Convert binary 1101111₂ into hexadecimal using correct nibble grouping from the right.
A 6E
B 7F
C 6F
D 5F
Explanation: Pad left: 0110 1111. 0110 is 6 and 1111 is F. So 1101111₂ equals 6F₁₆.
A 5-bit 2’s complement number is 10110₂. What decimal value does it represent?
A −9
B −10
C +22
D +10
Explanation: 10110 has MSB=1, so negative. Take 2’s complement: invert 01001, add 1 → 01010 = 10. So value is −10.
In 8-bit 2’s complement, the value 11100110₂ is interpreted as which decimal?
A −26
B −25
C +230
D +26
Explanation: MSB=1 so negative. Invert: 00011001, add 1 → 00011010 = 26. So the signed value is −26.
A CPU uses 8-bit signed integers. What happens when 01000000₂ is added to itself in 8-bit 2’s complement?
A No overflow
B Unsigned wrap only
C Signed overflow
D Always zero
Explanation: 01000000₂ is +64. +64 + +64 = +128, but 8-bit signed max is +127. Result becomes 10000000₂ (−128), indicating signed overflow.
Which statement about end-around carry is correct when using 1’s complement subtraction in fixed width?
A Always discard carry
B Add carry to LSB
C Store carry as sign
D Convert carry to XOR
Explanation: In 1’s complement arithmetic, any carry out of MSB is wrapped around and added to the least significant bit. This is essential for correct subtraction results.
In 2’s complement subtraction implemented as A + (2’s comp B), the final carry-out bit is generally
A Ignored
B Added back
C Stored as sign
D Used as parity
Explanation: In 2’s complement, carry out of MSB is discarded. The fixed-width sum gives the correct subtraction result without the end-around carry step.
A designer wants to convert an OR gate using only NAND gates. Which technique is primarily applied?
A Base conversion
B Parity setting
C BCD correction
D De Morgan law
Explanation: De Morgan’s laws allow converting OR into NAND form by complementing inputs and output appropriately. This “bubble pushing” enables NAND-only implementations.
The XOR of four bits (A⊕B⊕C⊕D) equals 1 when the number of 1s is
A Always odd
B Always even
C Always zero
D Always four
Explanation: XOR outputs 1 when there is an odd count of 1s among inputs. For multiple inputs, XOR acts like an odd-parity detector.
A 3-input XOR gate output is 1 for input combinations having
A Even ones
B Only zeros
C Odd ones
D Only ones
Explanation: For three inputs, XOR is 1 when exactly one or all three inputs are 1 (odd count). It is 0 when the count of ones is even.
For a 4-variable K-map, how many cells exist and why?
A 8, because 2³
B 16, because 2⁴
C 32, because 2⁵
D 64, because 2⁶
Explanation: A K-map has 2ⁿ cells for n variables. With 4 variables, 2⁴ = 16 cells, each representing one minterm combination.
In a K-map, a group of 8 adjacent 1s eliminates how many variables from the simplified product term?
A Eliminates 1
B Eliminates 2
C Eliminates 4
D Eliminates 3
Explanation: Group size 8 = 2³, so three variables change across the group and cancel. Only the remaining constant variable(s) appear in the simplified term.
A full adder carry-out is 1 when at least two inputs are 1. Which Boolean expression matches this carry logic?
A A⊕B⊕Cin
B (A+B+Cin)’
C AB + ACin + BCin
D A·B·Cin
Explanation: Carry-out is majority-of-three. It becomes 1 if any two inputs are 1. Expression AB + ACin + BCin covers all such cases.
A half adder sum output differs from carry output. Which pair correctly matches sum and carry equations?
A Sum=XOR, Carry=AND
B Sum=AND, Carry=XOR
C Sum=OR, Carry=NOR
D Sum=NAND, Carry=OR
Explanation: Half adder sum is XOR (difference), while carry is AND (both 1). This is the standard minimal-gate implementation.
A 4-to-1 multiplexer selects one input using 2 select lines. How many distinct select combinations are possible?
A 2
B 3
C 4
D 8
Explanation: Two select lines produce 2² = 4 combinations (00,01,10,11). Each combination selects one of the four data inputs.
A 3-to-8 decoder activates exactly one output line for each input code. How many input bits are required for 8 outputs?
A 3 bits
B 2 bits
C 4 bits
D 8 bits
Explanation: Decoder outputs are 2ⁿ. For 8 outputs, 2ⁿ=8 so n=3. Thus 3 input bits select one of eight outputs.
