401. Study the truth table.
| \(A\) | \(B\) | \(Y=\overline{A\cdot B}\) |
|---|
| \(0\) | \(0\) | \(1\) |
| \(0\) | \(1\) | \(1\) |
| \(1\) | \(0\) | \(1\) |
| \(1\) | \(1\) | \(0\) |
The logic gate represented is:
ⓐ. \(AND\)
ⓑ. \(NAND\)
ⓒ. \(XNOR\)
ⓓ. \(NOR\)
Correct Answer: \(NAND\)
Explanation: The expression \(Y=\overline{A\cdot B}\) is the direct Boolean expression of a \(NAND\) gate. The table also confirms this because the output is \(0\) only when both inputs are \(1\). For all other combinations, the \(AND\) output is \(0\), so the \(NAND\) output is \(1\). An \(AND\) gate would have the opposite output pattern in the last step. The bar over the product is the key symbol.
402. Study the truth table.
| \(A\) | \(B\) | \(Y=\overline{A+B}\) |
|---|
| \(0\) | \(0\) | \(1\) |
| \(0\) | \(1\) | \(0\) |
| \(1\) | \(0\) | \(0\) |
| \(1\) | \(1\) | \(0\) |
The logic gate represented is:
ⓐ. \(OR\)
ⓑ. \(NOR\)
ⓒ. \(AND\)
ⓓ. \(NAND\)
Correct Answer: \(NOR\)
Explanation: The expression \(Y=\overline{A+B}\) represents \(OR\) followed by inversion. The table shows output \(1\) only when both inputs are \(0\). If either input is \(1\), the \(OR\) output becomes \(1\), and the \(NOR\) output becomes \(0\). This output pattern is the defining truth table of a \(NOR\) gate. The gate should not be confused with \(NAND\), whose \(0\) output occurs only when both inputs are \(1\).
403. A logic circuit has output \(Y=\overline{A}\cdot B\). For inputs \(A=0\) and \(B=1\), the output is:
ⓐ. \(1\)
ⓑ. \(0\)
ⓒ. \(A+B\)
ⓓ. \(\overline{A+B}\)
Correct Answer: \(1\)
Explanation: \( \textbf{Given expression:} \) \(Y=\overline{A}\cdot B\).
\( \textbf{Input values:} \) \(A=0\) and \(B=1\).
\( \textbf{Complement of \(A\):} \)
\[
\overline{A}=\overline{0}=1
\]
\( \textbf{Substitution into expression:} \)
\[
Y=1\cdot1
\]
\( \textbf{Boolean multiplication:} \)
\[
1\cdot1=1
\]
\( \textbf{Final answer:} \) The output is \(1\).
404. A logic circuit has output \(Y=A+\overline{B}\). For inputs \(A=0\) and \(B=1\), the output is:
ⓐ. \(1\)
ⓑ. \(A\cdot B\)
ⓒ. \(\overline{A\cdot B}\)
ⓓ. \(0\)
Correct Answer: \(0\)
Explanation: \( \textbf{Given expression:} \) \(Y=A+\overline{B}\).
\( \textbf{Input values:} \) \(A=0\) and \(B=1\).
\( \textbf{Complement of \(B\):} \)
\[
\overline{B}=\overline{1}=0
\]
\( \textbf{Substitution into expression:} \)
\[
Y=0+0
\]
\( \textbf{Boolean addition:} \)
\[
0+0=0
\]
\( \textbf{Final answer:} \) The output is \(0\).
405. A two-input \(OR\) gate is followed by a \(NOT\) gate. The resulting combination behaves as:
ⓐ. \(AND\)
ⓑ. \(NAND\)
ⓒ. \(NOR\)
ⓓ. \(OR\)
Correct Answer: \(NOR\)
Explanation: A two-input \(OR\) gate first produces the output \(A+B\). When this output is passed through a \(NOT\) gate, it becomes \(\overline{A+B}\). This is exactly the Boolean expression of a \(NOR\) gate. The operation is not \(NAND\), because \(NAND\) would require inversion of \(A\cdot B\). The order “\(OR\) then \(NOT\)” directly means \(NOR\).
