How to perform in the new G2 SEC Mathematics examination when a question looks familiar begins with checking what has changed. Two questions can contain the same numbers and still require different methods. The difference may be one phrase: equal times rather than equal distances, a fraction of the total rather than a ratio of parts, or replacement rather than no replacement.
This G2 SEC Mathematics contrast-pair clinic gives you eight pairs of original questions. Each pair holds much of the situation steady while changing the relationship that matters. Your job is not only to calculate both answers. It is to explain why the first method should or should not carry into the second question.
The 2027 K210 Mathematics syllabus assesses selecting relevant mathematics, translating information and interpreting results. These exercises are teaching material for that selection work, not an official paper or mark scheme. Use the SEAB G2 syllabus directory to confirm the document for your examination year.
How to attempt a contrast pair
For every pair, read both questions before solving either. Underline the changed condition. Write one sentence describing what stays fixed. Then choose a representation: a fraction, equation, table, diagram or probability tree. Calculate only after the relationship is visible. Finish by explaining why the answers differ, even when both calculations are short.
This exercise separates recognition from execution. You may be able to perform a percentage calculation accurately but choose the wrong base. You may know the formula for average speed but fail to notice that travel times are unequal. A correct procedure used on the wrong model is still a wrong solution.
Keep an evidence record rather than a speed ranking. For each pair, note whether you identified the changed condition, chose the correct relationship, executed accurately and interpreted the result. These are teaching checks, not a conversion to an SEC grade. Work untimed first; introduce a realistic time limit only when the distinctions make sense.
Pair one: ratio of parts or fraction of the whole?
Question 1A: A box contains red and blue counters in the ratio 3:5. There are 40 counters altogether. How many are red? Question 1B: Three-fifths of the 40 counters in a different box are red. How many are red? The numbers 3, 5 and 40 appear in both, but they do not describe the same relationship.
In 1A, red and blue together make eight equal parts. One part is 40 ÷ 8 = 5 counters. Red counters occupy three parts, so there are 15 red counters. The fraction of all counters that are red is 3/8, not 3/5. The denominator in the original ratio names the blue part, not the total.
In 1B, the fraction already compares red counters with all counters. Calculate 3/5 × 40 = 24. The remaining 16 are not red. If the box contains only red and blue counters, its red-to-blue ratio would be 24:16, or 3:2. That check confirms that 1B does not describe the 3:5 ratio in 1A.
The decisive question is “three compared with what?” In 1A, three red parts are compared with five blue parts. In 1B, three red parts are compared with five total parts. Write labels beside a ratio before converting it into a fraction. The notation looks compact, but its meaning depends on the quantities attached to it.
A useful repair after an error is to draw eight equal boxes for 1A and shade three, then draw five equal boxes for 1B and shade three. This representation shows the different wholes. Once the difference is clear, return to symbolic working. The diagram is a bridge to understanding, not an extra picture required for every examination ratio question.
Transfer check: a class has boys and girls in the ratio 2:3, with 35 students altogether. There are 14 boys. In another class, two-thirds of 36 students are boys, giving 24 boys. Explain the denominator in each calculation before reading those answers as mere numerical facts.
Pair two: more machines, more output or less time?
Question 2A: Eight identical printers produce 2,400 pages in 30 minutes. At the same constant rate, how many pages do twelve printers produce in 30 minutes? Question 2B: At that rate, how long do twelve printers take to produce the original 2,400 pages? Assume printers work independently without setup delays, faults or shared capacity limits.
In 2A, time is fixed. Increasing the number of printers from eight to twelve multiplies output by 12/8. The result is 2,400 × 12/8 = 3,600 pages. Output and number of printers are directly proportional under the stated assumptions because each additional printer contributes the same amount during the same interval.
In 2B, the job size is fixed. Eight printers need 30 minutes, so the job requires 8 × 30 = 240 printer-minutes at this rate. Twelve printers complete it in 240 ÷ 12 = 20 minutes. Time decreases when the number of printers increases. Multiplying 30 by 12/8 would predict 45 minutes, contradicting the model’s intended relationship.
You can also derive a common rate: one printer produces 2,400 ÷ 8 ÷ 30 = 10 pages per minute. Then 2A is 12 × 10 × 30 and 2B is 2,400 ÷ (12 × 10). This route is especially useful when the direct-versus-inverse label feels uncertain. Build the relationship from the unit rate rather than guessing the label.