A priority encoder is preferred over a simple encoder because it can correctly handle what input condition?
A Single active input
B All inputs zero
C Clock missing
D Multiple active inputs
Explanation: If multiple inputs are 1, a simple encoder becomes ambiguous. A priority encoder outputs the code of the highest-priority active input, resolving conflicts.
A latch is called level-sensitive because its output can change while enable is
A Clock edge
B Power off
C Active level
D Grounded
Explanation: Latches follow input as long as enable is at its active level (often high). When enable becomes inactive, latch holds the last value.
A D flip-flop is often used to avoid invalid SR conditions because its single data input directly controls
A Parity state
B Next state
C Base state
D Hex state
Explanation: In a D flip-flop, Q(next) equals D at the clock edge. This avoids the forbidden S=R=1 condition found in SR latches, giving predictable behavior.
In binary division, shifting right by one bit for an unsigned number is equivalent to dividing by
A 2
B 4
C 8
D 16
Explanation: Right shift moves each bit to a lower weight, halving the value. For unsigned integers, a 1-bit right shift equals integer division by 2.
If the binary number 101000₂ is shifted left by 2 positions, what happens to its value in decimal terms?
A Multiplies by 2
B Multiplies by 8
C Multiplies by 4
D Divides by 4
Explanation: Each left shift multiplies by 2. Shifting left by 2 positions multiplies by 2² = 4, ignoring overflow beyond fixed bit width.
In unsigned arithmetic, overflow is best indicated by
A Carry-out bit
B Sign bit set
C Borrow bit only
D Parity mismatch
Explanation: In unsigned addition, if a carry out of MSB occurs, the true sum exceeds the representable range, indicating overflow for that fixed bit width.
In BCD, which range of 4-bit patterns is invalid for a single digit representation?
A 0000–1001
B 0001–0111
C 1010–1111
D 0010–1000
Explanation: BCD represents digits 0–9 using 0000 to 1001. Patterns 1010 to 1111 do not correspond to any decimal digit and are considered invalid.
Gray code is useful in encoders because adjacent codes differ by
A Two bits
B One bit
C Three bits
D Four bits
Explanation: Gray code changes only one bit between consecutive values, reducing error during transitions. This is important for mechanical or noisy sensors where bits may not switch simultaneously.
A logic function written as OR of AND terms is in which standard form used for implementation?
A SOP form
B POS form
C XOR form
D NOR form
Explanation: SOP stands for Sum of Products. It is implemented by AND gates producing product terms, followed by OR gates summing them.
A logic function written as AND of OR terms is in which standard form?
A SOP form
B XOR form
C NAND form
D POS form
Explanation: POS stands for Product of Sums. It is implemented by OR gates producing sum terms, followed by AND gates multiplying them.
The expression (A+B)’ equals A’·B’ which is used to change an OR followed by NOT into
A OR of NOTs
B AND of NOTs
C XOR of NOTs
D NAND of AND
Explanation: De Morgan’s law converts complement of OR into AND of complements. This helps when transforming OR-NOT structures into NAND/NOR equivalents.
A 4-bit unsigned number 1111₂ added to 0001₂ produces 1 0000₂. In 4-bit storage, what is stored?
A 0001
B 1111
C 0000
D 1000
Explanation: 15 + 1 = 16, which needs 5 bits. In 4-bit storage, the extra carry is discarded, leaving 0000₂, showing wrap-around overflow in unsigned arithmetic.
In an 8-bit signed system, adding 01100000₂ and 01100000₂ results in a negative pattern. What does this indicate?
A Signed overflow
B Valid sum
C No carry
D Parity error
Explanation: Both operands are positive (+96 each). Their true sum is +192, outside 8-bit signed range. When the result becomes negative, it signals signed overflow.
A binary-to-decimal conversion using weights requires multiplying each bit by powers of
A 10
B 8
C 2
D 16
Explanation: Binary is base-2. Each position weight is 2⁰, 2¹, 2², etc. Add weights for bits set to 1 to obtain the decimal value.
The hexadecimal number 0xFF corresponds to how many bits all set to 1?
A 4 bits
B 8 bits
C 12 bits
D 16 bits
Explanation: FF₁₆ equals 1111 1111₂, which is 8 bits all ones. Each hex digit corresponds to 4 bits, so two hex digits make 8 bits.
The octal digit 7 corresponds to which 3-bit binary group?
