Sec 1 G2 Maths Bukit Timah | Fractions of Remainder

Did you know? Spending one third of a budget and then one quarter of the remainder does not mean spending seven twelfths of the original. Why? The second quarter belongs to a smaller whole. That single change of reference explains many of the hardest-looking fraction word problems in Secondary 1 Mathematics.

For parents looking for a Secondary 1 G2 Mathematics tutor in Bukit Timah, the short method is: write the original amount, find the first remainder, apply the second fraction only to that remainder, then verify the three amounts add back to the starting total. For reverse problems, divide backwards through the remaining fractions.

The applications use the fractions, ratio and problem-solving ideas in the MOE G2 Mathematics syllabus. The unchanged eduKateSG Secondary 1 small-group reference describes weekly 90-minute sessions with three students near Sixth Avenue MRT, including focused feedback and independent practice.

The answer in sixty seconds

The first fractional removal normally refers to the original whole.
The words ‘of the remainder’ create a new reference amount.
Fraction left after removing p then q of remainder is (1-p)(1-q).
Fraction used in stage two, measured against original, is q(1-p).
Total fraction used is p+q(1-p), equal to 1-(1-p)(1-q).
If the final amount is known, divide by the remaining fraction to recover the original.
A bar model or start–used–left table makes the changing whole visible.
At every stage, previous amount = amount used + amount remaining.

The essential example: one third, then one quarter of the remainder

Start with a hypothetical $120. One third is spent, so $40 is used and $80 remains. The next instruction is to spend one quarter of the remainder. That is one quarter of $80, or $20. The final amount is $80-$20=$60. The total used is $40+$20=$60, exactly one half of the original.

There is a useful fraction-only route too. The first step leaves 2/3 of the original. The second leaves 3/4 of that. Therefore the final fraction is (2/3)(3/4)=1/2. The second removal itself is (2/3)(1/4)=1/6 of the original. The sum 1/3+1/6=1/2 matches the amount used.

Now change only the wording: spend one third of the original, then spend another quarter of the original. The second amount becomes $30 instead of $20, leaving $50. The fractions printed on the page are the same; the reference whole is different. That is why word reading is part of the Mathematics.

Twenty-eight principles that make the changing whole clear

1. What does ‘of the remainder’ mean?

The phrase changes the base of a fraction. In the first step the fraction applies to the original whole. After some amount has been removed, the second fraction applies only to what is left, unless the question explicitly says otherwise.

Worked connection: If one third of 120 is spent, 80 remains. One quarter of the remainder means one quarter of 80, or 20, not one quarter of 120.

Tutor’s checking question: Ask the child to name the whole for each stage before doing any arithmetic. Those reference quantities need not be equal.

2. The first remainder is one minus the fraction used

If p of the original whole is removed, 1-p of the original remains. This is a relationship, not an arbitrary subtraction trick.

Worked connection: Spending 2/5 of $180 leaves 3/5 of $180, or $108. That remaining amount becomes the base of the next calculation.

Tutor’s checking question: Write ‘fraction spent + fraction remaining = 1’ on the first bar diagram.

3. The second fraction is applied to a changed base

After the first stage, the remainder is itself a whole for the next local calculation. But it is only a fraction of the original whole. Mixing those two interpretations is the most common source of error.

Worked connection: From $180, spending 2/5 leaves $108. Spending 1/3 of the remainder means $36, so $72 remains.

Tutor’s checking question: Label each arrow ‘of original’ or ‘of what remains’ so the student cannot silently exchange the bases.

4. The final remaining fraction is a product

If p of the original is used and q of the remainder is then used, the remaining fraction of the original is (1-p)(1-q). Multiplication captures two successive proportional reductions.

Worked connection: Using 1/3 then 1/4 of the remainder leaves (2/3)(3/4)=1/2 of the starting quantity.

Tutor’s checking question: Explain the product with a bar that shrinks at each stage before using an algebraic shortcut.

5. The total fraction used can be found in two ways

The used portion is one minus the final remaining fraction, or it is the first used fraction plus the second used fraction expressed relative to the original. The two methods must agree.

Worked connection: For one third followed by one quarter of the remainder, total used is 1-1/2=1/2. Alternatively 1/3+(2/3)(1/4)=1/3+1/6=1/2.

Tutor’s checking question: A contradiction between the methods indicates a wrong reference whole or arithmetic slip.

6. The second removal is not simply q of the original

One quarter of a remainder is usually smaller than one quarter of the original, because some quantity has already been removed. Express the second removal as q(1-p) of the starting whole.

Worked connection: After removing 2/5, one third of the remainder is (1/3)(3/5)=1/5 of the original.

Tutor’s checking question: Ask ‘one third of what?’ before writing the second multiplication.

7. Tables can track each stage explicitly

A start–used–remaining table makes a changing whole visible. It is useful when both removed amounts and the final amount are requested.

Worked connection: Start $120; remove one third ($40); remainder $80; remove one quarter of $80 ($20); final $60. Every row satisfies ‘start minus used equals remain’.