The mathematical model is conditional. Real printers may share a network bottleneck or need setup time. Those possibilities are excluded by the question’s assumptions, so they should not prevent solving the stated task. In a question asking you to evaluate the model, however, they could become relevant limitations. Distinguish solving under assumptions from judging the assumptions.
Transfer check: six identical pumps fill a tank in 40 minutes at constant independent rates. Ten pumps take 24 minutes for the same tank. For a fixed 40-minute interval instead, ten pumps move 10/6 times as much water as six. The changed fixed quantity determines which relationship you need.
Pair three: a known price after a change or a new price before it?
Question 3A: A bag costs $96 after a 20% discount. Find its original price. Question 3B: A different bag currently costs $96 and its price is increased by 20%. Find its new price. Both involve $96 and 20%, but the known amount occupies a different place in the relationship.
For 3A, let the original price be P. The discounted price is 80% of P, so 0.8P = 96 and P = $120. The check is $120 − $24 = $96. The $24 discount is 20% of the original $120, not 20% of the sale price.
For 3B, $96 is already the starting base. The new price is 1.2 × 96 = $115.20. This time multiplication is correct because you are applying a change to a known original. The operation is not chosen from the percentage sign alone. It follows from which quantity is known and which is requested.
A common error is adding 20% to the sale price in 3A. That produces $115.20, the answer to a different question. The coincidence is useful: it shows exactly how an accurate calculation can solve the wrong task. Write “original”, “change” and “final” beside the values before deciding what to calculate.
Do not generalise this into “original means divide”. An original could be given in a question asking for something else, and not every change is multiplicative. The robust method is an equation linking the original and final amounts. Once the equation is correct, the algebra selects the operation for you.
Transfer check: an item costs $85 after a 15% discount, so its original price is $100. An item starting at $85 and increasing by 15% costs $97.75. In both cases, identify the quantity represented by 100% before performing any arithmetic.
Pair four: equal travel times or equal travel distances?
Question 4A: A cyclist travels for one hour at 30 km/h and one hour at 60 km/h. Find the average speed. Question 4B: A traveller covers 60 km at 30 km/h and another 60 km at 60 km/h. Find the average speed. Ignore stops. The listed speeds are identical, but the weighting differs.
Average speed is total distance divided by total time. In 4A, the distances are 30 km and 60 km, giving 90 km in two hours. The average is 45 km/h. The arithmetic mean of the speeds works here because each speed operates for the same duration.
In 4B, the first stage takes 60 ÷ 30 = 2 hours and the second takes 60 ÷ 60 = 1 hour. The whole journey is 120 km in three hours, giving 40 km/h. The lower speed operates for twice as long, so the average is pulled closer to 30 than an equal-time average would be.
The reliable formula does not change between the questions. What changes is how total distance and total time are obtained. This is an important form of method selection: sometimes you should not choose a new formula at all. You should preserve the general definition and change the inputs to match the new condition.
A stage table can prevent confusion. Label each stage’s distance, speed and time. Fill the missing quantity using the other two. Then total distances and total times separately. Never total the speeds as though they were distances. Units help keep the operations attached to their meanings.
Transfer check: equal half-hour stages at 20 km/h and 40 km/h give an average of 30 km/h. Equal 20 km stages at those speeds take one hour and half an hour, giving 40 ÷ 1.5 = 26.666… km/h. State the required rounding only at the final stage if the question specifies it.
Pair five: equal groups or unequal groups?
Question 5A: Eight students have a mean score of 12 and another eight have a mean score of 20. Find the combined mean. Question 5B: Twelve students have a mean score of 12 and another four have a mean score of 20. Find the combined mean. These are invented scores on the same assessment scale.
For 5A, the total scores are 8 × 12 = 96 and 8 × 20 = 160. The combined total is 256 across 16 students, giving 16. Because the groups are equal in size, averaging the group means also gives the correct result. That shortcut works because of the equal weighting, not because means can always be averaged directly.
For 5B, the totals are 12 × 12 = 144 and 4 × 20 = 80. The combined total is 224 across 16 students, giving 14. There are three times as many students in the lower-mean group, so the combined mean lies closer to 12. The simple mean 16 would ignore group size.
The general route is recover totals, add totals, divide by the combined count. This is more dependable than memorising a separate weighted-mean formula without understanding its parts. The product of group size and group mean reconstructs the group’s total, which is what must be combined.
