AQA A-Level Mathematics Paper 1, June 2025: Question 10

12 marks · Medium difficulty · Modelling

Rearrange an exponential cooling/warming model into linear logarithmic form, determine the parameters from a straight-line graph, and calculate initial temperature and time taken to reach a given temperature.

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Question

Question 10 presents an exponential model for temperature: theta = 21 - A * e^(-kt). Part (a) asks to show that this rearranges to ln(21 - theta) = -kt + ln A for 3 marks. Part (b) shows a straight-line graph with vertical axis ln(21 - theta) and horizontal axis t, intersecting the vertical axis at (0, 3.676) and horizontal axis at (19.98, 0). Sub-questions (i), (ii), and (iii) ask to find the values of A and k to three significant figures, and the initial temperature of the dessert (2 marks each). Part (c) asks to find the time taken for the dessert to reach 4 degrees Celsius to the nearest 10 minutes (3 marks).
Question text

10 A researcher working for a frozen‑food manufacturer uses the formula

θ = 21 – Ae–kt

to model the temperature of a dessert once it is taken out of a freezer.

In this model:

l θ is the temperature of the dessert in °C

l t is the time in hours since the dessert was removed from the freezer

l A and k are positive constants.

10 (a) Show how

θ = 21 – Ae–kt

can be rearranged to obtain

ln(21 – θ) = –kt + ln A

[3 marks]

10 (b) The researcher uses measurements they have recorded to plot the graph of

ln(21 – θ) against t as shown in the diagram below.

ln(21 – θ)

(0, 3.676)

(19.98, 0)

O t

(10)

10 (b) (i) Use the information on the graph to find the value of A

Give your answer to three significant figures.

[2 marks]

10 (b) (ii) Use the information on the graph to find the value of k

Give your answer to three significant figures.

[2 marks]

10 (b) (iii) Find the temperature of the dessert when it is initially removed from the freezer.

Give your answer to three significant figures.

[2 marks]

… U

(11)

10 (c) The dessert is ready to be eaten when its temperature reaches 4°C

Use the model to determine the time, after being removed from the freezer, for the

dessert to reach this temperature.

Give your answer to the nearest 10 minutes.

[3 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 10. 10(a) awards M1 for rearranging to isolate the exponential and taking logs, M1 for applying log laws to separate ln(A) and ln(e^(-kt)), and R1 for fully reasoned argument to achieve ln(21 - theta) = -kt + ln A. 10(b)(i) awards M1 for ln A = 3.676 and A1 for A = 39.5. 10(b)(ii) awards M1 for gradient calculation -3.676 / 19.98 and A1 for k = 0.184. 10(b)(iii) awards M1 for setting t = 0 and A1 for theta = -18.5 degrees Celsius. 10(c) awards M1 for setting theta = 4, A1 for t = 4.58 hours, and A1 for 4 hours 30 mins or 4 hours 40 mins (270 or 280 minutes).

Q Marking instructions AO Marks Typical solution

10(a) ln Ae−kt = ln 21( θ )

Obtains or θ = 21− Ae−kt

−kt 21 θ −kt

lne = ln or Ae = 21− θ

A 3.1a M1

ln Ae−kt = ln 21(− θ )

21 θ

−kt = ln −kt

A ln A + lne = ln 21(− θ)

ln Ae−kt = ln A + ln e−kt

Obtains ln A − kt = ln 21(− θ)

21 θ 1.1a M1

or ln = ln(21 θ) − ln A

A

Completes reasoned argument

to obtain ln 21(− θ) = −kt + ln A 2.1 R1

AG

Subtotal 3

10(b)(i) States ln A = 3.676 3.4 M1 ln A = 3.676

Obtains AWRT 39.5 A = e3.676 =

1.1b– A-LEVELA1 MATHEMATICS – 39.5 –

Subtotal 2

10(b)(ii) 3.676 3.676

Obtains −k = −

19.98 19.98

Or k = 0.1839...

