A simple inequality can be solved with correct algebra and still finish with the wrong answer because the learner forgets to decide whether the boundary value belongs. The difference between x < 6 and x ≤ 6 is one point, but that point can change a number line, an integer list and a real-world decision.
Use an endpoint audit. First solve the inequality while protecting its direction. Then substitute the boundary into the original condition, not only the rearranged line. Finally decide whether equality is allowed and show the correct open or closed endpoint, inequality sign and contextual answer.
This G2 Mathematics workshop is a narrow application of the K210 equations-and-inequalities topic. It extends earlier work on case splitting by focusing on boundary membership and final representation. The examples and practice questions are original.
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Understand the boundary (chapters 1–5) · Solve and test (chapters 6–11) · Interpret and diagnose (chapters 12–16) · Practise and transfer (chapters 17–30)
1. Locate inequalities in K210
2. What an endpoint means
3. Direction and membership are separate decisions
4. Why the original condition owns the check
5. Translate the four signs into words
6. Worked example with a positive coefficient
7. Worked example with a negative coefficient
8. Why multiplying by a negative reverses order
9. Number-line representation
10. Translate a number line back into symbols
11. Context example: a capacity condition
12. Discrete values can change the final statement
13. Boundary values created by measurement words
14. Common misconception: the filled circle means positive
15. Common misconception: greater symbol means shade right
16. Common misconception: the boundary test proves direction
17. An examination finishing routine
18. Practice set A
19. Practice set B
20. Review an error as evidence
21. For parents and tutors
22. Final audit: solve, substitute, interpret
23. Extended practice: inequalities and integer domains
24. Extended practice: negative values and ordering
25. Extended practice: diagnose three wrong answers
26. Extended practice: language-to-symbol contrasts
27. Extended practice: design a self-checking response
28. Mixed practice with explained answers
29. Error ledger for inequality performance
30. Reverse construction: build the inequality from its solution
1. Locate inequalities in K210
The 2027 K210 syllabus includes solving simple inequalities of the forms ax ≤ b, ax ≥ b, ax < b and ax > b, where a and b are integers. It also includes ordering numbers and using the four inequality symbols. This guide stays inside that scope while developing reliable examination performance.
An inequality describes a set of possible values rather than a single solution. The final response therefore needs both algebra and membership control. A learner must know which side of a boundary is permitted and whether the boundary itself is included.
The current question remains in charge. It may ask for an inequality, integers satisfying a condition, a number-line representation or a contextual decision. Do not force every answer into the same form.
During practice, say the set in words before drawing it: “all values less than six” or “all values at least negative two.” Language exposes direction and inclusion errors early.
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2. What an endpoint means
For x < 6, the number 6 is the boundary between permitted and non-permitted values, but 6 itself is excluded. For x ≤ 6, the same boundary is included. The endpoint audit asks one direct question: does substituting the boundary make the original statement true?
Substitute 6 into x < 6. The result 6 < 6 is false, so the endpoint is open. Substitute it into x ≤ 6. The result 6 ≤ 6 is true, so the endpoint is closed.
This test is simple enough to use as a check and powerful enough to repair memory confusion about empty and filled circles. The symbol is not decorative; it records membership.
On a number line, use the convention required by the question or your school materials. Whatever notation is used, ensure the drawing and inequality communicate the same set.
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3. Direction and membership are separate decisions
The direction tells which side of the boundary belongs: values greater than or values less than. Membership tells whether the boundary point joins that side. Learners often correct one feature while leaving the other wrong.
For x > 3, the permitted side is to the right and 3 is excluded. For x ≥ 3, the permitted side is still to the right but 3 is included. A filled endpoint cannot repair shading on the wrong side.
Use two checks: test the boundary and test one easy interior value. For x ≥ 3, boundary 3 works and interior 4 works. Value 2 fails. These three substitutions reconstruct the set.
This method also catches a copied sign. If your final answer says x ≤ 3 but the easy value 4 satisfied the original condition, the direction needs review.
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4. Why the original condition owns the check
Algebraic transformations can contain a sign error. Testing only the final line may confirm its internal consistency while missing the mistake. Substitute into the original inequality whenever practical.
Suppose −2x < 8. A learner writes x < −4, forgetting that division by a negative reverses the sign. The boundary −4 makes the original statement 8 < 8, false, so exclusion is correct; however an interior test such as x = −5 gives 10 < 8, false. The direction fails.
The correct solution is x > −4. Boundary −4 is excluded, and x = 0 gives 0 < 8, true. The pair of tests identifies both membership and direction.
