Question 1

Decimal→Binary conversion — “One valid method to convert a decimal integer to binary is successive division by 2 and collecting remainders.” Evaluate.

Accepted Answer

Introduction / Context: Converting between number bases is a common digital design and computer science task. For decimal integers, a standard manual technique is repeated division by the target base, recording remainders that become the digits in reverse order. This prompt asks whether that method applies to decimal→binary conversion. Given Data / Assumptions: We convert a nonnegative decimal integer to base 2. Division algorithm: divide by 2 until the quotient is 0; record each remainder (0 or 1). Binary digits read from last remainder to first remainder. Concept / Approach: In positional notation, the remainder sequence from successive division by the base yields the digits because each division peels off the least significant digit in that base. For base 2, remainders are bits. Reading them in reverse produces the correct binary representation. Step-by-Step Solution: Example: Convert 45.45/2 = 22 r1; 22/2 = 11 r0; 11/2 = 5 r1; 5/2 = 2 r1; 2/2 = 1 r0; 1/2 = 0 r1.Read remainders upward: 101101 (binary for 45). Verification / Alternative check: Multiply-and-add check: 132 + 016 + 18 + 14 + 02 + 11 = 45. Why Other Options Are Wrong: Incorrect: The method is standard for any base.Works only for even numbers: Odd numbers yield remainder 1; algorithm still works.Applicable to fractions only: Fractions use successive multiplication by base; integers use division. Common Pitfalls: Forgetting to reverse the remainder order.Mixing integer division (for integers) with fractional method (multiply by 2) for fraction parts. Final Answer: Correct

Question 2

BCD vs hexadecimal — Evaluate: “Hexadecimal is used to encode BCD numbers.” Clarify the distinction between 4-bit hex digits and BCD nibbles.

Accepted Answer

Introduction / Context: Hexadecimal (base 16) and Binary-Coded Decimal (BCD) both use 4-bit groups, which often leads to confusion. However, they encode different domains: hex digits represent values 0–15, while standard BCD encodes decimal digits 0–9 only. This question probes whether you can separate these encodings conceptually and practically. Given Data / Assumptions: Hex digits: 0–9, A–F mapping to 0–15. BCD digits: 0000–1001 valid; 1010–1111 are invalid in standard BCD. Use cases: hex for compact representation of arbitrary bit patterns; BCD for decimal-exact quantities. Concept / Approach: Although both use nibbles, BCD is not “encoded by hex.” BCD stores each decimal digit separately in 4 bits to preserve decimal semantics (e.g., 39 decimal → BCD 0011 1001). Hexadecimal is a base-16 numeral system; 0x27 equals decimal 39 but as a single base-16 number, not two decimal digits encoded. Treating hex and BCD as interchangeable causes value misinterpretations in registers and memory maps. Step-by-Step Solution: Represent 59 decimal in BCD: 0101 1001 (two nibbles).Represent 59 decimal in hex: 0x3B (0011 1011), a different bit pattern.Observe that BCD preserves decimal digits; hex represents a base-16 magnitude. Verification / Alternative check: Round-trip conversions confirm that hex and BCD agree numerically only after proper base interpretation, not by raw bit equality. Why Other Options Are Wrong: Correct: Would imply hex is the encoding mechanism for BCD, which is false.True only for values 0–9: While the hex symbols 0–9 exist, hex continues with A–F; BCD forbids A–F codes.Depends on signed magnitude: Sign representation is orthogonal to the radix/encoding distinction. Common Pitfalls: Reading a BCD byte as hex and getting the wrong decimal value.Using A–F nibbles in BCD fields, which are invalid. Final Answer: Incorrect

Question 3

Computer internal representation vs. character encoding: “A computer will use ASCII code to store information internally.” Decide whether this blanket statement is valid, considering that processors operate on binary data and that ASCII is only one …