406. A two-input \(AND\) gate is followed by a \(NOT\) gate. The resulting combination behaves as:
ⓐ. \(NOR\)
ⓑ. \(XNOR\)
ⓒ. \(NAND\)
ⓓ. \(AND\)
Correct Answer: \(NAND\)
Explanation: A two-input \(AND\) gate first gives output \(A\cdot B\). Passing this output through a \(NOT\) gate gives \(Y=\overline{A\cdot B}\). This is the Boolean expression of a \(NAND\) gate. The combination is not \(NOR\), because \(NOR\) is obtained by inverting \(A+B\). The gate name tells which basic operation is performed before inversion.
407. A circuit has output \(Y=\overline{A\cdot B}+C\). For \(A=1\), \(B=0\), and \(C=0\), the output is:
ⓐ. \(0\)
ⓑ. \(A\cdot B\)
ⓒ. \(\overline{C}\)
ⓓ. \(1\)
Correct Answer: \(1\)
Explanation: \( \textbf{Given expression:} \) \(Y=\overline{A\cdot B}+C\).
\( \textbf{Input values:} \) \(A=1\), \(B=0\), and \(C=0\).
\( \textbf{First operation:} \)
\[
A\cdot B=1\cdot0=0
\]
\( \textbf{Complement step:} \)
\[
\overline{A\cdot B}=\overline{0}=1
\]
\( \textbf{Substitution into final expression:} \)
\[
Y=1+0
\]
\( \textbf{Boolean addition:} \)
\[
1+0=1
\]
\( \textbf{Final answer:} \) The output is \(1\).
408. A logic circuit gives \(Y=(A+B)\cdot \overline{C}\). If \(A=0\), \(B=1\), and \(C=1\), then \(Y\) is:
ⓐ. \(1\)
ⓑ. \(A+B+C\)
ⓒ. \(\overline{A\cdot B}\)
ⓓ. \(0\)
Correct Answer: \(0\)
Explanation: \( \textbf{Given expression:} \) \(Y=(A+B)\cdot \overline{C}\).
\( \textbf{Input values:} \) \(A=0\), \(B=1\), and \(C=1\).
\( \textbf{OR operation:} \)
\[
A+B=0+1=1
\]
\( \textbf{Complement of \(C\):} \)
\[
\overline{C}=\overline{1}=0
\]
\( \textbf{Final product:} \)
\[
Y=1\cdot0
\]
\( \textbf{Boolean multiplication:} \)
\[
Y=0
\]
\( \textbf{Final answer:} \) The output is \(0\).
409. A logic network first sends \(A\) and \(B\) into an \(AND\) gate, then sends the result and \(C\) into an \(OR\) gate. The Boolean expression for the final output is:
ⓐ. \(Y=A+B\cdot C\)
ⓑ. \(Y=\overline{A+B+C}\)
ⓒ. \(Y=(A\cdot B)+C\)
ⓓ. \(Y=A\cdot(B+C)\)
Correct Answer: \(Y=(A\cdot B)+C\)
Explanation: The first gate is an \(AND\) gate with inputs \(A\) and \(B\), so its output is \(A\cdot B\). This intermediate output then becomes one input of an \(OR\) gate. The other input of that \(OR\) gate is \(C\). Therefore, the final output is \((A\cdot B)+C\). The order of gates matters because \(A\cdot B+C\) is not the same circuit as \(A\cdot(B+C)\) unless a specific Boolean identity is being applied.
410. A logic network first sends \(A\) and \(B\) into an \(OR\) gate, then inverts the output. The final expression and gate type are:
ⓐ. \(Y=A+B\), \(OR\)
ⓑ. \(Y=\overline{A\cdot B}\), \(NAND\)
ⓒ. \(Y=\overline{A+B}\), \(NOR\)
ⓓ. \(Y=A\cdot B\), \(AND\)
Correct Answer: \(Y=\overline{A+B}\), \(NOR\)
Explanation: The \(OR\) gate first produces the expression \(A+B\). The next operation is inversion, so the complete output becomes \(\overline{A+B}\). This is the Boolean expression of a \(NOR\) gate. The expression \(\overline{A\cdot B}\) would come from an \(AND\) gate followed by inversion. Reading the order of operations avoids confusing \(NOR\) with \(NAND\).
