A ratio question often looks easy until the question reverses direction. Instead of asking for the two masses in a known mixture, it might give the mass of one constituent, a percentage after a loss, or the desired composition after an addition. The reliable skill is knowing which quantity is the whole at each step. This guide builds that skill from simple equal parts through multi-stage reverse calculations.
For 2027 Singapore-Cambridge SEC G3 Mathematics, the official subject code is K310. Ratio, direct and inverse proportion, percentages and reverse percentages appear in the syllabus. Paper 1 and Paper 2 are each 2 hours 15 minutes, carry 90 marks and contribute 50% of the subject assessment. The official notes say that omission of essential working may lose marks and that non-exact answers normally follow the specified accuracy convention. Verify details via SEAB’s 2027 G3 directory and the K310 syllabus PDF. These are original mathematical practice contexts, not official examination questions.
First distinguish a part, the whole, and the base
Write three labels on rough working: ‘part’, ‘whole’, and ‘change’. If the ratio A:B is 3:2, there are five parts in the whole. A is three-fifths of that whole, not three-halves. If A is known but the whole is missing, divide by A’s share. If a mixture gains or loses one constituent, rebuild the new whole before calculating the new percentage.
For a clean first example, suppose A:B is 2:3 and the total is 250 g. One ratio part is 250/5 = 50 g, so A = 100 g and B = 150 g. Now reverse it. Suppose only A = 100 g is given. The same ratio tells us one part is 50 g and the total is still 250 g. The method does not change; only the starting evidence changes.
The five-line examination layout
Line one: define the quantity and unit. Line two: translate the wording into a ratio, fraction or equation. Line three: solve without early rounding. Line four: check that the component masses add to the stated whole and satisfy the ratio. Line five: write the answer in the requested units and accuracy. This layout may seem slower than a mental shortcut, but it exposes the reasoning when the arithmetic becomes complicated.
Some contexts use percentages greater than 100% or percentages applied repeatedly. Others compare percentage points with relative percentage change. Keep those questions separate. An increase from 15% to 18% is three percentage points, or a 20% relative increase in the percentage itself. It is not simply a three-percent increase in the physical quantity unless the complete context says so.
Twenty-four worked cases, from basic to advanced
1. A three-to-two batch
Problem. A prepared material mix contains material A and material B in the mass ratio 3:2. The total mass is 250 g. Write a short model before reaching for a calculator; use masses, percentages or ratios that refer to the same whole.
Worked route. Five equal parts make 250 g, so one part is 50 g. Material A is 3 times 50 = 150 g and B is 2 times 50 = 100 g. Notice which quantity serves as the denominator and how the answer connects to the original units.
Why the method transfers. The denominator must represent the whole to which the ratio refers. Write the number of equal parts before using any calculator. A ratio is not automatically a fraction of the total until all listed parts are added. This first move prevents the common error of dividing by the wrong number when a question describes one ingredient rather than the complete blend. For this problem, verify the calculated answer by substituting it into the original statement, not merely by repeating the calculation. A solution that satisfies the algebra but violates the described mixture is not a complete contextual answer.
Error to prevent. Taking 3/2 of the total for A would count the wrong whole; the denominator is five parts. Mark the exact step where the mistaken interpretation enters. Repairing the first wrong relationship is much more useful than checking only the final arithmetic.
Independent extension. Retest with a 400 g total and explain why each mass scales together. Once that answer is secure, state in one sentence why the chosen denominator is correct. A clear explanation is evidence that you can transfer the method to a new question.
2. Four parts to one
Problem. A school model uses four equal masses of clay A for every one mass of clay B. There are 180 g altogether. Write a short model before reaching for a calculator; use masses, percentages or ratios that refer to the same whole.
Worked route. There are five parts. One part equals 36 g, giving A = 144 g and B = 36 g. Notice which quantity serves as the denominator and how the answer connects to the original units.
Why the method transfers. A reverse problem starts from what is known, not from the largest printed number. Mark whether the given mass is the total, a single part, a difference, or a percentage of an unknown original. Then set a one-part value or an unknown x. The algebra should preserve the relationship described by the words. For this problem, verify the calculated answer by substituting it into the original statement, not merely by repeating the calculation. A solution that satisfies the algebra but violates the described mixture is not a complete contextual answer.