Tutor’s checking question: Recompute each row separately rather than jumping from the original to the final answer.

8. A bar model can show the original and later whole

Use a long rectangle for the original, shade the fraction removed, then partition only the unshaded remainder for the second fraction. The second partition’s pieces are not automatically the same size as those in the first partition.

Worked connection: For an original bar of twelve equal units, one third uses four units, leaving eight. One quarter of the remainder uses two units, leaving six.

Tutor’s checking question: This drawing proves that the final six of twelve parts equal one half of the original.

9. Use compatible denominators when drawing bars

A fraction bar becomes easy to count when the original is divided into a number of units that works with both stages. That may require multiplying the denominators or finding a convenient equivalent partition.

Worked connection: One third then one quarter of the remainder can be shown with twelve original units; a third uses four and a quarter of the eight remaining units uses two.

Tutor’s checking question: The diagram must preserve equal-sized original units even as the remainder is repartitioned.

10. Do not add percentages or fractions across different bases

The rates 1/3 and 1/4 cannot simply be added to claim 7/12 of the original was used. The second 1/4 applies to a smaller whole.

Worked connection: The true used fraction is 1/3+1/6=1/2. Adding 1/3+1/4=7/12 would overstate the removal.

Tutor’s checking question: Test the result on a simple original amount, such as $120, to show the difference.

11. Fractions of a remainder can be converted into percentages

A fraction such as three quarters corresponds to 75%. Changing format does not alter the relationship as long as the reference amount remains the same.

Worked connection: After a 20% reduction, 80% remains. Spending half of that remainder leaves 0.80×0.50=0.40, or 40% of the original.

Tutor’s checking question: Use fractions or decimals consistently; do not accidentally multiply percentage labels as whole numbers.

12. Reverse problems begin with the known final value

If the final remainder is given, work backwards by dividing by each remaining fraction, or by dividing once by their product. Both are equivalent when the stage rules are proportional.

Worked connection: If 80 remains after first using 1/4 and then 1/3 of the remainder, the remaining fraction is (3/4)(2/3)=1/2, so the original was 160.

Tutor’s checking question: Check by starting from 160 and applying both removals forward; you must recover 80.

13. Reversing one stage at a time improves explanation

When the final amount is known, the most transparent method may be to undo the second reduction first, then undo the first. This is especially helpful for students who find a product of fractions abstract.

Worked connection: If $90 remains after one quarter of a post-first-stage remainder was used, the earlier remainder was $90÷(3/4)=$120. If that was three fifths of the original, the original was $200.

Tutor’s checking question: The stage order of reverse working should be the opposite of the forward story.

14. A letter can represent the original unknown whole

Writing a simple equation helps with reverse problems and makes every relationship explicit. Students should define the variable and keep equality valid at each line.

Worked connection: If x is the original amount, 1/4 used then 1/3 of the remainder used gives (3/4)(2/3)x=80, or x/2=80, so x=160.

Tutor’s checking question: Substitute the solution into the original two-stage story rather than checking only the final rearranged line.

15. Sequential removals change the amount taken at each step

Even when two proportional reductions have a final multiplier that does not depend on the order, the amount removed at each stage can change. The story’s individual stage answers are therefore order-sensitive.

Worked connection: From 100, remove 20% then 50% of the remainder: first 20, then 40, final 40. Reversing the rates removes 50 then 10, final also 40.

Tutor’s checking question: Distinguish ‘final remaining total’ from ‘how much was removed in stage one or two’.

16. Not every two-stage story is multiplicative

A second step might remove a fixed number rather than a fraction. Then the mathematical model includes subtraction, not merely multiplying remaining fractions.

Worked connection: From 120, use one third, leaving 80, then use exactly 20 more. Final 60. The second step is not described as one quarter in this wording, even though it happens to equal one quarter of 80.

Tutor’s checking question: Underline whether the second stage states a fixed amount or a proportion.

17. Reading ‘another third of the original’ changes the model

A problem can intentionally keep the original reference whole for the second removal. Then the two fractions apply to the same base and can be added directly, provided they are distinct removals.

Worked connection: From 120, use one third of the original and later another quarter of the original. Total used is 40+30=70, leaving 50. This differs from one quarter of the remainder, which would leave 60.

Tutor’s checking question: Teach the student to underline the words ‘original’ and ‘remainder’.

18. The units should follow every stage

The original quantity may represent money, tokens, distance, liquid or objects. The fraction calculation itself is dimensionless, but the amount used and left must retain the original unit.

Worked connection: A tank contains 240 litres; removing one quarter leaves 180 litres; removing two thirds of that leaves 60 litres. Every amount is measured in litres.

Tutor’s checking question: Write ‘L’, ‘$’, ‘cm’ or ‘items’ after every stage amount so the answer stays connected to the story.

19. Counts of people or objects may need whole-number checks

Fraction-of-a-group questions about individuals need counts that are consistent with the stated fractions. Choosing a total divisible by appropriate denominators makes each stage an integer; a fractional person is not an acceptable ordinary count.

