AQA AS Paper 2 2018 June — Question 12 8 marks

Exam BoardAQA
ModuleAS Paper 2 (AS Paper 2)
Year2018
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Equations & Modelling
TypeFinding x from given y value
DifficultyStandard +0.3 This is a straightforward exponential modelling question requiring substitution of values to form simultaneous equations, basic algebraic manipulation, and interpretation. Part (a) involves routine algebra with exponentials, (b) is direct substitution, (c) requires solving e^{0.1t} = e^{0.2t} which simplifies to a basic logarithm, and (d) asks for a standard critique of exponential models. All techniques are standard AS-level with no novel insight required, making it slightly easier than average.
Spec1.06a Exponential function: a^x and e^x graphs and properties1.06g Equations with exponentials: solve a^x = b1.06i Exponential growth/decay: in modelling context

Trees in a forest may be affected by one of two types of fungal disease, but not by both. The number of trees affected by disease A, \(n_A\), can be modelled by the formula $$n_A = ae^{0.1t}$$ where \(t\) is the time in years after 1 January 2017. The number of trees affected by disease B, \(n_B\), can be modelled by the formula $$n_B = be^{0.2t}$$ On 1 January 2017 a total of 290 trees were affected by a fungal disease. On 1 January 2018 a total of 331 trees were affected by a fungal disease.
  1. Show that \(b = 90\), to the nearest integer, and find the value of \(a\). [3 marks]
  2. Estimate the total number of trees that will be affected by a fungal disease on 1 January 2020. [1 mark]
  3. Find the year in which the number of trees affected by disease B will first exceed the number affected by disease A. [3 marks]
  4. Comment on the long-term accuracy of the model. [1 mark]

Question 12:

AnswerMarks
12(a)Uses model to form one correct
equation (PI by a=200) (ACF)
AnswerMarks Guidance
Accept 1.105a + 1.221b = 331AO3.1b M1
331 = ae0.1 + be0.2
290e0.1 = ae0.1 + be0.1
(331βˆ’290𝑒0.1)
b = = 90.3 = 90 to the
(𝑒0.2βˆ’π‘’0.1)
nearest integer
so a = 200
Forms a second correct equation
AnswerMarks Guidance
(ACF)AO1.1a M1
Obtains correct a AWRT 200 and b
AWRT 90 (AG)
Only award if both previous M1’s
achieved. Do not award marks
retrospectively for correct values of
AnswerMarks Guidance
a and b used in part (b)AO1.1b A1
(b)Substitutes t = 3 and evaluates
CAOAO3.4 B1
(c)Forms inequality (accept < or >)
(condone use of equation) FT
AnswerMarks Guidance
β€˜their’ value of a, but b must be 90AO1.1a M1
e0.1t > 200
90
200
0.1t > ln ( )
90
200
t > 10 ln ( ) = 7.985
90
Just less than 8 so during 2024
Uses logs or calculator to solve
β€˜their’ inequality (or equation)
If using trial and error must see t=7
AnswerMarks Guidance
and t=8 testedAO1.1a M1
Interprets final result. (Do not
AnswerMarks Guidance
accept 2025)AO3.2a A1
(d)Gives one limitation of the model.
Eg. Model must break down as
both n and n will tend to infinity /
A B
model assumes nothing changes /
no attempt to control the diseases /
all the trees have died / finite
number of trees / cure for the
disease might be found / other
factors such as drought could
AnswerMarks Guidance
affect the model / etc.AO3.5b E1
so the model will no longer be
accurate.
AnswerMarks Guidance
Total8
QMarking Instructions AO
Question 12:
--- 12(a) ---
12(a) | Uses model to form one correct
equation (PI by a=200) (ACF)
Accept 1.105a + 1.221b = 331 | AO3.1b | M1 | 290 = a + b
331 = ae0.1 + be0.2
290e0.1 = ae0.1 + be0.1
(331βˆ’290𝑒0.1)
b = = 90.3 = 90 to the
(𝑒0.2βˆ’π‘’0.1)
nearest integer
so a = 200
Forms a second correct equation
(ACF) | AO1.1a | M1
Obtains correct a AWRT 200 and b
AWRT 90 (AG)
Only award if both previous M1’s
achieved. Do not award marks
retrospectively for correct values of
a and b used in part (b) | AO1.1b | A1
(b) | Substitutes t = 3 and evaluates
CAO | AO3.4 | B1 | 200e0.3 + 90e0.6 = 434
(c) | Forms inequality (accept < or >)
(condone use of equation) FT
β€˜their’ value of a, but b must be 90 | AO1.1a | M1 | 90e0.2t > 200e0.1t
e0.1t > 200
90
200
0.1t > ln ( )
90
200
t > 10 ln ( ) = 7.985
90
Just less than 8 so during 2024
Uses logs or calculator to solve
β€˜their’ inequality (or equation)
If using trial and error must see t=7
and t=8 tested | AO1.1a | M1
Interprets final result. (Do not
accept 2025) | AO3.2a | A1
(d) | Gives one limitation of the model.
Eg. Model must break down as
both n and n will tend to infinity /
A B
model assumes nothing changes /
no attempt to control the diseases /
all the trees have died / finite
number of trees / cure for the
disease might be found / other
factors such as drought could
affect the model / etc. | AO3.5b | E1 | Eventually all of the trees will die
so the model will no longer be
accurate.
Total | 8
Q | Marking Instructions | AO | Marks | Typical Solution
Trees in a forest may be affected by one of two types of fungal disease, but not by both.

The number of trees affected by disease A, $n_A$, can be modelled by the formula

$$n_A = ae^{0.1t}$$

where $t$ is the time in years after 1 January 2017.

The number of trees affected by disease B, $n_B$, can be modelled by the formula

$$n_B = be^{0.2t}$$

On 1 January 2017 a total of 290 trees were affected by a fungal disease.

On 1 January 2018 a total of 331 trees were affected by a fungal disease.

\begin{enumerate}[label=(\alph*)]
\item Show that $b = 90$, to the nearest integer, and find the value of $a$. [3 marks]

\item Estimate the total number of trees that will be affected by a fungal disease on 1 January 2020. [1 mark]

\item Find the year in which the number of trees affected by disease B will first exceed the number affected by disease A. [3 marks]

\item Comment on the long-term accuracy of the model. [1 mark]
\end{enumerate}

\hfill \mbox{\textit{AQA AS Paper 2 2018 Q12 [8]}}