Error to prevent. The ratio 4:1 does not mean A is four-fifths of B; it means A is four times B. Mark the exact step where the mistaken interpretation enters. Repairing the first wrong relationship is much more useful than checking only the final arithmetic.
Independent extension. Explain the fraction of the whole represented by B. Once that answer is secure, state in one sentence why the chosen denominator is correct. A clear explanation is evidence that you can transfer the method to a new question.
3. One part given
Problem. Two kinds of paper are stacked in the ratio 2:5. The first kind weighs 36 g. Write a short model before reaching for a calculator; use masses, percentages or ratios that refer to the same whole.
Worked route. Two parts equal 36 g. Each part is 18 g. The other paper weighs 5 times 18 = 90 g, and both weigh 126 g. Notice which quantity serves as the denominator and how the answer connects to the original units.
Why the method transfers. Percentages are comparisons with a particular base. Before every percentage operation, ask ‘percentage of what?’ The original quantity and the final quantity often have different denominators. If a 20% increase produces 120%, the inverse operation divides by 1.20. It does not subtract 20% of the new number. For this problem, verify the calculated answer by substituting it into the original statement, not merely by repeating the calculation. A solution that satisfies the algebra but violates the described mixture is not a complete contextual answer.
Error to prevent. Dividing 36 by seven uses the wrong unit; 36 describes two parts, not the entire stack. Mark the exact step where the mistaken interpretation enters. Repairing the first wrong relationship is much more useful than checking only the final arithmetic.
Independent extension. Find the new total if both stacks are doubled without changing their ratio. Once that answer is secure, state in one sentence why the chosen denominator is correct. A clear explanation is evidence that you can transfer the method to a new question.
4. Difference rather than total
Problem. A is to B as 7:3, and A exceeds B by 48 g. Neither individual mass is given. Write a short model before reaching for a calculator; use masses, percentages or ratios that refer to the same whole.
Worked route. The difference corresponds to four parts, so one part equals 12 g. Therefore A = 84 g, B = 36 g and the total = 120 g. Notice which quantity serves as the denominator and how the answer connects to the original units.
Why the method transfers. When blends are combined, first return to actual component masses. Percentages are useful descriptions but cannot normally be averaged without accounting for sample sizes. Add grams of A to grams of A and grams of B to grams of B; only then divide by the new total. This is the reliable route through unequal-batch questions. For this problem, verify the calculated answer by substituting it into the original statement, not merely by repeating the calculation. A solution that satisfies the algebra but violates the described mixture is not a complete contextual answer.
Error to prevent. Treating the difference as seven parts would incorrectly equate one component with the gap. Mark the exact step where the mistaken interpretation enters. Repairing the first wrong relationship is much more useful than checking only the final arithmetic.
Independent extension. Explain what a ‘difference of four parts’ means without a formula. Once that answer is secure, state in one sentence why the chosen denominator is correct. A clear explanation is evidence that you can transfer the method to a new question.
5. Three components
Problem. A 420 g blend has ingredients A:B:C in the ratio 1:4:5. Write a short model before reaching for a calculator; use masses, percentages or ratios that refer to the same whole.
Worked route. Ten parts weigh 420 g, so each part is 42 g. The component masses are 42 g, 168 g and 210 g respectively. Notice which quantity serves as the denominator and how the answer connects to the original units.
Why the method transfers. In selective addition or removal, identify what stays fixed. Adding pure A increases both A and the whole. Adding B changes the whole without changing A. Removing A reduces A and the whole. Writing those changes explicitly prevents the false assumption that the original ratio survives every operation. For this problem, verify the calculated answer by substituting it into the original statement, not merely by repeating the calculation. A solution that satisfies the algebra but violates the described mixture is not a complete contextual answer.
Error to prevent. Do not divide by the largest ratio number; divide by the sum of all parts. Mark the exact step where the mistaken interpretation enters. Repairing the first wrong relationship is much more useful than checking only the final arithmetic.
Independent extension. Check all three values against the original 420 g total. Once that answer is secure, state in one sentence why the chosen denominator is correct. A clear explanation is evidence that you can transfer the method to a new question.
6. Non-integer ratio written as integers
Problem. Two components are described in the ratio 2.5:1.5, and the total mass is 320 g. Write a short model before reaching for a calculator; use masses, percentages or ratios that refer to the same whole.
