AQA AS Level Mathematics Paper 1, June 2025: Question 13

9 marks · Medium difficulty · Modelling

Use an exponential decay model for points scored in a computer game to interpret parameters, linearise using logarithms, determine constants from a line of best fit, and solve for time given a specific score.

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Question

Question text

13 A computer game awards points to a player based on the time taken to complete

a level.

The points awarded decrease as the time taken to complete a level increases.

Rebekah believes that the points awarded, P, can be modelled by the equation

P = Aekt

where t is the time, in seconds, taken to complete the level and A and k are constants.

13 (a) Explain, in context, the meaning of the value of A

[1 mark]

13 (b) Show that

In P = In A + kt

[2 marks]

13 (c) Rebekah records the points and the time taken for her to complete each level.

She plots the values of ln P against t

Rebekah obtains a straight‑line graph with a gradient of –0.08 and a vertical intercept

of 5.30, as shown in the diagram.

In P

5.30

(17)

Find the value of A and the value of k

[3 marks]

13 (d) Rebekah scores 20 points for completing a particular level.

Find, to the nearest second, the time taken to complete this level.

[3 marks]

(18) …

END OF SECTION A

Mark scheme

Show the mark scheme

Q Marking instructions AO Marks Typical solution

13(a) Explains that A is the number of 3.2a E1 The points available at the start of

points at the start of the level. the level

Must be in context.

Accept ‘initial points’ or

‘maximum points’

Subtotal 1

13(b) Takes natural log of both sides 1.1a M1 P = Aekt

and correctly uses addition law kt

of logs ln P = ln Ae

Completes reasoned argument 2.1 R1 ln P = ln A + ln ekt

to show given result

ln P = ln A+ kt

Subtotal 2

13(c) States k = –0.08 2.2a B1 Gradient = k ⇒ k = −0 08.

Equates ln A to 5.30 2.2a M1

ln A = 5 30.

Obtains A = AWRT 200 1.1b A1

A = e5 30.

A = 200

Subtotal 3

13(d) Substitutes their values of A and 3.4 M1 200e−0 08.t = 20

k appropriately into the model

−0 08.t 1

PI by 5.3 – 0.08t e =

Obtains a correct equation in t 1.1b A1 10

Obtains t = 28 or AWRT 29 1.1b A1 1

−0 08.t = ln

t = 28 782.

∴time is 29 seconds

Subtotal 3

Question 13 Total 9

How to answer it

Exponential Modelling and Linear Reduction

📌 What this question tests

Contextual interpretation: Understanding the physical meaning of initial values (constants) in exponential models where t = 0.
Laws of logarithms: Applying natural logarithms ( ln ) to linearise an exponential equation of the form y = Ae^kt into Y = mX + c.
Connecting models to graphs: Relating the gradient and y-intercept of a straight line to algebraic constants.
Solving exponential equations: Rearranging equations, calculating values using logarithmic operations, and rounding to required units/accuracy.

Part (a) — Context of Parameter A

Interpreting Constant A

1 Mark • Assessment Objective 3.2a

✅ Model Answer

A represents the number of points available at the start of the level (or the initial points / maximum points).

💡 Key Knowledge

When t = 0 seconds (the start of the level):

P = Ae^k(0) = Ae⁰ = A × 1 = A

Hence, A is always the initial value of the dependent variable.

🧠 Exam Technique

The question specifies "in context". You will not receive marks for stating purely mathematical definitions such as "the y-intercept" or "the value when t is 0". You must explicitly refer to points and the start of the level.

❌ Common Errors

Writing "the rate of points decrease" (confusing A with the decay constant k).
Omitting context: writing only "the initial amount" without mentioning points or the game.

Mark Scheme Breakdown:
E1: Explains that A is the number of points at the start of the level (must be in context; accept 'initial points' or 'maximum points').

