Question 1

In digital electronics, convert the binary number 101000010111₂ to its equivalent hexadecimal (base-16) value.

Accepted Answer

Introduction / Context:Binary-to-hexadecimal conversion is a core skill in computer organization and digital electronics. It relies on grouping binary digits in sets of four (nibbles) because each nibble maps directly to one hexadecimal digit. This question checks your fluency with base conversions and nibble mapping.Given Data / Assumptions:Binary number: 101000010111₂No sign bit is implied; treat as an unsigned binary value.Hexadecimal digits map from 0000₂→0 to 1111₂→F.Concept / Approach:Pad the left of the binary string with zeros, if needed, so its length is a multiple of 4.Split into 4-bit groups (nibbles) from left to right.Convert each nibble to its hexadecimal digit and concatenate the results.Step-by-Step Solution:Start with 101000010111₂.Group into nibbles: 1010 0001 0111.Map each nibble: 1010₂ → A; 0001₂ → 1; 0111₂ → 7.Concatenate: A17₁₆.Verification / Alternative check:Convert A17₁₆ back to binary: A → 1010, 1 → 0001, 7 → 0111, yielding 1010 0001 0111, which matches the original.Why Other Options Are Wrong:D8F9₁₆ and D9F8₁₆: These would imply binary nibbles beginning with 1101 (D) and include F (1111) which are not present in the original grouping.A8B9₁₆: Would require nibbles 1010 1000 1011 1001, which do not match.None of the above: Incorrect because A17₁₆ exactly matches.Common Pitfalls:Grouping from the right but forgetting to pad on the left to complete the leftmost nibble.Confusing nibble values such as 1010 (A) versus 1001 (9).Dropping leading zeros within a nibble (e.g., 0001 must remain 0001).Final Answer:A17₁₆

Question 2

In digital logic families, which technology dissipates the minimum static power under typical operating conditions?

Accepted Answer

Introduction / Context:Different logic families—DTL, TTL, ECL, and CMOS—have distinct trade-offs in speed, noise margins, fan-out, and power dissipation. Power efficiency is critical in battery-operated devices and large-scale integration. This question tests recognition of which family minimizes static power consumption.Given Data / Assumptions:Logic families: DTL, TTL, ECL, and CMOS.Typical ambient conditions; nominal supply voltages.Focus on static (DC) power dissipation at logic 0/1, not dynamic at high switching rates.Concept / Approach:CMOS (Complementary Metal-Oxide-Semiconductor) uses complementary pairs of MOSFETs.In steady states, one device of the pair is OFF, leading to near-zero DC current through the pull-up/pull-down path.Other families (TTL/DTL) maintain bias currents; ECL uses differential pairs with constant current, hence higher power.Step-by-Step Solution:Identify static power mechanism: CMOS → negligible static current; TTL/DTL → bias currents; ECL → constant tail current.Compare orders of magnitude: ECL (highest), TTL/DTL (moderate), CMOS (lowest in static).Conclude: Minimum static power dissipation → CMOS.Verification / Alternative check:Cross-check by considering battery devices and microcontrollers which overwhelmingly use CMOS due to low idle consumption.Why Other Options Are Wrong:DTL: Diode-transistor logic still draws static current via biasing networks.TTL: Transistor-transistor logic consumes more static power than CMOS.ECL: Highest power due to constant current flow in differential pairs, traded for speed.None of the above: Incorrect because CMOS is correct.Common Pitfalls:Confusing static with dynamic power; CMOS dynamic power can be significant at high frequencies (P ≈ C * V^2 * f).Assuming ECL is efficient because it is fast; speed does not imply low power.Final Answer:CMOS

Question 3

Convert the decimal number 71₁₀ to its binary (base-2) equivalent.

