GELE Mathematics — Algebra and FundamentalsExam Answer Templates

Marks

1

Topic

Quadratic and Polynomial Equations

Difficulty

easy

Template Id

T1

Examiner Tip

Even in a 1-mark VSA, examiners expect ALL three discriminant conditions. Writing only 'two cases' forfeits the mark in many rubrics.

Model Answer

Quadratic formula: x = [–b ± √(b² – 4ac)] / (2a). The discriminant D = b² – 4ac: if D > 0, roots are real and distinct; if D = 0, roots are real and equal; if D < 0, roots are complex (conjugate pair).

Question Type

very_short_answer

Answer Structure

• Line 1: Write the quadratic formula in symbolic form [0.5 mark]
• Line 2: State the three conditions of the discriminant with their corresponding root nature [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Writing ±√(b² – 4ac) without the 2a denominator
• Stating only one or two discriminant cases instead of all three
• Confusing D < 0 with 'no solution' instead of 'complex roots'

Key Phrases To Include

• b² – 4ac
• real and distinct
• real and equal
• complex conjugate
• discriminant

Marks

1

Topic

Exponents and Radicals

Difficulty

easy

Template Id

T2

Examiner Tip

Name the law you are applying at each step (power rule, product rule). This demonstrates conceptual command and earns partial credit even if you make an arithmetic slip.

Model Answer

= a⁶b⁻⁴ · a⁻¹b³ [apply power rule: (xᵐ)ⁿ = xᵐⁿ] = a⁶⁻¹ · b⁻⁴⁺³ [apply product rule: aᵐaⁿ = aᵐ⁺ⁿ] = a⁵b⁻¹ = a⁵/b ← Answer

Question Type

numerical

Answer Structure

• Step 1: Apply the power rule to (a³b⁻²)² → a⁶b⁻⁴ [show exponent multiplication]
• Step 2: Multiply by a⁻¹b³ using the product rule [add exponents]
• Step 3: Write in positive-exponent form as the final simplified expression

Scoring Breakdown

Common Mark Deductions

• Multiplying exponents when the rule requires addition (product rule) — the most common error
• Leaving negative exponents in the final answer without converting to positive form
• Arithmetic errors in exponent addition (e.g., –4 + 3 = –2 instead of –1)

Key Phrases To Include

• power rule
• product rule
• aᵐaⁿ = aᵐ⁺ⁿ
• (aᵐ)ⁿ = aᵐⁿ

Marks

2

Topic

Logarithms

Difficulty

easy

Template Id

T3

Examiner Tip

Always write the rule in symbolic form before applying it numerically. Examiners award the 'method mark' for the correct rule even if arithmetic goes wrong.

Model Answer

Given: log₂ 32 + log₂ 4 – log₂ 16 Using logarithm product and quotient rules: = log₂ (32 × 4 / 16) = log₂ (128 / 16) = log₂ 8 = log₂ 2³ = 3 ∴ log₂ 32 + log₂ 4 – log₂ 16 = 3 ← Answer

Question Type

numerical

Answer Structure

• Line 1: State the log rules being applied (product and quotient) [1 mark]
• Line 2: Combine into a single logarithm and simplify the argument [0.5 mark]
• Line 3: Evaluate the resulting logarithm to get the numerical answer [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Incorrectly applying log(M + N) = log M + log N (reversing the rule)
• Not simplifying the argument to a power of 2 before evaluating
• Arithmetic error in 32 × 4 ÷ 16 = 8

Key Phrases To Include

• log(MN) = log M + log N
• log(M/N) = log M – log N
• log₂ 8 = 3
• log rules

Marks

2

Topic

Quadratic and Polynomial Equations

Difficulty

easy

Template Id

T4

Examiner Tip

The sign error on the sum formula (–b/a vs b/a) is the single most common mistake. Always rewrite the equation in standard form ax² + bx + c = 0 and carefully extract b with its sign before computing.