Worked connection: In a hypothetical group of 420 tokens, remove 2/7=120, leaving 300; remove 1/3 of 300=100; 200 remain. The same arithmetic could apply to people if the scenario and resulting counts are meaningful.

Tutor’s checking question: If a question implies a fractional number of indivisible objects, revisit the data or requested interpretation.

20. The fraction of the remainder is a new ‘whole’ locally

Students sometimes insist every fraction must use the initial total. A correct model can have a different whole at each step, with both reference points clearly labelled.

Worked connection: After spending 1/5 of $300, the $240 remainder is 100% of the amount available for stage two, even though it is only 80% of the original $300.

Tutor’s checking question: Use the words ‘stage-two whole’ and ‘original whole’ to prevent confusion.

21. A two-row comparison can expose an invalid shortcut

Compare two otherwise identical stories, changing only ‘one quarter of the remainder’ to ‘one quarter of the original’. The differing final results prove why an unexamined keyword shortcut is unreliable.

Worked connection: Original $120, spend one third, then one quarter of remainder: final $60. Change the second phrase to one quarter of original: final $50.

Tutor’s checking question: A learner should explain what changed in the model even though the visible fractions did not.

22. The final result must lie in a plausible range

When both removal fractions are positive and less than one, the final remainder must be nonnegative and no greater than the original. It must also be less than the remainder after stage one.

Worked connection: From 120, after first spending 40, only 80 remain; after further spending 20, the final 60 is below both 80 and 120.

Tutor’s checking question: An answer of 140 after two positive removals is an immediate signal to re-examine the operation.

23. Reverse answers should be large enough

If a problem gives a positive final amount after two reductions, the original should be at least as large as that final amount. Division by remaining fractions below one increases the recovered total.

Worked connection: If $90 is final after two proportional reductions, a claimed original of $75 cannot work because the story only removes value.

Tutor’s checking question: Before detailed calculation, estimate whether the reversed result should increase or decrease.

24. Check all three conservation relationships

At every stage, the quantity before that stage equals the amount removed plus the amount after it. The original equals first removal plus second removal plus final remainder.

Worked connection: In the $120 example, $40+$20+$60=$120. Also $40+$80=$120 and $20+$60=$80.

Tutor’s checking question: These checks catch stage-mixing errors even if the final answer looks reasonable.

25. The same structure appears in successive discounts

A discount is a fraction of a price, and the next discount applies to the newly reduced price. The mathematical structure matches a fraction-of-the-remainder story, although the real context is money.

Worked connection: A 20% discount followed by a 25% discount leaves 0.80×0.75=0.60 of original price, an overall 40% discount.

Tutor’s checking question: Do not add the two displayed rates and claim a 45% discount.

26. The same structure appears in resource planning

Multi-stage resource problems arise in simple budgeting, water use and stock distribution. The calculations are hypothetical learning models; actual real-world rates can vary.

Worked connection: From 600 tokens, 3/10 are allocated in one stage, leaving 420; 2/7 of the remainder is 120; 300 remain.

Tutor’s checking question: Ask which amounts change between stages and whether the rate applies to the remaining quantity.

27. A good first line is more valuable than a long answer

The first written step should name the relationship and reference whole: ‘1-1/3=2/3 remains’. A child who begins by adding every printed fraction may arrive at a plausible but wrong number.

Worked connection: For 1/3 then 1/4 of the remainder, write ‘fraction finally remaining = (2/3)(3/4)=1/2’, then multiply by the original total.

Tutor’s checking question: Use a bar model to explain the same first step when words are not sufficient.

28. Learn the reasoning before racing through worksheet sets

A tutor should determine whether the student has confused the reference whole, miscomputed equivalent fractions or ignored a fixed amount. Each mistake needs a different repair and a delayed independent retest.

Worked connection: A learner who gets 7/12 used in the first example needs a reference-base lesson, not simply another twenty similar arithmetic calculations.

Tutor’s checking question: Let the student explain the value of one stage-two unit and then solve a changed scenario without prompts.

Twenty-four forward worked clinics: two stages, one original whole

In these model stories the numerical quantities are hypothetical; they provide clean arithmetic for practising reference amounts. Each problem follows the same structure but varies the fractions, the original amount and the context. Cover the answers, draw the start–used–left table and calculate independently.

Forward clinic 1: 120 dollars of a fictional allowance, remove 1/3, then 1/4 of the remainder

Start with 120 dollars of a fictional allowance. First remove 1/3 of the original. Next remove 1/4 of what remains. How much is removed at each stage, and how much is left?

Answer: first removal 40; first remainder 80; second removal 20; final remainder 60. The remaining share of the original is 1/2, while the total share removed is 1/2.

Reason: multiply 120 by 1/3 to find the first removal. Subtract it from 120 before applying 1/4. The second removal is 1/6 of the original, not simply 1/4 of the original. Check: 40+20+60=120.