Accepted Answer

Introduction / Context:Converting from decimal (base-10) to binary (base-2) is foundational in computer architecture and number systems. The standard approach repeatedly divides the decimal number by 2, tracking remainders, or uses powers of 2 decomposition.Given Data / Assumptions:Decimal input: 71₁₀.We want an unsigned binary representation.Binary weights: 1, 2, 4, 8, 16, 32, 64, ...Concept / Approach:Power-of-two decomposition: represent 71 as a sum of distinct powers of two.Alternatively, use successive division by 2 and read remainders in reverse.Step-by-Step Solution:Find largest power of 2 ≤ 71 → 64 = 2^6.Compute remainder: 71 - 64 = 7.Express 7 as 4 + 2 + 1 → 2^2 + 2^1 + 2^0.Set bits at positions 6, 2, 1, 0 → 1 000 111.Final binary: 1000111₂.Verification / Alternative check:Convert back: 164 + 032 + 016 + 08 + 14 + 12 + 1*1 = 64 + 0 + 0 + 0 + 4 + 2 + 1 = 71₁₀.Why Other Options Are Wrong:110011₂: Equals 51₁₀, not 71.1110011₂: Equals 115₁₀, not 71.0110011₂: Equals 51₁₀ with a leading zero.None of the above: Incorrect because 1000111₂ is correct.Common Pitfalls:Reversing the order of remainders when using the division method.Dropping significant leading 1 in the highest place (2^6 here).Final Answer:1000111₂

Question 4

De Morgan's first theorem: a NOR gate is equivalent to a bubbled _____ gate. Choose the correct gate to complete the statement.

Accepted Answer

Introduction / Context:De Morgan's theorems provide equivalences between logic gate implementations, enabling circuit simplification and technology mapping. One classic identity states that a NOR operation can be realized using an AND gate with input inversion (bubbles) and output inversion representation.Given Data / Assumptions:NOR is the complement of OR: A NOR B = NOT(A OR B).Bubble notation indicates inversion on inputs or outputs of a logic symbol.Gate 'equivalence' is functional equivalence.Concept / Approach:De Morgan 1: NOT(A OR B) = (NOT A) AND (NOT B).This means a NOR gate behaves as an AND gate whose inputs are inverted (bubbles on inputs), and typically the symbol shows inversion at inputs rather than an explicit inverter block.Step-by-Step Solution:Start from NOR: Y = NOT(A OR B).Apply De Morgan: Y = (NOT A) AND (NOT B).Interpretation: Use an AND gate; place bubbles on both inputs to indicate NOT A and NOT B.Hence: A NOR B is equivalent to a bubbled AND gate.Verification / Alternative check:Truth table check for A, B ∈ {0,1} shows equality of outputs for NOR and bubbled-input AND.Why Other Options Are Wrong:XAND: Not a standard logic gate.XOR: NOT(A XOR B) is XNOR, unrelated to NOR equivalence above.NOR: Statement asks for gate equivalence with bubbling; the gate is AND with bubbles, not NOR itself.None of the above: Incorrect because AND is correct.Common Pitfalls:Placing the bubble on the output instead of inputs for this identity.Confusing NOR equivalence with NAND equivalence (which relates to OR with input bubbles).Final Answer:AND

Question 5

Convert the octal number 143₈ to its decimal (base-10) value.

Accepted Answer

Introduction / Context:Octal (base-8) numbers arise historically from grouping binary digits in sets of three. Converting octal to decimal reinforces positional-value understanding and prepares you for quick binary conversions as well.Given Data / Assumptions:Octal input: 143₈.Octal digits range from 0 to 7.Place values: 8^0, 8^1, 8^2, ...Concept / Approach:Use positional expansion: abc₈ = a8^2 + b8^1 + c8^0.Compute powers: 8^2 = 64, 8^1 = 8, 8^0 = 1.Step-by-Step Solution:143₈ = 18^2 + 48^1 + 38^0.Compute: 164 = 64; 48 = 32; 3*1 = 3.Sum: 64 + 32 + 3 = 99.Verification / Alternative check:Convert 143₈ → binary quickly: 1→001, 4→100, 3→011 giving 001 100 011₂ = 1100011₂ which equals 64 + 32 + 3 = 99₁₀.Why Other Options Are Wrong:90₁₀, 971₁₀, 107₁₀: Do not match the positional expansion result.None of the above: Incorrect because 99₁₀ is correct.Common Pitfalls:Accidentally treating octal digits as decimal during expansion.Miscomputing 8^2 or 8^1, which skews the final sum.Final Answer:99₁₀

Question 6

ASCII vs EBCDIC: these character encodings primarily differ in their collating sequences (ordering of characters) rather than storage size.