Checking is not a replacement for understanding the sign reversal. It is an independent route that can reveal when execution no longer matches the original condition.
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5. Translate the four signs into words
Less than means the boundary is excluded. Less than or equal to includes it. Greater than excludes it. Greater than or equal to includes it. The word “or equal to” is the membership signal.
Context words also carry signs. “Fewer than 20” usually translates to n < 20. “At most 20” translates to n ≤ 20. “More than 20” gives n > 20. “At least 20” gives n ≥ 20.
Do not memorise isolated pairs without checking the situation. “A lift can carry at most 600 kg” includes 600 kg if the stated limit is exact. “The temperature must be below 8°C” excludes 8°C.
Write one allowed boundary example beside each phrase. If the boundary is allowed, equality belongs. If it is not, use a strict inequality.
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6. Worked example with a positive coefficient
Solve 3x + 5 ≤ 20. Subtract 5 from both sides to obtain 3x ≤ 15. Divide by positive 3 to obtain x ≤ 5. The sign does not reverse because the division is by a positive number.
Endpoint audit: substitute x = 5 into the original expression. Then 3(5) + 5 = 20, and 20 ≤ 20 is true. The boundary belongs, so the endpoint is closed.
Interior test: x = 4 gives 17 ≤ 20, true. Outside test: x = 6 gives 23 ≤ 20, false. These checks confirm the side of the boundary.
If the question asks for integer values in a stated range, list only values satisfying both the inequality and that range. Do not invent a range that is not supplied.
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7. Worked example with a negative coefficient
Solve −4x ≥ 12. Dividing both sides by −4 reverses the inequality, giving x ≤ −3. The reversal preserves the set because multiplication or division by a negative reverses order on the number line.
Endpoint audit: x = −3 gives −4(−3) = 12, and 12 ≥ 12 is true. Equality is allowed, so −3 belongs.
Interior test: choose x = −4. Then 16 ≥ 12 is true. Value 0 gives 0 ≥ 12, false. The solution lies to the left of −3 and includes −3.
A memory phrase such as “flip the sign” is incomplete unless the learner knows when: only when multiplying or dividing both sides by a negative quantity in this simple process.
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8. Why multiplying by a negative reverses order
Begin with 2 < 5. Multiply both numbers by −1 to obtain −2 and −5. On the number line, −2 is greater than −5, so the true relationship is −2 > −5. The order reverses around zero.
The reversal is not a special punishment attached to algebra. It reflects the structure of negative numbers. Keeping this model reduces the risk of applying the rule during addition or subtraction, where no reversal is needed.
For example, subtracting 7 from x < 10 gives x − 7 < 3; the sign remains. Dividing −2x < 10 by −2 gives x > −5; the sign reverses.
Practise classifying each transformation before performing it: add, subtract, multiply by positive, divide by positive, multiply by negative or divide by negative.
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9. Number-line representation
After solving x > −4, place the boundary at −4, mark it as excluded and indicate values greater than −4. Check orientation using zero: since 0 > −4 is true, the indicated side must include zero.
For x ≤ −3, the boundary is included and the indicated side extends toward smaller values. Test −10: it satisfies −10 ≤ −3, so the side containing −10 must be selected.
Number lines expose sign errors because spatial direction becomes visible. They also expose endpoint errors because inclusion must be drawn rather than left implicit.
If your drawing and symbolic answer disagree, do not choose whichever looks more familiar. Return to the original inequality and test values.
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10. Translate a number line back into symbols
Suppose a number line has a closed endpoint at 2 and extends to the left. The values are at most 2, so x ≤ 2. Closed gives equality; left gives less than.
An open endpoint at −1 extending right represents x > −1. Open removes equality; right selects greater values. Use both features before writing the sign.
A frequent error is reading the arrow as an inequality arrow. Instead, interpret actual values on the line. Pick a marked-side value and compare it with the endpoint.
Practice with endpoints not centred at zero. The method should work for negative and fractional boundaries as well as positive integers, even when the syllabus examples use simple coefficients.
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11. Context example: a capacity condition
A container may hold a mass m satisfying 5m ≤ 40, where each identical package has mass 5 kg and total package mass cannot exceed 40 kg. Solving gives m ≤ 8 if m represents the number of packages.
The endpoint audit confirms that eight packages have mass 40 kg and are allowed because the condition says “cannot exceed”. Nine packages give 45 kg and fail.
Context adds a domain: the number of packages is a non-negative whole number. The algebraic solution x ≤ 8 alone also contains negative and non-integer values that make no sense as package counts.
State the contextual answer requested, such as “at most eight packages”. This is not extra decoration; it reconnects the set to the problem.
