Chapter 5: Number System and Digital Logic (Set-10)
A 10-bit 2’s complement register stores 1111000001₂. What is the correct decimal value of this signed number?
A −62
B +961
C −63
D +993
Explanation: MSB is 1, so it’s negative. Take 2’s complement: invert → 0000111110, add 1 → 0000111111 = 63. Hence stored value is −63.
A 10-bit unsigned register holds 1111000001₂. What decimal value does this represent as an unsigned magnitude?
A 945
B 993
C 1009
D 961
Explanation: Unsigned value equals sum of weights: 2⁹+2⁸+2⁷+2⁶ +2⁰ = 512+256+128+64+1 = 961. Same bits differ in meaning for signed vs unsigned.
A signed 8-bit system adds 10010000₂ and 10010000₂ using 2’s complement. What is the correct result interpretation?
A No overflow
B Signed overflow
C Unsigned overflow
D Result is zero
Explanation: 10010000₂ is negative (−112). −112 + −112 = −224, which is below −128, so it overflows. Same-sign negatives giving a positive/wrap indicates overflow.
In 8-bit 2’s complement, what is the decimal value of 10010000₂?
A −96
B +144
C −112
D +112
Explanation: Negative because MSB=1. 2’s complement magnitude: invert → 01101111, add 1 → 01110000 = 112. So value is −112.
A designer uses overflow flag = (carry into MSB) XOR (carry out of MSB). For which arithmetic is this overflow rule valid?
A Signed 2’s complement
B Unsigned addition
C BCD addition
D Gray code addition
Explanation: In 2’s complement signed addition, overflow occurs when carries into and out of sign bit differ. This rule does not indicate unsigned overflow, which uses carry-out only.
A 6-bit signed value is −7 in 2’s complement. Which bit pattern correctly represents −7?
A 111000
B 000111
C 100111
D 111001
Explanation: +7 is 000111. Invert → 111000, add 1 → 111001. So 111001₂ represents −7 in 6-bit 2’s complement.
In 6-bit 1’s complement, what is the representation of −7?
A 111001
B 000111
C 111000
D 100111
Explanation: In 1’s complement, negative is formed by inverting all bits of the positive number. +7 is 000111, so −7 is 111000. No “add 1” step is used.
A 4-bit 2’s complement number 0111₂ is added to 0101₂. What happens to the sign and overflow?
A Signed overflow
B No overflow
C Underflow only
D Becomes negative
Explanation: 0111 is +7 and 0101 is +5. Sum is +12, but 4-bit signed max is +7. Result wraps to 1100 (−4), indicating signed overflow.
A 4-bit 2’s complement result 1100₂ equals which decimal value?
A −3
B +12
C +4
D −4
Explanation: 1100 is negative. Invert → 0011, add 1 → 0100 = 4. So the signed value is −4.
A binary number 111111111111₂ (12 bits) is converted to hexadecimal. What is the correct hex result?
A F0F
B 0FF
C FFF
D EFF
Explanation: Group 12 bits into three nibbles: 1111 1111 1111. Each nibble is F. So the hex value is FFF₁₆.
Convert hexadecimal 7A3₁₆ to binary using correct 4-bit mapping per hex digit.
Convert octal 645₈ to binary using exact 3-bit groups per digit.
A 110110101
B 110100011
C 110100101
D 101100101
Explanation: Each octal digit maps to 3 bits: 6→110, 4→100, 5→101. Combine to get 110100101₂.
An 8-bit BCD number represents two decimal digits. Which pair is correctly valid BCD bytes for “59”?
A 0110 1001
B 0101 1010
C 0101 1111
D 0101 1001
Explanation: In BCD, 5 is 0101 and 9 is 1001. So “59” is 0101 1001. Values above 1001 are invalid for a BCD digit.
A BCD digit addition produces 1100₂ in the lower nibble without carry. What correction must be applied?
A Add 0011
B Add 0110
C Add 0101
D Add 1001
Explanation: 1100₂ is decimal 12, invalid in BCD. Add 6 (0110₂) to correct into valid BCD range and generate carry to next digit if required.
A K-map group of 8 cells in a 4-variable map simplifies to a term containing how many variables?
A 2 variables
B 3 variables
C 1 variable
D 4 variables
Explanation: A 4-variable map has 4 variables. Group size 8 = 2³ eliminates 3 variables, leaving 4−3 = 1 variable in the simplified term.
A K-map group of 2 cells in a 4-variable map leaves how many variables in the term?
A 3 variables
B 1 variable
C 2 variables
D 4 variables
Explanation: Group size 2 = 2¹ eliminates 1 variable. With 4 total variables, remaining variables are 4−1 = 3 in the simplified product term.
For a 4-variable K-map, which group shape is always invalid for simplification?
A Rectangle block
B Wrap-around block
C 2×2 square
D Diagonal pair
Explanation: K-map adjacency is horizontal/vertical with wrap-around edges, not diagonal. Diagonal cells differ in more than one bit, so grouping them gives incorrect simplification.
