This diagram represents a logic circuit. [Figure 6.1]
- (a)Complete the truth table for the given logic circuit. A B C D Working space Z 0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 0 0 0 1 0 0 1 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1[3]
- (b)Simplify the given Boolean expression using Boolean algebra. Show your working. Y = A.B.C.D + A.B.C.D + A.B.C.D + A.B.C.D[3]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.
- [3] One mark for every shaded block of rows for column Z correct (Max 3)
- Correct Z column: 1 0 1 0 0 1 0 1 0 1 0 1 0 0 0 0
- [2] One mark for correct working from points (Max 2), for example: (Y =) A.B.C.D + A.B.C.D + A.B.C.D + A.B.C.D (Y =) A.D.(B.C + B.C + B.C + B.C) (Y =) A.D.(B.(C + C) + B.(C + C)) (Y =) A.D.(B.(1) + B.(1)) (Y =) A.D.(B + B) (Y =) A.D.(1)
- [1] One mark for correct answer (Y =) A.D
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