Model Answer

Given: 3x² – 7x + 2 = 0 → a = 3, b = –7, c = 2 By Vieta's formulas: Sum of roots = –b/a = –(–7)/3 = 7/3 Product of roots = c/a = 2/3 ∴ Sum = 7/3, Product = 2/3 ← Answer

Question Type

numerical

Answer Structure

• Line 1: Identify coefficients a, b, c explicitly [0.5 mark]
• Line 2: State Vieta's formulas symbolically: sum = –b/a, product = c/a [0.5 mark]
• Line 3: Substitute and compute each value [0.5 mark]
• Line 4: State both answers clearly [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using b/a instead of –b/a for the sum (sign error — extremely common)
• Using b instead of –b because the equation is written with a minus sign (confusing sign of coefficient)
• Not identifying a, b, c before applying the formulas

Key Phrases To Include

• Vieta's formulas
• sum = –b/a
• product = c/a
• a = 3, b = –7, c = 2

Marks

3

Topic

Quadratic and Polynomial Equations

Difficulty

medium

Template Id

T5

Examiner Tip

Always compute the discriminant as a separate step before the formula. Examiners award a standalone mark for D = 49 even if you later make an arithmetic error in the roots.

Model Answer

Given: 2x² + 3x – 5 = 0 Required: x Identify coefficients: a = 2, b = 3, c = –5 Discriminant: D = b² – 4ac = (3)² – 4(2)(–5) = 9 + 40 = 49 > 0 → Two real distinct roots. Quadratic formula: x = [–b ± √D] / (2a) x = [–3 ± √49] / (2·2) x = [–3 ± 7] / 4 x₁ = (–3 + 7)/4 = 4/4 = 1 x₂ = (–3 – 7)/4 = –10/4 = –5/2 Verification (Vieta's): Sum = 1 + (–5/2) = –3/2 = –b/a = –3/2 ✓ Product = 1 × (–5/2) = –5/2 = c/a = –5/2 ✓ ∴ x = 1 or x = –5/2 ← Answer

Question Type

numerical

Answer Structure

• Step 1: Write Given and Required; identify a, b, c [0.5 mark]
• Step 2: Compute discriminant and state root nature [0.5 mark]
• Step 3: Write quadratic formula symbolically then substitute [1 mark]
• Step 4: Compute both roots x₁ and x₂ [0.5 mark]
• Step 5: Verify using Vieta's sum and product checks [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting the ± when applying the formula and giving only one root
• Arithmetic error inside the square root (e.g., computing 9 – 40 instead of 9 + 40 because c is negative)
• Omitting the Vieta's verification when it is specifically requested or when partial credit requires showing a check
• Not simplifying –10/4 to –5/2

Key Phrases To Include

• discriminant D = b² – 4ac
• real and distinct
• quadratic formula
• Vieta's verification
• x = 1
• x = –5/2

Marks

3

Topic

Arithmetic Progression

Difficulty

easy

Template Id

T6

Examiner Tip

Show both forms of the AP sum formula and verify with the second form. This demonstrates deeper understanding and protects you if you made an arithmetic error in one path.

Model Answer

Given: a₁ = 4, d = 6, n = 15 Required: S₁₅ Formula: S_n = (n/2)[2a₁ + (n – 1)d] Substitute: S₁₅ = (15/2)[2(4) + (15 – 1)(6)] = (15/2)[8 + 84] = (15/2)(92) = 15 × 46 = 690 ∴ S₁₅ = 690 ← Answer Alternative check using last term: a₁₅ = 4 + (14)(6) = 4 + 84 = 88 S₁₅ = (15/2)(4 + 88) = (15/2)(92) = 690 ✓