Accepted Answer

Introduction / Context: Computers fundamentally store and process binary patterns. ASCII is a widely used character encoding for representing text, but it does not define how all information is stored internally. This question tests understanding of the difference between raw binary representation, data types, and text encodings such as ASCII, UTF-8, and others. Given Data / Assumptions: The statement claims computers use ASCII for internal storage in general. Internal data include instructions, integers, floating-point numbers, pointers, images, audio, and text. ASCII is a 7-bit code for textual characters; extended 8-bit variants and Unicode encodings also exist. Concept / Approach: Distinguish “binary representation” (bits and machine formats) from “character encoding” (mapping bytes to symbols). Only data that are text strings may use a character encoding. Numeric values, instructions, and multimedia use other binary formats defined by the ISA, IEEE 754, file formats, and codecs. Therefore, saying computers “use ASCII internally” overgeneralizes a specific text use-case to all data. Step-by-Step Solution: Identify the domain of ASCII: text characters only.Note other internal formats: twos-complement integers, IEEE 754 floats, machine opcodes.Acknowledge alternative encodings for text: UTF-8/UTF-16 supersede ASCII in many systems.Conclude that the blanket statement is not valid because most internal data are not ASCII. Verification / Alternative check: Inspecting a process’s memory dump reveals a mixture of code, numeric data, pointers, and only some regions containing recognizable ASCII strings. Compilers and CPUs treat numbers and instructions as binary patterns unrelated to ASCII tables. Why Other Options Are Wrong: Correct: Overgeneralizes; only text uses encodings. Only for text stored as plain characters: This is true but does not rescue the original blanket claim. Only in legacy 7-bit terminals: Too narrow and still not “all internal information.” Only for floating-point numbers: Floats use IEEE 754, not ASCII. Common Pitfalls: Confusing “human-readable” storage with all machine storage; assuming that because text prints as letters, everything is ASCII. Ignoring Unicode’s prevalence also leads to overstatements about ASCII. Final Answer: Incorrect

Question 4

Even parity rule in digital communications: “In even parity, the sum (count) of 1-bits in the code group must be even.” Determine if this definition is accurate and complete.

Accepted Answer

Introduction / Context: Parity bits provide a simple, low-cost method to detect single-bit errors during storage or transmission. The question checks knowledge of the precise rule that defines even parity and how the parity bit relates to the data bits. Given Data / Assumptions: A “code group” is the set of data bits plus, when present, a parity bit. Even parity aims for an even count of logic 1s across the group after adding the parity bit. Scope includes serial links (UART), memory systems, and simple buses where parity may be used. Concept / Approach: Parity enforces a constraint on the count of 1s. With even parity, the transmitter sets the parity bit so that the total number of 1s in data + parity is even. At the receiver, the same count is performed; a mismatch indicates an odd number of flipped bits (most notably a single-bit error). Step-by-Step Solution: Count the 1s in the data bits.If the count is already even, set parity bit to 0; if odd, set parity bit to 1.Transmit data plus parity; receiver repeats the count check.If total 1s are odd at the receiver, flag a parity error. Verification / Alternative check: Worked example: data 1011001 has four 1s (even). Even parity bit = 0; total remains even. If noise flips one bit, total becomes odd; error is detected. If two bits flip, parity may miss it (even number of errors). Why Other Options Are Wrong: Incorrect: The definition provided is the standard rule. Applies only to 7-bit ASCII or odd parity: Parity rules are independent of the particular payload width and parity selection (odd/even). Depends on stop bits: Stop bits are UART framing parameters separate from parity logic. Common Pitfalls: Confusing parity with checksums/CRCs; assuming parity guarantees multi-bit error detection (it does not); mixing up the direction (even vs. odd). Final Answer: Correct

Question 5

Hexadecimal usage in digital systems: “Many digital electronic systems work with hexadecimal rather than binary.” Evaluate this statement by considering internal hardware operation versus human-friendly representation.

Accepted Answer

Introduction / Context: Digital hardware operates on binary signals. However, engineers and tools often use hexadecimal (base-16) to represent those binary values compactly. This question tests whether the learner recognizes the practical prevalence of hex in interfaces, documentation, and tooling, even though the physical logic remains binary. Given Data / Assumptions: Binary is the hardware-native representation. Hexadecimal is a compact textual/visual representation of binary data. Software tools, datasheets, and instruction set documents frequently present constants, addresses, and masks in hex. Concept / Approach: Each hexadecimal digit maps exactly to 4 binary bits (a nibble). This one-to-one nibble mapping makes hexadecimal ideal for representing bytes, words, and bitfields in a readable form. Thus, while systems operate in binary, many workflows “work in hexadecimal” from the user’s perspective: entering addresses, inspecting memory, writing constants, or configuring registers. Step-by-Step Solution: Relate hex to binary: 1 hex digit ↔ 4 bits.Recognize common use: memory maps (0x2000), masks (0xFF00), opcodes.Acknowledge that internal logic signals remain binary-level voltages.Conclude: The statement is practically correct in everyday engineering usage. Verification / Alternative check: Examine any microcontroller reference manual: register addresses and bit masks are provided in hex; debuggers display memory and disassemblies in hex; network protocols and file formats are commonly documented in hex dumps for clarity. Why Other Options Are Wrong: Incorrect: Ignores widespread hex use in tools and docs. Only for memory addresses or assemblers: Too narrow; hex is used across logs, protocols, and UI readouts. Depends on CPU word size: Hex usability is independent of 8/16/32/64-bit widths. Common Pitfalls: Equating representation with implementation; thinking “hexadecimal logic” exists in hardware. The logic is binary; hex is a human-friendly lens on those bits. Final Answer: Correct