Accepted Answer

Introduction / Context:ASCII and EBCDIC are legacy character encodings used to represent letters, digits, and symbols on computers. Understanding how they differ matters for sorting behavior, interoperability, and porting software between systems (e.g., mainframes vs microcomputers).Given Data / Assumptions:ASCII: 7-bit code commonly stored in 8-bit bytes.EBCDIC: 8-bit code used predominantly on IBM mainframes.Question focuses on what they differ in most fundamentally.Concept / Approach:Collating sequence is the order assigned to characters for comparison and sorting.ASCII and EBCDIC place letters, digits, and punctuation in different relative orders.Both are typically stored in single bytes on most systems, so storage size is not the core difference in practice.Step-by-Step Explanation:Review ASCII: digits '0'–'9' (0x30–0x39) occur before uppercase letters (0x41–0x5A) and lowercase (0x61–0x7A).Review EBCDIC: letters and digits are arranged in non-contiguous blocks, yielding different sort orders.Implication: programs that sort strings will yield different results across ASCII vs EBCDIC without translation tables.Verification / Alternative check:Consider sorting strings that mix digits and letters; ASCII puts digits before letters, while EBCDIC order is different, proving collating-sequence differences.Why Other Options Are Wrong:Efficiency in storing data: Both are byte-oriented in practice; difference is not fundamental here.Random/sequential access method: This pertains to storage/media, not character encoding.Number of bytes used: Typically one byte each; ASCII is 7-bit but stored in a byte.None of the above: Incorrect because collating sequence differences are well-known.Common Pitfalls:Assuming ASCII always uses fewer bytes; on most systems it still occupies one byte.Confusing encoding with file I/O access methods.Final Answer:their collating sequences

Question 7

Express −7 in 16-bit two’s complement (signed) binary form.

Accepted Answer

Introduction / Context:Two’s complement is the dominant representation for signed integers in computer systems because it yields a single representation for zero, simplifies arithmetic, and keeps addition/subtraction hardware uniform. This question asks for −7 in 16-bit two’s complement.Given Data / Assumptions:Word size: 16 bits.Target value: −7.Representation system: two’s complement.Concept / Approach:To obtain −N in two’s complement: write +N in binary, invert all bits (one’s complement), then add 1.Ensure the width (16 bits) is maintained throughout.Step-by-Step Solution:Write +7 in 16 bits: 0000 0000 0000 0111.Invert all bits (one’s complement): 1111 1111 1111 1000.Add 1: 1111 1111 1111 1000 + 0000 0000 0000 0001 = 1111 1111 1111 1001.Therefore, −7 = 1111 1111 1111 1001 (16-bit two’s complement).Verification / Alternative check:Add +7 and −7 in 16-bit two’s complement: 0000 0000 0000 0111 + 1111 1111 1111 1001 = 1 0000 0000 0000 0000 (discard carry) → 0000 0000 0000 0000, confirming correctness.Why Other Options Are Wrong:0000 0000 0000 0111: That is +7, not −7.1000 0000 0000 0111: Sign-magnitude-like form, not valid two’s complement for −7.1111 0000 0000 0001 and 0111 0000 0000 0000: Do not correspond to two’s complement of 7.None of the above: Correct here only because the accurate two’s complement (1111 1111 1111 1001) is not listed among the options.Common Pitfalls:Forgetting to fix the bit-width before inversion and addition.Confusing sign-magnitude or one’s complement with two’s complement.Final Answer:1111 1111 1111 1001 (not listed), so 'None of the above' is correct.

Question 8

Add the binary numbers 10₂ and 11₂ and express the sum in binary.

Accepted Answer

Introduction / Context:Binary addition is fundamental to arithmetic logic units (ALUs) and digital circuits. This item reinforces carry handling when summing two small binary numbers.Given Data / Assumptions:Operands: 10₂ and 11₂.No signed interpretation; simple unsigned addition.Concept / Approach:Binary addition rules: 0+0=0, 0+1=1, 1+0=1, 1+1=0 with carry 1.Align bits from least significant to most significant and add with carries.Step-by-Step Solution:Write vertically: 10₂ + 11₂.LSB column: 0 + 1 = 1 (carry 0).Next column: 1 + 1 = 0 with carry 1.Append carry to the left: result = 101₂.Verification / Alternative check:Convert to decimal: 10₂=2, 11₂=3; 2+3=5; 5 in binary is 101₂.Why Other Options Are Wrong:100₂: Equals 4, not 5.111₂: Equals 7, not 5.110₂: Equals 6, not 5.None of the above: Incorrect because 101₂ is correct.Common Pitfalls:Dropping the final carry when 1+1 occurs.Misaligning bit columns when adding.Final Answer:101₂

Question 9

The 8-bit binary value 1100 0101 corresponds to how many bytes of data?

