How to Perform in PSLE | Learner’s Guide Vol 0678 | Mathematics: In Multi-Step Fraction Problems, Update the Whole Before Taking the Next Fraction

CHAPTER 1 OF 12 · Orient · Back to contents

1. Why the reference whole changes

A multi-step fraction question can contain completely correct arithmetic and still produce a wrong answer. The hidden problem is often not multiplication; it is reference control. The first fraction acts on the original whole. After something is spent, removed, read, poured, sold or used, the remainder becomes a new whole for the next phrase.

Read “one-third of the remainder” as a complete mathematical instruction. The word of signals multiplication, but the noun phrase after it identifies the quantity to multiply. If the remainder is 90, then one-third means 30, even if the original amount was 120.

Your central habit is therefore: event, new state, next event. Never move directly from the first fraction to the second without recording the state between them. A bar model, arrow chain or state table can all make the changing base visible.

• Underline words that reset the base: remainder, left, reduced amount, balance.
• Write the original amount and unit.
• Compute the first change.
• Record the new state on its own line.
• Apply the next fraction to that state.
• Answer the requested amount or fraction, not a nearby quantity.

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CHAPTER 2 OF 12 · Model · Back to contents

2. The state-update model

Represent each event as a state transition. If an amount S loses a fraction f, the new state is S(1−f). If the next stage loses fraction g of the remainder, the final state is S(1−f)(1−g). The symbols are not required for every Primary 6 solution, but the relationship explains why successive fractions multiply rather than add.

A table can have three columns: starting state, change, ending state. The ending state of one row must equal the starting state of the next. This equality is a powerful audit. If stage two begins with the original amount again, the table exposes the mistake.

Use retained fractions when they simplify the reasoning. Removing 2/7 means retaining 5/7. Spending 1/4 of what remains means retaining 3/4 of that state. The product 5/7×3/4 describes the final fraction of the original.

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CHAPTER 3 OF 12 · Model · Back to contents

3. Bars, tables and equations

A bar model should be redrawn or relabelled after the first event. Suppose a bar of 12 units loses 3 units. The remaining 9 units now form the whole for the next phrase. If one-third of the remainder is removed, split those 9 units into three equal groups, not the original 12 into three groups.

An arrow chain is efficient under time pressure: 240 → 180 → 120. Write the operation above each arrow, such as ×3/4 or ×2/3. The chain records both the operation and the updated state.

An equation is useful for reverse questions. If $90 is left after retaining 3/4 and then 3/5, write original×3/4×3/5=90. Combine to original×9/20=90, then solve for the original. Always check by running the process forward.

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CHAPTER 4 OF 12 · Work examples · Back to contents

4. Worked foundations

Worked case 0: two fractions spent in sequence

Question. Mira has $120. She spends 1/4 of it, then spends 1/3 of the remainder. How much is left?

Tempting error. Taking both fractions of $120 and subtracting $30 + $40. The calculation may look tidy, but it uses the wrong reference quantity or fails to record the new state before the next operation.

Controlled solution. First spend 1/4 × 120 = $30, leaving $90. Then spend 1/3 × 90 = $30, leaving $60.

Mechanism. The phrase “of the remainder” changes the second reference whole from 120 to 90. A fraction is always a fraction of something. Naming that reference whole before multiplying prevents the numerator and denominator from floating free of meaning.

Transfer. If the second fraction becomes 2/5, keep the $90 state and take 2/5 of 90. Redraw the bar or state table after the first change. Then solve the variation without copying the original arithmetic.

Audit. Check the reference whole, remaining amount, fraction operation, unit and final demand. Estimate whether the result should be smaller than the current amount.

Worked case 1: fraction used, then fraction given away

Question. A tank contains 200 L. 3/10 is used. Then 1/4 of the remaining water is transferred. How much remains?

Tempting error. Subtracting 3/10 + 1/4 = 11/20 of the original 200 L. The calculation may look tidy, but it uses the wrong reference quantity or fails to record the new state before the next operation.

Controlled solution. After use: 200 − 60 = 140 L. Transfer 1/4 of 140 = 35 L. Remaining amount = 105 L.

Mechanism. The events are sequential, so the second denominator partitions 140 L, not 200 L. A fraction is always a fraction of something. Naming that reference whole before multiplying prevents the numerator and denominator from floating free of meaning.

