| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2018 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Finding constants from data |
| Difficulty | Challenging +1.2 This is a structured differential equations problem with clear scaffolding through parts (a) and (b). While it requires setting up a separable DE, integrating, and applying initial conditions, the question guides students through each step explicitly. Part (c)(i) involves substitution and solving a quadratic, which is routine. The conceptual insight needed (recognizing the model breaks down at t=0) is minimal. More challenging than average due to the DE context and multi-step nature, but the heavy scaffolding and standard techniques keep it accessible. |
| Spec | 1.02z Models in context: use functions in modelling1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks |
|---|---|
| 9(a) | Translates proportionality into a |
| Answer | Marks | Guidance |
|---|---|---|
| proportionality. | AO3.3 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| to find k | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| AG | AO2.1 | R1 |
| (b) | Integrates one side correctly | AO1.1a |
| Answer | Marks | Guidance |
|---|---|---|
| condone missing c | AO1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| result.AG | AO2.1 | R1 |
| (c)(i) | Translates rate of growth into |
| Answer | Marks | Guidance |
|---|---|---|
| of sales | AO3.3 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| equation or inequality in t | AO3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Need not be simplified | AO1.1b | A1 |
| Obtains t=5.17 | AO1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| minutes | AO3.4 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 09:30+their converted time | AO3.2a | R1F |
| (c)(ii) | Explains in context that when the | |
| stall opens there will be zero sales | AO3.5a | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| is undefined | AO3.5a | E1 |
| Total | 14 | |
| Q | Marking Instructions | AO |
Question 9:
--- 9(a) ---
9(a) | Translates proportionality into a
differential equation involving
(cid:3031)(cid:3051)
, t, x and a constant of
(cid:3031)(cid:3047)
proportionality. | AO3.3 | B1 | (cid:1856)(cid:1876) (cid:1863)(cid:4666)8(cid:3398)(cid:1872)(cid:4667)
(cid:3404)
(cid:1856)(cid:1872) (cid:1876)
(cid:3038)(cid:4666)(cid:2876)(cid:2879)(cid:2870)(cid:4667)
72 =
(cid:2871)(cid:2871)(cid:2874)
k = 4032
(cid:3031)(cid:3051) (cid:2872)(cid:2868)(cid:2871)(cid:2870)(cid:4666)(cid:2876)(cid:2879)(cid:3047)(cid:4667)
(cid:3404)
(cid:3031)(cid:3047) (cid:3051)
(cid:3031)(cid:3051)
(cid:1876) =4032(8 - t)
(cid:3031)(cid:3047)
(cid:3031)(cid:3051)
Substitutes t = 2, x = 336, (cid:3404) 72
(cid:3031)(cid:3047)
to find k | AO1.1a | M1
Obtains correct value of k shows
given result
AG | AO2.1 | R1
(b) | Integrates one side correctly | AO1.1a | M1 | xdx =4032(8t)dt
1 t2
x2 40328t c
2 2
1 22
3362 403282 c
2 2
c0
x2 64512t4032t2
x2 4032t16t
Integrates both sides correctly
condone missing c | AO1.1b | A1
Uses conditions to show c=0
and correctly obtains given
result.AG | AO2.1 | R1
(c)(i) | Translates rate of growth into
dx
24and uses in model for rate
dt
of sales | AO3.3 | M1 | 24x40328t
x1688t
1688t2 4032t16t
t²-16t+56 = 0
t = 5.171.. or 10.828…
5 hours 10 minutes
Earliest time 14:40
Eliminates x to form quadratic
equation or inequality in t | AO3.1a | M1
Obtains correct equation or
inequality in t
Need not be simplified | AO1.1b | A1
Obtains t=5.17 | AO1.1b | A1
Converts their t into hours and
minutes | AO3.4 | A1
Interprets the closing time as 14:40
09:30+their converted time | AO3.2a | R1F
(c)(ii) | Explains in context that when the
stall opens there will be zero sales | AO3.5a | E1 | When the stall opens there are
zero sales
(cid:3031)(cid:3051)
When x=0 is undefined as the
(cid:3031)(cid:3047)
denominator is zero
Explains that when x=0 the model
is undefined | AO3.5a | E1
Total | 14
Q | Marking Instructions | AO | Marks | Typical SolutionA market trader notices that daily sales are dependent on two variables:
number of hours, $t$, after the stall opens
total sales, $x$, in pounds since the stall opened.
The trader models the rate of sales as directly proportional to $\frac{8 - t}{x}$
After two hours the rate of sales is £72 per hour and total sales are £336
\begin{enumerate}[label=(\alph*)]
\item Show that
$$x \frac{dx}{dt} = 4032(8 - t)$$
[3 marks]
\item Hence, show that
$$x^2 = 4032t(16 - t)$$
[3 marks]
\item The stall opens at 09.30.
\begin{enumerate}[label=(\roman*)]
\item The trader closes the stall when the rate of sales falls below £24 per hour.
Using the results in parts (a) and (b), calculate the earliest time that the trader closes the stall.
[6 marks]
\item Explain why the model used by the trader is not valid at 09.30.
[2 marks]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2018 Q9 [14]}}