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12. Discrete values can change the final statement
Suppose an event needs more than 38 seats and each row has 6 seats. If r is a whole number of rows, 6r > 38 gives r > 38/6, or r > 6⅓. The smallest allowable whole number is 7.
The algebraic boundary 6⅓ is excluded. But because r is discrete, the practical decision is not drawn as every real number greater than 6⅓; it is 7, 8, 9 and so on, subject to any other limit.
Do not round 6⅓ to 6 or 7 without reference to the inequality. Test candidates in the original condition: 6 rows give 36 seats and fail; 7 give 42 and work.
Candidate testing is especially useful when a context forces whole numbers. It converts abstract membership into a defensible decision.
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13. Boundary values created by measurement words
Words such as minimum, maximum, below, above, no more than and no less than decide endpoint membership. Read them before calculating. The same numerical boundary can be included in one problem and excluded in another.
“A score of at least 50 qualifies” includes 50. “A score above 50 qualifies” excludes it. If only whole-number scores are possible, the practical qualifying sets begin at 50 and 51 respectively.
Avoid replacing precise phrases with casual synonyms. “Under” and “up to” do not express the same boundary. Copy the intended relationship into symbols while the wording is visible.
During review, highlight the phrase that authorises equality. If no phrase supports it, reconsider the endpoint.
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14. Common misconception: the filled circle means positive
A closed endpoint does not mean the values are positive. It means the boundary value is included. You can have x ≤ −3 with a closed point at a negative number.
Likewise, an open circle does not mean the answer is wrong or uncertain. It records exclusion. For x > 0, zero is excluded even though all permitted values are positive.
Repair by using membership language. Say “closed because equality is allowed” and “open because equality is not allowed.” This explanation attaches the drawing to the inequality.
Then test one negative-boundary example and one zero-boundary example. Variation prevents the learner from tying the symbol to visual position.
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15. Common misconception: greater symbol means shade right
The final solved form x > 4 does indicate values to the right of 4. But an unsolved inequality such as −2x > 8 cannot be drawn by following the visible greater-than symbol beside x before isolating x.
Solve first: divide by −2 and reverse the sign to obtain x < −4. The solution lies left. The coefficient changed the relationship.
If you choose to reason directly, test values rather than relying on symbol appearance. x = −5 makes 10 > 8 true; x = 0 makes 0 > 8 false.
A safe routine is isolate, audit boundary, test interior, represent. Each step has a different job.
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16. Common misconception: the boundary test proves direction
Testing only the boundary tells whether equality is included. Both x < 5 and x > 5 reject the value 5. Therefore the endpoint test alone cannot distinguish left from right.
Add an interior test. Choose a simple value on one proposed side, substitute it into the original condition and see whether it works. Use a value that avoids unnecessary arithmetic.
For 2x < 10, boundary 5 is excluded. Test 0: 0 < 10 is true, so values on the lower side are permitted. The solution is x < 5.
Name the two checks accurately: boundary membership and side direction. This prevents one successful check from being asked to do two jobs.
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17. An examination finishing routine
Before leaving an inequality, check five items: x is isolated; the sign was reversed if and only if a negative multiplication or division required it; the boundary has been tested; one interior value has been tested; the final form matches the question.
This routine should be brief. Most checks use simple substitution. If the working is long, the learner may be solving again rather than auditing the high-risk points.
Prioritise the original condition, especially after a negative coefficient or a contextual translation. Those are common first-divergence points.
If time is limited, a boundary and one easy interior value provide more targeted evidence than staring at the final sign.
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18. Practice set A
Solve and audit: 5x < 35. Dividing by positive 5 gives x < 7. Boundary 7 gives 35 < 35, false. Interior 0 gives 0 < 35, true. The endpoint is open and the permitted side is below 7.
Solve and audit: −3x ≤ 9. Dividing by −3 reverses the sign, giving x ≥ −3. Boundary −3 gives 9 ≤ 9, true. Interior 0 gives 0 ≤ 9, true.
Translate: “The temperature must be no higher than 4°C.” Let t be temperature. Then t ≤ 4; the boundary is included because 4 is allowed.
For each, write the inequality, one sentence explaining endpoint membership and the number-line direction.
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19. Practice set B
A hall admits fewer than 120 people. If n is attendance, n < 120. Because people are counted in whole numbers and negative attendance is impossible, valid practical values are non-negative whole numbers below 120.
A machine operates safely at a pressure of at most 8 units. Then p ≤ 8. The endpoint belongs. If the question provides a lower operating limit, intersect that condition rather than ignoring it.