Substitutes their value of A or 3.4 M1

= 0.184

ln A = 3.676 , t = 19.98 and

ln 21(−θ) = 0 to form an

equation for k

Obtains AWRT k = 0.184 1.1b A1

Subtotal 2

10(b)(iii) Uses the model θ = 21− Ae−kt θ = 21− 39.5e0

with their A or ln A = 3.676 = −18.5 C

and t = 0 3.4 M1

Or

Equates ln 21(− θ) to 3.676

Obtains AWRT −18.5 C

Condone missing units 1.1b A1

CAO

Subtotal 2

10(c) Uses either version of the model 11

with their A from (b)(i) or 4 = 21− 39.5e−0.184t

A = 3.4 M1

ln 3.676 , and their k from

t = 4.581996...

their (b)(ii) and θ = 4

Obtains AWRT 4.58 Time = 4 hrs 30 mins

PI by 4 hrs 30 mins (270min) or 1.1b A1

4 hrs 40 mins(280 min)

Obtains 4 hrs 30 mins or

4 hrs 40 mins

Accept 270 min or 280 min 3.2a A1

Answer must not come from an

incorrect value of t

Subtotal 3

Question 10 Total 12

How to answer it

Exponential Cooling Models & Linear Forms

📌 What this question tests

This question assesses your ability to manipulate exponential models and linearise non-linear data using natural logarithms (ln). Specifically, it covers:

  • Algebraic manipulation with logarithms: Applying laws of logarithms (product law and inverse functions) to convert a cooling model into the straight-line equation Y = mX + c .
  • Interpreting linearised graphs: Connecting the vertical intercept to ln A and the gradient to -k .
  • Model interpretation: Evaluating initial conditions ( t = 0 ) and solving for specific operational temperatures using logarithms.
  • Unit conversion & rounding: Converting decimal hours to hours and minutes rounded to the nearest 10 minutes.

Part (a) — Proof of Linear Form [3 Marks]

Show that θ = 21 - A e-kt rearranges to ln(21 - θ) = -kt + ln A

📐 Step-by-Step Proof

  1. Isolate the exponential term:
    θ = 21 - A e-kt
    A e-kt = 21 - θ
  2. Take the natural logarithm of both sides:
    ln(A e-kt) = ln(21 - θ)
  3. Apply the addition law of logs:
    ln(A) + ln(e-kt) = ln(21 - θ)
  4. Use ln(ex) = x and rearrange:
    ln A - kt = ln(21 - θ)
    ∴ ln(21 - θ) = -kt + ln A

💡 Key Laws Used

  • Product rule: ln(xy) = ln x + ln y
  • Inverse property: ln(ef(t)) = f(t)
  • Linear form comparison:
    Y = mX + c where:
    • Y = ln(21 - θ)
    • X = t
    • Gradient m = -k
    • Intercept c = ln A

🧠 Exam Technique (AG = Answer Given)

Because the target result is printed on the exam paper, you must show every single intermediate step clearly. Never skip the expansion of ln(A e-kt) into ln A + ln(e-kt) . Jumping straight to the final line loses the reasoning mark (R1).

❌ Common Errors

  • Taking logs term-by-term incorrectly: writing ln(θ) = ln(21) - ln(A e-kt) , which is mathematically invalid!
  • Forgetting to show how ln(e-kt) simplifies to -kt .
Mark Scheme Breakdown:
• M1: Rearranges correctly to obtain ln(A e-kt) = ln(21 ± θ) or divides by A first.
• M1: Correctly uses log laws to expand LHS to ln A + ln e-kt .
• R1: Completes fully reasoned argument to achieve the printed result.

Part (b)(i) & (b)(ii) — Finding Constants A and k [4 Marks]

Using the graph coordinates: (0, 3.676) and (19.98, 0)

📐 (b)(i) Finding Constant A

  1. The vertical intercept is at t = 0 , where ln(21 - θ) = 3.676 .
  2. From part (a), Intercept c = ln A :
    ln A = 3.676
  3. Solve for A by exponentiating:
    A = e3.676 = 39.488...
  4. Round to 3 significant figures:
    A = 39.5

