Algebra Basics

Overview

Algebra Basics appears about 1 times per paper across the last four AFCAT solved papers, placing it in the high yield band of Numerical Ability.

Algebra Basics carries roughly one question per AFCAT paper — three marks if correct, one mark lost if wrong. That is small in absolute weight, but the chapter is exceptionally high in return-on-time because almost every AFCAT algebra item is a one-step identity substitution, a two-line simultaneous-equation solve, or a direct application of a surd or log rule. There is no extended derivation work, no abstract proof, no calculus. The candidate who has memorised twelve identities, the discriminant rule and the surd-and-index table will solve the algebra item in under sixty seconds and walk away with three guaranteed marks.

This page builds the full toolkit. You get a comprehensive identity table covering twelve standard expansions, a polynomial vocabulary primer, the three methods for solving simultaneous linear equations, the condition trio (a₁/a₂, b₁/b₂, c₁/c₂) that tells you whether a system has one, none or infinite solutions, the quadratic discriminant and Vieta's formulas, a surd-and-index rule table with rationalisation worked through, the four logarithm laws with change-of-base, and the AP and GP formula sets. Twelve worked examples cover every pattern AFCAT has used in the last four papers.

The chapter is also a quiet feeder into other Numerical topics. The identity (a + b)² appears inside time-and-work mixture problems. The surd table appears inside geometry questions on diagonal length. The AP formula appears inside data-interpretation series. Mastering algebra basics pays a dividend across the entire Numerical Ability paper.

Why algebra carries reliable AFCAT marks

Across the last four solved AFCAT papers, algebra-basics items averaged 1.0 questions per paper. That is lower than percentage (~3.0) and time-speed-distance (~2.5), but algebra has a different property — it is almost completely formula-driven. Every AFCAT algebra item in recent papers has been solvable by recognising one of three structures:

• An algebraic identity that converts a given pair like a + b and ab into a target like a² + b².
• A two-variable linear system with small integer coefficients.
• A surd or index simplification with no more than three steps.

None of these require thinking — only recall. The candidate who has drilled the twelve identities and the surd rules walks in knowing every algebra item will close in under a minute. The candidate who has not drilled them spends two minutes on the same item, often misapplies an identity, and loses the mark.

A second reason this chapter matters: algebra is the quickest topic to revise. The full toolkit fits on one A4 sheet. Two days before the paper, a candidate can re-read the identity table, the discriminant rule and the surd table in twenty minutes and lock in three guaranteed marks. No other Numerical topic offers a revision-to-marks ratio this favourable.

Algebraic identities — comprehensive table

The twelve identities below cover every AFCAT algebra item in the last four papers. Memorise them in this exact form — most exam items hand you the right-hand side and ask for the left, or vice versa.

Polynomials — vocabulary and zero

A polynomial in one variable x is an expression of the form a_nxⁿ + a_n−1xⁿ⁻¹ + ... + a₁x + a₀, where the coefficients are real numbers and n is a non-negative integer.

A polynomial of degree n has at most n real zeros. A linear polynomial ax + b has exactly one zero, x = −b/a. A quadratic polynomial has zero, one or two real zeros depending on its discriminant (covered below).

Factor theorem. If p(a) = 0, then (x − a) is a factor of p(x). This lets you check candidate zeros by substitution rather than long division.

Remainder theorem. When a polynomial p(x) is divided by (x − a), the remainder equals p(a). AFCAT occasionally asks for the remainder when a cubic is divided by a linear factor — substitute the root, no long division required.

Linear equations — one and two variables

A linear equation in one variable has the form ax + b = 0 with a ≠ 0. Its unique solution is x = −b/a. AFCAT one-variable items usually wrap this skeleton inside a word problem.

A linear equation in two variables has the form ax + by + c = 0. By itself it has infinitely many solutions (a straight line in the xy-plane). Two such equations together — a simultaneous system — can have a unique solution, no solution or infinitely many.

Three solution methods for a 2 × 2 system.

1. Substitution. Solve one equation for one variable, substitute into the other. Best when one variable already has coefficient 1 or −1.
2. Elimination. Multiply one or both equations by suitable constants so that one variable's coefficients become equal in magnitude, then add or subtract. Best when coefficients are small integers.
3. Cross-multiplication. For a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, the solution is x / (b₁c₂ − b₂c₁) = y / (c₁a₂ − c₂a₁) = 1 / (a₁b₂ − a₂b₁). Memorise this only if it suits you — substitution and elimination handle every AFCAT case.

Worked. Solve 2x + 3y = 13 and 4x − y = 5.

1. From the second equation, y = 4x − 5.
2. Substitute: 2x + 3(4x − 5) = 13 → 2x + 12x − 15 = 13 → 14x = 28 → x = 2.
3. Then y = 4(2) − 5 = 3.