Accepted Answer

Introduction / Context:Computer memory and data movement are measured in bits and bytes. Knowing how many bits form a byte is fundamental for understanding storage sizes, data alignment, and instruction encoding. Here we translate an 8-bit sequence into the number of bytes it represents.Given Data / Assumptions:Binary sequence: 1100 0101 (8 bits shown, grouped 4+4 for readability).Definition: 1 byte = 8 bits.No parity, start/stop, or framing bits are implied.Concept / Approach:Count the number of bits in the sequence.Divide by 8 to obtain the number of bytes.Step-by-Step Solution:Count bits: '1100 0101' contains 8 bits.Apply definition: 8 bits = 1 byte.Therefore, the sequence occupies exactly 1 byte.Verification / Alternative check:Any 8-bit pattern—from 0000 0000 to 1111 1111—constitutes 1 byte; 1100 0101 is one such pattern.Why Other Options Are Wrong:2, 4, 8 bytes: Would correspond to 16, 32, 64 bits respectively, which are not shown.None of the above: Incorrect because 1 is exactly correct.Common Pitfalls:Confusing nibble (4 bits) with byte (8 bits).Assuming whitespace in grouped bits adds extra bits—spacing is for readability only.Final Answer:1

Question 10

In transistor circuits, when a BJT is driven into cutoff or saturation, transistor parameter variations have almost no effect on circuit operation.

Accepted Answer

Introduction / Context:In switching applications, bipolar junction transistors (BJTs) are commonly operated in two extreme regions: cutoff (no conduction) and saturation (maximum conduction). Designing for these regions reduces sensitivity to device parameter variations, improving robust digital switching behavior.Given Data / Assumptions:Operating regions considered: cutoff and saturation.Application focus: switching, not linear amplification.Parameters of concern include beta (current gain), V_BE, and V_CE(sat).Concept / Approach:Cutoff: Base-emitter junction is not forward-biased; collector current is ~0, largely independent of beta.Saturation: Both junctions forward-biased; transistor behaves like a low resistance from collector to emitter; small changes in beta have limited effect on logic-level outcome.Digital logic that drives transistors hard ON or hard OFF becomes resilient to parameter spread and temperature.Step-by-Step Explanation:At cutoff, base drive is below threshold → no conduction; output determined by external pull-ups/pull-downs, not beta.At saturation, ample base drive ensures V_CE is near V_CE(sat); collector current set mainly by external resistances and supply, not precise beta.Therefore, device-to-device 'variations' minimally affect the intended logic state in these extremes.Verification / Alternative check:Practical design rule: provide base overdrive (I_B ≥ I_C / beta_forced) to guarantee saturation regardless of beta spread.Why Other Options Are Wrong:wave: Not a relevant technical factor here.stage: Refers to circuit blocks, not parameter sensitivity.circuits: Too vague; statement concerns parameter variations.None of the above: Incorrect since 'variations' correctly captures the effect.Common Pitfalls:Assuming linear-region design; in linear mode beta variations matter far more.Under-driving the base, leaving the device in the active region instead of saturation.Final Answer:variations

Question 11

In TTL (Transistor–Transistor Logic), the maximum number of standard TTL inputs that a given TTL output can drive reliably across the specified temperature range is called the fanout.