Transfer. Reverse-check by adding the transferred 35 L to 105 L, then the first 60 L. Redraw the bar or state table after the first change. Then solve the variation without copying the original arithmetic.

Audit. Check the reference whole, remaining amount, fraction operation, unit and final demand. Estimate whether the result should be smaller than the current amount.

Worked case 2: fraction of a remainder retained

Question. A library lends 2/5 of 350 books. Of the books left, 3/7 are reserved. How many are neither lent nor reserved?

Tempting error. Multiplying 3/7 by 350 because 350 is the only stated total. The calculation may look tidy, but it uses the wrong reference quantity or fails to record the new state before the next operation.

Controlled solution. Lent = 140, leaving 210. Reserved = 3/7 × 210 = 90. Neither = 120 books.

Mechanism. The word left creates a new set of 210 books before reservation. A fraction is always a fraction of something. Naming that reference whole before multiplying prevents the numerator and denominator from floating free of meaning.

Transfer. Find the fraction of the original set that is neither lent nor reserved: 120/350 = 12/35. Redraw the bar or state table after the first change. Then solve the variation without copying the original arithmetic.

Audit. Check the reference whole, remaining amount, fraction operation, unit and final demand. Estimate whether the result should be smaller than the current amount.

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CHAPTER 5 OF 12 · Work examples · Back to contents

5. Mixed fraction sequences

Worked case 4: remaining fraction instead of remaining amount

Question. A pupil reads 3/8 of a book on Monday and 1/5 of the remainder on Tuesday. What fraction remains unread?

Tempting error. Subtracting 3/8 + 1/5 directly. The calculation may look tidy, but it uses the wrong reference quantity or fails to record the new state before the next operation.

Controlled solution. After Monday, 5/8 remains. Tuesday’s reading is 1/5 of 5/8 = 1/8 of the original. Unread = 5/8 − 1/8 = 1/2.

Mechanism. A multiplicative chain converts the Tuesday fraction to a fraction of the original only after multiplying by the current remainder. A fraction is always a fraction of something. Naming that reference whole before multiplying prevents the numerator and denominator from floating free of meaning.

Transfer. Solve by retained multipliers: (1−3/8)(1−1/5). Redraw the bar or state table after the first change. Then solve the variation without copying the original arithmetic.

Audit. Check the reference whole, remaining amount, fraction operation, unit and final demand. Estimate whether the result should be smaller than the current amount.

Worked case 5: three-stage reduction

Question. A store sells 1/3 of 540 tickets, then 1/4 of the remainder, then 1/5 of the new remainder. How many tickets remain?

Tempting error. Adding 1/3 + 1/4 + 1/5 and subtracting from 540. The calculation may look tidy, but it uses the wrong reference quantity or fails to record the new state before the next operation.

Controlled solution. 540 → 360 → 270 → 216. Each arrow applies the retained fraction 2/3, then 3/4, then 4/5.

Mechanism. Sequential fractional changes multiply retained proportions; their spent fractions do not share one base. A fraction is always a fraction of something. Naming that reference whole before multiplying prevents the numerator and denominator from floating free of meaning.

Transfer. Notice the telescoping product (2/3)(3/4)(4/5)=2/5. Redraw the bar or state table after the first change. Then solve the variation without copying the original arithmetic.

Audit. Check the reference whole, remaining amount, fraction operation, unit and final demand. Estimate whether the result should be smaller than the current amount.

Worked case 6: same fractions in a different order

Question. Compare spending 1/2 then 1/3 of the remainder with spending 1/3 then 1/2 of the remainder, starting from $90.

Tempting error. Assuming different order must always change the answer. The calculation may look tidy, but it uses the wrong reference quantity or fails to record the new state before the next operation.

Controlled solution. Both leave $30 because retained factors multiply: (1/2)(2/3)×90 = (2/3)(1/2)×90. The intermediate states differ, but the final retained product is equal.

Mechanism. Multiplication of fixed retained fractions is commutative, although the amounts spent at each stage are not identical. A fraction is always a fraction of something. Naming that reference whole before multiplying prevents the numerator and denominator from floating free of meaning.

Transfer. List each stage’s spent amount to show what changes even when the final amount does not. Redraw the bar or state table after the first change. Then solve the variation without copying the original arithmetic.