Question 6

Magnitude vs. representation: “When converting a decimal number to hexadecimal, the hex result will be a larger number than the original decimal.” Decide whether this claim makes sense in number-system conversions.

Accepted Answer

Introduction / Context: Conversions among bases (decimal, binary, hexadecimal) preserve numeric value. What changes is the representation—the symbols and their positions. The statement claims the hexadecimal result will be “larger,” which confuses value with how many digits or which symbols appear in a given base. Given Data / Assumptions: Decimal is base-10, hex is base-16. “Larger number” is interpreted as larger numeric value, not “more characters” or “visually higher symbols like A–F.” Standard positional weights apply: base^position. Concept / Approach: Base conversion is an isomorphism on natural numbers: value stays identical. For example, decimal 255 equals hex FF and binary 11111111; all three denote the same quantity. Digit count and glyph choices differ, but the magnitude does not change because we are only re-expressing the same value in a different radix. Step-by-Step Solution: Pick a number, e.g., 26.Convert to hex: 26 / 16 = 1 remainder 10 → 0x1A.Verify value: 1*16 + 10 = 26 (unchanged).Conclude that “larger number” is a misconception; only notation changes. Verification / Alternative check: Try another: decimal 12 = hex C. Although “C” looks like a single symbol, its value remains 12. Similarly, decimal 4096 = hex 1000; both represent the same magnitude with different digit counts due to base choice. Why Other Options Are Wrong: Correct: Would falsely imply value inflation through conversion. Only true above 255 / BCD / endianness: These factors do not alter mathematical equality across bases. Common Pitfalls: Equating string length or alphabetic ordering with numeric magnitude; assuming that “letters” in hex imply larger value than digits; confusing byte order (endianness) with base conversion (they are unrelated concepts). Final Answer: Incorrect

Question 7

Why hexadecimal is popular: “The primary advantage of the hexadecimal number system is the ease of conversion to and from binary.” Assess this rationale for using base-16 in digital design.

Accepted Answer

Introduction / Context: Engineers prefer hexadecimal because it maps neatly to binary, the native language of digital hardware. This question asks whether that mapping is the main reason hex dominates register maps, memory dumps, and low-level software tools. Given Data / Assumptions: Binary is base-2; hexadecimal is base-16. One hex digit corresponds to exactly four binary bits (a nibble). Digital artifacts are commonly byte (8-bit) aligned, making two hex digits per byte a perfect fit. Concept / Approach: The one-to-four mapping enables straightforward, error-resistant conversion without arithmetic. Group binary into nibbles and substitute hex digits; reverse by expanding each hex digit to four bits. This simplicity, combined with compactness, is precisely why hex is used in datasheets and debugging tools. Step-by-Step Solution: Take a binary value, e.g., 1011 0110.Group as nibbles: 1011 (B), 0110 (6) → 0xB6.Reverse easily: B → 1011, 6 → 0110.Note how this avoids repeated division or multiplication when converting. Verification / Alternative check: Memory dumps typically show bytes as two hex characters. Bitmask constants and register fields are expressed in hex to align with bit boundaries. Educational material emphasizes hex specifically because of this deterministic nibble mapping. Why Other Options Are Wrong: Incorrect: Dismisses the core mapping advantage. Only for signed integers: Sign does not affect base mapping. Only when bytes are 8 bits: Even with non-8-bit words, nibble groupings remain useful. Fewer digits than decimal: While true, compactness is secondary to the exact binary mapping. Common Pitfalls: Confusing ease of conversion with human familiarity; believing decimal would be better merely due to everyday use; overlooking that octal (3-bit groupings) was popular on systems with 12/36-bit words for the same mapping reason. Final Answer: Correct

Question 8

Mixed-base identity check: “12 (decimal) = 1101 (binary) = C (hexadecimal) = 0001 0010 (BCD).” Determine whether this chain of equalities is valid.