Solve −5x > 20. Division by −5 gives x < −4. Boundary −4 fails equality because the original sign is strict; x = −5 works.
Explain why “flip whenever there is a negative number” is wrong. The sign reverses when multiplying or dividing both sides by a negative, not merely because a negative appears.
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20. Review an error as evidence
If the direction is wrong after dividing by a negative, classify a transformation error. If the direction is correct but the endpoint symbol is wrong, classify membership. If the algebraic set is correct but the package count is impossible, classify context interpretation.
Keep the original response and add one diagnostic substitution. The substitution should reveal the exact failure rather than only show the corrected answer.
Then practise a contrast pair: one strict and one inclusive inequality with the same boundary, or one positive and one negative coefficient. Contrast prevents repair from becoming a single memorised example.
Retest later with an unfamiliar context and no label announcing the error type. Reliable performance requires selecting the check independently.
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21. For parents and tutors
Ask “Does the boundary itself work?” before telling the learner which circle to draw. Then ask for one value on the selected side. These two questions reveal whether the learner understands membership and direction.
Avoid teaching the endpoint as a picture-only code. Connect symbols, words, substitution and number line. Multiple representations make the concept easier to recover under pressure.
When context matters, ask what values the variable can represent. Whole packages, people and days differ from unrestricted real numbers. The domain should come from the problem, not from habit.
Move from untimed explanation to mixed practice only after the learner can solve and justify the endpoint calmly. Speed should not hide a sign misconception.
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22. Final audit: solve, substitute, interpret
An inequality answer is complete when the transformation is valid, the boundary membership is correct, the permitted side is correct and the final representation fits the question.
Use the original condition as the owner of the check. Substitute the endpoint to decide inclusion and a simple interior value to decide direction. For negative division, explain the reversal through order rather than relying on a fragile slogan.
In context, test the nearest practical candidates. A calculator decimal may sit between whole-number options, and the inequality determines which candidate is allowed.
This compact audit turns an easy-to-miss symbol into a reasoned decision. It also leaves visible evidence that can be reviewed when an answer fails.
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23. Extended practice: inequalities and integer domains
A club can spend at most $84 on identical tickets costing $7 each. Let n be the number of tickets. The condition is 7n ≤ 84, so n ≤ 12. Endpoint 12 gives 84 ≤ 84 and is allowed.
The algebraic set contains all real numbers at or below 12, but the context restricts n to non-negative whole numbers. The practical statement is that the club may buy at most 12 tickets, assuming no other costs.
Change “at most $84” to “less than $84”. Then 7n < 84 and n < 12. Twelve tickets are excluded because they cost exactly $84; the greatest whole-number choice is 11.
This contrast holds the numbers constant while changing membership. Test 11 and 12 in the original cost condition to make the decision visible.
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24. Extended practice: negative values and ordering
Solve −6x < 18. Divide by −6 and reverse the inequality: x > −3. Boundary −3 gives 18 < 18, false. Interior 0 gives 0 < 18, true. The solution lies to the right and excludes −3.
The common wrong answer x < −3 fails when x = −4: the original left side becomes 24, and 24 < 18 is false. One substitution reveals the direction error.
Now solve −6x ≤ 18. The result is x ≥ −3. Equality now includes the boundary. Holding coefficient and boundary constant isolates the strict-versus-inclusive distinction.
Use a number line including −5, −3 and 0. It shows why zero belongs to x ≥ −3 and why “greater” does not necessarily mean positive.
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25. Extended practice: diagnose three wrong answers
For −2x ≥ 10, the correct solution is x ≤ −5. Wrong Answer A, x ≥ −5, is a reversal failure. Wrong Answer B, x < −5, has correct direction but wrong membership. Wrong Answer C, x ≤ 5, has the wrong boundary sign.
Use different tests. Boundary −5 gives 10 ≥ 10, so equality belongs and B fails. Interior −6 gives 12 ≥ 10, so the lower side belongs and A fails. Substituting 5 gives −10 ≥ 10, false, defeating C.
No single check explains every failure. Boundary membership, side direction and sign copying are related but separable. Diagnostic feedback should name the broken component.
After correction, ask the learner to generate another inequality with solution x ≤ −5. Creating an example tests structure beyond this calculation.
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26. Extended practice: language-to-symbol contrasts
With boundary 30: fewer than 30 gives x < 30; no more than 30 gives x ≤ 30; more than 30 gives x > 30; no fewer than 30 gives x ≥ 30.
Test 30 verbally. Is exactly 30 permitted? That decides inclusion. Then test 29 and 31 to decide direction. The process is stronger than recalling a memorised pair.
