Question 1

Binary addition fact — evaluate the statement: “The addition 1 + 0 doesn’t generate a carry bit; one does not exist.” Consider one-bit addition rules and carry generation conditions.

Accepted Answer

Introduction / Context: Binary addition follows simple truth rules for single bits with an optional carry-in. Understanding when a carry is generated is fundamental to adder design (half adders, full adders) and to analyzing arithmetic overflows. Here, we assess whether adding 1 and 0 alone generates a carry. Given Data / Assumptions: One-bit addition without an incoming carry (carry-in = 0). Values being added: 1 and 0. We focus on carry generation in that minimal case. Concept / Approach: Binary single-bit addition tables state: 0+0 → sum 0, carry 0; 0+1 or 1+0 → sum 1, carry 0; 1+1 → sum 0, carry 1. Carries occur when the arithmetic sum is 2 or greater. The pair 1 and 0 sums to 1, which is below the threshold for carry generation. Step-by-Step Solution: Compute 1 + 0 with no carry-in.Sum = 1; carry-out = 0.Therefore, no carry bit is produced in this operation.If a carry-in of 1 existed, then 1 + 0 + carry-in would equal 2 → sum 0, carry 1 (different case). Verification / Alternative check: Truth tables for half adders confirm the result; full adder truth tables show the dependency on carry-in. Implementation in logic (XOR for sum, AND for carry in the 1+1 case) reflects the same rule. Why Other Options Are Wrong: Incorrect: Conflicts with standard adder logic. “Only correct if previous carry-in is zero”: The statement already assumes no carry-in; with carry-in = 1, the situation changes, but that is a different scenario. “Only correct for signed addition”: Signedness affects interpretation of overflow, not single-bit carry rules. Common Pitfalls: Confusing carry with overflow; in signed arithmetic, overflow detection has separate rules and is not identical to carry-out in two’s complement. Final Answer: Correct — 1 + 0 (with no carry-in) yields sum 1 and no carry.

Question 2

Hexadecimal to decimal — evaluate the statement: “64 hexadecimal equals 100 decimal.” Treat “64 hexadecimal” as the base-16 number 0x64 and verify its base-10 value.

Accepted Answer

Introduction / Context: Converting between hexadecimal (base 16) and decimal (base 10) is a routine skill in computer architecture, low-level programming, and digital electronics. This question asks you to verify whether the hex number “64” represents the decimal number 100. Given Data / Assumptions: Hexadecimal digits: 0–9, A(10), B(11), C(12), D(13), E(14), F(15). Number to convert: 0x64 (hex). Positional weights in base 16: 16^1 and 16^0 for the two digits. Concept / Approach: To convert hex to decimal, multiply each digit by its positional weight and sum. For 0x64, the left digit is 6 and the right digit is 4. Therefore, value = 6*16 + 4*1. Compute and compare with 100 in decimal to verify the statement. Step-by-Step Solution: Identify digits → 6 (sixteens place) and 4 (ones place).Compute sixteens place → 6 * 16 = 96.Compute ones place → 4 * 1 = 4.Add → 96 + 4 = 100 decimal. Verification / Alternative check: Use a mental cross-check: 0x64 is often remembered because 0x64 = 100, 0x32 = 50, and 0x0A = 10 in decimal. These are common anchor points for programmers. Why Other Options Are Wrong: Incorrect: The arithmetic confirms the statement. “Correct only for base-8” / “ignore the trailing 4”: Both are irrelevant or nonsensical to proper base-16 conversion. Common Pitfalls: Misinterpreting “64” as decimal 64 instead of hex 0x64; forgetting place values differ by powers of 16 in hexadecimal. Final Answer: Correct — 0x64 equals 100 in decimal.

Question 3

Terminology — evaluate the statement: “Base is the same as radix.” In number-system language, are the terms “base” and “radix” interchangeable?