For independent transfer, change the first fraction while holding the original total fixed, then predict whether the new final quantity should be larger or smaller. The student should justify the prediction before calculating. If the result goes in the opposite direction, re-examine which amount the second fraction was applied to.

Forward clinic 2: 180 practice tokens, remove 2/5, then 1/3 of the remainder

Start with 180 practice tokens. First remove 2/5 of the original. Next remove 1/3 of what remains. How much is removed at each stage, and how much is left?

Answer: first removal 72; first remainder 108; second removal 36; final remainder 72. The remaining share of the original is 2/5, while the total share removed is 3/5.

Reason: multiply 180 by 2/5 to find the first removal. Subtract it from 180 before applying 1/3. The second removal is 1/5 of the original, not simply 1/3 of the original. Check: 72+36+72=180.

Forward clinic 3: 240 litres of water in a hypothetical tank, remove 1/4, then 2/3 of the remainder

Start with 240 litres of water in a hypothetical tank. First remove 1/4 of the original. Next remove 2/3 of what remains. How much is removed at each stage, and how much is left?

Answer: first removal 60; first remainder 180; second removal 120; final remainder 60. The remaining share of the original is 1/4, while the total share removed is 3/4.

Reason: multiply 240 by 1/4 to find the first removal. Subtract it from 240 before applying 2/3. The second removal is 1/2 of the original, not simply 2/3 of the original. Check: 60+120+60=240.

Forward clinic 4: 300 cardboard tokens, remove 1/5, then 1/4 of the remainder

Start with 300 cardboard tokens. First remove 1/5 of the original. Next remove 1/4 of what remains. How much is removed at each stage, and how much is left?

Answer: first removal 60; first remainder 240; second removal 60; final remainder 180. The remaining share of the original is 3/5, while the total share removed is 2/5.

Reason: multiply 300 by 1/5 to find the first removal. Subtract it from 300 before applying 1/4. The second removal is 1/5 of the original, not simply 1/4 of the original. Check: 60+60+180=300.

Forward clinic 5: 160 tickets in a maths model, remove 3/8, then 1/5 of the remainder

Start with 160 tickets in a maths model. First remove 3/8 of the original. Next remove 1/5 of what remains. How much is removed at each stage, and how much is left?

Answer: first removal 60; first remainder 100; second removal 20; final remainder 80. The remaining share of the original is 1/2, while the total share removed is 1/2.

Reason: multiply 160 by 3/8 to find the first removal. Subtract it from 160 before applying 1/5. The second removal is 1/8 of the original, not simply 1/5 of the original. Check: 60+20+80=160.

Forward clinic 6: 360 cm of ribbon in a model, remove 1/6, then 2/5 of the remainder

Start with 360 cm of ribbon in a model. First remove 1/6 of the original. Next remove 2/5 of what remains. How much is removed at each stage, and how much is left?

Answer: first removal 60; first remainder 300; second removal 120; final remainder 180. The remaining share of the original is 1/2, while the total share removed is 1/2.

Reason: multiply 360 by 1/6 to find the first removal. Subtract it from 360 before applying 2/5. The second removal is 1/3 of the original, not simply 2/5 of the original. Check: 60+120+180=360.

Forward clinic 7: 420 paper squares, remove 2/7, then 1/3 of the remainder

Start with 420 paper squares. First remove 2/7 of the original. Next remove 1/3 of what remains. How much is removed at each stage, and how much is left?

Answer: first removal 120; first remainder 300; second removal 100; final remainder 200. The remaining share of the original is 10/21, while the total share removed is 11/21.

Reason: multiply 420 by 2/7 to find the first removal. Subtract it from 420 before applying 1/3. The second removal is 5/21 of the original, not simply 1/3 of the original. Check: 120+100+200=420.

Forward clinic 8: 280 counters, remove 3/7, then 1/4 of the remainder

Start with 280 counters. First remove 3/7 of the original. Next remove 1/4 of what remains. How much is removed at each stage, and how much is left?

Answer: first removal 120; first remainder 160; second removal 40; final remainder 120. The remaining share of the original is 3/7, while the total share removed is 4/7.

Reason: multiply 280 by 3/7 to find the first removal. Subtract it from 280 before applying 1/4. The second removal is 1/7 of the original, not simply 1/4 of the original. Check: 120+40+120=280.

Forward clinic 9: 150 dollars in a fictional budget, remove 1/5, then 1/2 of the remainder

Start with 150 dollars in a fictional budget. First remove 1/5 of the original. Next remove 1/2 of what remains. How much is removed at each stage, and how much is left?

Answer: first removal 30; first remainder 120; second removal 60; final remainder 60. The remaining share of the original is 2/5, while the total share removed is 3/5.

Reason: multiply 150 by 1/5 to find the first removal. Subtract it from 150 before applying 1/2. The second removal is 2/5 of the original, not simply 1/2 of the original. Check: 30+60+60=150.

Forward clinic 10: 200 cards, remove 1/4, then 1/5 of the remainder

Start with 200 cards. First remove 1/4 of the original. Next remove 1/5 of what remains. How much is removed at each stage, and how much is left?

