| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Rational inequality algebraically |
| Difficulty | Easy -1.2 This is a straightforward inverse proportion question requiring basic algebraic manipulation. Part (a) is trivial verification (1 mark), part (b) involves finding a constant for a second inverse proportion then solving a simple inequality (standard AS technique), and part (c) is interpretation. The question is entirely procedural with no problem-solving insight required, making it easier than average for AS level. |
| Spec | 1.02r Proportional relationships: and their graphs |
| Answer | Marks |
|---|---|
| 9(a) | Forms inverse proportion |
| Answer | Marks | Guidance |
|---|---|---|
| AG | 2.1 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 9(b) | Forms inverse proportion |
| Answer | Marks | Guidance |
|---|---|---|
| inequalities | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| inequalities | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| only | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| CAO | 1.1b | A1 |
| Subtotal | 4 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 9(c) | Identifies correctly the number |
| Answer | Marks | Guidance |
|---|---|---|
| Providing n > 0 OE | 3.5a | E1F |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 1 | |
| Question 9 Total | 6 | |
| Q | Marking instructions | AO |
Question 9:
--- 9(a) ---
9(a) | Forms inverse proportion
equation for P and substitutes
given values to obtain given
result
AG | 2.1 | B1 | k
P =
n
k
24 =
10
k = 240
240
P =
n
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 9(b) ---
9(b) | Forms inverse proportion
equation for C and substitutes
given value
PI by sight of 25 with or without
inequalities | 1.1a | M1 | l
C =
n2
l
60 =
100
l = 6000
240 6000
>
n n2
240n > 6000
n > 25
Obtains correct value of
constant of proportionality
PI by sight of 25 with or without
inequalities | 1.1b | A1
Forms inequality linking P and
their C. Condone equality at
this stage.
Or
Shows an attempt at trial and
error to solve the inequality
Condone equality at this stage.
Or
States n > 25 n = 25 or n ≥ 25
only | 1.1a | M1
Obtains n > 25 ignore any extra
inequality containing 0
CAO | 1.1b | A1
Subtotal | 4
Q | Marking instructions | AO | Marks | Typical solution
--- 9(c) ---
9(c) | Identifies correctly the number
of items that need to be sold to
make a profit corresponding to
their range of n from part (b)
Providing n > 0 OE | 3.5a | E1F | The artist makes a profit if they sell
more than 25 items
Subtotal | 1
Question 9 Total | 6
Q | Marking instructions | AO | Marks | Typical solutionA craft artist is producing items and selling them in a local market.
The selling price, £P, of an item is inversely proportional to the number of items produced, $n$
\begin{enumerate}[label=(\alph*)]
\item When $n = 10$, $P = 24$
Show that $P = \frac{240}{n}$ [1 mark]
\item The production cost, £C, of an item is inversely proportional to the square of the number of items produced, $n$
When $n = 10$, $C = 60$
Find the set of values of $n$ for which $P > C$ [4 marks]
\item Explain the significance to the craft artist of the range of values found in part (b). [1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 2023 Q9 [6]}}