Edexcel Paper 2 2024 June — Question 13 9 marks

Exam BoardEdexcel
ModulePaper 2 (Paper 2)
Year2024
SessionJune
Marks9
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Mark schemeDownload PDF ↗
TopicExponential Equations & Modelling
Typelog(y) vs x: convert and interpret
DifficultyStandard +0.3 This is a standard log-linear modelling question requiring students to convert between exponential and linear forms using logarithms. The steps are routine: use intercept to find a (10^0.81), use gradient to find b (10^0.0054), interpret constants, and substitute t=26. Part (d) requires basic understanding of model limitations. Slightly easier than average due to clear scaffolding and standard technique.
Spec1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form

  1. The world human population, \(P\) billions, is modelled by the equation
$$P = a b ^ { t }$$ where \(a\) and \(b\) are constants and \(t\) is the number of years after 2004
Using the estimated population figures for the years from 2004 to 2007, a graph is plotted of \(\log _ { 10 } P\) against \(t\). The points lie approximately on a straight line with
  • gradient 0.0054
  • intercept 0.81 on the \(\log _ { 10 } P\) axis
    1. Estimate, to 3 decimal places, the value of \(a\) and the value of \(b\).
In the context of the model,
    1. interpret the value of the constant \(a\),
    2. interpret the value of the constant \(b\).
  • Use the model to estimate the world human population in 2030
  • Comment on the reliability of the answer to part (c).

  • Question 13:
    Part (a):
    AnswerMarks Guidance
    Answer/WorkingMark Guidance
    \(\log_{10} b = 0.0054 \Rightarrow b = 10^{0.0054}\) or \(\log_{10} a = 0.81 \Rightarrow a = 10^{0.81}\)M1 Must use base 10; may be implied by \(a \approx\) awrt 6.46 or \(b \approx\) awrt 1.01
    \(b = 1.01\) or \(a = 6.46\)A1 Allow 3 sf
    \(\log_{10} b = 0.0054 \Rightarrow b = 10^{0.0054}\) and \(\log_{10} a = 0.81 \Rightarrow a = 10^{0.81}\)M1 Both values; may be implied if no incorrect work seen
    \(b = 1.013\) and \(a = 6.457\)A1 Requires \(a =\) awrt 6.457 and \(b =\) awrt 1.013; isw once correct answers seen
    Special case - constants wrong way round: \(a = 1.013\) and \(b = 6.457\) scores M1A1M1A0 unless equation formed correctly.
    Part (b)(i):
    AnswerMarks Guidance
    Answer/WorkingMark Guidance
    e.g. The world population in billions in 2004B1ft Follow through their \(a\); must reference "billions"; allow "original/initial population in billions"
    Part (b)(ii):
    AnswerMarks Guidance
    Answer/WorkingMark Guidance
    \(b = 1.013\) represents the scale factor of the yearly increase in the world populationB1ft Follow through their \(b\); must reference "each year" or "yearly"; allow "proportional increase/change in each year", "population rises by 1.3% each year", "multiplier representing year on year increase"
    Do not accept: "the amount it is rising", "how much it is rising", "the rate the population increases", "the percentage increase each year", "the rate of increase in billions annually"
    Part (c):
    AnswerMarks Guidance
    Answer/WorkingMark Guidance
    \(P = 6.457...(1.013...)^{26}\) or \(\log P = 0.81 + 26\times 0.0054 \Rightarrow P = ...\)M1 Substitutes \(t = 25, 26\) or \(27\); must use their \(a\) and \(b\) correctly in \(P = ab^t\)
    awrt 9 billionA1 From correct model; condone incorrect/premature rounding leading to awrt 9 billion; allow \(9\,000\,000\,000\) or \(9\times10^9\); just "9" without "billions" is A0
    Part (d):
    AnswerMarks Guidance
    Answer/WorkingMark Guidance
    Not reliable since data used for model covered 2004–2007 and it would not be sensible to assume the model still holds in 2030B1 Must refer to unreliability and reference data being a long way from 2030
    Accept: "Not good as 2030 is a long way from 2004–2007", "unreliable as based on old data", "questionable as extrapolated over a long time", "not reliable due to how far we have extrapolated"
    Do not accept: "unreliable, extrapolation" (alone), "not good as outside the range", "disease may happen", "reliable as based on old data"
    ## Question 13:
    
