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Grade 10 · Exam & test walkthrough

Mathematics Paper 1 November 2025

About this video

Grade 10 Mathematics Paper 1, November 2025 (Nelson Mandela Metro, Eastern Cape) worked through in full: surds and recurring decimals, factorising and simplifying, equations and inequalities, simultaneous equations, number patterns, finance and exchange rates, functions and graphs, and probability.

What it covers

  • Between which integers a root lies; a recurring decimal as a fraction
  • Factorising, exponents with mixed bases, algebraic fractions
  • Equations: zero product rule, exponential, making x the subject, inequalities
  • Simultaneous equations by substitution
  • Linear patterns: the general term, the first negative term, a sum
  • Finance: present value, hire purchase, exchange rates
  • Hyperbola, parabola and exponential graphs; probability and Venn diagrams

Extra worked examples

✍️ Extra example: write 0,2̇7̇ as a fraction
  1. 1
    x = 0,2727…
    Call the recurring decimal x.
  2. 2
    100x = 27,2727…
    × 100 lines up the repeating part (× 10 gives 2,727… — not the same decimals).
  3. 3
    99x = 27 → x = 27/99 = 3/11
    Subtract: the recurring decimals cancel. Check: 3 ÷ 11 = 0,2727… ✓
✍️ Extra example: solve −6 < 3x − 9 ≤ 6 for x ∈ ℤ
  1. 1
    3 < 3x ≤ 15
    Add 9 to all three sides.
  2. 2
    1 < x ≤ 5
    Divide by 3. A POSITIVE number keeps the signs; a negative one would flip them.
  3. 3
    x ∈ {2; 3; 4; 5}
    Integers only: 1 is left out, 5 is included.

Exercises to try

Work each one out, then click to reveal the answer.

  1. 1
    Between which two consecutive integers does √90 lie?
    Show answer ▾
    81 < 90 < 100, so 9 < √90 < 10
  2. 2
    A linear pattern is 50; 46; 42; … Find the general term and the first negative term.
    Show answer ▾
    Tn = −4n + 54; −4n + 54 < 0 → n > 13,5, so T14 = −2 is the first negative term
  3. 3
    Solve simultaneously: y = 2x − 1 and 3x + y = 9.
    Show answer ▾
    Substitute: 3x + (2x − 1) = 9 → 5x = 10 → x = 2 and y = 3
  4. 4
    How much must be invested now at 7% per year, compounded annually, to have R30 000 in 4 years?
    Show answer ▾
    P = (30 000)/(1 + 7/100)4 = R22 886,86
  5. 5
    P(A) = 0,4, P(B) = 0,5 and P(A and B) = 0,1. Find P(A or B).
    Show answer ▾
    P(A or B) = 0,4 + 0,5 − 0,1 = 0,8

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