Accepted Answer

Introduction / Context: Verifying equivalence across bases is a common skill in digital fundamentals. This item intentionally mixes correct and incorrect conversions to test careful checking of each representation: binary, hexadecimal, and packed BCD for the same decimal value 12. Given Data / Assumptions: Target value: 12 decimal. Binary conversion must reflect base-2 positional weights. Hexadecimal C corresponds to 12 decimal. Packed BCD encodes each decimal digit in 4 bits. Concept / Approach: Evaluate each equality independently. Decimal 12 in binary is 1100 (8+4+0+0), not 1101 (which equals 13). Hexadecimal C equals 12, which is correct. Packed BCD for 12 is 0001 0010 (digit “1” → 0001; digit “2” → 0010), which is also correct. Therefore, because the binary term is wrong, the full chain is not valid. Step-by-Step Solution: Compute 12 in binary: 12 = 8 + 4 = 1100.Check given binary 1101: 8 + 4 + 1 = 13 → mismatch.Confirm hex: C = 12 → correct.Confirm BCD: 12 → 0001 0010 → correct. Verification / Alternative check: Convert the allegedly equal 1101 back to decimal: 18 + 14 + 02 + 11 = 13. Since this differs from 12, the chain fails. Two correct items (hex and BCD) cannot fix one wrong item (binary). Why Other Options Are Wrong: Correct: Would require all representations to match 12; they do not. Correct only if parity/rounding: Parity and rounding are irrelevant to base conversion. Correct except for BCD: BCD part is actually correct; the binary is not. Common Pitfalls: Mistaking 1101 for 12 due to visual similarity; forgetting that BCD encodes decimal digits, not values directly; neglecting to recheck each piece independently. Final Answer: Incorrect

Question 9

Division-by-two method nuance: “When converting decimal to binary using repeated division by 2, the first remainder obtained becomes the most significant bit (MSB).” Judge this statement.

Accepted Answer

Introduction / Context: The repeated division-by-two algorithm is a staple for manual decimal-to-binary conversion. Correct bit ordering is crucial. This question probes whether the learner knows which remainder becomes the least significant bit (LSB) and which becomes the most significant bit (MSB). Given Data / Assumptions: We repeatedly divide the decimal number by 2 and record remainders (0 or 1). Remainders are produced from least significant position upward. The final remainder corresponds to the MSB when read in reverse order. Concept / Approach: Each division by 2 peels off the current LSB. Thus, the very first remainder is the LSB, not the MSB. The sequence of remainders, when read from last to first, yields the correct binary digits from MSB to LSB. Step-by-Step Solution: Start with N; compute r0 = N % 2 (first remainder) → this is the LSB.Update N = floor(N / 2) and repeat to get r1, r2, ...Stop when N = 0; the last non-zero division produces the MSB remainder.Write bits in reverse remainder order: rn ... r2 r1 r0. Verification / Alternative check: Example: 13. Divisions yield remainders 1, 0, 1, 1. Writing them in reverse gives 1101, which is the binary for 13. The first remainder (1) is clearly the LSB at the rightmost end, not the MSB. Why Other Options Are Wrong: Correct: Misstates the ordering; the first remainder is LSB. Other distractors (hex, signed magnitude, Gray code): Unrelated to the division-by-two remainder ordering. Common Pitfalls: Listing remainders in the order computed without reversing; confusing modulus operation output position; mixing methods (e.g., subtraction of powers of two) with division-by-two and conflating their steps. Final Answer: Incorrect

Question 10

Binary-to-decimal conversion method: “A binary number can be converted to decimal by summing the decimal weights of all positions that contain a 1.” Decide if this description is correct.