    ### Part (a):
    
    | Answer/Working | Mark | Guidance |
    |---|---|---|
    | $\log_{10} b = 0.0054 \Rightarrow b = 10^{0.0054}$ **or** $\log_{10} a = 0.81 \Rightarrow a = 10^{0.81}$ | M1 | Must use base 10; may be implied by $a \approx$ awrt 6.46 or $b \approx$ awrt 1.01 |
    | $b = 1.01$ **or** $a = 6.46$ | A1 | Allow 3 sf |
    | $\log_{10} b = 0.0054 \Rightarrow b = 10^{0.0054}$ **and** $\log_{10} a = 0.81 \Rightarrow a = 10^{0.81}$ | M1 | Both values; may be implied if no incorrect work seen |
    | $b = 1.013$ **and** $a = 6.457$ | A1 | Requires $a =$ awrt 6.457 and $b =$ awrt 1.013; isw once correct answers seen |
    
    **Special case - constants wrong way round:** $a = 1.013$ **and** $b = 6.457$ scores M1A1M1A0 unless equation formed correctly.
    
    ---
    
    ### Part (b)(i):
    
    | Answer/Working | Mark | Guidance |
    |---|---|---|
    | e.g. The world population **in billions** in 2004 | B1ft | Follow through their $a$; must reference "billions"; allow "original/initial population in billions" |
    
    ---
    
    ### Part (b)(ii):
    
    | Answer/Working | Mark | Guidance |
    |---|---|---|
    | $b = 1.013$ represents the scale factor of the yearly increase in the world population | B1ft | Follow through their $b$; must reference "each year" or "yearly"; allow "proportional increase/change in each year", "population rises by 1.3% each year", "multiplier representing year on year increase" |
    
    Do **not** accept: "the amount it is rising", "how much it is rising", "the rate the population increases", "the percentage increase each year", "the rate of increase in billions annually"
    
    ---
    
    ### Part (c):
    
    | Answer/Working | Mark | Guidance |
    |---|---|---|
    | $P = 6.457...(1.013...)^{26}$ or $\log P = 0.81 + 26\times 0.0054 \Rightarrow P = ...$ | M1 | Substitutes $t = 25, 26$ or $27$; must use **their** $a$ and $b$ correctly in $P = ab^t$ |
    | awrt 9 billion | A1 | From correct model; condone incorrect/premature rounding leading to awrt 9 billion; allow $9\,000\,000\,000$ or $9\times10^9$; just "9" without "billions" is A0 |
    
    ---
    
    ### Part (d):
    
    | Answer/Working | Mark | Guidance |
    |---|---|---|
    | Not reliable since data used for model covered 2004–2007 and it would not be sensible to assume the model still holds in 2030 | B1 | Must refer to unreliability **and** reference data being a long way from 2030 |
    
    Accept: "Not good as 2030 is a long way from 2004–2007", "unreliable as based on old data", "questionable as extrapolated over a long time", "not reliable due to how far we have extrapolated"
    
    Do **not** accept: "unreliable, extrapolation" (alone), "not good as outside the range", "disease may happen", "reliable as based on old data"
    
    ---
    \begin{enumerate}
      \item The world human population, $P$ billions, is modelled by the equation
    \end{enumerate}
    
    $$P = a b ^ { t }$$
    
    where $a$ and $b$ are constants and $t$ is the number of years after 2004\\
    Using the estimated population figures for the years from 2004 to 2007, a graph is plotted of $\log _ { 10 } P$ against $t$.
    
    The points lie approximately on a straight line with
    
    \begin{itemize}
      \item gradient 0.0054
      \item intercept 0.81 on the $\log _ { 10 } P$ axis\\
    (a) Estimate, to 3 decimal places, the value of $a$ and the value of $b$.
    \end{itemize}
    
    In the context of the model,\\
    (b) (i) interpret the value of the constant $a$,\\
    (ii) interpret the value of the constant $b$.\\
    (c) Use the model to estimate the world human population in 2030\\
    (d) Comment on the reliability of the answer to part (c).
    
    \hfill \mbox{\textit{Edexcel Paper 2 2024 Q13 [9]}}