Accepted Answer

Introduction / Context: In number systems, the size of the digit set and the positional weighting depend on a fundamental parameter often called the “base” or “radix.” Students encounter both terms in digital electronics, computer science, and mathematics. This question clarifies whether the two words are synonymous in this context. Given Data / Assumptions: Digit sets correspond to base b (or radix r), yielding digits 0..b−1. Common examples: binary (base/radix 2), octal (8), decimal (10), hexadecimal (16). The terms are used in textbooks and standards. Concept / Approach: “Base” and “radix” are interchangeable names for the same concept: the number of unique digit symbols and the factor by which place value increases when moving left in a positional numeral system. Each position represents powers of the base/radix: b^0, b^1, b^2, etc. The term “radix” is often used in more formal mathematical or computer-science contexts, but it does not denote a different quantity from “base.” Step-by-Step Solution: Define base/radix → number of symbols and place-value multiplier.Provide examples → base 2 (0–1), base 10 (0–9), base 16 (0–9, A–F).Conclude → the two terms name the same parameter.Hence the statement is correct. Verification / Alternative check: Documentation for programming languages (e.g., integer parsing functions) and mathematics texts routinely use “radix/base” interchangeably when describing conversions and representations. Why Other Options Are Wrong: Incorrect: Contradicts standard usage. “Only correct for binary/decimal” or “only in floating-point” adds unnecessary constraints; base/radix applies across all positional systems. Common Pitfalls: Confusing “radix point” (generalized decimal point) with the radix value itself; both concepts are related but not identical. Final Answer: Correct — “base” and “radix” refer to the same concept in positional number systems.

Question 4

Binary Coded Decimal (BCD) — in digital electronics, BCD represents each decimal digit 0–9 independently. Each single decimal digit is encoded as which fixed-length binary pattern?

Accepted Answer

Introduction / Context: Binary Coded Decimal (BCD) is a widely used number representation in digital systems where each individual decimal digit is encoded separately. Unlike pure binary, BCD preserves the decimal digit boundaries, which simplifies human-readable I/O in calculators, digital meters, and financial devices. Given Data / Assumptions: We are encoding a single decimal digit 0–9. Standard BCD uses “8421” weighting per digit position. We are not discussing packed formats beyond the per-digit code width. Concept / Approach: In BCD, each decimal digit is represented by a 4-bit binary code (weights 8, 4, 2, 1). Thus 0→0000, 1→0001, …, 9→1001. Multi-digit decimal numbers are formed by concatenating these 4-bit nibbles. This differs from straight binary, where the entire number is encoded as a single binary value with place weights 2^0, 2^1, 2^2, etc. BCD keeps conversion to and from decimal trivial at the digit level and avoids rounding issues in some financial computations. Step-by-Step Solution: Identify the question scope: one decimal digit (0–9).Recall the standard BCD mapping: 0–9 use 4-bit codes from 0000 to 1001.Conclude that the fixed length per digit in standard BCD is 4 bits.Reject longer widths (8-bit, 16-bit) since they are unnecessary for a single digit. Verification / Alternative check: Check the digit 9: 1001 is a 4-bit pattern. Digits 10–15 do not exist in decimal, so the six remaining 4-bit combinations (1010–1111) are “invalid” in basic BCD and may be unused or reserved in packed formats. Why Other Options Are Wrong: 8-bit/16-bit: excessive width for a single decimal digit; used for bytes/words, not BCD per digit.ASCII: encodes characters; the ASCII code for '3' is 0x33, not a BCD digit code.Binary floating-point: represents real numbers with mantissa/exponent, unrelated to BCD digit encoding. Common Pitfalls: Confusing “binary representation of the whole number” with “BCD per digit”; forgetting that some 4-bit patterns are invalid in basic BCD. Final Answer: 4-bit binary code

Question 5

Unit-distance coding — which binary code progresses so that only one bit changes between any two successive code words (minimizing transition errors)?

