| Exam Board | AQA |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2008 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sorting Algorithms |
| Type | Algorithm Tracing |
| Difficulty | Easy -1.8 This is a straightforward algorithm tracing exercise requiring only systematic substitution of values into a simple loop structure. The algorithm is clearly written with no complex logic, and part (b) requires only basic observation that division/leading term issues arise when A=0. This is significantly easier than average A-level maths, being purely mechanical execution with no mathematical insight or problem-solving required. |
| Spec | 7.03c Working with algorithms: trace, interpret, adapt |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| SCA with \(A=1, B=-6, C=11, D=-6\) | M1 | Must use at least 3 variables |
| \(K=1, N=0, X=1, Y=0\) (1st pass) | A1 | 1st pass |
| \(K=2, N=1, X=2, Y=0\) (2nd pass) | A1 | 2nd pass |
| \(K=3, N=2, X=3, Y=0\) (3rd pass, final answer \(N=3\)) | A1 | All correct — 4 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A=1, B=-10, C=29, D=-20\), SCA | M1 | 1st pass, must use at least 3 variables |
| \(K=2, X=2, Y=6\) (2nd pass) | A1 | 2nd pass |
| \(K=3, X=3, Y=4\) (3rd pass) | A1 | 3rd pass |
| \(K=5, X=5, Y=0\), final \(N=3\) | A1 | All correct — 4 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Line 90 | B1 | |
| Never ending or \(N \neq 3\) | B1 | 2 marks total |
# Question 6(a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| SCA with $A=1, B=-6, C=11, D=-6$ | M1 | Must use at least 3 variables |
| $K=1, N=0, X=1, Y=0$ (1st pass) | A1 | 1st pass |
| $K=2, N=1, X=2, Y=0$ (2nd pass) | A1 | 2nd pass |
| $K=3, N=2, X=3, Y=0$ (3rd pass, final answer $N=3$) | A1 | All correct — 4 marks total |
# Question 6(a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A=1, B=-10, C=29, D=-20$, SCA | M1 | 1st pass, must use at least 3 variables |
| $K=2, X=2, Y=6$ (2nd pass) | A1 | 2nd pass |
| $K=3, X=3, Y=4$ (3rd pass) | A1 | 3rd pass |
| $K=5, X=5, Y=0$, final $N=3$ | A1 | All correct — 4 marks total |
# Question 6(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Line 90 | B1 | |
| Never ending or $N \neq 3$ | B1 | 2 marks total |
---6 A student is solving cubic equations that have three different positive integer solutions.\\
The algorithm that the student is using is as follows:\\
Line 10 Input $A , B , C , D$\\
Line $20 \quad$ Let $K = 1$\\
Line $30 \quad$ Let $N = 0$\\
Line $40 \quad$ Let $X = K$\\
Line 50 Let $Y = A X ^ { 3 } + B X ^ { 2 } + C X + D$\\
Line 60 If $Y \neq 0$ then go to Line 100\\
Line $70 \quad$ Print $X$, "is a solution"\\
Line $80 \quad$ Let $N = N + 1$\\
Line 90 If $N = 3$ then go to Line 120\\
Line $100 \quad$ Let $K = K + 1$\\
Line 110 Go to Line 40\\
Line 120 End
\begin{enumerate}[label=(\alph*)]
\item Trace the algorithm in the case where the input values are:
\begin{enumerate}[label=(\roman*)]
\item $A = 1 , B = - 6 , C = 11$ and $D = - 6$;
\item $A = 1 , B = - 10 , C = 29$ and $D = - 20$.
\end{enumerate}\item Explain where and why this algorithm will fail if $A = 0$.
\end{enumerate}
\hfill \mbox{\textit{AQA D1 2008 Q6 [10]}}