Worked route. Multiplying both ratio entries by two gives 5:3. Eight parts weigh 320 g; the masses are 200 g and 120 g. Notice which quantity serves as the denominator and how the answer connects to the original units.
Why the method transfers. Check units, feasibility and rounding. An answer in grams cannot be silently compared with a value in kilograms, and a negative mass would be physically impossible in this context. Where a syllabus answer calls for three significant figures, keep exact values or enough calculator digits during working and round only when reporting the final numerical result. For this problem, verify the calculated answer by substituting it into the original statement, not merely by repeating the calculation. A solution that satisfies the algebra but violates the described mixture is not a complete contextual answer.
Error to prevent. Dividing only the first part by 0.5 changes the ratio; scale both entries equally. Mark the exact step where the mistaken interpretation enters. Repairing the first wrong relationship is much more useful than checking only the final arithmetic.
Independent extension. Give another equivalent ratio and explain why the physical mix is unchanged. Once that answer is secure, state in one sentence why the chosen denominator is correct. A clear explanation is evidence that you can transfer the method to a new question.
7. Same ratio at different scales
Problem. A workshop has two batches with A:B = 3:2, weighing 100 g and 150 g. Write a short model before reaching for a calculator; use masses, percentages or ratios that refer to the same whole.
Worked route. Batch one has 60 g A and 40 g B. Batch two has 90 g A and 60 g B. Combined, A = 150 g and B = 100 g. Notice which quantity serves as the denominator and how the answer connects to the original units.
Why the method transfers. The denominator must represent the whole to which the ratio refers. Write the number of equal parts before using any calculator. A ratio is not automatically a fraction of the total until all listed parts are added. This first move prevents the common error of dividing by the wrong number when a question describes one ingredient rather than the complete blend. For this problem, verify the calculated answer by substituting it into the original statement, not merely by repeating the calculation. A solution that satisfies the algebra but violates the described mixture is not a complete contextual answer.
Error to prevent. Adding ratios directly is unsafe unless the part size and total quantities are accounted for. Mark the exact step where the mistaken interpretation enters. Repairing the first wrong relationship is much more useful than checking only the final arithmetic.
Independent extension. Prove that the combined ratio remains 3:2 in this special case. Once that answer is secure, state in one sentence why the chosen denominator is correct. A clear explanation is evidence that you can transfer the method to a new question.
8. A percentage of the whole
Problem. A 250 g manufactured item contains 12% of a named component by mass. Write a short model before reaching for a calculator; use masses, percentages or ratios that refer to the same whole.
Worked route. Component mass = 0.12 times 250 = 30 g. The remainder is 220 g. The percentage is a share of the entire item. Notice which quantity serves as the denominator and how the answer connects to the original units.
Why the method transfers. A reverse problem starts from what is known, not from the largest printed number. Mark whether the given mass is the total, a single part, a difference, or a percentage of an unknown original. Then set a one-part value or an unknown x. The algebra should preserve the relationship described by the words. For this problem, verify the calculated answer by substituting it into the original statement, not merely by repeating the calculation. A solution that satisfies the algebra but violates the described mixture is not a complete contextual answer.
Error to prevent. Reading 12% as 12 g confuses a proportion with an amount. Mark the exact step where the mistaken interpretation enters. Repairing the first wrong relationship is much more useful than checking only the final arithmetic.
Independent extension. Repeat for a 500 g item with the same percentage and identify the scaling. Once that answer is secure, state in one sentence why the chosen denominator is correct. A clear explanation is evidence that you can transfer the method to a new question.
9. Mass to percentage
Problem. An object has total mass 500 g, of which 80 g is material A. Write a short model before reaching for a calculator; use masses, percentages or ratios that refer to the same whole.
Worked route. The percentage of A is 80 divided by 500 multiplied by 100 = 16%. This is the fraction of the whole expressed per hundred. Notice which quantity serves as the denominator and how the answer connects to the original units.
Why the method transfers. Percentages are comparisons with a particular base. Before every percentage operation, ask ‘percentage of what?’ The original quantity and the final quantity often have different denominators. If a 20% increase produces 120%, the inverse operation divides by 1.20. It does not subtract 20% of the new number. For this problem, verify the calculated answer by substituting it into the original statement, not merely by repeating the calculation. A solution that satisfies the algebra but violates the described mixture is not a complete contextual answer.