Accepted Answer

Introduction / Context: Understanding positional weights is foundational in number systems. The conversion from binary to decimal hinges on the idea that each bit represents a power of 2, and that the decimal value is the sum of the weights for bits set to 1. This question validates mastery of that principle. Given Data / Assumptions: Binary digits (bits) have weights 2^0, 2^1, 2^2, ... from right to left. Only positions with bit value 1 contribute to the sum. Unsigned interpretation is assumed unless specified otherwise. Concept / Approach: For any binary string b_n ... b_2 b_1 b_0, the decimal value is Σ(b_k * 2^k). When b_k = 1, that power of 2 adds to the total; when b_k = 0, it contributes nothing. This direct sum is the standard, exact method for conversion. Step-by-Step Solution: Label bit positions with powers: 2^0, 2^1, 2^2, ...Identify which positions are set to 1.Add corresponding powers of 2 to form the decimal value.Verify result by alternative methods if desired (e.g., Horner’s rule). Verification / Alternative check: Example: 101101₂ = 12^5 + 02^4 + 12^3 + 12^2 + 02^1 + 12^0 = 32 + 8 + 4 + 1 = 45. This matches the decimal interpretation, confirming the rule. Why Other Options Are Wrong: Incorrect: Denies the fundamental positional weight definition. Only for unsigned / up to 8 bits / endianness: These caveats do not change how binary-to-decimal conversion is computed; sign conventions affect interpretation, not the arithmetic of weights. Common Pitfalls: Mislabeling bit positions (starting at 1 instead of 0), forgetting to include a weight, or confusing endianness (byte order) with numeric value; endianness is a storage layout issue, not a mathematical one. Final Answer: Correct

Question 11

Acronym expansion: “ASCII stands for American Standard Code for Information Interchange.” Confirm whether this expansion is correct.

Accepted Answer

Introduction / Context: Acronyms are pervasive in computing, and expanding them correctly is essential for clear communication. ASCII is one of the most historical and influential encodings for text in computers and telecommunications. The question checks recall of the exact expansion of ASCII. Given Data / Assumptions: ASCII defines a mapping from numerical codes to characters and control symbols. Original ASCII is a 7-bit code (0–127), with many 8-bit extensions historically used. Unicode and UTF encodings supersede ASCII for multilingual text, but ASCII remains a subset and a baseline. Concept / Approach: Confirm the words behind each letter: A = American, S = Standard, C = Code, I = Information, I = Interchange. This exact phrase is canonical in standards documentation and textbooks. Step-by-Step Solution: Identify the five-word expansion commonly cited.Cross-check typical confusions (Character vs. Code, Communication vs. Information).Affirm the correct, historically standardized expansion.Recognize ASCII’s role as the foundation for many later encodings. Verification / Alternative check: Standards references and technical glossaries consistently give “American Standard Code for Information Interchange.” It appears in documentation for serial communications, file formats, and programming language references. Why Other Options Are Wrong: Incorrect expansions replace “Code” with “Character” or “Communication,” or change “Information Interchange” to other phrases; these do not match the standard. Common Pitfalls: Swapping “Code” and “Character,” or using “International” instead of “Information”; these are common memory slips but are not accurate for ASCII. Final Answer: Correct

Question 12

ASCII hex decoding exercise: “In ASCII, the hexadecimal byte sequence 53 54 55 44 45 4E 54 corresponds to the text string ‘STUDENT’.” Evaluate this statement.

Accepted Answer

Introduction / Context: Interpreting hex byte sequences as ASCII characters is a common task when reading protocol traces, memory dumps, or file headers. This question checks the learner’s ability to map each hex byte to its ASCII character and recognize the resulting string. Given Data / Assumptions: ASCII is a 7-bit code commonly stored in bytes (the high bit often zero). Hex bytes map directly to characters: 0x41–0x5A are uppercase A–Z, 0x61–0x7A are lowercase a–z. Given bytes: 53 54 55 44 45 4E 54 (hex). Concept / Approach: Decode each byte: 0x53→‘S’, 0x54→‘T’, 0x55→‘U’, 0x44→‘D’, 0x45→‘E’, 0x4E→‘N’, 0x54→‘T’. Concatenate to form the string. No endianness issues arise for byte-wise character sequences; endianness affects multi-byte numeric fields, not the order of bytes already presented in a stream. Step-by-Step Solution: Map 0x53 → S.Map 0x54 → T.Map 0x55 → U.Map 0x44 → D.Map 0x45 → E.Map 0x4E → N.Map 0x54 → T. Verification / Alternative check: Consult any ASCII table: hexadecimal column for uppercase letters confirms the mappings. Many hex editors display an ASCII panel that would show “STUDENT” for these bytes. Why Other Options Are Wrong: Incorrect: Conflicts with the standard ASCII table. EBCDIC: A different encoding used on mainframes; the same hex bytes would map differently. Endianness and parity: Irrelevant to interpreting individual bytes as ASCII characters. Common Pitfalls: Reading hex without verifying with a table; confusing case (0x53 is uppercase S, not lowercase s which is 0x73); assuming CPU endianness changes pre-ordered byte streams. Final Answer: Correct

Question 13

Parity check with odd parity: For the 9-bit data pattern 100011010, would this word pass an odd-parity check at the receiver (i.e., total count of 1s should be odd)? Choose the best evaluation.