Question Type

numerical

Answer Structure

• Step 1: Write Given, Required; state AP sum formula [0.5 mark]
• Step 2: Substitute n = 15, a₁ = 4, d = 6 [0.5 mark]
• Step 3: Simplify inside the brackets [0.5 mark]
• Step 4: Compute final numerical answer [1 mark]
• Step 5: Verify using alternate formula S_n = n/2 (a₁ + aₙ) [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using (n/2)[a₁ + aₙ] without first finding the last term aₙ correctly
• Substituting (n) instead of (n – 1) in the formula: writing (15)(6) instead of (14)(6)
• Arithmetic error: 8 + 84 = 90 instead of 92

Key Phrases To Include

• S_n = (n/2)[2a₁ + (n–1)d]
• arithmetic progression
• common difference d
• S₁₅ = 690

Marks

3

Topic

Geometric Progression

Difficulty

medium

Template Id

T7

Examiner Tip

The convergence check '|r| < 1' is a mandatory sentence for full marks. Examiners are specifically instructed to deduct marks if this condition is not explicitly stated.

Model Answer

Given: a₁ = 12, r = 4/12 = 1/3 Required: S∞ Check convergence: |r| = 1/3 < 1 → Series converges. Formula: S∞ = a₁ / (1 – r) Substitute: S∞ = 12 / (1 – 1/3) = 12 / (2/3) = 12 × (3/2) = 18 ∴ S∞ = 18 ← Answer

Question Type

numerical

Answer Structure

• Step 1: Identify a₁ and compute r = a₂/a₁ [0.5 mark]
• Step 2: Check convergence condition |r| < 1 [0.5 mark]
• Step 3: State the infinite GP sum formula [0.5 mark]
• Step 4: Substitute and simplify to get the numerical answer [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Skipping the convergence check — even if the answer is correct, omitting this step costs marks
• Using S∞ = a₁/(1 + r) instead of (1 – r) — sign error
• Computing r = a₁/a₂ = 12/4 = 3 instead of a₂/a₁ = 1/3 (inverted ratio)
• Forgetting to verify |r| < 1 before applying the infinite sum formula

Key Phrases To Include

• S∞ = a₁/(1 – r)
• common ratio r = 1/3
• |r| < 1
• converges
• geometric progression

Marks

3

Topic

Work Problems

Difficulty

medium

Template Id

T8

Examiner Tip

In all work-rate problems, the golden rule is: ADD RATES, NOT TIMES. Write this to yourself on scratch paper every exam. One sentence defining your variable earns the problem-setup mark.

Model Answer

Given: Pipe A fills in 3 h, Pipe B fills in 5 h, Drain C empties in 4 h. Required: t = fill time with all three open (in hours). Let the full tank = 1 job. Rates (jobs per hour): Pipe A: +1/3 Pipe B: +1/5 Drain C: –1/4 (negative because it empties) Net rate: R_net = 1/3 + 1/5 – 1/4 LCM of 3, 5, 4 = 60: R_net = 20/60 + 12/60 – 15/60 = 17/60 jobs/hour Time to fill: t = 1 / R_net = 1 / (17/60) = 60/17 ≈ 3.53 hours ∴ t = 60/17 h ≈ 3.53 hours ← Answer

Question Type

numerical

Answer Structure

• Step 1: Define the job and assign rates with correct signs (+ fill, – drain) [0.5 mark]
• Step 2: Write the net rate equation and find the LCD [1 mark]
• Step 3: Compute R_net = 17/60 [0.5 mark]
• Step 4: Compute t = 1/R_net = 60/17 and express in decimal [1 mark]

Scoring Breakdown

Common Mark Deductions

• Adding times instead of rates: t = 3 + 5 – 4 = 4 h (the most common and fatal error)
• Treating the drain rate as positive instead of negative
• Arithmetic error in finding LCD or adding fractions
• Not defining the variable t explicitly before solving

Key Phrases To Include

• rate = 1/time
• net rate
• LCD
• jobs per hour
• drain rate is negative
• t = 1/R_net

Marks

3

Topic

Binomial Theorem

Difficulty

medium

Template Id

T9

Examiner Tip

Always write '6th term → r + 1 = 6 → r = 5' as a separate line. This single sentence shows the examiner you understand the theorem's indexing, and earns a mark even if the computation has errors.