Accepted Answer

Introduction / Context: When digital systems encode ordered states (such as shaft angles or counter positions), changing multiple bits at once can create transient errors if not all lines switch simultaneously. A “unit-distance” code mitigates this by ensuring that adjacent values differ by only one bit. Given Data / Assumptions: We need a code where successive values differ by Hamming distance 1. Use case includes encoders, ADCs, and state machines where transition glitches must be minimized. We consider common codes: weighted BCD, complements, and Gray. Concept / Approach: Gray code (reflected binary code) is constructed so that each successive word differs from the previous one by exactly one bit. This property reduces ambiguity during transitions because even if lines switch at slightly different times, at most one bit is in flux, which minimizes decoding errors. Weighted codes like 8421 BCD can change multiple bits between successive numbers (e.g., 7→8 toggles four bits), increasing the risk of transient misreads. Step-by-Step Solution: Define unit distance: Hamming distance of 1 between neighbors.Recall Gray code sequence: 00, 01, 11, 10 for 2-bit; generalizes by reflection.Check weighted codes (8421): observe multi-bit toggles between sequential decimal values.Conclude that Gray code uniquely satisfies the one-bit-change criterion. Verification / Alternative check: Create a small 3-bit table and compare transitions: Gray code shows single-bit steps while straight binary shows multi-bit steps such as 3 (011) to 4 (100), which toggles three bits. Why Other Options Are Wrong: 9's complement/excess-3: intended for decimal arithmetic and self-complementing properties, not unit-distance transitions.8421 BCD: weighted; adjacent decimal digits often change several bits.ASCII: character code, not numeric progression code. Common Pitfalls: Confusing “nonweighted” with “unit distance”; some codes are nonweighted yet not strictly single-bit transition for all sequences. Final Answer: Gray code

Question 6

Bit significance — in a fixed-width binary number, changing which bit causes the largest change in numeric value for a single-bit toggle?

Accepted Answer

Introduction / Context: Binary numbers assign positional weights that are powers of 2. Understanding bit significance is essential for error analysis, resolution calculations, and designing counters or DACs/ADCs where bit toggles directly map to step sizes in value or voltage. Given Data / Assumptions: A fixed binary word length is used (e.g., 8-bit). Each position weight doubles as you move toward more significant bits. We toggle exactly one bit and compare its impact on numeric value. Concept / Approach: The weight of each bit position is 2^n, with n increasing toward the MSB. Therefore, toggling the MSB changes the number by the largest magnitude among single-bit changes since its increment or decrement equals 2^(N−1) for an N-bit unsigned word. Conversely, toggling the LSB changes the value by only 1. This logic holds for signed magnitude or two’s complement as well, where the MSB influences sign or large magnitude changes. Step-by-Step Solution: Identify bit weights: LSB = 2^0 = 1; next = 2; …; MSB = 2^(N−1).Compare magnitudes: 2^(N−1) is greater than any other single-bit weight.Therefore, toggling the MSB produces the largest value step.Confirm with an example: in 8-bit, MSB toggle changes value by 128. Verification / Alternative check: Consider DAC resolution: the MSB contributes half of the full-scale range, reinforcing its dominant effect on output changes. Why Other Options Are Wrong: LSB: smallest unit change (1) only.Duty cycle/frequency/parity bits: not positional weights in numeric representation; they are control/metadata concepts. Common Pitfalls: Confusing bit roles in data frames (parity, control) with positional weights in numeric binary representation; mixing two’s complement sign effects with magnitude analysis without considering absolute value change. Final Answer: Most Significant Bit (MSB)

Question 7

Foundational base — digital electronics, logic gates, and computer arithmetic are fundamentally built on which numbering system?

Accepted Answer

Introduction / Context: Modern digital hardware represents information using discrete voltage levels mapped to symbols. The most robust and economical choice is two stable states, enabling noise-immune switching and straightforward implementation with transistors. This naturally aligns with the binary numbering system. Given Data / Assumptions: Logic circuits use two distinct states (low/high). Voltage thresholds separate 0 from 1 to provide noise margins. Other radices may be used for human representation but not for core hardware operation. Concept / Approach: The binary system (base 2) uses digits 0 and 1. In hardware, these correspond to two voltage ranges (e.g., 0 V and +V). Boolean algebra maps logic operations (AND, OR, NOT) to simple transistor networks. While hexadecimal and octal are convenient shorthand for grouping bits (4 and 3 respectively), they are not the physical basis; they simply compress binary for readability. Decimal is used in human interfaces and BCD storage when necessary, but internal logic remains binary. Step-by-Step Solution: Identify physical implementation: two-state devices (transistors, flip-flops).Relate states to symbols: assign 0 and 1 to voltage ranges.Conclude the fundamental numbering system is binary.Note that other bases are derived not native for hardware switching. Verification / Alternative check: Observe that memory cells, registers, and buses store and transmit bits. Groupings (nibbles/bytes) are powers of 2, reinforcing the binary foundation. Why Other Options Are Wrong: Decimal: human-centric; inefficient for hardware implementation.Hex/Octal: compact encodings of binary, not the physical basis.Balanced ternary: interesting academically but not standard in modern hardware. Common Pitfalls: Assuming hex is “native” because of its prevalence in programming; it merely maps to 4-bit groups. Final Answer: Binary (base 2)

Question 8

Nonweighted, unit-distance identification — which of the following is a nonweighted binary code that also changes only one bit between successive values (a unit-distance code)?