Answer: first removal 50; first remainder 150; second removal 30; final remainder 120. The remaining share of the original is 3/5, while the total share removed is 2/5.

Reason: multiply 200 by 1/4 to find the first removal. Subtract it from 200 before applying 1/5. The second removal is 3/20 of the original, not simply 1/5 of the original. Check: 50+30+120=200.

Forward clinic 11: 900 practice points, remove 2/9, then 1/7 of the remainder

Start with 900 practice points. First remove 2/9 of the original. Next remove 1/7 of what remains. How much is removed at each stage, and how much is left?

Answer: first removal 200; first remainder 700; second removal 100; final remainder 600. The remaining share of the original is 2/3, while the total share removed is 1/3.

Reason: multiply 900 by 2/9 to find the first removal. Subtract it from 900 before applying 1/7. The second removal is 1/9 of the original, not simply 1/7 of the original. Check: 200+100+600=900.

Forward clinic 12: 500 ml of hypothetical liquid, remove 3/10, then 2/7 of the remainder

Start with 500 ml of hypothetical liquid. First remove 3/10 of the original. Next remove 2/7 of what remains. How much is removed at each stage, and how much is left?

Answer: first removal 150; first remainder 350; second removal 100; final remainder 250. The remaining share of the original is 1/2, while the total share removed is 1/2.

Reason: multiply 500 by 3/10 to find the first removal. Subtract it from 500 before applying 2/7. The second removal is 1/5 of the original, not simply 2/7 of the original. Check: 150+100+250=500.

Forward clinic 13: 264 buttons, remove 1/3, then 3/8 of the remainder

Start with 264 buttons. First remove 1/3 of the original. Next remove 3/8 of what remains. How much is removed at each stage, and how much is left?

Answer: first removal 88; first remainder 176; second removal 66; final remainder 110. The remaining share of the original is 5/12, while the total share removed is 7/12.

Reason: multiply 264 by 1/3 to find the first removal. Subtract it from 264 before applying 3/8. The second removal is 1/4 of the original, not simply 3/8 of the original. Check: 88+66+110=264.

Forward clinic 14: 315 paper clips, remove 2/7, then 1/5 of the remainder

Start with 315 paper clips. First remove 2/7 of the original. Next remove 1/5 of what remains. How much is removed at each stage, and how much is left?

Answer: first removal 90; first remainder 225; second removal 45; final remainder 180. The remaining share of the original is 4/7, while the total share removed is 3/7.

Reason: multiply 315 by 2/7 to find the first removal. Subtract it from 315 before applying 1/5. The second removal is 1/7 of the original, not simply 1/5 of the original. Check: 90+45+180=315.

Forward clinic 15: 768 cm of material, remove 1/4, then 1/3 of the remainder

Start with 768 cm of material. First remove 1/4 of the original. Next remove 1/3 of what remains. How much is removed at each stage, and how much is left?

Answer: first removal 192; first remainder 576; second removal 192; final remainder 384. The remaining share of the original is 1/2, while the total share removed is 1/2.

Reason: multiply 768 by 1/4 to find the first removal. Subtract it from 768 before applying 1/3. The second removal is 1/4 of the original, not simply 1/3 of the original. Check: 192+192+384=768.

Forward clinic 16: 288 markers, remove 1/6, then 1/5 of the remainder

Start with 288 markers. First remove 1/6 of the original. Next remove 1/5 of what remains. How much is removed at each stage, and how much is left?

Answer: first removal 48; first remainder 240; second removal 48; final remainder 192. The remaining share of the original is 2/3, while the total share removed is 1/3.

Reason: multiply 288 by 1/6 to find the first removal. Subtract it from 288 before applying 1/5. The second removal is 1/6 of the original, not simply 1/5 of the original. Check: 48+48+192=288.

Forward clinic 17: 840 sample tokens, remove 2/7, then 1/5 of the remainder

Start with 840 sample tokens. First remove 2/7 of the original. Next remove 1/5 of what remains. How much is removed at each stage, and how much is left?

Answer: first removal 240; first remainder 600; second removal 120; final remainder 480. The remaining share of the original is 4/7, while the total share removed is 3/7.

Reason: multiply 840 by 2/7 to find the first removal. Subtract it from 840 before applying 1/5. The second removal is 1/7 of the original, not simply 1/5 of the original. Check: 240+120+480=840.

Forward clinic 18: 216 counters, remove 1/4, then 1/3 of the remainder

Start with 216 counters. First remove 1/4 of the original. Next remove 1/3 of what remains. How much is removed at each stage, and how much is left?

Answer: first removal 54; first remainder 162; second removal 54; final remainder 108. The remaining share of the original is 1/2, while the total share removed is 1/2.

Reason: multiply 216 by 1/4 to find the first removal. Subtract it from 216 before applying 1/3. The second removal is 1/4 of the original, not simply 1/3 of the original. Check: 54+54+108=216.