Accepted Answer

Introduction / Context: Parity is one of the simplest error-detection methods used in digital communications and storage. With odd parity, the total number of 1 bits in the transmitted unit (data plus any parity bit) must be odd. The receiver recomputes the number of 1s and checks whether the rule holds. This question asks you to determine whether the 9-bit word 100011010 would pass an odd-parity check when evaluated as received data (no extra parity bit shown). Given Data / Assumptions: Word under test: 100011010 (9 bits shown explicitly). Odd parity rule: total count of 1s must be odd. Assume these 9 bits constitute the unit being checked (no hidden parity bit beyond what is shown). Noise may flip bits; parity aims only to detect, not correct, such flips. Concept / Approach: To test odd parity, count the number of 1s. If the count is odd, the unit passes; if the count is even, it fails. Parity is a lightweight integrity check and primarily detects any odd number of bit errors; it cannot detect all multi-bit error patterns (for example, two-bit flips can go undetected). Step-by-Step Solution: List bits: 1 0 0 0 1 1 0 1 0.Count 1s: positions with 1 are the 1st, 5th, 6th, and 8th bits → total 4 ones.Evaluate odd parity: 4 is even, so the odd-parity condition is not satisfied.Conclusion: the word fails an odd-parity check when considered as the complete unit. Verification / Alternative check: To make this unit satisfy odd parity, you would append a parity bit of 1 (increasing the total number of 1s from 4 to 5). The placement (MSB or LSB) depends on protocol, but the value required is 1 to achieve an odd count. Why Other Options Are Wrong: “Passes” is wrong because the count of 1s (4) is even.“Passes only if parity bit 1 appended” describes a possible fix, not the stated word as-is.“Indeterminate” is unnecessary here because the count decision is direct on the given unit. Common Pitfalls: Confusing odd with even parity, miscounting 1s, or assuming an implicit parity bit when none is shown. Always count exactly the bits under test and apply the specified parity rule. Final Answer: Incorrect — it fails the odd-parity check

Question 14

Number base conversion principle: Is “repeated division-by-10” the standard manual technique to convert a decimal (base-10) number to its binary (base-2) representation?

Accepted Answer

Introduction / Context: Converting numbers between bases is a foundational digital-logic and computer-architecture skill. For decimal (base 10) to binary (base 2) conversion, there is a well-known manual “short division” procedure. This question checks whether repeated division by 10 is the correct approach for decimal-to-binary conversion. Given Data / Assumptions: Source base: 10 (decimal). Target base: 2 (binary). We are discussing the standard manual algorithm for integer conversion. No calculators or programming, just algorithmic steps. Concept / Approach: The standard method for converting an integer N from decimal to binary uses repeated division by the target base (2). At each step, you note the remainder (0 or 1), then divide the quotient by 2 again. After the quotient becomes zero, the binary digits are the remainders read in reverse order. Division by 10 is unrelated to producing base-2 digits and would not yield correct binary bits. Step-by-Step Solution: Initialize with N (decimal integer).Compute r0 = N mod 2; set N0 = floor(N / 2).Repeat: ri = Ni mod 2; Ni+1 = floor(Ni / 2) until Ni = 0.Binary result = sequence r_last … r2 r1 r0 (remainders read backwards). Verification / Alternative check: Try N = 13: 13/2 → q=6 r=1; 6/2 → q=3 r=0; 3/2 → q=1 r=1; 1/2 → q=0 r=1. Reading remainders backward yields 1101, which is correct for 13. Any attempt to divide by 10 produces decimal digits, not binary bits. Why Other Options Are Wrong: “Division by 10”: Produces decimal digits, not binary bits.“Repeated subtraction of powers of 10”: Also tied to base 10, not base 2.“Only repeated multiplication by 2 works”: Multiplication by 2 is used for fractional binary; integers use division by 2. Common Pitfalls: Mixing up methods for integer vs. fractional parts; forgetting to reverse the remainder sequence; dropping leading zeros in fixed-width formats without considering word size. Final Answer: Incorrect — use repeated division by 2 for decimal→binary

Question 15

Two’s complement rule: Is the two’s complement of a binary number obtained by forming the one’s complement and then adding 1 to the result?

