Question 9 - A-Level Maths

AQA Paper 1 2021 June — Question 9 15 marks

Exam Board	AQA
Module	Paper 1 (Paper 1)
Year	2021
Session	June
Marks	15
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Exponential Equations & Modelling
Type	log(y) vs x: convert and interpret
Difficulty	Moderate -0.3 This is a standard log-linear modelling question requiring routine techniques: taking logs to linearize (part a is just algebraic manipulation), calculating log values, plotting points, finding gradient and intercept from a graph, and making predictions. While multi-part with several marks, each step follows a well-practiced procedure with no novel problem-solving required, making it slightly easier than average.
Spec	1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form 1.06i Exponential growth/decay: in modelling context 2.02c Scatter diagrams and regression lines 2.02d Informal interpretation of correlation

9 The table below shows the annual global production of plastics, $P$, measured in millions of tonnes per year, for six selected years.

Year	1980	1985	1990	1995	2000	2005
$\boldsymbol { P }$	75	94	120	156	206	260

It is thought that $P$ can be modelled by $$P = A \times 10 ^ { k t }$$ where $t$ is the number of years after 1980 and $A$ and $k$ are constants.
9

Show algebraically that the graph of $\log _ { 10 } P$ against $t$ should be linear.
9
(i) Complete the table below.
$\boldsymbol { t }$ 0 5 10 15 20 25
$\boldsymbol { \operatorname { l o g } } _ { \mathbf { 1 0 } } \boldsymbol { P }$ 1.88 1.97 2.08 2.31
9 (b) (ii) Plot $\log _ { 10 } P$ against $t$, and draw a line of best fit for the data. \includegraphics[max width=\textwidth, alt={}, center]{042e248a-9efa-4844-957d-f05715900ffc-13_1203_1308_360_367} 9
(i) Hence, show that $k$ is approximately 0.02
9 (c) (ii) Find the value of $A$.
9
Using the model with $k = 0.02$ predict the number of tonnes of annual global production of plastics in 2030. 9
Using the model with $k = 0.02$ predict the year in which $P$ first exceeds 8000
9
Give a reason why it may be inappropriate to use the model to make predictions about future annual global production of plastics. \includegraphics[max width=\textwidth, alt={}, center]{042e248a-9efa-4844-957d-f05715900ffc-15_2488_1716_219_153}

Show mark scheme Show mark scheme source

Question 9(a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Takes $\log_{10}$ of both sides to obtain $\log_{10} P = \log_{10}(A \times 10^{kt})$	M1 (1.1a)	Or states that $A = 10^c$
$\log_{10} P = \log_{10} A + \log_{10} 10^{kt}$, or $P = 10^{kt+c}$	A1 (1.1b)
Completes rigorous argument to show $\log_{10} P = \log_{10} A + kt$, or $\log_{10} P = kt + c$	R1 (2.1)

Subtotal: 3 marks

Question 9(b)(i):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Completes table: $t$: 0, 5, 10, 15, 20, 25; $\log_{10} P$: 1.88, 1.97, 2.08, 2.19, 2.31, 2.41	B1 (1.1b)

Subtotal: 1 mark

Question 9(b)(ii):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Plots at least four points correctly	M1 (1.1a)	Allow $\pm$ one small square
Draws a ruled line of best fit from $t=0$ to $t=25$ or better	A1 (1.1b)	CSO

Subtotal: 2 marks

Question 9(c)(i):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Calculates gradient: $k = \frac{2.41 - 1.88}{25} = 0.0212 \approx 0.02$	M1 (1.1a)	Using line of best fit or two points from table of values
Obtains a value of $k$ which rounds to $0.02$	R1 (1.1b)

Subtotal: 2 marks

Question 9(c)(ii):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$A = 75$	B1F (2.2b)	Infers value of $A$; uses 75 from data or uses $10^{\text{their intercept}}$

Subtotal: 1 mark

Question 9(d):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Substitutes $t = 50$ into $P = A \times 10^{0.02t}$: $P = 75 \times 10^{0.02 \times 50}$	M1 (3.4)	PI by 750
750 million tonnes	A1F (3.2a)	Follow through on their $70 < A < 90$

Subtotal: 2 marks

Question 9(e):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Forms equation: $8000 = 75 \times 10^{0.02t}$, $t = 101.401$	M1 (3.4)	Using their model $P = A \times 10^{0.02t}$ and $P = 8000$
$t = 101.4$; AWFW [97.44, 102.90]	A1F (1.1b)
2082	A1F (3.2a)	Interprets as year: integer part of $t$ + 1980 + 1, provided their $t > 50$

Subtotal: 3 marks

Question 9(f):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
E.g. "The world will produce less plastics to be more environmentally friendly"	E1 (3.5b)	Gives a reason in context why the model for production of plastics will be inappropriate; e.g. it is not appropriate to extrapolate future global production from data provided; global production may decrease in future

