AQA Paper 3 2022 June — Question 7 7 marks

Exam BoardAQA
ModulePaper 3 (Paper 3)
Year2022
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Equations & Modelling
Typeln(y) vs ln(x) linear graph
DifficultyStandard +0.3 This is a standard log-linear regression question requiring students to find the equation of a line from a graph, convert between logarithmic and exponential forms, and apply the model. While it involves multiple steps (7 marks total), each step uses routine A-level techniques: reading coordinates from a graph, calculating gradient/intercept, using log laws (10^(log T) = T), and substituting into a formula. No novel insight or complex problem-solving is required—this is slightly easier than average due to its formulaic nature.
Spec1.06f Laws of logarithms: addition, subtraction, power rules1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form
A planet takes \(T\) days to complete one orbit of the Sun. \(T\) is known to be related to the planet's average distance \(d\), in millions of kilometres, from the Sun. A graph of \(\log_{10} T\) against \(\log_{10} d\) is shown with data for Mercury and Uranus labelled. \includegraphics{figure_7}
    1. Find the equation of the straight line in the form $$\log_{10} T = a + b \log_{10} d$$ where \(a\) and \(b\) are constants to be found. [3 marks]
    2. Show that $$T = K d^n$$ where K and n are constants to be found. [2 marks]
  2. Neptune takes approximately 60 000 days to complete one orbit of the Sun. Use your answer to 7(a)(ii) to find an estimate for the average distance of Neptune from the Sun. [2 marks]
Question 7:
AnswerMarks
7(a)(i)Forms a correct expression for
the gradient or sets up two
correct simultaneous equations
PI by a = 0.7 or b = 1.5
Ignore missing labels
AnswerMarks Guidance
−1.1a M1
=1.5
3.46−1.76
log T −1.94 =1.5 ( log d −1.76 )
10 10
log T = −0.7+1.5log d
10 10
Obtains a = 0.7 or b = 1.5
OE
−
AnswerMarks Guidance
Ignore miss i ng labels1.1b A1
Obtains a = 0.7 and b = 1.5 or
seen in the logarithmic equation
−
AnswerMarks Guidance
ISW1.1b A1
Subtotal3
QMarking instructions AO
AnswerMarks
7(a)(ii)Uses one law of logarithm
correctly
Allow use of original equation
without values for a and b
AnswerMarks Guidance
If values are used, a ≠ 03.3 M1
10 10
 T 
log   =−0.7
10 d1.5 
T
=10 −0.7
d1.5
T =10 −0.7×d1.5
Completes reasoned argument
to obtain T = Kdn with
K = 10-0.7 or AWRT 0.2 and
n = 1.5
ISW
AnswerMarks Guidance
Must come from correct working2.1 R1
Subtotal2
QMarking instructions AO
AnswerMarks
7(b)Forms an equation using their
answer to (a)(ii) with K > 0 and
n > 0 and T =60000
Must only have unknown d in
AnswerMarks Guidance
the equation3.4 M1
d =4488.5
Average distance is approximately
4500 million kilometres
Obtains AWRT 4500 million
kilometres
ACF with units
For example
• 4.5 × 109 kilometres
• 4500 × 106 kilometres
• 4.5 × 1012 metres
AnswerMarks Guidance
• 4500 × 109 metres3.2a A1
Subtotal2
Question 7 Total7
QMarking instructions AO 
Question 7:
--- 7(a)(i) ---
7(a)(i) | Forms a correct expression for
the gradient or sets up two
correct simultaneous equations
PI by a = 0.7 or b = 1.5
Ignore missing labels
− | 1.1a | M1 | 4.49−1.94
=1.5
3.46−1.76
log T −1.94 =1.5 ( log d −1.76 )
10 10
log T = −0.7+1.5log d
10 10
Obtains a = 0.7 or b = 1.5
OE
−
Ignore miss i ng labels | 1.1b | A1
Obtains a = 0.7 and b = 1.5 or
seen in the logarithmic equation
−
ISW | 1.1b | A1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 7(a)(ii) ---
7(a)(ii) | Uses one law of logarithm
correctly
Allow use of original equation
without values for a and b
If values are used, a ≠ 0 | 3.3 | M1 | log T −log d1.5 =−0.7
10 10
 T 
log   =−0.7
10 d1.5 
T
=10 −0.7
d1.5
T =10 −0.7×d1.5
Completes reasoned argument
to obtain T = Kdn with
K = 10-0.7 or AWRT 0.2 and
n = 1.5
ISW
Must come from correct working | 2.1 | R1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 7(b) ---
7(b) | Forms an equation using their
answer to (a)(ii) with K > 0 and
n > 0 and T =60000
Must only have unknown d in
the equation | 3.4 | M1 | 60000=0.2×d1.5
d =4488.5
Average distance is approximately
4500 million kilometres
Obtains AWRT 4500 million
kilometres
ACF with units
For example
• 4.5 × 109 kilometres
• 4500 × 106 kilometres
• 4.5 × 1012 metres
• 4500 × 109 metres | 3.2a | A1
Subtotal | 2
Question 7 Total | 7
Q | Marking instructions | AO | Marks | Typical solution
A planet takes $T$ days to complete one orbit of the Sun.

$T$ is known to be related to the planet's average distance $d$, in millions of kilometres, from the Sun.

A graph of $\log_{10} T$ against $\log_{10} d$ is shown with data for Mercury and Uranus labelled.

\includegraphics{figure_7}

\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the equation of the straight line in the form
$$\log_{10} T = a + b \log_{10} d$$
where $a$ and $b$ are constants to be found.
[3 marks]

\item Show that
$$T = K d^n$$
where K and n are constants to be found.
[2 marks]
\end{enumerate}

\item Neptune takes approximately 60 000 days to complete one orbit of the Sun.

Use your answer to 7(a)(ii) to find an estimate for the average distance of Neptune from the Sun.
[2 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 3 2022 Q7 [7]}}
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