A further boundary matters: the two means must refer to comparable measurements. Combining a mean distance with a mean time would not produce a meaningful overall mean of one quantity. In these questions, the scores share a scale. Keep that comparability visible when the context changes to costs, masses or durations.
Transfer check: five items have mean mass 10 g and fifteen have mean mass 18 g. The combined mean is (50 + 270) ÷ 20 = 16 g. It lies between the group means and nearer 18 g because the larger group has that mean. This is a useful plausibility check, not an alternative to the total calculation.
Pair six: a map length or a map area?
A map has scale 1:20,000. Question 6A: A route measures 4 cm on the map. Find its actual length in metres. Question 6B: A region measures 4 cm² on the map. Find its actual area in square kilometres. The numeral 4 is unchanged; the dimension is not.
For 6A, 1 cm represents 20,000 cm, or 200 m. Therefore 4 cm represents 800 m. The scale factor acts once because a length has one dimension. State the unit conversion explicitly so that centimetres do not silently become metres.
For 6B, imagine a 1 cm by 1 cm square on the map. Its actual dimensions are 200 m by 200 m, so its area is 40,000 m². A map area of 4 cm² therefore represents 160,000 m², or 0.16 km². The linear factor has affected both dimensions.
Multiplying 4 cm² by 20,000 only once is not enough. Area scales with the square of a linear scale factor. Working through a unit square makes that relationship visible without relying on a memorised warning to “square it”. The same reasoning explains why conversion from square metres to square kilometres uses one million, not one thousand.
Do not treat 4 cm and 4 cm² as interchangeable just because both are map measurements. A unit is part of the mathematical information. This pair rewards careful reading before calculation and demonstrates why dimensional checks can catch an otherwise neat-looking numerical answer.
Transfer check: a 1:10,000 map shows a route 3 cm long and a park 3 cm² in area. The route is 300 m; the park is 30,000 m², or 0.03 km². Explain the unit-square reasoning before reducing the solution to a short scale calculation.
Pair seven: replacement or no replacement?
A bag contains three red counters and two blue counters. Draw one counter at random, then draw another. Question 7A: The first counter is returned to the bag and the bag is mixed before the second draw. Find the probability that both are red. Question 7B: The first counter is not returned. Find the probability that both are red.
For 7A, the first red probability is 3/5. Replacement restores the original contents, so the second red probability is also 3/5. The probability of two reds is 3/5 × 3/5 = 9/25. The model assumes each draw is random from the stated bag contents.
For 7B, start with the same 3/5. If the first counter is red, two red counters remain among four counters. The second red probability on that branch is 2/4. Multiply 3/5 × 2/4 = 3/10. The condition of the bag after the first result has changed.
The phrase “if the first counter is red” is not optional reasoning. It identifies the branch on which the second probability is calculated. A tree diagram can make this clear. Label each branch with both the colour and its probability rather than writing a second 3/5 by habit.
Another check is to compare the answers. With no replacement after a red, obtaining another red should become less likely, so 3/10 should be smaller than 9/25. This comparison fits: 0.30 is smaller than 0.36. A plausible direction is helpful evidence, though it does not prove every fraction in the working is correct.
Transfer check: a bag has four green and two yellow counters. The probability of two greens is 4/9 with replacement and 2/5 without replacement. State the new bag contents after the first green before calculating the second draw. That sentence is the point at which the methods separate.
Pair eight: simplifying an expression or solving an equation?
Question 8A: Simplify 3(x − 2). Question 8B: Solve 3(x − 2) = 15. The left-hand form is identical. The extra equality in 8B changes the job. One task asks for an equivalent expression; the other asks for values of x satisfying a condition.
For 8A, distribute the multiplier to obtain 3x − 6. There is no single value of x to find because no condition has been provided. Setting the expression equal to zero invents a new problem. Simplification preserves value for every allowed input; it does not choose one input.
For 8B, divide both sides by three to get x − 2 = 5, then add two to get x = 7. Alternatively, expand and solve 3x − 6 = 15. Check in the original equation: 3(7 − 2) = 15. Both methods are valid because they preserve the equality condition.
An answer of 3x − 6 to 8B is unfinished. An answer of x = 2 to 8A introduces an unsupported equation. These are task-type errors, not arithmetic errors. Before calculating, name the mathematical object: expression, equation, formula, inequality or another structure. Similar symbols do not guarantee the same operation.