Error to prevent. Dividing 500 by 80 produces a multiplier, not the percentage of A. Mark the exact step where the mistaken interpretation enters. Repairing the first wrong relationship is much more useful than checking only the final arithmetic.
Independent extension. Explain how the percentage would change if another 100 g of material B were added. Once that answer is secure, state in one sentence why the chosen denominator is correct. A clear explanation is evidence that you can transfer the method to a new question.
10. Reversing a mass loss
Problem. After losing 25% of its original mass, a dry sample weighs 300 g. Write a short model before reaching for a calculator; use masses, percentages or ratios that refer to the same whole.
Worked route. The remaining 300 g represents 75% of the original. Original mass = 300 / 0.75 = 400 g, so the mass lost was 100 g. Notice which quantity serves as the denominator and how the answer connects to the original units.
Why the method transfers. When blends are combined, first return to actual component masses. Percentages are useful descriptions but cannot normally be averaged without accounting for sample sizes. Add grams of A to grams of A and grams of B to grams of B; only then divide by the new total. This is the reliable route through unequal-batch questions. For this problem, verify the calculated answer by substituting it into the original statement, not merely by repeating the calculation. A solution that satisfies the algebra but violates the described mixture is not a complete contextual answer.
Error to prevent. Multiplying 300 by 1.25 gives 375 g, because the 25% refers to the original whole. Mark the exact step where the mistaken interpretation enters. Repairing the first wrong relationship is much more useful than checking only the final arithmetic.
Independent extension. Check by finding 25% of 400 g and subtracting it. Once that answer is secure, state in one sentence why the chosen denominator is correct. A clear explanation is evidence that you can transfer the method to a new question.
11. Reversing an increase
Problem. A plastic model weighs 120 g after an added coating raises its original mass by 20%. Write a short model before reaching for a calculator; use masses, percentages or ratios that refer to the same whole.
Worked route. The final 120 g represents 120% of the original. Original mass = 120 / 1.20 = 100 g; coating mass = 20 g. Notice which quantity serves as the denominator and how the answer connects to the original units.
Why the method transfers. In selective addition or removal, identify what stays fixed. Adding pure A increases both A and the whole. Adding B changes the whole without changing A. Removing A reduces A and the whole. Writing those changes explicitly prevents the false assumption that the original ratio survives every operation. For this problem, verify the calculated answer by substituting it into the original statement, not merely by repeating the calculation. A solution that satisfies the algebra but violates the described mixture is not a complete contextual answer.
Error to prevent. Subtracting 20% of 120 treats the final mass as the wrong base. Mark the exact step where the mistaken interpretation enters. Repairing the first wrong relationship is much more useful than checking only the final arithmetic.
Independent extension. Write an equation using x for the original mass. Once that answer is secure, state in one sentence why the chosen denominator is correct. A clear explanation is evidence that you can transfer the method to a new question.
12. Unknown whole from remainder
Problem. Material A forms 60% of a mix. Material B makes up all the remaining 90 g. Write a short model before reaching for a calculator; use masses, percentages or ratios that refer to the same whole.
Worked route. B forms 40% of the total. Total = 90 / 0.40 = 225 g, and A = 135 g. Notice which quantity serves as the denominator and how the answer connects to the original units.
Why the method transfers. Check units, feasibility and rounding. An answer in grams cannot be silently compared with a value in kilograms, and a negative mass would be physically impossible in this context. Where a syllabus answer calls for three significant figures, keep exact values or enough calculator digits during working and round only when reporting the final numerical result. For this problem, verify the calculated answer by substituting it into the original statement, not merely by repeating the calculation. A solution that satisfies the algebra but violates the described mixture is not a complete contextual answer.
Error to prevent. Do not set 90 g equal to 60%; it is the unnamed remaining portion. Mark the exact step where the mistaken interpretation enters. Repairing the first wrong relationship is much more useful than checking only the final arithmetic.
Independent extension. Rebuild the answer from the ratio 3:2 and check it agrees. Once that answer is secure, state in one sentence why the chosen denominator is correct. A clear explanation is evidence that you can transfer the method to a new question.
13. Ratio and a fixed additive
Problem. A: B = 4:1 and the smaller component B has mass 45 g. Write a short model before reaching for a calculator; use masses, percentages or ratios that refer to the same whole.