Model Answer

Given: (2x – 1)⁹; required: 6th term (r + 1 = 6, so r = 5) General term (Binomial Theorem): T_{r+1} = C(n, r) · aⁿ⁻ʳ · bʳ Here: n = 9, a = 2x, b = –1, r = 5 T₆ = C(9, 5) · (2x)⁹⁻⁵ · (–1)⁵ = C(9, 5) · (2x)⁴ · (–1) = 126 · 16x⁴ · (–1) = –2016x⁴ ∴ 6th term = –2016x⁴ ← Answer [Note: C(9,5) = 9!/(5!4!) = 126]

Question Type

numerical

Answer Structure

• Step 1: Identify n, a, b, and determine r from 'r + 1 = 6 → r = 5' [0.5 mark]
• Step 2: Write the general term formula T_{r+1} = C(n,r) · aⁿ⁻ʳ · bʳ [0.5 mark]
• Step 3: Substitute all values including signs [1 mark]
• Step 4: Compute C(9,5) = 126 and (2x)⁴ = 16x⁴; multiply to get final term [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using r = 6 instead of r = 5 (confusing the 'r+1' relationship)
• Forgetting that b = –1 carries a negative sign, giving +2016x⁴ instead of –2016x⁴
• Computing (2)⁴ = 8 instead of 16
• Miscalculating C(9,5): confusing with C(9,4)

Key Phrases To Include

• T_{r+1} = C(n,r) · aⁿ⁻ʳ · bʳ
• r = 5 for the 6th term
• C(9,5) = 126
• (–1)⁵ = –1
• Binomial Theorem

Marks

5

Topic

Logarithms

Difficulty

hard

Template Id

T10

Examiner Tip

The domain check is not optional — it is a mandatory final step in any logarithmic equation. In every board exam logarithm problem, at least one extraneous root is planted by the question setter specifically to test whether you do this check.

Model Answer

Given: log₃(x + 2) + log₃(x – 4) = 3 Required: x Step 1 — Combine using the product rule: log₃[(x + 2)(x – 4)] = 3 Step 2 — Convert to exponential form (definition of logarithm: logᵦ M = N ↔ bᴺ = M): (x + 2)(x – 4) = 3³ = 27 Step 3 — Expand and form a quadratic: x² – 4x + 2x – 8 = 27 x² – 2x – 8 = 27 x² – 2x – 35 = 0 Step 4 — Solve the quadratic (factor or formula): (x – 7)(x + 5) = 0 x = 7 or x = –5 Step 5 — Check domain (arguments of log must be > 0): x = 7: (7 + 2) = 9 > 0 ✓ and (7 – 4) = 3 > 0 ✓ x = –5: (–5 + 2) = –3 < 0 ✗ → REJECTED ∴ x = 7 ← Answer

Question Type

numerical

Answer Structure

• Step 1: Apply log product rule to combine LHS into a single logarithm [1 mark]
• Step 2: Convert logarithmic to exponential form; evaluate 3³ = 27 [1 mark]
• Step 3: Expand the product and rearrange into standard quadratic form [1 mark]
• Step 4: Solve the quadratic equation (factoring preferred) [1 mark]
• Step 5: Domain check — reject any solution that makes an argument ≤ 0 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Accepting x = –5 without performing a domain check — losing the final mark
• Applying the product rule incorrectly: log₃(x+2) + log₃(x–4) ≠ log₃(x+2+x–4)
• Arithmetic error when expanding (x+2)(x–4): commonly getting –x² instead of x²
• Forgetting to set the quadratic equal to zero before factoring

Key Phrases To Include

• product rule
• log₃[(x+2)(x–4)] = 3
• exponential form
• 3³ = 27
• domain check
• argument > 0
• rejected

Marks

5

Topic

Motion (Distance-Rate-Time) Problems

Difficulty

medium

Template Id

T11

Examiner Tip

For motion problems, always draw a simple sketch: mark A, B, and the directions of travel. This visual setup prevents errors and signals to the examiner that you've modeled the problem correctly.