A 111
B 110
C 101
D 100
Explanation: Octal digits map to 3-bit binary. The digit 7 equals binary 111, representing decimal value 7.
A 2-input XNOR gate can be used as a basic comparator because it outputs 1 when inputs are
A Different
B Both 1 only
C Same
D Both 0 only
Explanation: XNOR outputs 1 for equal inputs (00 or 11). This equality detection is used as the basic element for comparing bits in digital comparators.
The complement of a Boolean variable is also called its
A Inverse
B Identity
C Product
D Sum
Explanation: Complement means logical NOT. It is also called inverse because it reverses the truth value: A becomes 0 when A is 1, and becomes 1 when A is 0.
In Boolean algebra, the distributive law allows rewriting A(B+C) as
A AB+BC
B A+B+C
C (AB)’
D AB+AC
Explanation: Distributive law states A·(B+C)=A·B + A·C. This supports expansion and simplification when converting expressions into SOP or factoring them.
The associative law for AND is correctly written as
A A·B=B·A
B A+A=A
C A·(B·C)=(A·B)·C
D A·0=0
Explanation: Associative law means grouping does not change result. For AND, you can group as A·(B·C) or (A·B)·C, giving the same output.
A function expressed in canonical SOP includes terms that each correspond to
A Output 1 rows
B Output 0 rows
C Random rows
D Don’t-care rows
Explanation: Canonical SOP is sum of minterms. Each minterm corresponds to an input combination where the function output is 1, ensuring full coverage of all 1-cases.
A function expressed in canonical POS includes terms that each correspond to
A Output 1 rows
B Output 0 rows
C Only MSB rows
D Only LSB rows
Explanation: Canonical POS is product of maxterms. Each maxterm corresponds to an input combination where output is 0, and multiplying them recreates the function.
A binary code used for text representation where ‘A’ and digits have assigned numeric codes is
A BCD
B Gray code
C ASCII
D Excess-3
Explanation: ASCII maps characters to numeric codes, enabling storage and transfer of text. Standard ASCII uses 7 bits, covering 128 symbols including letters, digits, and control codes.
A parity bit is mainly used for which type of basic checking during transmission?
A Error detection
B Error correction
C Data encryption
D Data sorting
Explanation: Parity checks detect many single-bit errors by verifying whether total 1s are even or odd. It does not reliably correct errors and may miss some even-bit errors.
A simple parity check can miss errors most commonly when
A One bit flips
B Two bits flip
C No bits flip
D LSB is 1
Explanation: Parity detects odd numbers of flips. Two flipped bits keep parity unchanged, so receiver may not detect the error using only parity.
A binary adder that adds two bits and a carry-in uses which circuit?
A Half adder
B Decoder
C Encoder
D Full adder
Explanation: Full adder has three inputs (A, B, Cin) and produces Sum and Carry-out. It is the building block for multi-bit adders in CPUs and ALUs.
A flip-flop is part of sequential logic because it requires a
A Memory element
B Only OR gates
C Only NAND gates
D Only XOR gates
Explanation: Sequential logic depends on past state, which is stored in memory elements. Flip-flops store one bit and update on clock events, enabling counters and registers.
In sign extension for a negative 2’s complement number, new bits added on the left are
A Always 0
B Alternating
C Always 1
D Random
Explanation: Negative numbers have MSB=1. Sign extension repeats the sign bit into higher bits to preserve the value. So for negative, added bits are 1s.
The binary 00101100₂ equals which decimal value using weights?
A 42
B 44
C 45
D 46
Explanation: 00101100₂ = 32 + 8 + 4 = 44. Add the place values where bits are 1 (2⁵, 2³, 2²) to get decimal.
Converting binary 11110000₂ to hex gives which result?
A F0
B 0F
C FF
D F8
Explanation: Group into nibbles: 1111 0000. 1111 is F and 0000 is 0, so hex is F0₁₆.
In a 2-input AND gate, output becomes 1 for how many input combinations?
A Two cases
B Three cases
C One case
D Four cases
Explanation: AND output is 1 only when both inputs are 1 (11). For 00, 01, 10 output is 0, so only one combination produces 1.
A NOR gate is often used to build other functions because it is universal. Which condition makes NOR output 1?
A Both inputs 1
B Inputs different
C Any input 1
D Both inputs 0
Explanation: NOR is NOT(OR). OR outputs 1 if any input is 1. Therefore NOR outputs 1 only when both inputs are 0, otherwise output is 0.