Worked route. B is one part, so A weighs 180 g and the complete mix weighs 225 g. Notice which quantity serves as the denominator and how the answer connects to the original units.
Why the method transfers. The denominator must represent the whole to which the ratio refers. Write the number of equal parts before using any calculator. A ratio is not automatically a fraction of the total until all listed parts are added. This first move prevents the common error of dividing by the wrong number when a question describes one ingredient rather than the complete blend. For this problem, verify the calculated answer by substituting it into the original statement, not merely by repeating the calculation. A solution that satisfies the algebra but violates the described mixture is not a complete contextual answer.
Error to prevent. Do not use 45 g as one fifth of A; it is one fifth of the total. Mark the exact step where the mistaken interpretation enters. Repairing the first wrong relationship is much more useful than checking only the final arithmetic.
Independent extension. Calculate the percentage of B in the combined mix. Once that answer is secure, state in one sentence why the chosen denominator is correct. A clear explanation is evidence that you can transfer the method to a new question.
14. Percentage points versus relative growth
Problem. The same 500 g item rises from 15% A to 18% A in a revised design. Write a short model before reaching for a calculator; use masses, percentages or ratios that refer to the same whole.
Worked route. The masses are 75 g and 90 g, a difference of 15 g. The composition rises by three percentage points, while the relative percentage increase is 20%. Notice which quantity serves as the denominator and how the answer connects to the original units.
Why the method transfers. A reverse problem starts from what is known, not from the largest printed number. Mark whether the given mass is the total, a single part, a difference, or a percentage of an unknown original. Then set a one-part value or an unknown x. The algebra should preserve the relationship described by the words. For this problem, verify the calculated answer by substituting it into the original statement, not merely by repeating the calculation. A solution that satisfies the algebra but violates the described mixture is not a complete contextual answer.
Error to prevent. A three-percentage-point increase is not the same as a three-percent increase. Mark the exact step where the mistaken interpretation enters. Repairing the first wrong relationship is much more useful than checking only the final arithmetic.
Independent extension. Explain which change a manufacturer should use when comparing masses. Once that answer is secure, state in one sentence why the chosen denominator is correct. A clear explanation is evidence that you can transfer the method to a new question.
15. Combining unequal concentrations
Problem. A 200 g batch is 70% A, while a 100 g batch is 80% A. They are mixed. Write a short model before reaching for a calculator; use masses, percentages or ratios that refer to the same whole.
Worked route. The first contains 140 g A; the second contains 80 g A. The combined 300 g contains 220 g A, or about 73.3%. Notice which quantity serves as the denominator and how the answer connects to the original units.
Why the method transfers. Percentages are comparisons with a particular base. Before every percentage operation, ask ‘percentage of what?’ The original quantity and the final quantity often have different denominators. If a 20% increase produces 120%, the inverse operation divides by 1.20. It does not subtract 20% of the new number. For this problem, verify the calculated answer by substituting it into the original statement, not merely by repeating the calculation. A solution that satisfies the algebra but violates the described mixture is not a complete contextual answer.
Error to prevent. A simple average of 70% and 80% gives 75% and wrongly treats the batches as equal sizes. Mark the exact step where the mistaken interpretation enters. Repairing the first wrong relationship is much more useful than checking only the final arithmetic.
Independent extension. Check whether the combined percentage lies between the two starting percentages. Once that answer is secure, state in one sentence why the chosen denominator is correct. A clear explanation is evidence that you can transfer the method to a new question.
16. Remove only one constituent
Problem. A 750 g blend is A:B = 3:7. Then 75 g of B is removed without removing any A. Write a short model before reaching for a calculator; use masses, percentages or ratios that refer to the same whole.
Worked route. Originally A = 225 g and B = 525 g. Afterwards A = 225 g, B = 450 g and total = 675 g. The new A share is one-third. Notice which quantity serves as the denominator and how the answer connects to the original units.
Why the method transfers. When blends are combined, first return to actual component masses. Percentages are useful descriptions but cannot normally be averaged without accounting for sample sizes. Add grams of A to grams of A and grams of B to grams of B; only then divide by the new total. This is the reliable route through unequal-batch questions. For this problem, verify the calculated answer by substituting it into the original statement, not merely by repeating the calculation. A solution that satisfies the algebra but violates the described mixture is not a complete contextual answer.