Simultaneous equations — conditions for solution count

For two equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, the number of solutions depends on the three ratios a₁/a₂, b₁/b₂, c₁/c₂.

Worked. For what value of k does the system 2x + 3y = 7 and 4x + ky = 14 have infinite solutions? Apply the third condition: 2/4 = 3/k = 7/14. From 1/2 = 3/k, we get k = 6. Check the third ratio: 7/14 = 1/2. All three match, so k = 6.

Quadratic equations — discriminant and Vieta

A quadratic equation has the standard form ax² + bx + c = 0 with a ≠ 0. The two roots are given by the quadratic formula x = (−b ± √(b² − 4ac)) / (2a).

Discriminant. The quantity D = b² − 4ac decides the nature of the roots.

Vieta's formulas (sum and product of roots). If α and β are the roots of ax² + bx + c = 0, then:

• Sum of roots: α + β = −b / a
• Product of roots: αβ = c / a

These two relations let you read off both root statistics directly from the coefficients — no need to solve the equation. AFCAT uses this heavily.

Forming a quadratic from given roots. If the roots are α and β, the equation can be written as x² − (α + β)x + αβ = 0.

Worked. Find the equation whose roots are 3 and −5. Sum = 3 + (−5) = −2. Product = 3 × (−5) = −15. Equation: x² − (−2)x + (−15) = 0, i.e. x² + 2x − 15 = 0.

Quadratic factorisation — splitting the middle term

For a quadratic ax² + bx + c with small integer coefficients, factorisation by splitting the middle term is faster than the quadratic formula. The method has three steps.

1. Compute the product a × c.
2. Find two numbers whose product is a × c and whose sum is b.
3. Split b into those two numbers and factor in pairs.

Worked. Factor x² − 7x + 12.

1. a × c = 1 × 12 = 12.
2. Two numbers with product 12 and sum −7: that is −3 and −4.
3. Rewrite: x² − 3x − 4x + 12 = x(x − 3) − 4(x − 3) = (x − 3)(x − 4).

So the roots are x = 3 and x = 4.

Worked. Factor 2x² + 7x + 3.

1. a × c = 2 × 3 = 6.
2. Two numbers with product 6 and sum 7: that is 1 and 6.
3. Rewrite: 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3).

Roots: x = −1/2 and x = −3.

When splitting fails. If the discriminant is not a perfect square, no integer split exists. Switch to the quadratic formula. AFCAT keeps coefficients clean — almost every quadratic item factors over the integers.

Inequalities — linear and modulus

A linear inequality looks like ax + b > 0 (or ≥, <, ≤). The solution method mirrors a linear equation, with one rule: when you multiply or divide both sides by a negative number, the inequality flips.

Worked. Solve −3x + 7 > 1.

1. Subtract 7: −3x > −6.
2. Divide by −3 (flip the inequality): x < 2.

Modulus inequalities. The absolute value |x| is the distance of x from zero on the number line.

• |x| < a (with a > 0) means −a < x < a.
• |x| > a (with a > 0) means x < −a or x > a.
• |x − k| < a means k − a < x < k + a (a window of width 2a centred at k).

Worked. Solve |x − 4| ≤ 3. This gives 4 − 3 ≤ x ≤ 4 + 3, i.e. 1 ≤ x ≤ 7.

AFCAT inequality items are rare (one item every few papers) and almost always linear. The modulus form occasionally appears wrapped inside a number-line interpretation question.

Surds and indices — rules and rationalisation

An index (or exponent) tells you how many times the base is multiplied by itself. A surd is an irrational root, typically √2, √3 or √5. AFCAT mixes the two in roughly one item per paper.

Rationalising the denominator. A fraction with a surd in the denominator is simplified by multiplying numerator and denominator by the conjugate of the denominator. The conjugate of (a + √b) is (a − √b), and vice versa.

Worked. Rationalise 1 / (3 + √2).

1. Multiply top and bottom by (3 − √2): (3 − √2) / ((3 + √2)(3 − √2)).
2. Denominator = 3² − (√2)² = 9 − 2 = 7.
3. Result: (3 − √2) / 7.

Logarithms — definition and four laws

The logarithm log_a(x) answers the question: to what power must a be raised to give x? Formally, if a^y = x, then log_a(x) = y, where a > 0, a ≠ 1, and x > 0.

Common log. log₁₀(x) is written simply as log(x) on most AFCAT items. Natural log uses base e (≈ 2.718) and is written ln(x), but AFCAT rarely uses it.

Worked. Find log₂(32).

1. 32 = 2⁵.
2. log₂(2⁵) = 5 × log₂(2) = 5 × 1 = 5.

Worked. Simplify log(8) + log(125) (base 10).

1. log(8) + log(125) = log(8 × 125) = log(1000).
2. log(1000) = log(10³) = 3.