Forward clinic 19: 600 pieces of paper, remove 3/10, then 2/7 of the remainder

Start with 600 pieces of paper. First remove 3/10 of the original. Next remove 2/7 of what remains. How much is removed at each stage, and how much is left?

Answer: first removal 180; first remainder 420; second removal 120; final remainder 300. The remaining share of the original is 1/2, while the total share removed is 1/2.

Reason: multiply 600 by 3/10 to find the first removal. Subtract it from 600 before applying 2/7. The second removal is 1/5 of the original, not simply 2/7 of the original. Check: 180+120+300=600.

Forward clinic 20: 450 virtual reward points, remove 1/5, then 1/4 of the remainder

Start with 450 virtual reward points. First remove 1/5 of the original. Next remove 1/4 of what remains. How much is removed at each stage, and how much is left?

Answer: first removal 90; first remainder 360; second removal 90; final remainder 270. The remaining share of the original is 3/5, while the total share removed is 2/5.

Reason: multiply 450 by 1/5 to find the first removal. Subtract it from 450 before applying 1/4. The second removal is 1/5 of the original, not simply 1/4 of the original. Check: 90+90+270=450.

Forward clinic 21: 250 dollars in a fictional fund, remove 1/5, then 1/4 of the remainder

Start with 250 dollars in a fictional fund. First remove 1/5 of the original. Next remove 1/4 of what remains. How much is removed at each stage, and how much is left?

Answer: first removal 50; first remainder 200; second removal 50; final remainder 150. The remaining share of the original is 3/5, while the total share removed is 2/5.

Reason: multiply 250 by 1/5 to find the first removal. Subtract it from 250 before applying 1/4. The second removal is 1/5 of the original, not simply 1/4 of the original. Check: 50+50+150=250.

Forward clinic 22: 1000 identical blocks, remove 1/4, then 1/5 of the remainder

Start with 1000 identical blocks. First remove 1/4 of the original. Next remove 1/5 of what remains. How much is removed at each stage, and how much is left?

Answer: first removal 250; first remainder 750; second removal 150; final remainder 600. The remaining share of the original is 3/5, while the total share removed is 2/5.

Reason: multiply 1000 by 1/4 to find the first removal. Subtract it from 1000 before applying 1/5. The second removal is 3/20 of the original, not simply 1/5 of the original. Check: 250+150+600=1000.

Forward clinic 23: 540 tickets in an arithmetic story, remove 1/3, then 1/5 of the remainder

Start with 540 tickets in an arithmetic story. First remove 1/3 of the original. Next remove 1/5 of what remains. How much is removed at each stage, and how much is left?

Answer: first removal 180; first remainder 360; second removal 72; final remainder 288. The remaining share of the original is 8/15, while the total share removed is 7/15.

Reason: multiply 540 by 1/3 to find the first removal. Subtract it from 540 before applying 1/5. The second removal is 2/15 of the original, not simply 1/5 of the original. Check: 180+72+288=540.

Forward clinic 24: 312 cm of strip material, remove 1/4, then 2/3 of the remainder

Start with 312 cm of strip material. First remove 1/4 of the original. Next remove 2/3 of what remains. How much is removed at each stage, and how much is left?

Answer: first removal 78; first remainder 234; second removal 156; final remainder 78. The remaining share of the original is 1/4, while the total share removed is 3/4.

Reason: multiply 312 by 1/4 to find the first removal. Subtract it from 312 before applying 2/3. The second removal is 1/2 of the original, not simply 2/3 of the original. Check: 78+156+78=312.

Eight reverse problems: recover the starting total

Reverse questions are best checked twice: first by undoing the second stage, then the first, and finally by running the entire forward process. The answer is not secure until it reproduces the given final amount.

Reverse clinic 1: 60 remains after 1/3 and then 1/4

An unknown initial quantity loses 1/3 of itself, then loses 1/4 of the amount remaining. The final amount is 60 units. What was the original quantity?

Answer: 120 units. The amount left after the first stage is 80, so reversing the second removal gives 60÷(3/4)=80. Reversing the first gives 80÷(2/3)=120.

Forward check: start at 120; take away 40, then 20, and the result is 60. The second fraction is applied only after the first removal. Explain why both reverse divisions increase the candidate earlier amount rather than reduce it.

Reverse clinic 2: 72 remains after 2/5 and then 1/3

An unknown initial quantity loses 2/5 of itself, then loses 1/3 of the amount remaining. The final amount is 72 units. What was the original quantity?

Answer: 180 units. The amount left after the first stage is 108, so reversing the second removal gives 72÷(2/3)=108. Reversing the first gives 108÷(3/5)=180.

Forward check: start at 180; take away 72, then 36, and the result is 72. The second fraction is applied only after the first removal. Explain why both reverse divisions increase the candidate earlier amount rather than reduce it.

Reverse clinic 3: 180 remains after 1/5 and then 1/4

An unknown initial quantity loses 1/5 of itself, then loses 1/4 of the amount remaining. The final amount is 180 units. What was the original quantity?

Answer: 300 units. The amount left after the first stage is 240, so reversing the second removal gives 180÷(3/4)=240. Reversing the first gives 240÷(4/5)=300.

