OCR MEI D2 — Question 1 16 marks

Exam BoardOCR MEI
ModuleD2 (Decision Mathematics 2)
Marks16
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DifficultyEasy -1.2 This is a straightforward D2 question testing basic Boolean algebra and circuit representation. All parts involve standard recall and routine application: describing circuit conditions, writing Boolean expressions, completing a truth table, drawing standard circuits, and applying the distributive law. No novel problem-solving or insight required—purely mechanical application of well-practiced techniques.
The switching circuit in Fig. 1.1 shows switches, s for a car's sidelights, h for its dipped headlights and f for its high-intensity rear foglights. It also shows the three sets of lights. \includegraphics{figure_1} (Note: s and h are each "ganged" switches. A ganged switch consists of two connected switches sharing a single switch control, so that both are either on or off together.)
    1. Describe in words the conditions under which the foglights will come on. [2]
    Fig. 1.2 shows a combinatorial circuit. \includegraphics{figure_2}
    1. Write the output in terms of a Boolean expression involving s, h and f. [2]
    2. Use a truth table to prove that \(s \wedge h \wedge f = \sim (\sim s \vee \sim h) \wedge f\). [3]
  2. A car's first gear can be engaged (g) if either both the road speed is low (r) and the clutch is depressed (d), or if both the road speed is low (r) and the engine speed is the correct multiple of the road speed (m).
    1. Draw a switching circuit to represent the conditions under which first gear can be engaged. Use two ganged switches to represent r, and single switches to represent each of d, m and g. [2]
    2. Draw a combinatorial circuit to represent the Boolean expression \(r \wedge (d \vee m) \wedge g\). [4]
    3. Use Boolean algebra to prove that \(r \wedge (d \vee m) \wedge g = ((r \wedge d) \vee (r \wedge m)) \wedge g\). [2]
    4. Draw another switching circuit to represent the conditions under which first gear can be selected, but without using a ganged switch. [1]
Question 1:
AnswerMarks Guidance
16 3
12 2
122 14
13 2
144 22
12 2
1∞ 15
11 2
114 13
15 5
1−6 −3 
Question 1:
1 | 6 | 3 | 10 | 5
1 | 2 | 2 | 2 | 2
1 | 22 | 14 | 11 | 23
1 | 3 | 2 | 3 | 3
1 | 44 | 22 | 42 | 15 | 15
1 | 2 | 2 | 2 | 4 | 5
1 | ∞ | 15 | ∞ | ∞ | 7 | 8
1 | 1 | 2 | 3 | 4 | 5 | 6
1 | 14 | 13 | 18 | 15 | 7 | 8
1 | 5 | 5 | 5 | 5 | 5 | 6
1 | −6 | −3 | −7 | 0 | 0 | 0 | 0
The switching circuit in Fig. 1.1 shows switches, s for a car's sidelights, h for its dipped headlights and f for its high-intensity rear foglights. It also shows the three sets of lights.

\includegraphics{figure_1}

(Note: s and h are each "ganged" switches. A ganged switch consists of two connected switches sharing a single switch control, so that both are either on or off together.)

\begin{enumerate}[label=(\alph*)]
\item 
\begin{enumerate}[label=(\roman*)]
\item Describe in words the conditions under which the foglights will come on. [2]
\end{enumerate}

Fig. 1.2 shows a combinatorial circuit.

\includegraphics{figure_2}

\begin{enumerate}[label=(\roman*), start=2]
\item Write the output in terms of a Boolean expression involving s, h and f. [2]

\item Use a truth table to prove that $s \wedge h \wedge f = \sim (\sim s \vee \sim h) \wedge f$. [3]
\end{enumerate}

\item A car's first gear can be engaged (g) if either both the road speed is low (r) and the clutch is depressed (d), or if both the road speed is low (r) and the engine speed is the correct multiple of the road speed (m).

\begin{enumerate}[label=(\roman*)]
\item Draw a switching circuit to represent the conditions under which first gear can be engaged. Use two ganged switches to represent r, and single switches to represent each of d, m and g. [2]

\item Draw a combinatorial circuit to represent the Boolean expression $r \wedge (d \vee m) \wedge g$. [4]

\item Use Boolean algebra to prove that $r \wedge (d \vee m) \wedge g = ((r \wedge d) \vee (r \wedge m)) \wedge g$. [2]

\item Draw another switching circuit to represent the conditions under which first gear can be selected, but without using a ganged switch. [1]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{OCR MEI D2  Q1 [16]}}
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Q1 16 Q2 16 Q3 20 Q4 20