Accepted Answer

Introduction / Context: Binary codes are often categorized by whether they are weighted (each position has a fixed weight) or nonweighted (no simple positional weights). In applications like position encoders and certain counters, a “unit-distance” property—successive code words differ by one bit—is highly desirable to minimize transition errors. Given Data / Assumptions: We need a code that is nonweighted. We also need only a single-bit change between successive values. Codes under consideration include Gray and Excess-3. Concept / Approach: Gray code (reflected binary) is both nonweighted and unit-distance. It is constructed recursively to ensure that adjacent code words differ by exactly one bit, which reduces decoding ambiguity during changes. Excess-3, while nonweighted and self-complementing, does not guarantee single-bit transitions between successive decimal digits and is designed for decimal arithmetic, not transition robustness. Weighted codes like 8421 BCD explicitly use positional weights and are therefore not nonweighted. Step-by-Step Solution: List properties: Gray → nonweighted + unit-distance.Check Excess-3: nonweighted, but not unit-distance for general sequences.Eliminate weighted code (8421) immediately.Select Gray code as the only option satisfying both criteria. Verification / Alternative check: Construct a short Gray sequence (3-bit): 000, 001, 011, 010, 110, 111, 101, 100 — observe one-bit transitions throughout. Compare to Excess-3 of decimal digits and note multiple-bit changes frequently occur. Why Other Options Are Wrong: Excess-3: nonweighted, but not unit-distance by design goal.Multi-bit parity/minival: not standard nonweighted unit-distance codes for ordered numeric progression.8421 BCD: weighted, so fails the nonweighted requirement. Common Pitfalls: Equating “nonweighted” with “unit-distance”; they are independent properties. Gray happens to possess both, which is why it is common in encoders. Final Answer: Gray code (reflected binary)

Question 9

Excess-3 (XS-3) BCD — which code represents each decimal digit by its 4-bit binary value after adding 3 to that digit (i.e., code = binary(digit + 3))?

Accepted Answer

Introduction / Context: Several BCD variants exist to simplify arithmetic and error detection. Excess-3 (also written XS-3) is a self-complementing code that shifts each decimal digit by 3 before encoding it in 4-bit binary. This property eases 9’s complement and subtraction operations in certain hardware designs. Given Data / Assumptions: Per-digit BCD with 4-bit fields. Mapping formula: codeword = binary(digit + 3). Valid for digits 0–9 (so 3–12 in binary before mapping; 10–12 codes are still 4-bit patterns). Concept / Approach: By adding 3 to the decimal digit, Excess-3 ensures that the 9’s complement of a number corresponds to bitwise complement of its code, simplifying subtraction via complements. For example, digit 0 maps to 0011, 1→0100, …, 9→1100. This is not a weighted code and differs from 8421 BCD, which maps 0→0000 through 9→1001 directly without the +3 offset. Step-by-Step Solution: Start with digit d in 0–9.Compute d + 3.Encode the result as a 4-bit binary value.Thus obtain the Excess-3 codeword for that digit. Verification / Alternative check: Take d = 5: 5 + 3 = 8 → 1000; d = 9: 9 + 3 = 12 → 1100. The mapping matches standard XS-3 tables. Why Other Options Are Wrong: 9's complement: arithmetic concept, not the code itself.8421: direct weighted BCD without the +3 shift.Gray: unit-distance code for sequencing, not decimal encoding.Two's complement: signed binary integer representation, not per-digit BCD. Common Pitfalls: Confusing Excess-3 with 8421 due to both using 4 bits; overlooking its self-complementing advantage for subtraction operations. Final Answer: Excess-3 code

Number Systems and Codes Questions