Forward check: start at 300; take away 60, then 60, and the result is 180. The second fraction is applied only after the first removal. Explain why both reverse divisions increase the candidate earlier amount rather than reduce it.

Reverse clinic 4: 180 remains after 1/6 and then 2/5

An unknown initial quantity loses 1/6 of itself, then loses 2/5 of the amount remaining. The final amount is 180 units. What was the original quantity?

Answer: 360 units. The amount left after the first stage is 300, so reversing the second removal gives 180÷(3/5)=300. Reversing the first gives 300÷(5/6)=360.

Forward check: start at 360; take away 60, then 120, and the result is 180. The second fraction is applied only after the first removal. Explain why both reverse divisions increase the candidate earlier amount rather than reduce it.

Reverse clinic 5: 120 remains after 1/4 and then 1/5

An unknown initial quantity loses 1/4 of itself, then loses 1/5 of the amount remaining. The final amount is 120 units. What was the original quantity?

Answer: 200 units. The amount left after the first stage is 150, so reversing the second removal gives 120÷(4/5)=150. Reversing the first gives 150÷(3/4)=200.

Forward check: start at 200; take away 50, then 30, and the result is 120. The second fraction is applied only after the first removal. Explain why both reverse divisions increase the candidate earlier amount rather than reduce it.

Reverse clinic 6: 600 remains after 2/9 and then 1/7

An unknown initial quantity loses 2/9 of itself, then loses 1/7 of the amount remaining. The final amount is 600 units. What was the original quantity?

Answer: 900 units. The amount left after the first stage is 700, so reversing the second removal gives 600÷(6/7)=700. Reversing the first gives 700÷(7/9)=900.

Forward check: start at 900; take away 200, then 100, and the result is 600. The second fraction is applied only after the first removal. Explain why both reverse divisions increase the candidate earlier amount rather than reduce it.

Reverse clinic 7: 480 remains after 2/7 and then 1/5

An unknown initial quantity loses 2/7 of itself, then loses 1/5 of the amount remaining. The final amount is 480 units. What was the original quantity?

Answer: 840 units. The amount left after the first stage is 600, so reversing the second removal gives 480÷(4/5)=600. Reversing the first gives 600÷(5/7)=840.

Forward check: start at 840; take away 240, then 120, and the result is 480. The second fraction is applied only after the first removal. Explain why both reverse divisions increase the candidate earlier amount rather than reduce it.

Reverse clinic 8: 270 remains after 1/5 and then 1/4

An unknown initial quantity loses 1/5 of itself, then loses 1/4 of the amount remaining. The final amount is 270 units. What was the original quantity?

Answer: 450 units. The amount left after the first stage is 360, so reversing the second removal gives 270÷(3/4)=360. Reversing the first gives 360÷(4/5)=450.

Forward check: start at 450; take away 90, then 90, and the result is 270. The second fraction is applied only after the first removal. Explain why both reverse divisions increase the candidate earlier amount rather than reduce it.

Eight quick interpretation clinics

Interpretation clinic 1: One third followed by one quarter of the remainder: fraction finally left?

Answer: 1/2. Reason: The remaining share is (2/3)(3/4)=1/2 of original.

Tell the story aloud again, emphasising which base the fraction uses. A learner should be able to draw or label the stage-one and stage-two wholes before relying on the calculation shortcut. Try rewriting ‘remainder’ as ‘original’ to see why the answer changes.

Interpretation clinic 2: Same story: fraction removed in stage two relative to original?

Answer: 1/6. Reason: Second removal is (2/3)(1/4)=1/6 of the original.

Interpretation clinic 3: Same story: fraction removed in both stages together?

Answer: 1/2. Reason: First 1/3 plus second 1/6 equals 1/2.

Interpretation clinic 4: If one third is removed then one quarter of original is removed, fraction left?

Answer: 5/12. Reason: Both removals use the original base: 1-1/3-1/4=5/12.

Interpretation clinic 5: If one third is removed then a fixed 20 out of 120 is removed, how much remains?

Answer: 60. Reason: After first use 80 remains; subtract fixed 20.

Interpretation clinic 6: Which is larger from 120: quarter of original or quarter of 80 remaining?

Answer: 30 versus 20. Reason: A quarter of the original is 30; a quarter of remainder is 20.

Interpretation clinic 7: 20% off then 25% off the new price: fraction of original left?

Answer: 3/5. Reason: 0.8×0.75=0.6=3/5.

Interpretation clinic 8: Is it correct to add 1/3 and 1/4 when they apply to different wholes?

Answer: No. Reason: The second fraction must be expressed relative to the original base first.

Three activities that make the reference amount visible

A two-colour bar diagram

Draw twelve equal blocks for a starting amount of $120, so each original block is worth $10. Colour four blocks as the first third used. Only eight uncoloured blocks remain. A quarter of those eight is two blocks, so the second removal is $20. Six blocks remain, or $60. Ask which printed fraction refers to the twelve-block whole and which refers to the eight-block remainder.