Error to prevent. Keeping the original 3:7 ratio after selective removal ignores the changed denominator. Mark the exact step where the mistaken interpretation enters. Repairing the first wrong relationship is much more useful than checking only the final arithmetic.
Independent extension. Work out the new B-to-A mass ratio. Once that answer is secure, state in one sentence why the chosen denominator is correct. A clear explanation is evidence that you can transfer the method to a new question.
17. Add pure A to reach a target
Problem. A 200 g blend starts at 20% A. How much pure A must be added to make the final share 30%? Write a short model before reaching for a calculator; use masses, percentages or ratios that refer to the same whole.
Worked route. Initially there are 40 g A. Let x g of pure A be added. (40 + x) / (200 + x) = 0.30, giving x = 200/7, approximately 28.6 g. Notice which quantity serves as the denominator and how the answer connects to the original units.
Why the method transfers. In selective addition or removal, identify what stays fixed. Adding pure A increases both A and the whole. Adding B changes the whole without changing A. Removing A reduces A and the whole. Writing those changes explicitly prevents the false assumption that the original ratio survives every operation. For this problem, verify the calculated answer by substituting it into the original statement, not merely by repeating the calculation. A solution that satisfies the algebra but violates the described mixture is not a complete contextual answer.
Error to prevent. Adding 20 g because 30 minus 20 is ten percentage points uses the wrong denominator. Mark the exact step where the mistaken interpretation enters. Repairing the first wrong relationship is much more useful than checking only the final arithmetic.
Independent extension. Substitute the calculated x into both the final component mass and final total. Once that answer is secure, state in one sentence why the chosen denominator is correct. A clear explanation is evidence that you can transfer the method to a new question.
18. Dilute with the other component
Problem. A 240 g mixture contains 25% A. Then 60 g of pure B is added. Write a short model before reaching for a calculator; use masses, percentages or ratios that refer to the same whole.
Worked route. Initially A = 60 g, B = 180 g. Afterwards A remains 60 g, total rises to 300 g, and A’s share becomes 20%. Notice which quantity serves as the denominator and how the answer connects to the original units.
Why the method transfers. Check units, feasibility and rounding. An answer in grams cannot be silently compared with a value in kilograms, and a negative mass would be physically impossible in this context. Where a syllabus answer calls for three significant figures, keep exact values or enough calculator digits during working and round only when reporting the final numerical result. For this problem, verify the calculated answer by substituting it into the original statement, not merely by repeating the calculation. A solution that satisfies the algebra but violates the described mixture is not a complete contextual answer.
Error to prevent. A’s mass did not decrease; its percentage declined because the whole became larger. Mark the exact step where the mistaken interpretation enters. Repairing the first wrong relationship is much more useful than checking only the final arithmetic.
Independent extension. Explain this difference to someone who thinks a lower percentage always means material was lost. Once that answer is secure, state in one sentence why the chosen denominator is correct. A clear explanation is evidence that you can transfer the method to a new question.
19. A ratio becomes equal after an addition
Problem. A 300 g blend has A:B = 2:3. Then 60 g of A is added. Write a short model before reaching for a calculator; use masses, percentages or ratios that refer to the same whole.
Worked route. Initially A = 120 g and B = 180 g. The new amounts are 180 g each; A:B becomes 1:1 and the total 360 g. Notice which quantity serves as the denominator and how the answer connects to the original units.
Why the method transfers. The denominator must represent the whole to which the ratio refers. Write the number of equal parts before using any calculator. A ratio is not automatically a fraction of the total until all listed parts are added. This first move prevents the common error of dividing by the wrong number when a question describes one ingredient rather than the complete blend. For this problem, verify the calculated answer by substituting it into the original statement, not merely by repeating the calculation. A solution that satisfies the algebra but violates the described mixture is not a complete contextual answer.
Error to prevent. The original 2:3 does not continue after a selective addition. Mark the exact step where the mistaken interpretation enters. Repairing the first wrong relationship is much more useful than checking only the final arithmetic.
Independent extension. Find the new percentage of each component. Once that answer is secure, state in one sentence why the chosen denominator is correct. A clear explanation is evidence that you can transfer the method to a new question.
20. Find an unseen second mass
Problem. There are 120 g of pure A. Material A must be one quarter of the final mix, with only B added. Write a short model before reaching for a calculator; use masses, percentages or ratios that refer to the same whole.