Accepted Answer

Introduction / Context:Digital designers must know how many inputs a logic output can drive without violating voltage levels or timing. In classic TTL families, this capability is standardized and captured by a parameter known as fanout. Correctly sizing fanout prevents loading-induced logic errors and excessive propagation delays.Given Data / Assumptions:We are discussing TTL logic families and standard loads.Temperature and supply variations are within data-sheet ranges.All driven inputs are comparable TTL loads.Concept / Approach:Fanout is defined as the number of standard inputs that a single output can drive while maintaining valid logic levels (VOH/VOL) and noise margins. It relates to output source/sink current limits and input currents (IIH/IIL) of the loads. Designers sum load currents and compare against output current capabilities, also considering worst-case data-sheet values and temperature derating.Step-by-Step Solution:Identify the relevant TTL parameters: IOH, IOL, IIH, IIL.Compute allowable loads: fanout = min( IOH/IIH_max , IOL/IIL_max ).Ensure margins across temperature/supply extremes.Name the parameter that captures this count: fanout.Verification / Alternative check:Manufacturer data sheets list nominal and worst-case fanout numbers for popular families (e.g., 74LS, 74HC with CMOS equivalents), confirming the term and its usage.Why Other Options Are Wrong:Bipolar: describes device technology (BJT), not the load count.Chip: generic term for integrated circuit.Universal logic circuit: informal phrase, not a standard parameter.None of the above: incorrect because the correct term is fanout.Common Pitfalls:Ignoring mixed-family loading (TTL to CMOS requires level considerations); overlooking dynamic loading and rise-time effects; failing to count pull-ups/pull-downs that affect current budgets.Final Answer:Fanout

Question 12

Integrated circuits (ICs) pack more circuitry into less area. What are the practical advantages of ICs over equivalent discrete-device circuits due to this higher functional density?

Accepted Answer

Introduction / Context:Integration places many transistors and passive elements on a single die, replacing assemblies of discrete components. This shift revolutionized electronics by shrinking products, increasing robustness, and driving down cost per function. Understanding why these gains occur helps in technology selection and system economics.Given Data / Assumptions:ICs implement the same functions as discrete circuits in far less physical space.Manufacturing uses highly automated wafer fabrication and packaging.Yield, reliability, and economies of scale affect cost and quality.Concept / Approach:Smaller size: integration eliminates interconnects and bulky passives, enabling compact products. Higher reliability: fewer solder joints and interconnects reduce failure points; controlled semiconductor processes improve consistency. Lower cost: mass production and learning curves decrease cost per transistor; assembly and logistics are simpler than managing many discrete parts. These benefits reinforce each other—miniaturization reduces materials and improves thermal and signal integrity, further boosting reliability.Step-by-Step Solution:Compare discrete assembly (many parts, many joints) to a single-chip IC (few joints).Relate interconnect reduction to reliability (lower MTBF degradation from joints).Connect wafer-scale economies to cost reductions per function.Conclude that all listed advantages apply.Verification / Alternative check:Historical cost curves show exponential drops in cost/transistor and rising reliability metrics as integration increased from SSI to VLSI, validating the multi-faceted advantage of ICs.Why Other Options Are Wrong:Each single benefit is correct but incomplete alone.None of the above contradicts decades of semiconductor progress.Common Pitfalls:Assuming ICs are always cheaper regardless of volume—at very low quantities or with exotic processes, discrete may be economical; ignoring heat dissipation and ESD handling needs for dense ICs.Final Answer:All of the above

Question 13

In synchronous digital design with positive clocking, the flip-flop responds (captures input and changes state) only when the clock signal is at its active high condition (e.g., during a high level or a rising edge, depending on design).

Accepted Answer

Introduction / Context:Flip-flops are the building blocks of synchronous sequential circuits. Their state changes in response to a clock signal, which defines when data is sampled and propagated. Positive clocking means the active portion of the clock is the high condition (often specifically the rising edge in edge-triggered devices). Understanding the active clock condition is crucial for correct timing design.Given Data / Assumptions:The term “positive clocking” is used to denote an active-high clocking convention.We are considering standard level- or edge-sensitive flip-flops.Only one answer identifies the required clock condition.Concept / Approach:With positive clocking, level-sensitive latches are transparent when clock is high; edge-triggered flip-flops capture data on the positive-going (rising) edge. Either interpretation maps to an active-high condition. This ensures consistent data capture and avoids hazards when combined with proper setup and hold timing.Step-by-Step Solution:Interpret “positive clocking” as active-high behavior.Relate to device behavior: transparent or sampling when high/rising.Select “High” as the proper clock condition.Verification / Alternative check:Data sheets for standard D flip-flops show a triangle symbol at the clock input for positive-edge devices, indicating sampling on the rising edge (active-high clocking), confirming the principle.Why Other Options Are Wrong:Low: corresponds to negative/active-low clocking.Set: refers to asynchronous set input, not clock level.Race: undesired condition, not a valid clock requirement.None of the above: incorrect because “High” is correct.Common Pitfalls:Mixing level-sensitive and edge-triggered interpretations; ignoring setup/hold times; failing to align clock domains leading to metastability.Final Answer:High

Question 14

ASCII (American Standard Code for Information Interchange) is a 7-bit character code used to represent letters, numbers, and a wide range of control and punctuation symbols.