Audit. Check the reference whole, remaining amount, fraction operation, unit and final demand. Estimate whether the result should be smaller than the current amount.

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CHAPTER 6 OF 12 · Work examples · Back to contents

6. Reverse and comparison cases

Worked case 7: fraction added to the remainder

Question. A container has 160 mL. 3/8 is removed. Then water equal to 1/4 of the remainder is added. What is the final volume?

Tempting error. Treating addition as another removal or taking 1/4 of 160. The calculation may look tidy, but it uses the wrong reference quantity or fails to record the new state before the next operation.

Controlled solution. Remove 60 mL, leaving 100 mL. Add 1/4 of 100 = 25 mL. Final volume = 125 mL.

Mechanism. The base updates after removal, and the operation sign must follow the verb added. A fraction is always a fraction of something. Naming that reference whole before multiplying prevents the numerator and denominator from floating free of meaning.

Transfer. Represent the final step with multiplier 5/4 applied to the 100 mL state. Redraw the bar or state table after the first change. Then solve the variation without copying the original arithmetic.

Audit. Check the reference whole, remaining amount, fraction operation, unit and final demand. Estimate whether the result should be smaller than the current amount.

Worked case 8: find the original from final remainder

Question. After 1/4 of a sum is spent and then 2/5 of the remainder is spent, $90 remains. Find the original sum.

Tempting error. Adding spent fractions and treating 90 as the missing part of one original partition. The calculation may look tidy, but it uses the wrong reference quantity or fails to record the new state before the next operation.

Controlled solution. Retained fraction = 3/4 × 3/5 = 9/20. If 9/20 is $90, then 1/20 is $10 and the original is $200.

Mechanism. Reverse problems require identifying what fraction of the original the final state represents. A fraction is always a fraction of something. Naming that reference whole before multiplying prevents the numerator and denominator from floating free of meaning.

Transfer. Check forward: 200 → 150 → 90. Redraw the bar or state table after the first change. Then solve the variation without copying the original arithmetic.

Audit. Check the reference whole, remaining amount, fraction operation, unit and final demand. Estimate whether the result should be smaller than the current amount.

Worked case 9: fraction of remainder versus fraction of original

Question. A statement says 1/3 of the remainder is donated. Another says 1/3 of the original amount is donated. Explain the difference after 1/4 has already been used from 240.

Tempting error. Treating both donations as 80. The calculation may look tidy, but it uses the wrong reference quantity or fails to record the new state before the next operation.

Controlled solution. After use, 180 remains. One-third of the remainder is 60; one-third of the original is 80. The phrases produce different reference wholes.

Mechanism. The words following of name the base. Reference language is mathematical data. A fraction is always a fraction of something. Naming that reference whole before multiplying prevents the numerator and denominator from floating free of meaning.

Transfer. Write equations 1/3×180 and 1/3×240 beside the phrases. Redraw the bar or state table after the first change. Then solve the variation without copying the original arithmetic.

Audit. Check the reference whole, remaining amount, fraction operation, unit and final demand. Estimate whether the result should be smaller than the current amount.

Worked case 10: percentage form of the same structure

Question. A price is reduced by 20%, then by 25% of the reduced price. Starting from $80, find the final price.

Tempting error. Adding 20% and 25% to claim a 45% reduction. The calculation may look tidy, but it uses the wrong reference quantity or fails to record the new state before the next operation.

Controlled solution. $80 × 0.80 = $64; $64 × 0.75 = $48. The overall reduction is 40%, not 45%.

Mechanism. Successive percentages are sequential fractions with changing bases. A fraction is always a fraction of something. Naming that reference whole before multiplying prevents the numerator and denominator from floating free of meaning.

Transfer. Reverse-check that 48/80=0.60, so 60% is retained. Redraw the bar or state table after the first change. Then solve the variation without copying the original arithmetic.

Audit. Check the reference whole, remaining amount, fraction operation, unit and final demand. Estimate whether the result should be smaller than the current amount.

Worked case 11: objects require divisibility

Question. A class has some whole badges. After giving away 1/4, it gives 1/3 of the remainder. Could it have started with 10 badges if no badge is divided?

Tempting error. Applying decimal answers and ignoring whole-object constraints. The calculation may look tidy, but it uses the wrong reference quantity or fails to record the new state before the next operation.