Model Answer

Given: Distance AB = 360 km; speed of Train X (from A) = 80 km/h; speed of Train Y (from B) = 100 km/h; both depart simultaneously. Required: (a) time t when they meet; (b) distance from A. Step 1 — Define variable: Let t = time (in hours) until the trains meet. Step 2 — Write the distance equation (sum of distances = total separation): Distance by X + Distance by Y = 360 80t + 100t = 360 180t = 360 t = 2 hours Step 3 — Distance from Station A: d_A = 80 × t = 80 × 2 = 160 km Step 4 — Verification: Distance by Y = 100 × 2 = 200 km 160 + 200 = 360 km ✓ ∴ (a) The trains meet after t = 2 hours. (b) The meeting point is 160 km from Station A. ← Answer

Question Type

numerical

Answer Structure

• Step 1: Define variable t and state the physical meaning [0.5 mark]
• Step 2: Set up the equation d_X + d_Y = 360 and solve for t [2 marks]
• Step 3: Compute distance from Station A = 80t [1 mark]
• Step 4: Verify by checking both distances sum to 360 km [0.5 mark]
• Step 5: State both answers clearly with units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Setting up d_X = d_Y (as if trains have equal distances) instead of d_X + d_Y = 360
• Using different time variables for each train when they depart simultaneously
• Computing distance from B instead of A as required
• Omitting units in the final answer

Key Phrases To Include

• d = vt
• sum of distances
• 180t = 360
• t = 2 hours
• 160 km from Station A
• verification

Marks

3

Topic

Arithmetic Progression

Difficulty

medium

Template Id

T12

Examiner Tip

The system-of-equations approach earns marks at every step. Even if you know d by inspection, show the formal setup — examiners reward structured methodology.

Model Answer

Given: a₅ = 22, a₁₀ = 42 Required: a₁ and d Using the AP general term formula: aₙ = a₁ + (n – 1)d Equation 1 (n = 5): a₁ + 4d = 22 … (1) Equation 2 (n = 10): a₁ + 9d = 42 … (2) Subtract (1) from (2): (a₁ + 9d) – (a₁ + 4d) = 42 – 22 5d = 20 d = 4 Substitute d = 4 into (1): a₁ + 4(4) = 22 a₁ + 16 = 22 a₁ = 6 Verification: a₁₀ = 6 + 9(4) = 6 + 36 = 42 ✓ ∴ a₁ = 6, d = 4 ← Answer

Question Type

numerical

Answer Structure

• Step 1: Write the AP general term formula [0.5 mark]
• Step 2: Form two simultaneous equations from the given terms [1 mark]
• Step 3: Subtract equations to find d [0.5 mark]
• Step 4: Back-substitute to find a₁ and verify [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using n = 5 but writing (5)d instead of (4)d — forgetting the (n – 1) factor
• Not showing the subtraction step explicitly (jumping from two equations to d)
• Arithmetic error in back-substitution

Key Phrases To Include

• aₙ = a₁ + (n–1)d
• simultaneous equations
• elimination method
• d = 4
• a₁ = 6
• verification

Marks

5

Topic

Mixture Problems

Difficulty

medium

Template Id

T13

Examiner Tip

Mixture problems hinge on one key principle: the AMOUNT of the solute (alcohol, salt, etc.) is conserved across the operation. Always write the equation in terms of amounts, not percentages alone.