The point of the activity is to identify the changing whole, not merely to produce the numerical answer. After the model is removed, use an unfamiliar set of fractions and record whether the learner can start the question independently.

Compare two nearly identical stories

Story A: remove one third of a 120-token supply, then one quarter of the remainder; sixty remain. Story B: remove one third, then one quarter of the original; fifty remain. The student should explain that the second instruction changes the base. Give a new starting supply of 240 and let the student predict which story leaves more before calculating.

Find the original in reverse

A card box contains an unknown number of cards. One quarter are used, then one third of what remains are used, leaving eighty cards. Draw the final amount as two thirds of the first remainder; recover the first remainder 80÷(2/3)=120. That is three quarters of the original, so original 120÷(3/4)=160. Finally perform both removals forward to verify eighty.

A four-week fraction-word-problem improvement plan

Week 1: identify every reference whole

Begin with one-step fractions of a quantity and simple bar models. Highlight ‘of original’, ‘of remaining’ and ‘fixed amount’. Ask the student to write the first remaining fraction before any second-stage calculation.

Short, focused practice with an explanation and a delayed independent check is usually more useful than a long worksheet copied from a familiar model. The aim is to make the child’s choice of reference amount reliable in new contexts.

Week 2: use two-stage tables and fraction products

Introduce two proportional removals and practise start–used–left tables. Solve each using actual amounts and using the remaining-fraction product, then verify the routes agree.

Week 3: reverse from a known final amount

Work backwards using division by remaining fractions. Add a simple linear equation for students ready to express the unknown original symbolically. Check every answer by applying both reductions forward.

Week 4: mixed wording and independent transfer

Mix ‘fraction of the original’, ‘fraction of the remainder’ and ‘remove a fixed amount’ questions. Remove chapter labels and ask the child to select a model and write a clear first line. Retest persistent errors after several days.

How a tutor can identify the first wrong assumption

Three students may all reach the wrong final total for different reasons. One adds 1/3 and 1/4 directly, another finds one quarter of the original when the wording says remainder, and a third models the stages correctly but mishandles the fraction multiplication. Those are different teaching targets. A useful tutor listens to the child’s first sentence before correcting the arithmetic.

Once the reference model is secure, the student can solve a changed example without hints. During another lesson, a short delayed question tests whether the idea survived outside the immediate worked demonstration. The result should be recorded as independent or assisted, not merely correct or incorrect.

The eduKateSG Secondary 1 small-group reference describes three-pupil weekly 90-minute sessions with guided correction near Sixth Avenue MRT. Current availability and individual suitability should be confirmed directly.

Frequently asked questions

What does ‘one third of the remainder’ mean?

It means calculate one third of the amount remaining after the previous stage, not one third of the original total.

Why can’t I add 1/3 and 1/4 directly?

Because one third may refer to the original while one quarter refers to the smaller remainder. Fractions with different reference wholes cannot be added as though they describe the same whole.

How do I calculate the final fraction left?

Multiply the remaining fractions from each proportional stage, such as (1-1/3)(1-1/4)=1/2.

How do I find the original amount if only the final amount is given?

Divide the final amount by the product of the remaining fractions, or reverse each proportional reduction one stage at a time.

Can I solve these questions using a bar model?

Yes. Partition the original bar, mark the used part, then partition only the remainder for the second stage. This is especially useful for explaining the changing reference whole.

What if the second step removes a fixed number?

Then subtract that fixed amount from the remainder rather than multiplying by another remaining fraction.

Does the order of successive percentages matter?

The final multiplier for two pure percentage reductions commutes, but the amount removed at each named stage generally changes with the order. Read what the question specifically asks.

Do fractions of a group always produce whole people?

Not for arbitrary starting data. Classroom problems involving individual people normally use compatible totals or instruct how to interpret counts; check the wording and arithmetic.

What is a useful check of my answer?

Verify that first removal plus second removal plus the final remainder equals the original, and that each stage uses its correct reference amount.

Does this belong in Secondary 1 G2 Mathematics?

It applies the Secondary 1 G2 number, ratio, fractions and percentage reasoning strands in a multi-step problem-solving setting. The student’s school may present these applications in a different sequence.

Why does my child solve short fraction sums but struggle with these stories?

They may understand fractional arithmetic but misidentify the reference whole. Teach the story and model before assigning more computation.

What can a small-group tutor do differently?

A tutor can identify whether the error began in reading, modelling, a common-denominator step or arithmetic, then test the repair on a fresh story without hints.

The MOE G2 Mathematics syllabus provides the national number and ratio context, and the unchanged eduKateSG tutoring reference describes the small-group setting. Continue with the following guides for fraction operations, ratios and modelling:

The takeaway is one question you can ask before every calculation: ‘A fraction of which amount?’ When the student keeps the changing whole visible, the rest becomes a sequence of understandable, checkable steps.

Secondary 1 G2 Mathematics Tutor Bukit Timah | Fraction of the Remainder Word Problems