Controlled solution. No. One quarter of 10 is not a whole badge. A valid starting count must make each stage integral; multiples of 4 are candidates, then the remainder must be divisible by 3.

Mechanism. A context constraint can reject arithmetic that is formally expressible but physically impossible. A fraction is always a fraction of something. Naming that reference whole before multiplying prevents the numerator and denominator from floating free of meaning.

Transfer. Test 12, 16 and 24 and state which complete both stages with whole badges. Redraw the bar or state table after the first change. Then solve the variation without copying the original arithmetic.

Audit. Check the reference whole, remaining amount, fraction operation, unit and final demand. Estimate whether the result should be smaller than the current amount.

Worked case 12: comparison after sequential use

Question. A and B each start with 300 g. A uses 1/5 then 1/4 of the remainder. B uses 2/5 once. Who has more left?

Tempting error. Comparing 1/5 + 1/4 with 2/5. The calculation may look tidy, but it uses the wrong reference quantity or fails to record the new state before the next operation.

Controlled solution. A retains 300×4/5×3/4 = 180 g. B retains 300×3/5 = 180 g. They have equal amounts left.

Mechanism. Convert each process into a retained multiplier before comparing. A fraction is always a fraction of something. Naming that reference whole before multiplying prevents the numerator and denominator from floating free of meaning.

Transfer. Change A’s second fraction to 1/3 and predict the direction before calculating. Redraw the bar or state table after the first change. Then solve the variation without copying the original arithmetic.

Audit. Check the reference whole, remaining amount, fraction operation, unit and final demand. Estimate whether the result should be smaller than the current amount.

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CHAPTER 7 OF 12 · Repair errors · Back to contents

7. Misconceptions and repairs

0. Add sequential fractions immediately

Why it fails. The fractions usually refer to different wholes, so adding them combines unlike quantities. The repair is to write “current whole = …” before applying the next fraction. This short statement is more reliable than trying to hold the changing base in memory.

Diagnostic. After the first event, what exact amount is the second denominator partitioning? Change the order or one fraction and ask whether the base changes. A learner who only repeats a procedure has not yet transferred the idea.

1. Use the original total at every stage

Why it fails. Words such as remainder, left, reduced amount explicitly reset the base. The repair is to write “current whole = …” before applying the next fraction. This short statement is more reliable than trying to hold the changing base in memory.

Diagnostic. Which noun phrase follows the word of? Change the order or one fraction and ask whether the base changes. A learner who only repeats a procedure has not yet transferred the idea.

2. Track only what was used

Why it fails. A state table needs both change and new remaining state; otherwise the next base is hidden. The repair is to write “current whole = …” before applying the next fraction. This short statement is more reliable than trying to hold the changing base in memory.

Diagnostic. Can you point to the current whole before stage two? Change the order or one fraction and ask whether the base changes. A learner who only repeats a procedure has not yet transferred the idea.

3. Assume order always changes the final result

Why it fails. For pure successive proportional reductions, retained factors may commute, though stage amounts differ. The repair is to write “current whole = …” before applying the next fraction. This short statement is more reliable than trying to hold the changing base in memory.

Diagnostic. Are the operations all fixed multipliers, or is there a fixed amount or new condition? Change the order or one fraction and ask whether the base changes. A learner who only repeats a procedure has not yet transferred the idea.

4. Assume order never matters

Why it fails. Fixed additions, caps, rounding or context constraints can make order matter. The repair is to write “current whole = …” before applying the next fraction. This short statement is more reliable than trying to hold the changing base in memory.

Diagnostic. What feature breaks a simple product of multipliers? Change the order or one fraction and ask whether the base changes. A learner who only repeats a procedure has not yet transferred the idea.

5. Ignore whole-object conditions

Why it fails. A fraction of a set of indivisible objects must yield a possible count. The repair is to write “current whole = …” before applying the next fraction. This short statement is more reliable than trying to hold the changing base in memory.

Diagnostic. Which stages require divisibility? Change the order or one fraction and ask whether the base changes. A learner who only repeats a procedure has not yet transferred the idea.

6. Reverse by undoing only the last subtraction

Why it fails. The final amount represents a product of retained fractions of the original. The repair is to write “current whole = …” before applying the next fraction. This short statement is more reliable than trying to hold the changing base in memory.