Model Answer

Given: Initial volume = 20 L at 30% alcohol; pure alcohol added = x litres; final concentration = 50%. Required: x Step 1 — Alcohol content in original mixture: Alcohol_initial = 0.30 × 20 = 6 L Step 2 — Set up the mixture equation: (Alcohol initially) + (pure alcohol added) = (50% of new total volume) 6 + x = 0.50 × (20 + x) Step 3 — Solve for x: 6 + x = 10 + 0.50x x – 0.50x = 10 – 6 0.50x = 4 x = 8 litres Step 4 — Verification: New total = 20 + 8 = 28 L Total alcohol = 6 + 8 = 14 L Concentration = 14/28 = 0.50 = 50% ✓ ∴ x = 8 litres of pure alcohol must be added. ← Answer

Question Type

numerical

Answer Structure

• Step 1: Define x and compute initial alcohol content [0.5 mark]
• Step 2: Write the mixture equation balancing alcohol amounts [1.5 marks]
• Step 3: Solve the linear equation for x [1.5 marks]
• Step 4: Verify by computing the new concentration [1 mark]
• Step 5: State the answer with correct unit (litres) [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Writing 0.50 × 20 instead of 0.50 × (20 + x) — not accounting for the increased total volume
• Setting up the equation in terms of the new volume without including x in the denominator/total
• Omitting the verification step when 5 marks require explicit checking
• Not defining x clearly before writing the equation

Key Phrases To Include

• mixture equation
• alcohol content = concentration × volume
• 0.30 × 20 = 6
• 6 + x = 0.50(20 + x)
• x = 8 litres
• verification

Marks

2

Topic

Logarithms

Difficulty

easy

Template Id

T14

Examiner Tip

Always write the formula before the numbers. If you invert numerator and denominator, writing the formula first allows the examiner to identify a substitution error rather than a conceptual error — and may preserve a method mark.

Model Answer

Given: log₇ 200 Required: Numerical value (4 decimal places) Change-of-base formula: logᵦ x = log x / log b (using common log, base 10) log₇ 200 = log 200 / log 7 = 2.3010 / 0.8451 = 2.7237 ∴ log₇ 200 ≈ 2.7237 ← Answer

Question Type

numerical

Answer Structure

• Line 1: State the change-of-base formula explicitly [0.5 mark]
• Line 2: Substitute log 200 and log 7 with their values [0.5 mark]
• Line 3: Divide and state the answer to 4 decimal places [1 mark]

Scoring Breakdown

Common Mark Deductions

• Inverting the formula: writing log 7 / log 200 instead of log 200 / log 7
• Using ln incorrectly or mixing log and ln in the same calculation
• Rounding intermediate values too early, leading to an inaccurate final answer

Key Phrases To Include

• change-of-base formula
• logᵦ x = log x / log b
• log 200 = 2.3010
• log 7 = 0.8451

Marks

2

Topic

Ratio and Word Problems

Difficulty

easy

Template Id

T15

Examiner Tip

The ratio-variable method (3k, 5k) is the cleanest and fastest approach for ratio problems. It instantly sets up a one-equation, one-unknown system and shows the examiner a structured approach.

Model Answer

Given: Ratio of two numbers = 3:5; sum = 96. Required: The two numbers. Let the numbers be 3k and 5k. Sum: 3k + 5k = 96 8k = 96 k = 12 Numbers: 3k = 3(12) = 36 and 5k = 5(12) = 60 Verification: 36 + 60 = 96 ✓; 36:60 = 3:5 ✓ ∴ The two numbers are 36 and 60. ← Answer

Question Type

numerical

Answer Structure

• Line 1: Define the two numbers as 3k and 5k using a ratio variable [0.5 mark]
• Line 2: Write and solve the sum equation for k [0.5 mark]
• Line 3: Compute both numbers and state a double verification [1 mark]

Scoring Breakdown

Common Mark Deductions

• Setting up the equation incorrectly as (3/5)x = 96 — misinterpreting the ratio
• Finding k but not computing the actual numbers
• Providing only one number instead of both

Key Phrases To Include

• ratio variable k
• 3k and 5k
• 8k = 96
• k = 12
• verification