Worked route. Final total = 120 / 0.25 = 480 g. The required B mass is 480 – 120 = 360 g. Notice which quantity serves as the denominator and how the answer connects to the original units.
Why the method transfers. A reverse problem starts from what is known, not from the largest printed number. Mark whether the given mass is the total, a single part, a difference, or a percentage of an unknown original. Then set a one-part value or an unknown x. The algebra should preserve the relationship described by the words. For this problem, verify the calculated answer by substituting it into the original statement, not merely by repeating the calculation. A solution that satisfies the algebra but violates the described mixture is not a complete contextual answer.
Error to prevent. Adding 25% of 120 misunderstands the desired share of the completed blend. Mark the exact step where the mistaken interpretation enters. Repairing the first wrong relationship is much more useful than checking only the final arithmetic.
Independent extension. Show the final ratio A:B and check against one-quarter. Once that answer is secure, state in one sentence why the chosen denominator is correct. A clear explanation is evidence that you can transfer the method to a new question.
21. 60 percent corresponds to a known quantity
Problem. In a set of components, 60% of the total weighs 150 g. Find the complete mass. Write a short model before reaching for a calculator; use masses, percentages or ratios that refer to the same whole.
Worked route. The total is 150 / 0.60 = 250 g, while the other 40% weighs 100 g. Notice which quantity serves as the denominator and how the answer connects to the original units.
Why the method transfers. Percentages are comparisons with a particular base. Before every percentage operation, ask ‘percentage of what?’ The original quantity and the final quantity often have different denominators. If a 20% increase produces 120%, the inverse operation divides by 1.20. It does not subtract 20% of the new number. For this problem, verify the calculated answer by substituting it into the original statement, not merely by repeating the calculation. A solution that satisfies the algebra but violates the described mixture is not a complete contextual answer.
Error to prevent. Do not multiply 150 by 0.60: that would find a fraction of a part. Mark the exact step where the mistaken interpretation enters. Repairing the first wrong relationship is much more useful than checking only the final arithmetic.
Independent extension. Represent this relationship with an equation. Once that answer is secure, state in one sentence why the chosen denominator is correct. A clear explanation is evidence that you can transfer the method to a new question.
22. Selective sample removal
Problem. A 500 g blend has A:B = 2:3. A 10 g sample is taken entirely from A. Write a short model before reaching for a calculator; use masses, percentages or ratios that refer to the same whole.
Worked route. Originally A = 200 g and B = 300 g. Afterwards A = 190 g, B = 300 g, total = 490 g. The new percentage of A is about 38.8%. Notice which quantity serves as the denominator and how the answer connects to the original units.
Why the method transfers. When blends are combined, first return to actual component masses. Percentages are useful descriptions but cannot normally be averaged without accounting for sample sizes. Add grams of A to grams of A and grams of B to grams of B; only then divide by the new total. This is the reliable route through unequal-batch questions. For this problem, verify the calculated answer by substituting it into the original statement, not merely by repeating the calculation. A solution that satisfies the algebra but violates the described mixture is not a complete contextual answer.
Error to prevent. Removing 10 g of whole blend would preserve the ratio only if the sample was representative; this sample is all A. Mark the exact step where the mistaken interpretation enters. Repairing the first wrong relationship is much more useful than checking only the final arithmetic.
Independent extension. Compute the new B percentage and check the two shares add to 100. Once that answer is secure, state in one sentence why the chosen denominator is correct. A clear explanation is evidence that you can transfer the method to a new question.
23. Two-stage reduction
Problem. A 640 g mix has C:D = 3:5. Then 20% of C alone is discarded. Write a short model before reaching for a calculator; use masses, percentages or ratios that refer to the same whole.
Worked route. C initially weighs 240 g and D 400 g. Discarding 48 g C leaves 192 g C and 400 g D, total 592 g. C’s new share is about 32.4%. Notice which quantity serves as the denominator and how the answer connects to the original units.
Why the method transfers. In selective addition or removal, identify what stays fixed. Adding pure A increases both A and the whole. Adding B changes the whole without changing A. Removing A reduces A and the whole. Writing those changes explicitly prevents the false assumption that the original ratio survives every operation. For this problem, verify the calculated answer by substituting it into the original statement, not merely by repeating the calculation. A solution that satisfies the algebra but violates the described mixture is not a complete contextual answer.