Part (b) — Proof / Logarithmic Laws

Linearising the Exponential Model

2 Marks • Assessment Objectives 1.1a, 2.1

📐 Step-by-Step Proof

Take natural logarithms of both sides:
ln P = ln(Ae^kt)
Apply the addition law of logs, ln(xy) = ln x + ln y:
ln P = ln A + ln(e^kt)
Use the inverse property ln(e^x) = x:
ln P = ln A + kt

🧠 Exam Technique

In a "Show that" question, you cannot skip steps. You must clearly show the intermediate step ln A + ln(e^kt) before writing the final result. Going directly from ln(Ae^kt) to ln A + kt will lose the final reasoning mark.

❌ Common Errors

Writing ln(Ae^kt) = kt ln(A) , incorrectly applying the power rule before separating factors.
Forgetting brackets around Ae^kt when taking the log of the right-hand side.

Mark Scheme Breakdown:
M1 (1.1a): Takes natural logs of both sides and correctly applies the addition law of logarithms.
R1 (2.1): Completes a fully reasoned argument to arrive cleanly at the given equation.

Part (c) — Finding Parameters from Graph

Determining Constants A and k

3 Marks • Assessment Objectives 2.2a, 1.1b

📐 Step-by-Step Calculation

Match the linearised equation to y = mx + c:
Here, the vertical axis is ln P (corresponding to y) and the horizontal axis is t (corresponding to x):
ln P = kt + ln A &longleftrightarrow; y = mx + c
Identify the gradient:
Gradient m = k
Given gradient = −0.08 ⇒ k = -0.08
Identify the vertical intercept:
Vertical intercept c = ln A
Given intercept = 5.30 ⇒ ln A = 5.30
Solve for A using the inverse of ln (base e):
A = e^5.30 = 200.3368... ⇒ A = 200 (to 3 s.f.)

✅ Final Values

k = −0.08

A = 200 (accept values rounding to 200, e.g. 200.3)

❌ Common Calculation Traps

Sign Error: Forgetting the negative sign on k. Since points decrease as time increases, k must be negative ( k = -0.08 ).
Base confusion: Calculating 10^5.30 instead of e^5.30 because of confusing ln with log₁₀ .

Mark Scheme Breakdown:
B1 (2.2a): Correctly states k = -0.08 .
M1 (2.2a): Equates ln A to 5.30.
A1 (1.1b): Obtains A = 200 (AWRT 200).

Part (d) — Problem Solving with the Model

Calculating Completion Time for 20 Points

3 Marks • Assessment Objectives 3.4, 1.1b

📐 Step-by-Step Calculation

Method 1: Using the exponential form P = Ae^kt

Substitute known values into the equation:
P = 20 , A = 200 , k = -0.08
20 = 200e^-0.08t
Rearrange to isolate the exponential term:
e^-0.08t = 20 / 200 = 0.1 (or 1/10)
Take natural logarithms of both sides:
-0.08t = ln(0.1)
Solve for t:
t = ln(0.1) / (-0.08) = -2.302585... / (-0.08) = 28.7823...
Round to the nearest second:
t = 29 seconds

Method 2: Using the linearised equation ln P = ln A + kt

Substitute: ln(20) = 5.30 - 0.08t
Rearrange: 0.08t = 5.30 - ln(20) = 5.30 - 2.99573... = 2.30426...
Solve: t = 2.30426... / 0.08 = 28.803... ⇒ t = 29 seconds

✅ Final Answer

Time taken = 29 seconds (or 28 seconds depending on rounding in intermediate values; AWRT 29).

🧠 Exam Technique

Always re-read the final line of the question to confirm the required level of precision. Here, it explicitly says "to the nearest second". Leaving your answer as 28.8 s would lose the final mark!

Mark Scheme Breakdown:
M1 (3.4): Substitutes values of A and k appropriately into the model (implied by 5.3 - 0.08t or 200e^-0.08t = 20 ).
A1 (1.1b): Obtains a correct unsimplified or simplified equation in t.
A1 (1.1b): Obtains t = 28 or AWRT 29 (final value rounded to nearest integer = 29).

Topics

Pure Mathematics · F: Exponentials and logarithms

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