Accepted Answer

Introduction / Context:ASCII is a foundational character encoding that maps binary values to textual characters. Although modern systems often use Unicode, understanding ASCII’s scope is essential for data interchange, protocols, and legacy systems. ASCII uses 7 bits, allowing 128 code points including control characters and printable symbols.Given Data / Assumptions:ASCII includes uppercase and lowercase letters.ASCII includes digits 0–9.ASCII includes punctuation and control codes (e.g., CR, LF, ESC).Concept / Approach:The 7-bit range (0–127) covers control characters (0–31, 127) and printable characters (32–126): space, digits, uppercase A–Z, lowercase a–z, and punctuation. Therefore, ASCII clearly represents letters, numbers, and other symbols, not just one category. Extended 8-bit “ASCII” variants add locale-specific symbols but the core remains 7-bit.Step-by-Step Solution:List ASCII categories: control, digits, letters, punctuation.Confirm inclusion of all three categories in the options.Select the comprehensive choice: all of the above.Verification / Alternative check:ASCII tables show, for example, 48–57 for digits, 65–90 for uppercase letters, 97–122 for lowercase letters, and various punctuation codes—verifying coverage.Why Other Options Are Wrong:Letters only or numbers only: incomplete and thus incorrect.Other symbols only: incomplete and misses alphanumerics.None of the above: false because ASCII includes all categories.Common Pitfalls:Confusing ASCII with Unicode; assuming ASCII is 8-bit; forgetting control characters and whitespace are part of ASCII and crucial to protocols and file formats.Final Answer:All of the above

Question 15

Convert the binary number 110001 (base 2) to its decimal (base 10) value. Show the place-value reasoning clearly.

Accepted Answer

Introduction / Context:Converting from binary (base 2) to decimal (base 10) reinforces understanding of positional number systems. Each bit represents a power of 2 based on its position from the rightmost bit (least significant bit). We will convert 110001 (base 2) to decimal using place-value expansion.Given Data / Assumptions:Binary digits are 1 1 0 0 0 1 from left to right.Positions from rightmost: 2^0, 2^1, 2^2, 2^3, 2^4, 2^5.Only positions with bit 1 contribute to the sum.Concept / Approach:Interpret the binary as a weighted sum: value = sum(bit_i * 2^i). For 110001, bits set to 1 are at positions 5, 4, and 0 (counting from 0 on the right). Therefore, decimal value = 2^5 + 2^4 + 2^0 = 32 + 16 + 1.Step-by-Step Solution:Write powers of two: 2^0=1, 2^1=2, 2^2=4, 2^3=8, 2^4=16, 2^5=32.Map bits: 1 1 0 0 0 1 corresponds to 2^5 + 2^4 + 02^3 + 02^2 + 0*2^1 + 2^0.Compute the sum: 32 + 16 + 1 = 49.State the decimal value: 49.Verification / Alternative check:Group as 110000 + 000001 (base 2). 110000 (base 2) = 48, and 000001 (base 2) = 1. Sum = 49. Both methods agree.Why Other Options Are Wrong:27: would correspond to 11011 (base 2), not 110001.39: corresponds to 100111 (base 2), not 110001.50: would be 110010 (base 2), one more than our number.None of the above: incorrect because 49 is exactly correct.Common Pitfalls:Misaligning bit positions; forgetting that any 0 bit contributes nothing; reading from left instead of right when assigning powers; arithmetic slips when adding powers of two quickly.Final Answer:49

Question 16

A microcomputer with 64K of addressable memory has hexadecimal addresses ranging from the first location 0000 to the last location FFFF (inclusive). Identify the correct first and last addresses.