📐 (b)(ii) Finding Constant k

  1. Method 1 (Gradient):
    Gradient m = (y₂ - y₁) / (x₂ - x₁)
    -k = (0 - 3.676) / (19.98 - 0)
    -k = -0.18398... ⇒ k = 0.184 (3 s.f.)
  2. Method 2 (Substitution):
    Substitute (19.98, 0) into the line equation:
    0 = -k(19.98) + 3.676
    19.98k = 3.676 ⇒ k = 0.184

✅ Correct Answers

  • (b)(i): A = 39.5 (AWRT 39.5)
  • (b)(ii): k = 0.184 (AWRT 0.184)

❌ Common Errors

  • Thinking the intercept directly gives A (i.e. stating A = 3.676 instead of ln A = 3.676 ).
  • Sign errors: stating k = -0.184 . The question notes that k is a positive constant, and the gradient is -k .
  • Premature rounding during intermediate steps.
Mark Scheme Breakdown:
• (b)(i) M1: Sets ln A = 3.676 | A1: A = 39.5 (AWRT)
• (b)(ii) M1: Calculates ± 3.676 / 19.98 or sets up equation at (19.98, 0) | A1: k = 0.184 (AWRT)

Part (b)(iii) — Initial Temperature [2 Marks]

Find the temperature of the dessert when initially removed from the freezer

📐 Step-by-Step Calculation

  1. "Initially removed" means at time t = 0.
  2. Substitute t = 0 into the original model:
    θ = 21 - A e-k(0) = 21 - A(1) = 21 - A
  3. Using A = 39.488... (or 39.5):
    θ = 21 - 39.488... = -18.488... °C
  4. Alternative method:
    At t = 0 , ln(21 - θ) = 3.676
    21 - θ = e3.676 = 39.488...
    θ = 21 - 39.488... = -18.5 °C

✅ Final Answer

Initial temperature = -18.5 °C (to 3 s.f.)

Note: Mark scheme condones missing °C units, but signs must be correct.

Mark Scheme Breakdown:
• M1: Uses model with t = 0 and their value of A, or equates ln(21 - θ) = 3.676 .
• A1: Obtains AWRT -18.5 °C (Correct Answer Only).

Part (c) — Time to Reach Serving Temperature [3 Marks]

Find the time taken to reach 4 °C, to the nearest 10 minutes

📐 Step-by-Step Solution

  1. Substitute θ = 4 into the formula:
    4 = 21 - 39.49e-0.1840t
  2. Rearrange to isolate the exponential:
    39.49e-0.1840t = 21 - 4 = 17
    e-0.1840t = 17 / 39.49 = 0.4305...
  3. Take logs and solve for t:
    -0.1840t = ln(17 / 39.49) = -0.8428...
    t = -0.8428... / -0.1840 = 4.581996... hours
  4. Convert decimal hours to minutes:
    0.581996... × 60 = 34.92 minutes
    Total time = 4 hours 34.9 minutes (or 274.9 minutes)
  5. Round to the nearest 10 minutes:
    34.9 mins rounds to 30 or 40 mins depending on rounding convention.
    • 4 hours 30 mins (270 mins) OR
    • 4 hours 40 mins (280 mins)

❌ Common Calculation Traps

  • Decimal Time Trap: Thinking 4.58 hours is 4 hours 58 minutes! Always multiply the decimal part by 60: 0.58 × 60 ≈ 35 minutes .
  • Premature Rounding: Using A = 39.5 and k = 0.184 directly gives t ≈ 4.579 hours , which still leads to 275 minutes. Keep full precision in calculator memory where possible.
  • Ignoring the Target Accuracy: Leaving the answer as 4.58 hours loses the final accuracy mark (A1). You must round to the nearest 10 minutes.
Mark Scheme Breakdown:
• M1: Sets up equation with θ = 4 , their A, and their k.
• A1: Obtains t = AWRT 4.58 hours (implied by 4 hrs 30 mins or 4 hrs 40 mins).
• A1: Final answer stated as 4 hours 30 mins, 4 hours 40 mins, 270 mins, or 280 mins.

Topics

Pure Mathematics · C: Coordinate geometry in the (x, y) plane · F: Exponentials and logarithms

Question and mark scheme from the AQA A-Level Mathematics examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.