Error to prevent. Twenty percent of C is not twenty percent of the whole mix. Mark the exact step where the mistaken interpretation enters. Repairing the first wrong relationship is much more useful than checking only the final arithmetic.
Independent extension. Work backwards from 192 g remaining C to reconstruct its original mass. Once that answer is secure, state in one sentence why the chosen denominator is correct. A clear explanation is evidence that you can transfer the method to a new question.
24. Conservation check with round-off
Problem. A sample has A:B = 5:7 and a total of 389 g, with answers requested to 3 significant figures. Write a short model before reaching for a calculator; use masses, percentages or ratios that refer to the same whole.
Worked route. A = (5/12) × 389 = 162.083… g and B = (7/12) × 389 = 226.916… g. Report 162 g and 227 g, and note they total 389 g. Notice which quantity serves as the denominator and how the answer connects to the original units.
Why the method transfers. Check units, feasibility and rounding. An answer in grams cannot be silently compared with a value in kilograms, and a negative mass would be physically impossible in this context. Where a syllabus answer calls for three significant figures, keep exact values or enough calculator digits during working and round only when reporting the final numerical result. For this problem, verify the calculated answer by substituting it into the original statement, not merely by repeating the calculation. A solution that satisfies the algebra but violates the described mixture is not a complete contextual answer.
Error to prevent. Rounding the ratio parts early may create avoidable errors before the final arithmetic. Mark the exact step where the mistaken interpretation enters. Repairing the first wrong relationship is much more useful than checking only the final arithmetic.
Independent extension. Repeat keeping exact fractions through the working before applying the answer convention. Once that answer is secure, state in one sentence why the chosen denominator is correct. A clear explanation is evidence that you can transfer the method to a new question.
A realistic multi-step example: a revised manufacturing blend
A fictional school design brief proposes an initial 600 g mix with A:B = 2:3. Before the model is completed, 60 g of material B is removed, and 30 g of pure A is added. The task asks for the final percentage by mass of A. Begin with the original masses: A = 240 g, B = 360 g. Remove 60 g from B, leaving 300 g. Add 30 g to A, giving 270 g. The final total is 570 g. The new percentage is 270/570 × 100, approximately 47.4%.
The answer is plausible because A’s share was originally 40%. Removing B and adding A both increase A’s fraction of the whole. A useful sanity check is that the final percentage must therefore exceed 40%. It also cannot exceed 100%. These comparisons catch many mistakes before the written answer is complete.
For a reverse challenge, imagine that the target is instead a final mixture with A equal to exactly half the total after B has been reduced to 300 g. With A initially 240 g, let x grams of A be added. Then (240+x)/(540+x) = 1/2. Solving gives 480+2x=540+x, so x=60 g. Substitution gives 300 g A and 300 g B. This is a direct bridge from percentage composition to an equation.
Diagnose the first weak link
When the answer is wrong, classify the failure. A whole error means the wrong base was chosen. A ratio error means the part count or order was mistaken. A reversal error means a final percentage was treated as if it were an original percentage. A conservation error means mass was invented or removed twice. A rounding error means the learner approximated too soon. A reasoning error means the arithmetic result was never checked against the story.
Repair one category at a time. For a wrong base, draw a box representing the full mass and label the ingredient portion. For a wrong ratio, use five or ten equal bars. For a reverse percentage, write 100%, then the multiplier after the change. For a multi-stage mixture, keep a two-column ledger of component A and component B through each operation. Use a second example with different numbers to confirm that the method, not the answer, was learned.
Connect to earlier mathematics and science foundations
Learners who still confuse fractions, ratios and percentages can return to the PSLE Learning Guide and Primary 6 Mathematics Learning Hub. For G3 continuity, revisit Volume 0347 on recovering missing data and Volume 0351 on comparing statistics. Science contexts in this guide are illustrative; the chemical meaning of a formula must be checked independently against the Science syllabus.
A final five-question check
What is the whole? Which part is known? Is the percentage measured before or after the change? Did the operation change one component, both components or neither? Does the final answer satisfy the original wording, keep the correct units, and use the requested accuracy? Students who can answer these five questions without help usually make fewer errors even when the next question changes the numbers, the story or the direction of calculation.
