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Themed maths · 5 November

Bonfire Night maths activities for IGCSE and GCSE

A ready-to-teach lesson for the week of 5 November: a 5-minute starter, a 35-minute main activity, an extension and full worked answers. It practises quadratics, sequences and cones.

Level
IGCSE and GCSE (Higher and Extended)
Time
40 minutes, plus a 10-minute extension
Topics
Quadratic graphs and completing the square; Linear and quadratic sequences; Exponential patterns; Volume and surface area of a cone
Equipment
The starter is non-calculator. A calculator helps with the cone task.

Download student sheet (PDF)Answers (PDF)

Suggested timings

PartTimeWhat
Starter5 minQuick questions on the board
Main: task A12 minThe rocket
Main: task B12 minRemember, remember: sparkler patterns
Main: task C11 minThe bonfire
Extension10 minFast finishers or homework

Starter (5 minutes)

No calculator.

  1. Write down the next two terms: 5, 11, 17, 23, …
  2. Find the nth term of 5, 11, 17, 23, …
  3. Solve x2 − 9x + 20 = 0.
  4. h = 30t − 5t2. Find h when t = 2.

Main activity (35 minutes)

Task A: The rocket (12 min)

The height of a firework rocket, h metres, t seconds after launch is modelled by h = 40t − 5t2.

  1. Find the height after 3 seconds.
  2. If the rocket did not explode, after how many seconds would it land?
  3. Complete the square to find the greatest height and when it happens.
  4. For how long is the rocket more than 60 m above the ground?

Task B: Remember, remember: sparkler patterns (12 min)

Pattern 1 uses 5 sparklers, pattern 2 uses 9 and pattern 3 uses 13.

  1. Find the nth term for the number of sparklers.
  2. How many sparklers are in pattern 50?
  3. Can a pattern use exactly 100 sparklers? Explain.
  4. In a firework display, the numbers of stars in each burst are 3, 8, 15, 24, 35, … Find the next term and the nth term.
  5. Which burst has 168 stars?

Task C: The bonfire (11 min)

A bonfire is built as a cone with base radius 1.5 m and height 2 m. Volume of a cone = 1⁄3πr2h; curved surface area = πrl.

  1. Find the volume of the bonfire, to 3 significant figures.
  2. Find the slant height and the curved surface area, to 3 significant figures.

Extension (10 minutes)

Two rockets.

  1. A second rocket is launched 1 second after the first. Its height is h = 50(t − 1) − 5(t − 1)2, where t is the time since the first launch. When are the two rockets at the same height, and how high are they? Give the height to 3 significant figures.

For teachers

Teacher notes and full worked answers
  • Enjoy fireworks at an organised display; this pack is about the maths only.
  • In the rocket task, ask why t = 0 is also a solution of h = 0 and what it means.
  • For the quadratic sequence, students often halve the first difference instead of the second difference.

Starter

  1. 29, 35
    • The terms go up by 6 each time.
  2. 6n − 1
    • The difference is 6, so start with 6n. The first term is 5 = 6 − 1, so 6n − 1.
  3. x = 4 or 5
    • (x − 4)(x − 5) = 0
  4. 40
    • 60 − 20 = 40

Task A: The rocket

  1. 75 m
    • 40 × 3 − 5 × 9 = 120 − 45 = 75
  2. 8 seconds
    • 40t − 5t2 = 0 gives 5t(8 − t) = 0, so t = 0 or 8.
  3. 80 m after 4 seconds
    • −5(t2 − 8t) = −5[(t − 4)2 − 16] = 80 − 5(t − 4)2.
    • The greatest value is 80, when t = 4.
  4. 4 seconds
    • 40t − 5t2 = 60 gives t2 − 8t + 12 = 0, so (t − 2)(t − 6) = 0.
    • It is above 60 m from t = 2 to t = 6: 4 seconds.

Task B: Remember, remember: sparkler patterns

  1. 4n + 1
    • The difference is 4 and 5 = 4 + 1.
  2. 201
    • 4 × 50 + 1 = 201
  3. No
    • 4n + 1 = 100 gives n = 24.75, which is not a whole number.
  4. 48; n2 + 2n
    • First differences 5, 7, 9, 11; second difference 2, so start with n2.
    • Subtract n2: 2, 4, 6, 8, 10, which is 2n. So n2 + 2n.
    • Next term: 62 + 12 = 48.
  5. The 12th
    • n2 + 2n = 168 gives (n + 14)(n − 12) = 0, so n = 12.

Task C: The bonfire

  1. 4.71 m3
    • 1⁄3 × π × 1.52 × 2 = 1.5π = 4.712…
  2. 2.5 m; 11.8 m2
    • l = √(1.52 + 22) = √6.25 = 2.5
    • π × 1.5 × 2.5 = 3.75π = 11.78…

Extension

  1. t = 2.75 s, at 72.2 m
    • Expand: 50t − 50 − 5t2 + 10t − 5 = 60t − 55 − 5t2.
    • Set equal to 40t − 5t2: the t2 terms cancel, so 20t = 55 and t = 2.75.
    • Height = 40 × 2.75 − 5 × 2.752 = 72.1875, so 72.2 m.

The same answers are in the answers PDF. Every answer was recomputed by computer before publishing.

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