Place the phrases in attendance, temperature, storage and score contexts. Some quantities are discrete while temperature may be continuous. Domain changes practical values but not the boundary logic.
A minimum of 30 means values cannot be lower, so x ≥ 30. A maximum means values cannot be higher, so x ≤ 30. Defend each sign in one sentence.
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27. Extended practice: design a self-checking response
Arrange working so another reader can audit it: original condition, transformation isolating the variable, any negative division clearly shown, solution statement and a compact boundary or interior check where useful.
Do not turn every answer into an essay. A check can be “x = −3 gives equality, so included” or a substituted line. Its purpose is to protect a high-risk decision.
When a number line is requested, place the symbolic result and drawing together. When context is requested, add the practical statement after the algebra. Layout should make the final set easy to locate.
In review, cover the final answer and reconstruct the set from the checks. Clear checks should identify membership and side without relying on memory.
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28. Mixed practice with explained answers
1. Solve 4x > 28. Answer: x > 7. Seven is excluded because 4(7) > 28 is false; eight is a valid interior integer. 2. Solve 4x ≥ 28. Answer: x ≥ 7. Only the endpoint membership changes.
3. Solve −7x < 21. Answer: x > −3. Division by −7 reverses the sign. Boundary −3 gives 21 < 21, false; zero gives 0 < 21, true. 4. Solve −7x ≤ 21. Answer: x ≥ −3, with the endpoint included.
5. Translate “a parcel must weigh less than 5 kg”: w < 5. Exactly 5 kg is excluded. 6. Translate “a parcel may weigh up to 5 kg”: w ≤ 5. Exactly 5 kg is included. The context phrase decides equality.
7. A van carries boxes of 12 kg and the box load cannot exceed 96 kg. Let b be a whole-number count. Then 12b ≤ 96, so b ≤ 8 and the practical maximum is eight boxes.
8. A team needs more than 96 bottles supplied in packs of 12. Then 12p > 96, so p > 8; the smallest whole-number pack count is nine. “More than” excludes the exact boundary.
For every item, write two tiny checks: boundary and one interior value. If both checks agree with the symbolic result and context, the answer has independent support.
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29. Error ledger for inequality performance
Use five ledger codes: T for translation, R for negative reversal, B for boundary membership, D for direction and C for context/domain. One wrong answer may have more than one code, but identify the earliest failure first.
Example: “at most 40” becomes x < 40. The first failure is T because equality was lost during translation. A later open circle is a consequence. Repair the phrase-to-symbol link before drilling number lines.
Example: −3x > 12 becomes x > −4. The first failure is R. The learner may understand endpoint exclusion yet still choose the wrong side. Practise negative-order transformations with simple numerical checks.
Example: n ≤ 6 is reported as 6.4 items. The algebra may be irrelevant or incomplete because n is a count. Code C and return to variable meaning. A domain is part of the mathematical model.
After three practice sets, count recurring first-failure codes. Choose the next exercise family from the pattern, then retest inside mixed questions so the repair must be selected rather than announced.
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30. Reverse construction: build the inequality from its solution
Given solution x < 4, construct a simple inequality with a positive coefficient: 3x < 12. Construct one with a negative coefficient: −2x > −8. Solving the second requires reversal and returns x < 4.
Given solution x ≥ −5, one positive-coefficient example is 2x ≥ −10. A negative-coefficient example is −3x ≤ 15. Boundary −5 makes equality true in both.
Reverse construction reveals whether the learner controls equivalence or only follows a forward recipe. It also demonstrates that many original inequalities can describe the same solution set.
Check each construction with the boundary and an interior value. For x ≥ −5, choose zero as interior. The original condition should accept −5 and zero but reject a lower value such as −6.
Then write a verbal condition with the same set, such as “the temperature is no lower than −5°C.” Linking algebra, number line and language strengthens retrieval.
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Continue your learner route
Build from G2 Vol 0098: Mathematics Case Splitting and use eduKateSG Mathematics Start Here Hub for the underlying algebra and number-line learning.
Use Examination Craft for paper-level control and PSLE Learner’s Guide Vol 0080 for the earlier complete-then-stop habit.
Continue within this new batch: G2 Vol 0180: Instruction Authority Ladder · G2 Vol 0181: English Viewpoint-Knowledge Boundary.
Official references and scope
Framework checked on 13 October 2026: SEAB 2027 G2 syllabuses lists Mathematics as K210. Read the 2027 K210 G2 Mathematics syllabus.
K210 includes simple inequalities of the forms ax ≤ b, ax ≥ b, ax < b and ax > b, with integer a and b. These examples are original practice and do not predict examination questions.