AQA Paper 1 2018 June — Question 10 8 marks

Exam BoardAQA
ModulePaper 1 (Paper 1)
Year2018
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Functions
TypeTime to reach target in exponential model
DifficultyModerate -0.3 This is a straightforward application of the exponential decay model requiring students to find k from the half-life, then solve m = m₀e^(-kt) for t. While it involves logarithms and multi-step working, it follows a standard textbook pattern with clear given information and no conceptual surprises, making it slightly easier than average.
Spec1.06i Exponential growth/decay: in modelling context

10 A scientist is researching the effects of caffeine. She models the mass of caffeine in the body using $$m = m _ { 0 } \mathrm { e } ^ { - k t }$$ where \(m _ { 0 }\) milligrams is the initial mass of caffeine in the body and \(m\) milligrams is the mass of caffeine in the body after \(t\) hours. On average, it takes 5.7 hours for the mass of caffeine in the body to halve.
One cup of strong coffee contains 200 mg of caffeine.
10
  1. The scientist drinks two strong cups of coffee at 8 am. Use the model to estimate the mass of caffeine in the scientist's body at midday.
    10
  2. The scientist wants the mass of caffeine in her body to stay below 480 mg
    10 (b)
    Use the model to find the earliest time
    coffee.
    Give your answer to the nearest minute

Question 10:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(200 = 400e^{-kx5.7}\)M1 (AO3.4) Uses model to form equation with \(t=5.7\), \(m = \frac{1}{2}m_0\)
\(k = 0.1216047\ldots\)A1 (AO1.1b) Obtains correct value of \(k\)
\(m = 400e^{-0.1216\ldots \times 4}\)M1 (AO3.4) Uses model to find \(m\) with \(t=4\), \(m_0=400\) and their \(k\). Condone \(m_0=200\)
\(m = 250\)A1 (AO1.1b) Obtains correct value of \(m\). CAO. AWRT 250. (245.9296...)
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(400e^{-0.1216t} \leq 280\)M1 (AO3.1b) Uses model to set up inequality or equation using their \(k\) and 280
\(e^{-0.1216t} \leq 0.7\)
\(-0.1216t \leq \ln(0.7)\)
\(t \geq 2.933\)A1F (AO1.1b) Solves their inequality or equation to find \(t\). Follow through their \(k\) only. (2.933067)
10:56 amA1F (AO3.2a) Interprets their solution. Only follow through if time is earlier than 1:42 pm
Part (c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
e.g. Different people eliminate caffeine at different rates; The model is based on an average person; The length of time taken to drink two cups of coffee may have been significant; The amount of caffeine in a "strong cup of coffee" may varyB1 (AO3.5b) States any sensible reason
## Question 10:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $200 = 400e^{-kx5.7}$ | M1 (AO3.4) | Uses model to form equation with $t=5.7$, $m = \frac{1}{2}m_0$ |
| $k = 0.1216047\ldots$ | A1 (AO1.1b) | Obtains correct value of $k$ |
| $m = 400e^{-0.1216\ldots \times 4}$ | M1 (AO3.4) | Uses model to find $m$ with $t=4$, $m_0=400$ and their $k$. Condone $m_0=200$ |
| $m = 250$ | A1 (AO1.1b) | Obtains correct value of $m$. CAO. AWRT 250. (245.9296...) |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $400e^{-0.1216t} \leq 280$ | M1 (AO3.1b) | Uses model to set up inequality or equation using their $k$ and 280 |
| $e^{-0.1216t} \leq 0.7$ | | |
| $-0.1216t \leq \ln(0.7)$ | | |
| $t \geq 2.933$ | A1F (AO1.1b) | Solves their inequality or equation to find $t$. Follow through their $k$ only. (2.933067) |
| 10:56 am | A1F (AO3.2a) | Interprets their solution. Only follow through if time is earlier than 1:42 pm |

### Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. Different people eliminate caffeine at different rates; The model is based on an average person; The length of time taken to drink two cups of coffee may have been significant; The amount of caffeine in a "strong cup of coffee" may vary | B1 (AO3.5b) | States any sensible reason |

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10 A scientist is researching the effects of caffeine. She models the mass of caffeine in the body using

$$m = m _ { 0 } \mathrm { e } ^ { - k t }$$

where $m _ { 0 }$ milligrams is the initial mass of caffeine in the body and $m$ milligrams is the mass of caffeine in the body after $t$ hours.

On average, it takes 5.7 hours for the mass of caffeine in the body to halve.\\
One cup of strong coffee contains 200 mg of caffeine.\\
10
\begin{enumerate}[label=(\alph*)]
\item The scientist drinks two strong cups of coffee at 8 am. Use the model to estimate the mass of caffeine in the scientist's body at midday.\\

10
\item The scientist wants the mass of caffeine in her body to stay below 480 mg

\begin{center}
\begin{tabular}{ | l | }
\hline
10 (b) \\
 \\
Use the model to find the earliest time \\
coffee. \\
Give your answer to the nearest minute \\
\end{tabular}
\end{center}

□
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 1 2018 Q10 [8]}}