| Exam Board | OCR MEI |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof |
| Type | Logical statements and converses |
| Difficulty | Easy -1.2 This is a Decision Mathematics question on basic logical reasoning and truth tables. Parts (a) and (b) involve straightforward application of logical connectives and contrapositive reasoning with everyday contexts. Part (c) requires understanding circuit diagrams as logical statements, but the reasoning is mechanical once the setup is understood. All parts are accessible with standard D2 knowledge and require minimal problem-solving insight—mostly pattern recognition and direct application of logical rules. |
| Spec | 4.01b Complex proofs: conjecture and demanding proofs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Emelia's condition is: NOT dry OR NOT warm \(\equiv\) NOT(dry AND warm) | M1 | Correct interpretation |
| Gemma says it's warm but not that it's dry, so Emelia is not convinced | A1 | Need both dry and warm |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| The dashboard lights are not dimmed | B1 B1 | Two marks for correct deduction with justification |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Let D = daytime, H = headlights on, R = raining, L = dashboard dimmed | B1 | Defining propositions |
| \(D \wedge H \Rightarrow R\); \(L \Rightarrow H\); \(D\); \(\neg R\) | B1 | Correct statement of premises |
| From \(D\) and \(D \wedge H \Rightarrow R\) and \(\neg R\): by contrapositive \(\neg R \Rightarrow \neg(D \wedge H) \equiv \neg D \vee \neg H\) | M1 | |
| Since \(D\) is true, \(\neg H\) must be true, i.e. headlights are off | A1 | |
| From \(L \Rightarrow H\) and \(\neg H\): by modus tollens, \(\neg L\) | M1 A1 | Dashboard lights not dimmed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| A and B in parallel (current flows if A up OR B up) in series with C (current flows if C down, i.e. \(\neg C\)) | B1 B1 | |
| \((A \vee B)\) models the parallel section; series with \(\neg C\) gives \((A \vee B) \wedge \neg C\) | M1 | |
| \((A \wedge B) \Rightarrow C \equiv \neg(A \wedge B) \vee C \equiv \neg A \vee \neg B \vee C\); contrapositive checking gives correct equivalence | A1 | Full explanation required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| A is down so A doesn't conduct; C is up so C conducts in lower path; current flows so B1 must conduct meaning B1 is up, hence B2 is up (ganged); D must conduct so D is up | B1 B1 |
# Question 2:
## Part (a) [2 marks]
| Answer | Marks | Guidance |
|--------|-------|----------|
| Emelia's condition is: NOT dry OR NOT warm $\equiv$ NOT(dry AND warm) | M1 | Correct interpretation |
| Gemma says it's warm but not that it's dry, so Emelia is not convinced | A1 | Need both dry and warm |
## Part (b)(i) [2 marks]
| Answer | Marks | Guidance |
|--------|-------|----------|
| The dashboard lights are not dimmed | B1 B1 | Two marks for correct deduction with justification |
## Part (b)(ii) [6 marks]
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let D = daytime, H = headlights on, R = raining, L = dashboard dimmed | B1 | Defining propositions |
| $D \wedge H \Rightarrow R$; $L \Rightarrow H$; $D$; $\neg R$ | B1 | Correct statement of premises |
| From $D$ and $D \wedge H \Rightarrow R$ and $\neg R$: by contrapositive $\neg R \Rightarrow \neg(D \wedge H) \equiv \neg D \vee \neg H$ | M1 | |
| Since $D$ is true, $\neg H$ must be true, i.e. headlights are off | A1 | |
| From $L \Rightarrow H$ and $\neg H$: by modus tollens, $\neg L$ | M1 A1 | Dashboard lights not dimmed |
## Part (c)(i) [4 marks]
| Answer | Marks | Guidance |
|--------|-------|----------|
| A and B in parallel (current flows if A up OR B up) in series with C (current flows if C down, i.e. $\neg C$) | B1 B1 | |
| $(A \vee B)$ models the parallel section; series with $\neg C$ gives $(A \vee B) \wedge \neg C$ | M1 | |
| $(A \wedge B) \Rightarrow C \equiv \neg(A \wedge B) \vee C \equiv \neg A \vee \neg B \vee C$; contrapositive checking gives correct equivalence | A1 | Full explanation required |
## Part (c)(ii) [2 marks]
| Answer | Marks | Guidance |
|--------|-------|----------|
| A is down so A doesn't conduct; C is up so C conducts in lower path; current flows so B1 must conduct meaning B1 is up, hence B2 is up (ganged); D must conduct so D is up | B1 B1 | |
---2
\begin{enumerate}[label=(\alph*)]
\item Emelia: 'I won't go out for a walk if it's not dry or not warm.'\\
Gemma: 'It's warm. Let's go!'\\
Will what Gemma has said convince Emelia, and if not, why not?
\item If it is daytime and the car headlights are on, then it is raining.
If the dashboard lights are dimmed then the car headlights are on.\\
It is daytime.\\
It is not raining.
\begin{enumerate}[label=(\roman*)]
\item What can you deduce?
\item Prove your deduction.
\end{enumerate}\item In this part of the question the switch X is represented by\\
\includegraphics[max width=\textwidth, alt={}, center]{d254fbd2-7443-4b6d-87ba-f0d71fce5e17-3_104_138_824_1226}
The switch can be wired into a circuit so that current flows when\\
the switch is up\\
\includegraphics[max width=\textwidth, alt={}, center]{d254fbd2-7443-4b6d-87ba-f0d71fce5e17-3_103_177_1005_593}\\
but does not flow when it is down\\
\includegraphics[max width=\textwidth, alt={}, center]{d254fbd2-7443-4b6d-87ba-f0d71fce5e17-3_111_167_1000_1334}
Or the switch can be wired so that current flows when\\
the switch is down\\
\includegraphics[max width=\textwidth, alt={}, center]{d254fbd2-7443-4b6d-87ba-f0d71fce5e17-3_109_172_1228_639}\\
but does not flow when it is up\\
\includegraphics[max width=\textwidth, alt={}, center]{d254fbd2-7443-4b6d-87ba-f0d71fce5e17-3_109_174_1228_1327}
\begin{enumerate}[label=(\roman*)]
\item Explain how the following circuit models $( \mathrm { A } \wedge \mathrm { B } ) \Rightarrow \mathrm { C }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{d254fbd2-7443-4b6d-87ba-f0d71fce5e17-3_365_682_1484_694}
In the following circuit B1 and B2 represent 'ganged' switches. This means that the two switches are either both up or both down.\\
\includegraphics[max width=\textwidth, alt={}, center]{d254fbd2-7443-4b6d-87ba-f0d71fce5e17-3_364_1278_2042_397}
\item Given that A is down, C is up and current is flowing, what can you deduce?
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI D2 2016 Q2 [16]}}