A NAND-only implementation of NOT(A) is created by connecting NAND inputs
A To different signals
B One to 0
C Both to A
D One floating
Explanation: NAND(A,A) = (A·A)’ = A’. Tying both inputs makes NAND behave as an inverter, useful when only NAND gates are available.
Which expression is the correct De Morgan equivalent of (A + B + C)’ ?
A A’·B’·C’
B A’ + B’ + C’
C (A·B·C)’
D A·B·C
Explanation: De Morgan’s law generalizes: complement of OR becomes AND of complements. So (A+B+C)’ = A’·B’·C’.
Which expression is the correct De Morgan equivalent of (A·B·C)’ ?
A A’·B’·C’
B A’ + B’ + C’
C (A+B+C)’
D A+B+C
Explanation: Complement of AND becomes OR of complements. So (A·B·C)’ = A’ + B’ + C’, useful in converting between NAND/NOR forms.
A full adder carry-out equation AB + AC + BC is an example of which logic concept?
A Parity function
B Identity function
C Null function
D Majority function
Explanation: Carry-out is 1 when at least two inputs are 1, which is exactly the majority-of-three behavior. This is why the carry expression is often called a majority function.
A 3-input parity generator wants even parity. Which logic operation naturally helps compute parity across bits?
A AND chain
B OR chain
C XOR chain
D NOR chain
Explanation: XOR across bits produces 1 when odd number of ones exists. Using XOR and then choosing parity bit appropriately generates even parity efficiently in hardware.
A 2-to-4 decoder has how many selectable output lines and why?
A 2 outputs
B 4 outputs
C 6 outputs
D 8 outputs
Explanation: With 2 input bits, total combinations are 2² = 4. A 2-to-4 decoder provides one-hot outputs where exactly one of four lines is active per input code.
A 1-to-16 demultiplexer requires how many select lines to choose a single output?
A 2 lines
B 3 lines
C 5 lines
D 4 lines
Explanation: Number of outputs equals 2^select. For 16 outputs, select lines must be 4 because 2⁴ = 16. These select bits decide which output gets the input.
The ASCII standard is called 7-bit because it can represent exactly how many codes?
A 64 codes
B 96 codes
C 128 codes
D 256 codes
Explanation: 7 bits produce 2⁷ = 128 unique patterns. ASCII assigns these patterns to control codes, letters, digits, punctuation, and other symbols.
In 2’s complement, why is the negative range one larger than the positive range in fixed width?
A Pattern 100…0
B Two zeros exist
C MSB reserved
D Hex mapping
Explanation: The most negative pattern 1000…0 represents −2^(n−1) and has no positive counterpart. Positive max is 2^(n−1)−1, making negative range one larger.
A binary number 10000000₂ equals 128 in unsigned but −128 in signed 2’s complement. This difference happens due to
A Base change
B Parity addition
C Sign bit meaning
D BCD correction
Explanation: In signed 2’s complement, MSB indicates sign and weight changes interpretation. Same bit pattern is interpreted differently depending on whether number is treated as signed or unsigned.
Which condition guarantees unsigned addition overflow in n-bit arithmetic?
A Result MSB is 1
B Inputs same sign
C XOR is 1
D Carry out occurs
Explanation: Unsigned overflow occurs when sum exceeds 2ⁿ−1. This is exactly indicated by a carry out of the MSB in fixed-width addition.
In a logic circuit, a “don’t care” condition in K-map is used mainly to
A Aid simplification
B Force output 0
C Force output 1
D Remove variables
Explanation: Don’t cares can be treated as 0 or 1 to form larger groups in K-map, helping reduce the expression and minimize gate count without changing required outputs.
A 4-bit signed 2’s complement value 1000₂ equals
A −7
B +8
C −8
D +0
Explanation: In 4-bit 2’s complement, range is −8 to +7. Pattern 1000₂ is the most negative value, equal to −8.
The binary subtraction 10000₂ − 00001₂ equals 01111₂. This result is correct because borrowing in binary effectively adds
A 1 to bit
B 8 to bit
C 16 to bit
D 2 to bit
Explanation: Borrowing from the next higher bit gives 1 unit of that place, which equals 2 units in the current lower place in binary. That enables the subtraction to proceed correctly.
Convert decimal 4095 to hexadecimal. Which is correct?
A FFFF
B FFF
C 0FFF
D EFF
Explanation: 4095 equals 2¹²−1, so in hex it is FFF. That is 3 hex digits. Correct answer is A.
Convert hexadecimal 0x8000 (16-bit) to decimal unsigned value.
A 16384
B 65535
C 32768
D 8192
Explanation: 0x8000 has 1 in the 2¹⁵ position. Unsigned value is 2¹⁵ = 32768, since all other bits are 0.
Convert decimal 1023 to binary.