411. The output of a two-input \(NAND\) gate is connected to both inputs of another \(NAND\) gate. The overall circuit behaves as:
ⓐ. \(AND\) gate
ⓑ. \(OR\) gate
ⓒ. \(NOR\) gate
ⓓ. \(NOT\) gate with input \(A\) only
Correct Answer: \(AND\) gate
Explanation: The first \(NAND\) gate gives \(X=\overline{A\cdot B}\). The second \(NAND\) gate has both of its inputs connected to \(X\). A \(NAND\) gate with tied inputs works as an inverter, so its output is \(Y=\overline{X}\). Therefore, \(Y=\overline{\overline{A\cdot B}}\). Double complementation gives \(Y=A\cdot B\), which is the \(AND\) operation.
412. Use the circuit description below.
Two \(NAND\) gates are first used as inverters to produce \(\overline{A}\) and \(\overline{B}\). These two inverted signals are then applied to a third \(NAND\) gate.
The final output is:
ⓐ. \(A\cdot B\)
ⓑ. \(A+B\)
ⓒ. \(\overline{A+B}\)
ⓓ. \(\overline{A\cdot B}\)
Correct Answer: \(A+B\)
Explanation: The first two \(NAND\) gates act as \(NOT\) gates, so their outputs are \(\overline{A}\) and \(\overline{B}\). These signals are then applied to a \(NAND\) gate. The output is \(Y=\overline{\overline{A}\cdot\overline{B}}\). By De Morgan's theorem, \(\overline{\overline{A}\cdot\overline{B}}=A+B\). Thus, the described circuit builds an \(OR\) gate using only \(NAND\) gates.
413. Use the circuit description below.
Two \(NOR\) gates are first used as inverters to produce \(\overline{A}\) and \(\overline{B}\). These two inverted signals are then applied to a third \(NOR\) gate.
The final output is:
ⓐ. \(A+B\)
ⓑ. \(A\cdot B\)
ⓒ. \(\overline{A+B}\)
ⓓ. \(\overline{A\cdot B}\)
Correct Answer: \(A\cdot B\)
Explanation: A \(NOR\) gate with both inputs tied together behaves as an inverter. The first two \(NOR\) gates therefore produce \(\overline{A}\) and \(\overline{B}\). Feeding these into another \(NOR\) gate gives \(Y=\overline{\overline{A}+\overline{B}}\). By De Morgan's theorem, this simplifies to \(A\cdot B\). This shows how an \(AND\) operation can be built using only \(NOR\) gates.
414. Study the truth table for a logic circuit.
| \(A\) | \(B\) | \(Y\) |
|---|
| \(0\) | \(0\) | \(0\) |
| \(0\) | \(1\) | \(1\) |
| \(1\) | \(0\) | \(1\) |
| \(1\) | \(1\) | \(1\) |
The simplest two-input gate represented is:
ⓐ. \(OR\)
ⓑ. \(AND\)
ⓒ. \(NAND\)
ⓓ. \(NOR\)
Correct Answer: \(OR\)
Explanation: The output is \(0\) only when both inputs are \(0\). In every case where at least one input is \(1\), the output becomes \(1\). This is the defining truth table of a two-input \(OR\) gate. An \(AND\) gate would give \(1\) only in the \(A=1\), \(B=1\) row. The table is identified by the pattern of output values, not by the order in which the input rows are listed.
415. A logic circuit has output \(Y=\overline{A}+B\). The output for \(A=1\) and \(B=0\) is:
ⓐ. \(0\)
ⓑ. \(1\)
ⓒ. \(A\cdot B\)
ⓓ. \(\overline{A+B}\)
Correct Answer: \(0\)
Explanation: \( \textbf{Given expression:} \) \(Y=\overline{A}+B\).
\( \textbf{Input values:} \) \(A=1\) and \(B=0\).
\( \textbf{Complement of \(A\):} \)
\[
\overline{A}=\overline{1}=0
\]
\( \textbf{Substitution:} \)
\[
Y=0+0
\]
\( \textbf{Boolean addition:} \)
\[
0+0=0
\]
\( \textbf{Final answer:} \) The output is \(0\).