Subtotal: 1 mark

Question 9 Total: 15 marks

## Question 9(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Takes $\log_{10}$ of both sides to obtain $\log_{10} P = \log_{10}(A \times 10^{kt})$ | M1 (1.1a) | Or states that $A = 10^c$ |
| $\log_{10} P = \log_{10} A + \log_{10} 10^{kt}$, or $P = 10^{kt+c}$ | A1 (1.1b) | |
| Completes rigorous argument to show $\log_{10} P = \log_{10} A + kt$, or $\log_{10} P = kt + c$ | R1 (2.1) | |

**Subtotal: 3 marks**

---

## Question 9(b)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Completes table: $t$: 0, 5, 10, 15, 20, 25; $\log_{10} P$: 1.88, 1.97, 2.08, 2.19, 2.31, 2.41 | B1 (1.1b) | |

**Subtotal: 1 mark**

---

## Question 9(b)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Plots at least four points correctly | M1 (1.1a) | Allow $\pm$ one small square |
| Draws a ruled line of best fit from $t=0$ to $t=25$ or better | A1 (1.1b) | CSO |

**Subtotal: 2 marks**

---

## Question 9(c)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Calculates gradient: $k = \frac{2.41 - 1.88}{25} = 0.0212 \approx 0.02$ | M1 (1.1a) | Using line of best fit or two points from table of values |
| Obtains a value of $k$ which rounds to $0.02$ | R1 (1.1b) | |

**Subtotal: 2 marks**

---

## Question 9(c)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $A = 75$ | B1F (2.2b) | Infers value of $A$; uses 75 from data or uses $10^{\text{their intercept}}$ |

**Subtotal: 1 mark**

---

## Question 9(d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $t = 50$ into $P = A \times 10^{0.02t}$: $P = 75 \times 10^{0.02 \times 50}$ | M1 (3.4) | PI by 750 |
| 750 million tonnes | A1F (3.2a) | Follow through on their $70 < A < 90$ |

**Subtotal: 2 marks**

---

## Question 9(e):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Forms equation: $8000 = 75 \times 10^{0.02t}$, $t = 101.401$ | M1 (3.4) | Using their model $P = A \times 10^{0.02t}$ and $P = 8000$ |
| $t = 101.4$; AWFW [97.44, 102.90] | A1F (1.1b) | |
| 2082 | A1F (3.2a) | Interprets as year: integer part of $t$ + 1980 + 1, provided their $t > 50$ |

**Subtotal: 3 marks**

---

## Question 9(f):

| Answer/Working | Mark | Guidance |
|---|---|---|
| E.g. "The world will produce less plastics to be more environmentally friendly" | E1 (3.5b) | Gives a reason in context why the model for production of plastics will be inappropriate; e.g. it is not appropriate to extrapolate future global production from data provided; global production may decrease in future |

**Subtotal: 1 mark**

**Question 9 Total: 15 marks**

Show LaTeX source

9 The table below shows the annual global production of plastics, $P$, measured in millions of tonnes per year, for six selected years.

\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
Year & 1980 & 1985 & 1990 & 1995 & 2000 & 2005 \\
\hline
$\boldsymbol { P }$ & 75 & 94 & 120 & 156 & 206 & 260 \\
\hline
\end{tabular}
\end{center}

It is thought that $P$ can be modelled by

$$P = A \times 10 ^ { k t }$$

where $t$ is the number of years after 1980 and $A$ and $k$ are constants.\\
9
\begin{enumerate}[label=(\alph*)]
\item Show algebraically that the graph of $\log _ { 10 } P$ against $t$ should be linear.\\

9
\item (i) Complete the table below.

\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
$\boldsymbol { t }$ & 0 & 5 & 10 & 15 & 20 & 25 \\
\hline
$\boldsymbol { \operatorname { l o g } } _ { \mathbf { 1 0 } } \boldsymbol { P }$ & 1.88 & 1.97 & 2.08 &  & 2.31 &  \\
\hline
\end{tabular}
\end{center}

9 (b) (ii) Plot $\log _ { 10 } P$ against $t$, and draw a line of best fit for the data.\\
\includegraphics[max width=\textwidth, alt={}, center]{042e248a-9efa-4844-957d-f05715900ffc-13_1203_1308_360_367}

9
\item (i) Hence, show that $k$ is approximately 0.02\\

9 (c) (ii) Find the value of $A$.\\

9
\item Using the model with $k = 0.02$ predict the number of tonnes of annual global production of plastics in 2030.

9
\item Using the model with $k = 0.02$ predict the year in which $P$ first exceeds 8000\\

9
\item Give a reason why it may be inappropriate to use the model to make predictions about future annual global production of plastics.\\

\includegraphics[max width=\textwidth, alt={}, center]{042e248a-9efa-4844-957d-f05715900ffc-15_2488_1716_219_153}
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 1 2021 Q9 [15]}}

This paper (15 questions)

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Q1 1 Q2 1 Q3 1 Q4 1 Q5 6 Q6 7 Q7 7 Q8 9 Q9 15 Q10 8 Q11 8 Q12 8 Q13 8 Q14 10 Q15 10

\(\boldsymbol { t }\)	0	5	10	15	20	25
\(\boldsymbol { \operatorname { l o g } } _ { \mathbf { 1 0 } } \boldsymbol { P }\)	1.88	1.97	2.08		2.31