Accepted Answer

Introduction / Context:Address ranges in microcomputers are often expressed in hexadecimal. A “64K” memory space contains 65,536 addresses, typically numbered from 0000 to FFFF in hex, where each address points to one byte. Recognizing these limits is foundational for memory maps, stack placement, and I/O decoding.Given Data / Assumptions:Total size is 64K bytes = 65,536 decimal addresses.Hexadecimal digits span 0–F, so four hex digits represent 0–65,535.Addressing is byte-addressable and contiguous.Concept / Approach:Four hex digits cover 16^4 = 65,536 values. The first value is 0000 (decimal 0). The last value is FFFF (decimal 65,535). Therefore, a 64K space exactly fits within 0000–FFFF. Other pairs either undercount or are not valid hex ranges for 64K.Step-by-Step Solution:Compute range size: 16^4 = 65,536 addresses.Map first address: 0000.Map last address: FFFF.Select the pair that matches this mapping.Verification / Alternative check:Decimal check: 0000 = 0, FFFF = 65,535; count of addresses = 65,536 = 64 * 1024 = 64K, confirming correctness.Why Other Options Are Wrong:0000, EEEE: ends at 61,182 decimal, not 64K.0, 64: mixes decimal and lacks 4-digit hex formatting.0000, 9999: stops at 39,321 decimal; not the 64K boundary.None of the above: incorrect since 0000, FFFF is correct.Common Pitfalls:Confusing K as 1000 vs 1024; forgetting that hex FFFF equals 65,535; mixing numeric bases when reading address ranges.Final Answer:0000, FFFF

Question 17

An EXCLUSIVE–NOR (XNOR) gate is logically equivalent to an EXCLUSIVE–OR (XOR) gate followed by an inverter (logical NOT). Select the correct equivalence.

Accepted Answer

Introduction / Context:Composite logic design often relies on equivalences between gates. Understanding that XNOR is the complement of XOR simplifies synthesis and analysis of digital circuits, especially when implementing parity checks, equality comparators, and error-detecting logic.Given Data / Assumptions:XOR outputs 1 when inputs differ and 0 when they are equal.XNOR outputs 1 when inputs are equal and 0 when they differ.An inverter flips a logic value: 0 becomes 1, 1 becomes 0.Concept / Approach:If Y = XOR(A, B), then XNOR(A, B) = NOT(Y). Therefore, cascading an XOR gate with a NOT gate yields XNOR behavior. This equivalence is used in hardware to build equality comparators by XNORing bits and then ANDing the results to confirm all bits are equal.Step-by-Step Solution:Define XOR truth: 0⊕0=0, 0⊕1=1, 1⊕0=1, 1⊕1=0.Apply inversion: NOT(0)=1, NOT(1)=0 to each case.Observe the result matches XNOR truth: 1 when inputs equal, 0 otherwise.Conclude that “XOR followed by inverter” implements XNOR.Verification / Alternative check:Boolean algebra: XNOR = A·B + A'·B' = (A ⊕ B)'; reading “double prime” as NOT of XOR, confirming the cascade equivalence.Why Other Options Are Wrong:OR: cannot produce equality detection by simple inversion.AND/NAND: do not match the “equal inputs are 1” truth table when inverted.None of the above: incorrect because XOR is the correct precursor to an inverter for XNOR.Common Pitfalls:Confusing XOR with OR; forgetting that XNOR is “equivalence” (outputs 1 when inputs are the same). In multi-bit comparisons, remember to combine XNOR results with AND to detect equality across all bits.Final Answer:XOR

Question 18

In computer architecture, the unit that integrates the control section and the arithmetic-logic section—coordinating instruction sequencing, data movement, and arithmetic/logic operations—is commonly referred to as the:

Accepted Answer

Introduction / Context:A modern digital computer is organized into functional subsystems. Two of the most central are the control section, which fetches and sequences instructions, and the arithmetic-logic section, which performs calculations and logical decisions. Understanding the umbrella term that encapsulates both is foundational for computer organization and systems design.Given Data / Assumptions:The control section orchestrates instruction fetch, decode, and control signal generation.The arithmetic-logic section (ALU) executes arithmetic and logic operations on data.We seek the common name for the component that comprises both sections together.Concept / Approach:In standard nomenclature, the Central Processing Unit (CPU) is the component that combines the control unit (CU) and the arithmetic-logic unit (ALU). The CPU executes program instructions, coordinating with memory and I/O subsystems. While a “control unit” is part of the CPU, it does not include the ALU by itself; a block diagram is only a drawing; and the Input/Output unit handles peripherals, not core instruction execution.Step-by-Step Solution:Identify roles: CU controls sequencing; ALU computes.Recognize the standard grouping: CU + ALU = CPU.Exclude distractors: a diagram is not a functional unit; I/O is peripheral handling.Conclude the integrated name is “Central Processing Unit.”Verification / Alternative check:Introductory architecture texts explicitly define the CPU as the union of CU and ALU, sometimes also including registers and immediate datapaths as internal CPU resources.Why Other Options Are Wrong:Block diagram: a representation, not a device.Control Unit: only one half of the pair.Input / Output Unit: interfaces with peripherals, not core processing.None of the above: incorrect, a standard term exists.Common Pitfalls:Confusing “processor” with the entire computer; conflating CPU with GPU or specialized accelerators; assuming I/O control is part of the CPU rather than coordinated by it.Final Answer:Central Processing Unit

Question 19

In binary fixed-point arithmetic, compute the 2’s complement of the binary number 010111.1100 (invert all bits and add 1 across the fixed-point boundary). Which option matches the correct 2’s complement representation?

Accepted Answer

Introduction / Context:Two’s complement is the most common representation for signed binary numbers. For fixed-point values (with a binary point), the method is the same as for integers: invert every bit and then add 1 to the least significant position, carrying across the fractional boundary if necessary. This question reinforces precise bitwise manipulation for fractional binaries.Given Data / Assumptions:Original number: 010111.1100 (six integer bits and four fractional bits).We apply two’s complement over the entire fixed-width pattern, including the fractional part.Binary point location remains fixed.Concept / Approach:Compute two’s complement in two steps: (1) one’s complement (bitwise invert) and (2) add 1 at the least significant fractional bit. Care must be taken to preserve the bit positions on both sides of the binary point. The procedure mirrors integer two’s complement but with the fixed-point layout retained.Step-by-Step Solution:Start with 010111.1100.Invert all bits (one’s complement): 101000.0011.Add 1 to the least significant bit position: 101000.0011 + 0.0001 = 101000.0100.Keep the binary point fixed; no overflow occurs beyond the fixed width.Match with options: 101000.0100.Verification / Alternative check:Reverse the process: two’s complement again of 101000.0100 should return the original. Invert to 010111.1011 and add 1 → 010111.1100, confirming correctness.Why Other Options Are Wrong:101001.1100: incorrect bit pattern; not the add-1 result after inversion.010111.0011: this is not a two’s complement; looks like partial inversion.101000.0011: only one’s complement, missing the +1 step.None of the above: incorrect since a correct option exists.Common Pitfalls:Forgetting to add 1 after inversion; shifting the binary point; performing addition only to integer bits and ignoring fractional carries.Final Answer:101000.0100

Question 20

Number base conversion: convert the binary number 1010000₂ into its equivalent octal representation by grouping bits in sets of three (from the right).

Accepted Answer

Introduction / Context:Binary–octal conversion is efficient because 8 = 2^3, so every octal digit corresponds to exactly three binary bits. Grouping the binary string into triples from the least significant side (rightmost) yields a quick, error-resistant method to obtain the octal value without intermediate decimal conversion.Given Data / Assumptions:Binary input: 1010000₂.We may pad leading zeros to complete a leftmost group of three bits.No sign bit is involved; this is an unsigned magnitude conversion.Concept / Approach:Partition the binary sequence into 3-bit groups from right to left. Translate each group to its octal digit using the mapping 000→0, 001→1, 010→2, 011→3, 100→4, 101→5, 110→6, 111→7. Concatenate the resulting octal digits.Step-by-Step Solution:Binary: 1 010 000 (visual grouping from right).Pad leading zeros if desired: 001 010 000.Map groups: 001→1, 010→2, 000→0.Combine digits: 120 in base 8.Answer: 120₈.Verification / Alternative check:Convert to decimal as a check: 1010000₂ = 64 + 16 = 80. Now 120₈ = 164 + 28 + 0 = 64 + 16 = 80. Both match.Why Other Options Are Wrong:119₈, 101₈, 111₈: do not evaluate to 80 in decimal.None of the above: incorrect because 120₈ is correct.Common Pitfalls:Grouping from the left instead of the right; forgetting to pad leading zeros; misreading 3-bit groups when zeros occur.Final Answer:120₈

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