Diagnostic. What fraction of the original survives all stages? Change the order or one fraction and ask whether the base changes. A learner who only repeats a procedure has not yet transferred the idea.

7. Confuse amount remaining with fraction remaining

Why it fails. A fraction has no unit; an amount carries dollars, litres, grams or objects. The repair is to write “current whole = …” before applying the next fraction. This short statement is more reliable than trying to hold the changing base in memory.

Diagnostic. What exactly does the final demand request? Change the order or one fraction and ask whether the base changes. A learner who only repeats a procedure has not yet transferred the idea.

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CHAPTER 8 OF 12 · Examination control · Back to contents

8. Examination control

Use a compact state routine: mark the original, circle every reset word, draw one arrow per event and write the new amount after each arrow. Estimate direction before calculating: after two removals, the result must be smaller than after the first removal.

• Keep units on every state.
• Do not combine fractions until you have proved they share a whole.
• For reverse problems, label the final fraction of the original.
• For objects, test divisibility at each stage.
• If the question asks total used, add stage changes only after computing them separately.
• Stop when a forward check reproduces the stated final condition.

Time-control rule. If your second fraction is being multiplied by the original amount, point to the exact words that justify that base. If the sentence says “of the remainder,” restart from the recorded remainder.

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CHAPTER 9 OF 12 · Practice · Back to contents

9. Independent practice I

Practice 0: From 240 stickers, 1/6 are used and 1/5 of the remainder are shared. How many remain?

Explained answer. 240→200→160, so 160 remain. Show the amount after each event. The final number alone cannot reveal whether the correct whole was used at the second stage.

Delayed variation. Find the overall retained fraction and confirm 160/240. Return to this later and use a different representation—bar, state table or reverse check—to confirm the same structure.

Practice 1: A rope is cut by 1/4 of its length, then 1/3 of what remains. What fraction remains?

Explained answer. Retained fraction = 3/4×2/3=1/2. Show the amount after each event. The final number alone cannot reveal whether the correct whole was used at the second stage.

Delayed variation. Use an initial length of 120 cm and confirm the amount. Return to this later and use a different representation—bar, state table or reverse check—to confirm the same structure.

Practice 2: After 2/7 of 350 mL is poured out, 1/5 of the remainder evaporates. Find the final amount.

Explained answer. 350→250→200 mL. Show the amount after each event. The final number alone cannot reveal whether the correct whole was used at the second stage.

Delayed variation. Change evaporation to 20 mL and decide why a multiplier no longer applies. Return to this later and use a different representation—bar, state table or reverse check—to confirm the same structure.

Practice 3: A pupil saves 3/5 of $150, then spends 1/3 of the saved amount. How much of the original money remains saved?

Explained answer. Saved $90; one-third spent is $30; $60 remains saved. Show the amount after each event. The final number alone cannot reveal whether the correct whole was used at the second stage.

Delayed variation. Express $60 as a fraction of $150. Return to this later and use a different representation—bar, state table or reverse check—to confirm the same structure.

Practice 4: A box has 480 beads. 1/8 are red and removed; 2/7 of the remainder are blue and removed. How many stay?

Explained answer. Remove 60, leaving 420; remove 120, leaving 300. Show the amount after each event. The final number alone cannot reveal whether the correct whole was used at the second stage.

Delayed variation. Check that the second denominator divides the updated 420. Return to this later and use a different representation—bar, state table or reverse check—to confirm the same structure.

Practice 5: Why is 1/2 of the remainder not necessarily 1/2 of the original?

Explained answer. Because an earlier event may have changed the base. Half of a smaller current whole is a smaller amount. Show the amount after each event. The final number alone cannot reveal whether the correct whole was used at the second stage.

Delayed variation. Give a numerical counterexample starting from 100. Return to this later and use a different representation—bar, state table or reverse check—to confirm the same structure.

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CHAPTER 10 OF 12 · Practice · Back to contents

10. Independent practice II

Practice 7: A quantity is reduced by 30%, then increased by 10% of the reduced amount. Start at 200.

Explained answer. 200×0.70=140; add 14; final 154. Show the amount after each event. The final number alone cannot reveal whether the correct whole was used at the second stage.