Arithmetic and geometric progressions

An arithmetic progression (AP) is a sequence where each term is obtained by adding a fixed number (the common difference d) to the previous term. A geometric progression (GP) is a sequence where each term is obtained by multiplying the previous by a fixed number (the common ratio r).

AM ≥ GM. For any two non-negative reals a and b, the arithmetic mean is at least as large as the geometric mean. Equality holds only when a = b. This inequality occasionally features in AFCAT max-min items.

Worked AP. Find the 20-th term of 3, 7, 11, 15, ...

1. a = 3, d = 4.
2. a₂₀ = 3 + (20 − 1) × 4 = 3 + 76 = 79.

Worked GP. Find the sum of the first 6 terms of 2, 4, 8, 16, ...

1. a = 2, r = 2, n = 6.
2. S₆ = 2 × (2⁶ − 1) / (2 − 1) = 2 × 63 = 126.

Mensuration meets algebra — scaling rules

Whenever a linear dimension of a figure is multiplied by k, the area scales by k² and the volume scales by k³. This is pure algebra applied to geometry and AFCAT loves the trick.

The percentage effect on area for a small percentage change p on side is approximately 2p + p²/100, derived from (1 + p/100)² − 1. For volume it is 3p + 3p²/100 + p³/10000, derived from (1 + p/100)³ − 1. AFCAT uses these as one-line shortcuts.

Worked. If the radius of a circle is increased by 50%, by what percentage does the area increase?

1. k = 1.5. New area / old area = 1.5² = 2.25.
2. Increase = 1.25, i.e. 125%.

AFCAT trap patterns to memorise

• Identity confusion. (a − b)² equals a² − 2ab + b², not a² − b². A skim reader who mixes these loses three marks on the first identity item.
• Missing cross-terms in (a + b + c)². The expansion has six cross terms, not three. The correct form is a² + b² + c² + 2(ab + bc + ca). Forgetting the factor of 2 is the single most common error.
• Surd simplification slip. √(a + b) is not √a + √b. Only √(ab) = √a × √b. Many candidates apply the multiplicative rule to a sum and lose the mark.
• Log base assumption. If the question writes log(x) without a base, assume base 10 (common log) unless the question context explicitly suggests natural log. Mixing the bases gives a wrong numerical answer even if every law is applied correctly.
• Sign error in Vieta. For ax² + bx + c = 0, the sum of roots is −b/a, not b/a. A sign slip flips every answer in a question that asks for the sum.
• Discriminant zero versus negative. D = 0 gives two equal real roots, not no real roots. Only D < 0 means no real roots. A common item is ‘equation has equal roots, find k’ — set D = 0, not D < 0.
• Infinite-solution ratio. A 2 × 2 system has infinite solutions only when all three ratios a₁/a₂, b₁/b₂, c₁/c₂ are equal. Many candidates check only the first two and pick the wrong k.
• GP sum to infinity restriction. The formula a / (1 − r) holds only when |r| < 1. If r ≥ 1 or r ≤ −1, the series diverges and the question has no finite answer. AFCAT occasionally sets a trap by giving r = 2.
• Quadratic that is not a quadratic. An equation like (k − 1)x² + 3x + 2 = 0 is quadratic only when k ≠ 1. If k = 1, the equation collapses to a linear form. Always check the leading coefficient.

Time budget per item

AFCAT gives 72 seconds per question on average. For algebra-basics items, target the following bands.

If an algebra item exceeds 90 seconds, mark it for review and move on. The opportunity cost of a stuck algebra item is one or two faster percentage or time-speed items lost later in the paper.

Worked AFCAT-style examples

Exam-day strategy

1. Lock the twelve identities in week one. Most AFCAT algebra items reduce to one identity substitution. The reciprocal pair (x + 1/x) deserves its own drill — it appears in roughly one paper out of three.
2. For simultaneous linear systems, prefer substitution when one variable has coefficient 1 or −1; otherwise use elimination. Reach for cross-multiplication only if you have drilled the formula to instinct.
3. For solution-count questions, write the three ratios a₁/a₂, b₁/b₂, c₁/c₂ in a vertical stack and compare. The geometry interpretation (intersecting / parallel / coincident) is a useful memory aid.
4. For quadratics, use splitting-the-middle-term first; switch to the quadratic formula only when the discriminant is not a perfect square. Always state sum and product of roots by Vieta before computing the roots — saves time when the question asks for either.
5. For surds, factor radicands into the largest perfect square inside before simplifying. To rationalise, multiply by the conjugate.
6. For logarithms, treat log(x) as base 10 unless told otherwise. Combine logs into a single log expression before computing.
7. For AP and GP, write the four primitives (a, d or r, n, S) at the top of your scratch space and plug into the right formula. Never derive on paper — use the table directly.
8. Aim for 30–60 seconds on direct identity items, 45–75 seconds on simultaneous systems and quadratics. Mark anything past 90 seconds and move on.

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