A Level Math Revision Free diagnostic
Themed maths · 31 October

Halloween maths activities for A Level

A ready-to-teach Halloween lesson for A Level Maths: a 5-minute starter, a 35-minute main activity, an extension and full worked answers. Tasks A and C suit Year 12; the logistic and rates-of-change parts are Year 13.

Level
A Level Maths (Year 12 and Year 13)
Time
40 minutes, plus a 10-minute extension
Topics
Exponential models and logarithms; Logistic growth from a differential equation (Year 13); Connected rates of change (Year 13); Conditional probability and the binomial distribution; Equations of circles
Equipment
The starter is non-calculator. A calculator is needed for the main activity.

Download student sheet (PDF)Answers (PDF)

Suggested timings

PartTimeWhat
Starter5 minQuick questions on the board
Main: task A12 minZombie outbreak
Main: task B8 minThe growing pumpkin (Year 13)
Main: task C8 minTrick or treat
Main: task D7 minThe haunted room
Extension10 minFast finishers or homework

Starter (5 minutes)

No calculator. Exact answers.

  1. Solve ex = 7.
  2. Differentiate 5e2x.
  3. P(A) = 0.4 and P(B | A) = 0.25. Find P(A ∩ B).
  4. Find the centre and radius of the circle x2 + y2 − 6x + 4y − 12 = 0.

Main activity (35 minutes)

Task A: Zombie outbreak (12 min)

The number of zombies t hours after midnight on 31 October is first modelled by Z = 30ekt. After 4 hours there are 90 zombies.

  1. Find the exact value of k.
  2. Find the time when the model predicts 1000 zombies, to 3 significant figures.
  3. (Year 13) A town has 1000 people. A better model is dZ/dt = 0.5Z(1 − Z/1000) with Z = 10 when t = 0. Use partial fractions to show that Z = 1000 ÷ (1 + 99e−0.5t).
  4. Using the logistic model, find when 900 people are zombies, to 3 significant figures. What is the long-term number of zombies?

Task B: The growing pumpkin (Year 13) (8 min)

A prize pumpkin is a sphere of radius r cm. Its volume grows at a constant 50 cm3 per day.

  1. Find the rate at which the radius is increasing when r = 10. Give an exact answer and a decimal to 3 significant figures.
  2. Find the rate at which the surface area is increasing at the same moment.

Task C: Trick or treat (8 min)

60% of houses have a pumpkin outside. At a house with a pumpkin, P(treat) = 0.9; without one, P(treat) = 0.4.

  1. Find P(treat) and P(pumpkin | treat).
  2. Are the events ‘pumpkin’ and ‘treat’ independent? Give a reason.
  3. You visit 12 houses chosen at random. Find the probability of at most 6 treats, to 3 significant figures.

Task D: The haunted room (7 min)

The floor of a round haunted room is the circle x2 + y2 − 10x − 4y + 4 = 0. A ghost walks along the line y = x + 2.

  1. Find the centre and radius of the room.
  2. Find where the ghost enters and leaves the room, and the length of its path inside.
  3. Find the equation of the tangent to the room at (5, 7).

Extension (10 minutes)

Year 13 stretch.

  1. For dZ/dt = 0.5Z(1 − Z/1000), show that the zombies spread fastest when Z = 500. Find this greatest rate and the time it happens, to 3 significant figures.

For teachers

Teacher notes and full worked answers
  • Year 12 classes can do the starter, tasks A (a)–(b), C and D, and leave the differential equation and rates of change for a Year 13 lesson.
  • In A (c), check that students keep the modulus signs when they integrate and explain why they can drop them here (0 < Z < 1000).
  • In C (b), the common error is to compare P(treat | pumpkin) with P(treat) using the wrong numbers. Insist on one clear test.

Starter

  1. x = ln 7
    • Take natural logs of both sides: x = ln 7.
  2. 10e2x
    • d/dx e2x = 2e2x, so the answer is 10e2x.
  3. 0.1
    • P(A ∩ B) = P(A) × P(B | A) = 0.4 × 0.25 = 0.1
  4. Centre (3, −2), radius 5
    • Complete the square: (x − 3)2 + (y + 2)2 = 12 + 9 + 4 = 25.

Task A: Zombie outbreak

  1. k = ¼ ln 3 ≈ 0.275
    • 30e4k = 90, so e4k = 3.
    • 4k = ln 3, so k = ¼ ln 3 = 0.2746…
  2. 12.8 hours
    • ekt = 100/3, so t = ln(100/3) ÷ k = 4 ln(100/3) ÷ ln 3 = 12.767…
    • 12.8 hours
  3. Z = 1000 ÷ (1 + 99e−0.5t)
    • Separate: ∫ 1000 ÷ (Z(1000 − Z)) dZ = ∫ 0.5 dt.
    • Partial fractions: 1000 ÷ (Z(1000 − Z)) = 1/Z + 1/(1000 − Z), so ln Z − ln(1000 − Z) = 0.5t + c.
    • At t = 0: c = ln(10/990) = −ln 99. So Z ÷ (1000 − Z) = e0.5t ÷ 99.
    • Rearrange: Z = 1000 ÷ (1 + 99e−0.5t).
  4. 13.6 hours; 1000
    • 1 + 99e−0.5t = 1000/900 = 10/9, so e−0.5t = 1/891.
    • t = 2 ln 891 = 13.58…, so 13.6 hours.
    • As t → ∞, Z → 1000.

Task B: The growing pumpkin (Year 13)

  1. 1/(8π) ≈ 0.0398 cm per day
    • dV/dr = 4πr2 = 400π at r = 10.
    • dr/dt = (dV/dt) ÷ (dV/dr) = 50 ÷ 400π = 1/(8π) = 0.0398…
  2. 10 cm2 per day
    • S = 4πr2, so dS/dr = 8πr = 80π.
    • dS/dt = 80π × 1/(8π) = 10

Task C: Trick or treat

  1. 0.7 and 27/35 ≈ 0.771
    • P(treat) = 0.54 + 0.16 = 0.7.
    • P(pumpkin | treat) = 0.54 ÷ 0.7 = 27/35.
  2. No
    • P(pumpkin) × P(treat) = 0.6 × 0.7 = 0.42, but P(pumpkin ∩ treat) = 0.54.
    • 0.42 ≠ 0.54, so they are not independent.
  3. 0.118
    • X ~ B(12, 0.7). Use the cumulative binomial function.
    • P(X ≤ 6) = 0.1178…, so 0.118.

Task D: The haunted room

  1. Centre (5, 2), radius 5
    • (x − 5)2 + (y − 2)2 = 25 + 4 − 4 = 25.
  2. (0, 2) and (5, 7); 5√2
    • Substitute: x2 + (x + 2)2 − 10x − 4(x + 2) + 4 = 0 gives 2x2 − 10x = 0.
    • x = 0 or 5, so (0, 2) and (5, 7).
    • Length = √(52 + 52) = 5√2
  3. y = 7
    • The radius to (5, 7) goes from (5, 2) straight up, so it is vertical.
    • The tangent is perpendicular to it, so it is horizontal: y = 7.

Extension

  1. 125 zombies per hour, at t = 2 ln 99 ≈ 9.19 hours
    • dZ/dt = 0.5Z − 0.0005Z2 is a quadratic in Z with its maximum at Z = 0.5 ÷ 0.001 = 500.
    • Greatest rate = 0.5 × 500 × 0.5 = 125.
    • 1000 ÷ (1 + 99e−0.5t) = 500 gives t = 2 ln 99 = 9.19 hours.

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

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