A 1111111111
B 1000000000
C 1111111110
D 0111111111
Explanation: 1023 equals 2¹⁰−1, so it is 10 bits all ones: 1111111111₂.
A 10-bit unsigned maximum value is
A 511
B 1024
C 2047
D 1023
Explanation: Unsigned max for n bits is 2ⁿ−1. For 10 bits, 2¹⁰−1 = 1024−1 = 1023.
In 10-bit 2’s complement, the minimum value equals
A −511
B −1023
C −512
D 0
Explanation: For n-bit 2’s complement, minimum is −2^(n−1). For 10 bits, minimum is −2⁹ = −512.
In 10-bit 2’s complement, the maximum positive equals
A 512
B 511
C 1023
D 1024
Explanation: Max positive in n-bit 2’s complement is 2^(n−1)−1. For 10 bits, 2⁹−1 = 512−1 = 511.
The XOR of bits is commonly used in checksums because it is sensitive to
A Even flips only
B Any change always
C No change
D Odd flips only
Explanation: XOR parity changes when an odd number of bits flip. This makes XOR useful for parity generation, but it may miss errors involving an even number of bit flips.
A 2-input XOR gate can be built from basic gates by combining AND/OR/NOT. Which is a correct identity?
A A⊕B=AB + A’B’
B A⊕B=(A+B)’
C A⊕B=A’B + AB’
D A⊕B=AB
Explanation: XOR is true when inputs differ. That occurs for A=0,B=1 giving A’B, and for A=1,B=0 giving AB’. OR-ing these gives XOR.
A 2-input XNOR identity can be written as
A AB + A’B’
B AB’ + A’B
C (A+B)’
D A’ + B’
Explanation: XNOR is 1 when inputs are equal. That occurs when both are 1 (AB) or both are 0 (A’B’). OR of these terms gives XNOR.
A K-map wrap-around grouping is allowed because map variables are arranged in
A Decimal order
B Gray code order
C Random order
D ASCII order
Explanation: K-map uses Gray code arrangement so adjacent cells differ by one variable. This includes edges being adjacent, enabling wrap-around grouping and better simplification.
When converting a Boolean expression to NAND-only, the step of moving inversion bubbles across gates is popularly called
A Carry chaining
B Bit stuffing
C Nibble grouping
D Bubble pushing
Explanation: “Bubble pushing” uses De Morgan’s laws to move inversions across AND/OR structures, transforming a circuit into an equivalent NAND-only or NOR-only implementation.
A 4-variable function has minterms m(0,1,2,3,8,9,10,11). This set forms which simple simplified term?
A A’
B C’
C B’
D D’
Explanation: Indices 0–3 and 8–11 correspond to B=0 while other variables vary. That means function is 1 whenever B=0, so simplified expression is B’.
In binary, subtracting using 2’s complement works because adding a number and its 2’s complement yields
A Zero with carry
B All ones
C Random bits
D Invalid code
Explanation: A + (2’s complement of A) equals 2ⁿ, which appears as 0 in n bits with a carry out. This property enables subtraction using only addition circuitry.
A 2-to-1 multiplexer can implement NOT(A) by selecting between constants. Which setup works?
A S=A, I0=0, I1=1
B S=0, I0=A, I1=A
C S=1, I0=A, I1=A
D S=A, I0=1, I1=0
Explanation: Output of 2:1 MUX is I0 when S=0 and I1 when S=1. With S=A, output becomes 1 when A=0 and 0 when A=1, i.e., NOT(A).
A demultiplexer can act like a decoder when its data input is fixed to
A 0 constant
B Alternating
C 1 constant
D Floating
Explanation: With data input tied to 1, the DEMUX routes a logic 1 to one selected output line, producing one-hot outputs similar to a decoder for the select code.
In a 4-bit BCD, adding 8 (1000) and 7 (0111) gives 1111. After correction, what is the valid BCD digit and carry?
A 0100 carry1
B 0101 carry1
C 0011 carry1
D 1111 carry0
Explanation: 8+7=15 invalid. Add 6: 1111+0110=1 0101. Lower nibble 0101 (=5) and carry 1 goes to next digit, representing 15.
A binary number is said to be normalized in a given base when its most significant digit is
A Always 0
B Always 1
C Always even
D Non-zero
Explanation: Normalization means the leading digit is not zero, ensuring a unique representation and maximum use of available digits. For binary, normalized numbers typically start with 1.
A logic circuit that remembers state and uses feedback paths is typically
A Sequential
B Combinational
C Pure arithmetic
D Encoding-only
Explanation: Sequential circuits include memory, often through feedback and flip-flops/latches. Their output depends on past states, unlike combinational circuits that depend only on current inputs.
A 4-bit binary 1010 is treated as signed 2’s complement. What decimal value does it represent?
A −5
B +10
C −6
D +6
Explanation: 1010 has MSB=1 so negative. Invert → 0101, add 1 → 0110 (=6). Therefore 1010₂ represents −6 in 4-bit 2’s complement.