For a fresh pair, simplify 2(y + 4), then solve 2(y + 4) = 18. The first answer is 2y + 8; the second is y = 5. Explain why only the second has a unique value in this case. This is a compact test of whether the learner has preserved the difference between rewriting and solving.
What the eight pairs are really testing
The pairs do not reward memorising eight traps. They train a repeated decision: identify the quantity being compared and the condition being held fixed. In ratio, it is the reference whole. In proportionality, it is time or job size. In averages, it is the weight of each part. In scale, it is dimension. In probability, it is the updated sample space.
The final pair adds another distinction: the object and task can change even when the visible expression does not. That lesson extends beyond algebra. A graph may be given for reading a value, describing a trend or solving an intersection. The command and conditions determine what you do with the representation.
Vol 0015: Method Selection gives the broader recognition framework. This clinic supplies contrast cases in which the decision can be inspected. The aim is a learner who can explain a method choice before relying on speed.
Classify the error before choosing the repair
If you missed the changed phrase, repair reading. If you noticed the phrase but could not model it, repair the concept or representation. If the model was correct but arithmetic failed, repair execution and checking. If the number was correct but the unit or final decision was wrong, repair interpretation. These causes should not all be called carelessness.
For example, using 3/5 of 40 in Question 1A indicates a reference-whole problem. Writing 3/8 of 40 but obtaining 18 indicates an execution problem. Both produce a wrong count, but the next lesson should be different. The written setup lets you separate those causes without guessing what the learner intended.
Similarly, a student may know that no replacement changes the bag yet subtract from the wrong colour count. That is not the same as assuming the two draws use identical probabilities. Ask the learner to state the bag contents at the branch. One short explanation often reveals the precise misunderstanding.
A compact second-session test
On a later day, use only one question from each pair, with changed numbers and without a topic heading. Before solving, write the relationship in one line. Examples include “known amount is 85% of original”, “same job, more pumps” and “second draw uses updated bag contents”. These statements are temporary scaffolds for method recognition.
After several accurate attempts, stop requiring a written sentence for every routine question. The objective is independent recognition, not permanent extra writing. Retain the sentence on difficult or unfamiliar questions where it prevents a wrong launch. A useful tool should reduce confusion more than it consumes time.
Do not immediately repeat the full clinic from memory and interpret the higher score as transfer. Use the fresh numerical checks or ask a teacher to supply a new context. Success on a remembered answer is different from recognising the same relationship inside new information. Keep these two forms of practice evidence separate.
Make your own contrast pair carefully
Choose a question whose method you understand. Change one structural condition and predict how the method should change. For example, change equal durations to equal distances, or change a part-to-part ratio into a fraction of the total. Solve both and verify that the contrast is genuine. Merely changing numbers does not necessarily change the mathematical decision.
Check the new question for sufficient information. A poorly designed contrast may accidentally remove a necessary condition and make the answer indeterminate. That can be a useful separate exercise, but it should be intentional. State what is known, what is unknown and what assumptions are allowed before asking someone else to solve it.
Explain the pair to another learner without revealing the answers first. Ask them which phrase controls the method. If they identify a different ambiguity, inspect the wording rather than assuming they misunderstood. Writing a clear problem is itself a test of whether you understand the relationship you want it to represent.
Move from the clinic to examination work
The official K210 notes require essential working, so do not confuse fast recognition with a bare final answer. Show the equation, ratio, stage times or event structure that carries the method. For non-exact numerical answers, follow the question and the syllabus accuracy conventions; avoid rounding intermediate values before they have finished their job.
Use Vol 0031: Representation Switching when the right relationship is difficult to see in prose. Use Vol 0035: Constraints and Feasibility when a result needs a final practical decision. These are supporting skills, not alternative ways to avoid reading the changed condition.
For underlying topic teaching, return to the Complete Mathematics Index. For whole-paper decisions, use Examination Craft. The PSLE bridge remains Represent Before You Calculate. At G2, add a question before representation: what has changed, and does that change the relationship?
Similar numbers are not permission to reuse a method. Find the reference whole, fixed quantity, weighting, dimension, sample space or equation condition. Then solve the problem that is actually present. The strongest result of this clinic is not eight memorised warnings. It is a habit of checking why a familiar method belongs here.
