detail would not be enough to change the reasoning.
Challenge. Write the equality that justifies the route, then use a second form as an independent check. This comparison teaches selection rather than memorisation: divisibility by eight is a reason to choose the nineteen-eighths form, while a multiplicand near a friendly ten may favour a nearby-multiple strategy. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 21: Complete audit
Case. Equality, route, scale, factors, bound and units agree.
Analysis. The method is transparent. Notice that the answer depends on the exact representation of 2375, divisibility of the multiplicand, distribution to every term, preservation of scale and an independent route or bound; changing a decorative detail would not be enough to change the reasoning.
Challenge. Apply to an unfamiliar n. 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: Use seventeen eighths
What the learner may think. Copy the 2125 rule.
Why it fails. 2375 has nineteen groups of 125. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Derive the numerator. Use this short test immediately: How many 125s?
Misconception 2: Drop 1000
What the learner may think. Write 2375=19/8.
Why it fails. Scale is lost. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Use (19/8)×1000. Use this short test immediately: Do magnitudes match?
Misconception 3: Add 375 once
What the learner may think. Write 2000n+375.
Why it fails. n distributes to both terms. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Use 2000n+375n. Use this short test immediately: Can n factor back?
Misconception 4: Wrong sign
What the learner may think. Treat 2375 as 2500+125.
Why it fails. That equals 2625. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Use 2500−125. Use this short test immediately: Is 2375 below 2500?
Misconception 5: Round eighths
What the learner may think. Round n÷8.
Why it fails. Exactness is lost. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Choose another decomposition. Use this short test immediately: Can 2500−125 help?
Misconception 6: Ignore bound
What the learner may think. Accept above 2500n.
Why it fails. 2375 is below 2500. 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 corridor. Use this short test immediately: What is the upper bound?
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: Anchor
Question. Find 8×2375.
Explained answer. 19,000.
Why this answer earns trust. Nineteen eighths of 1000. Transfer variation: Show 19×125.
Practice 2: Multiple of eight
Question. Find 56×2375.
Explained answer. 133,000.
Why this answer earns trust. 56÷8=7; 7×19×1000. Transfer variation: Verify with subtraction.
Practice 3: Subtraction
Question. Find 12×2375.
Explained answer. 28,500.
Why this answer earns trust. 30,000−1,500. Transfer variation: Check with addition.
Practice 4: Near ten
Question. Find 9×2375.
Explained answer. 21,375.
Why this answer earns trust. Ten groups minus one. Transfer variation: Check final digits.
Practice 5: Decimal
Question. Find 3.2×2375.
Explained answer. 7,600.
Why this answer earns trust. 32 groups give 76,000; divide by ten. Transfer variation: Use 3+0.2.
Practice 6: Reverse
Question. What times 2375 equals 152,000?
Explained answer. 64.
Why this answer earns trust. 152,000÷1000×8÷19. Transfer variation: Multiply back.
Practice 7: Repair
Question. Repair 24×2375=48,000+375.
Explained answer. 48,000+24×375=57,000.
Why this answer earns trust. Distribute 24. Transfer variation: Factor back.
Practice 8: Routine
Question. State the routine.
Explained answer. Derive forms, inspect divisibility, choose exact work, retain factors and verify independently.
Why this answer earns trust. The routine protects accuracy. Transfer variation: Apply to 48.
Delayed transfer: prove that the learning lasts
Transfer test 1: Representation race
Write two exact setups for twelve multiplicands. Do this after a delay, without rereading the worked examples. The purpose is to test retrieval and decision-making, not immediate imitation.
Success criterion. Selection follows number structure. If the criterion is not met, return to the smallest missing decision rather than repeating the entire chapter.
Transfer test 2: Family shuffle
Mix 1875, 2125 and 2375 after a delay. 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 derives 15, 17 and 19 eighths. If the criterion is not met, return to the smallest missing decision rather than repeating the entire chapter.
Transfer test 3: Error repair
Repair missing scale, one-sided distribution and wrong signs. 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 repair names the broken equality. 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
Build 2375 from nineteen groups of 125 before showing shortcuts.
Alternate multiples of eight with values better suited to 2500−125.
Require a scaled bound and second route so speed remains accountable.
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 is 2375 nineteen eighths of 1000?
Because 1000÷8=125 and 19×125=2375.
What other forms are useful?
2000+375 and 2500−125.
When should I divide by eight first?
When the multiplicand is divisible by eight.
How do I avoid confusing 2125 and 2375?
Derive 17 or 19 groups of 125.
What bound works?
For positive n, between 2250n and 2500n.
Is long multiplication acceptable?
Yes; decomposition is an exact alternative and check.
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.