416. A logic circuit has \(Y=A\cdot\overline{B}+C\). For \(A=1\), \(B=1\), and \(C=0\), the output is:
ⓐ. \(1\)
ⓑ. \(0\)
ⓒ. \(A+B\)
ⓓ. \(\overline{C}\)
Correct Answer: \(0\)
Explanation: \( \textbf{Given expression:} \) \(Y=A\cdot\overline{B}+C\).
\( \textbf{Input values:} \) \(A=1\), \(B=1\), and \(C=0\).
\( \textbf{Complement of \(B\):} \)
\[
\overline{B}=\overline{1}=0
\]
\( \textbf{Product term:} \)
\[
A\cdot\overline{B}=1\cdot0=0
\]
\( \textbf{Final expression:} \)
\[
Y=0+0
\]
\( \textbf{Boolean result:} \)
\[
Y=0
\]
\( \textbf{Final answer:} \) The output is \(0\).
417. A claim says: “A \(NAND\) gate is the same as an \(OR\) gate because both usually give output \(1\) in many input cases.” The best evaluation is:
ⓐ. The claim is suitable because the number of \(1\) outputs is the only important feature
ⓑ. The claim is suitable only when both inputs are \(1\)
ⓒ. The claim is unsuitable because \(NAND\) gates cannot have two inputs
ⓓ. The claim is unsuitable; truth tables and Boolean expressions differ.
Correct Answer: The claim is unsuitable; truth tables and Boolean expressions differ.
Explanation: A two-input \(NAND\) gate has output \(Y=\overline{A\cdot B}\). A two-input \(OR\) gate has output \(Y=A+B\). Although both gates give output \(1\) for some of the same input combinations, they differ when \(A=0\), \(B=0\) and when \(A=1\), \(B=1\). A gate is identified by its complete truth table, not by a rough count of output \(1\) cases. Comparing all rows prevents treating different gates as identical.
418. Consider the following statements about universal gates.
I. A \(NAND\) gate with tied inputs can act as a \(NOT\) gate.
II. A \(NOR\) gate with tied inputs can act as a \(NOT\) gate.
III. Universal gates cannot be used to construct \(AND\) or \(OR\) gates.
The valid statements are:
ⓐ. II and III only
ⓑ. I and III only
ⓒ. I and II only
ⓓ. I, II, and III
Correct Answer: I and II only
Explanation: Statement I is valid because a tied-input \(NAND\) gate gives \(Y=\overline{A\cdot A}=\overline{A}\). Statement II is valid because a tied-input \(NOR\) gate gives \(Y=\overline{A+A}=\overline{A}\). Statement III is not valid because \(NAND\) and \(NOR\) are called universal precisely because other basic gates can be built from either one of them. The inverter construction is the first step in many universal-gate realizations. Once inversion is available, \(AND\) and \(OR\) can be constructed using De Morgan's theorem.
419. A logic system uses \(1\) to mean “switch closed” and \(0\) to mean “switch open.” Two switches must both be closed for a lamp to glow. The suitable logic gate model is:
ⓐ. \(AND\)
ⓑ. \(OR\)
ⓒ. \(NOT\)
ⓓ. \(NOR\)
Correct Answer: \(AND\)
Explanation: The lamp glows only when both conditions are satisfied. If either switch is open, one required condition is missing and the lamp remains off. This matches the \(AND\) rule: output \(1\) only when all inputs are \(1\). The \(OR\) rule would allow the lamp to glow when either one switch alone is closed. The phrase “both must be closed” is the physical condition corresponding to logical multiplication.
420. In another switching system, a warning light should turn on if either sensor \(P\) or sensor \(Q\) detects a fault. The suitable logic gate is:
ⓐ. \(AND\)
ⓑ. \(NOT\)
ⓒ. \(NAND\) only
ⓓ. \(OR\)
Correct Answer: \(OR\)
Explanation: The warning light should turn on when sensor \(P\) detects a fault, or when sensor \(Q\) detects a fault, or when both detect a fault. This is exactly the \(OR\) condition. An \(AND\) gate would require both sensors to detect faults before the light turns on, which is not the stated requirement. A \(NOT\) gate would invert a single input rather than combine two sensor signals. The word “either” naturally represents logical addition.
