Question 5
A logic gate circuit is shown.
For which combination of inputs is the output X equal to 1?
Not quite. You read the circuit as a NOR gate, which gives 1 only when both inputs are 0.
Not quite. You applied the NOT to A instead of B, which gives $\overline{A}\cdot B$.
That is it. Step through the hints below to see why.
Not quite. You ignored the NOT gate and treated the circuit as a plain AND, $A\cdot B$.
The printed circuit redrawn. A wire carrying 1 is drawn thick and orange with moving dashes. A wire carrying 0 stays thin and black. The number on each wire is its signal.
A = 0, so the AND gate has a 0 on its top input and X = 0. Not this one.
A = 0 and NOT B = 0, so both AND inputs are 0 and X = 0. Not this one.
A = 1 and NOT B = 1, so both AND inputs are 1 and X = 1. This is the one.
A = 1, but the NOT gate turns B = 1 into 0, so the AND gate sees a 0 and X = 0. Not this one.
Input A goes straight into the AND gate. Input B goes through the NOT gate first.
$$X = A \cdot \overline{B}$$
$X = 1$ only when $A = 1$ and $\overline{B} = 1$, that is, $A = 1$ and $B = 0$.
An AND gate outputs 1 only when both its inputs are 1, and the NOT gate inverts B.
$$X = A \cdot \overline{B}$$
$X = 1$ only when $A = 1$ and $\overline{B} = 1$, that is, $A = 1$ and $B = 0$.
The past QCAA questions this one was modelled on
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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