Delayed variation. Explain why the final is not 160. Return to this later and use a different representation—bar, state table or reverse check—to confirm the same structure.

Practice 8: After two stages, 3/8 of the original remains and equals 75. Find the original.

Explained answer. 1/8=25, so the original is 200. Show the amount after each event. The final number alone cannot reveal whether the correct whole was used at the second stage.

Delayed variation. Construct two retained fractions whose product is 3/8. Return to this later and use a different representation—bar, state table or reverse check—to confirm the same structure.

Practice 9: A has 5/6 left, then keeps 3/5 of that. What fraction of the original is left?

Explained answer. 5/6×3/5=1/2. Show the amount after each event. The final number alone cannot reveal whether the correct whole was used at the second stage.

Delayed variation. Draw six equal original units and regroup if useful. Return to this later and use a different representation—bar, state table or reverse check—to confirm the same structure.

Practice 10: A shop has 96 mugs. It sells 1/4, then 1/3 of the remainder. How many are sold altogether?

Explained answer. Sell 24, then 24; total sold 48. Show the amount after each event. The final number alone cannot reveal whether the correct whole was used at the second stage.

Delayed variation. Also find and verify the final remainder. Return to this later and use a different representation—bar, state table or reverse check—to confirm the same structure.

Practice 11: A tank loses 1/5, then a fixed 10 L. Why can you not combine these into one fraction without knowing the initial amount?

Explained answer. The 10 L is not a fixed proportion of an unknown initial state. The process mixes a multiplier and an additive change. Show the amount after each event. The final number alone cannot reveal whether the correct whole was used at the second stage.

Delayed variation. Choose an initial amount and write the two-stage equation. Return to this later and use a different representation—bar, state table or reverse check—to confirm the same structure.

Practice 12: Sam says 1/4 then 1/3 means 7/12 used. Give the correct general fraction used.

Explained answer. Retained = 3/4×2/3=1/2, so 1/2 is used, not 7/12. Show the amount after each event. The final number alone cannot reveal whether the correct whole was used at the second stage.

Delayed variation. Explain which 1/3 Sam treated as if it had the wrong whole. Return to this later and use a different representation—bar, state table or reverse check—to confirm the same structure.

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CHAPTER 11 OF 12 · Transfer · Back to contents

11. Delayed transfer and proof

Delay practice so that the diagram is not supplied. On day one, use bars. On day three, use arrow chains. On day seven, explain the multiplier method in words. The same answer through three representations is strong evidence of transfer.

Prove a result by reverse checking. Add back the second-stage change to recover the first remainder, then add back the first-stage change to recover the original. For a multiplier solution, run the original amount forward and confirm every stated condition.

Mix sequential fraction questions with questions where both fractions are explicitly of the original. The learner must decide the base from wording, not from expecting one familiar method.

Seven-day test. Solve one amount-left question, one total-used question, one fraction-left question and one reverse-original question. For each, write the reference whole beside every occurrence of of.

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CHAPTER 12 OF 12 · Transfer · Back to contents

12. Family guidance, FAQs and next routes

Guidance for parents and tutors

Ask “one-third of what?” before correcting arithmetic. If the learner can name the current whole, allow the calculation to proceed. If not, return to the state diagram. This keeps support diagnostic and brief.

Fade the diagram only after the learner independently records intermediate states. Speed should come from compressed representation, not from skipping the state update.

Frequently asked questions

Why can’t I just add the fractions?

Because the fractions often refer to different wholes. Addition is valid only after quantities are expressed relative to a common base.

Does order matter?

For pure successive proportional changes, the final retained product may be the same in either order. Intermediate amounts and context can still differ, and fixed additions or rounding can make order matter.

What does “of” mean?

In these problems, it signals multiplication, but you must also identify the quantity following it: original amount, remainder, saved amount or another current set.

Should I use bars or equations?

Use the representation that keeps the changing whole visible. Bars are concrete, state tables are auditable, and equations are efficient for reverse problems.

How do I know the answer is reasonable?

After removals, each state should not exceed the previous one. Fractions of a remainder should be smaller than the same fraction of a larger original whole.

What if the answer is not a whole object?

Check whether the data are possible for indivisible objects. A fraction of a set must produce whole counts unless the context allows division.

Next routes

Continue with the Mathematics tuition hub or return to the series directory.

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