Accepted Answer

Introduction / Context: Two’s complement is the dominant representation for signed integers in digital systems. It simplifies hardware for addition and subtraction, supports a unique zero, and makes sign extension straightforward. This question confirms the canonical construction rule for two’s complement. Given Data / Assumptions: We are dealing with fixed-width binary words. One’s complement is defined as bitwise inversion of all bits. Two’s complement is defined relative to a fixed width n. No BCD or floating-point encodings are under consideration. Concept / Approach: To obtain the two’s complement of an n-bit binary number X, first invert every bit to get the one’s complement (~X), then add 1: (two’s complement of X) = (~X) + 1, computed modulo 2^n. This operation is central for negation: the negative of a number A is its two’s complement with respect to the chosen width. Step-by-Step Solution: Start with X = b_{n−1}…b_1 b_0.Compute one’s complement: ~X flips each bit (0 ↔ 1).Add 1: (~X) + 1 modulo 2^n to form two’s complement.Verify by addition: X + (two’s complement of X) = 0 modulo 2^n. Verification / Alternative check: Example (8-bit): X = 00010110 (22). One’s complement = 11101001. Add 1 → 11101010, which represents −22 in two’s complement. Adding X and this result yields 00000000 with a carry out discarded, validating the rule. Why Other Options Are Wrong: “One’s complement − 1”: That procedure gives a different value and breaks the negation identity.BCD/floating-point claims: Two’s complement applies to binary integers, not BCD or floating-point mantissas. Common Pitfalls: Forgetting fixed width during inversion, mishandling carry out, or mixing sign-magnitude with two’s complement rules. Final Answer: Correct — two’s complement = one’s complement + 1

Question 16

BCD definition check: Is a decimal number converted to BCD by replacing each decimal digit with a 3-bit binary code?

Accepted Answer

Introduction / Context: Binary-Coded Decimal (BCD) is a family of encodings that map each decimal digit 0–9 to its own binary pattern. BCD is widely used in digital displays, financial arithmetic, and interfaces where precise decimal digit handling is required. The question probes whether BCD uses 3-bit groups per digit. Given Data / Assumptions: We refer to standard 8421 BCD unless otherwise specified. Digits: 0–9 must be representable individually. We are not discussing octal or specialized ternary systems. Concept / Approach: Since decimal has 10 symbols (0–9), at least 4 bits are required to encode all digits because 3 bits yield only 8 distinct combinations (0–7). Standard BCD uses 4-bit nibbles, typically 0000–1001 for digits 0–9; the six remaining 4-bit patterns 1010–1111 are invalid in plain BCD (or used for special purposes in some variants). Step-by-Step Solution: Count capacity: 3 bits → 2^3 = 8 codes < 10 digits → insufficient.Choose 4 bits: 2^4 = 16 codes → enough to represent 10 digits plus unused values.Conclude: standard BCD requires 4-bit codes per decimal digit. Verification / Alternative check: Example: 79 in BCD is 0111 1001 (7 → 0111, 9 → 1001). Any attempt to use 3-bit groups cannot represent digits 8 or 9. Why Other Options Are Wrong: “3-bit BCD”: Impossible to cover digits 8 and 9.“Packed BCD exception”: Packed/unpacked refers to storage layout, still 4 bits per digit.“Only true when digits 0–7”: That would be octal, not decimal BCD. Common Pitfalls: Confusing BCD with binary, octal, or hexadecimal; mishandling invalid BCD codes (A–F) that appear in hex dumps but are not valid decimal digits. Final Answer: Incorrect — standard BCD uses a 4-bit code per decimal digit

Question 17

Translate BCD to pure binary: Given the BCD-coded sequence 0111 1001 0011 (which represents decimal digits 7, 9, and 3), what is the equivalent binary bit string written contiguously?

Accepted Answer

Introduction / Context: Binary-Coded Decimal (BCD) stores each decimal digit as an independent 4-bit group. Converting a BCD representation to “binary” may mean two things: (1) concatenating the BCD nibbles into a single bit string (still digit-wise coded), or (2) translating the decimal value into a single pure binary number. Many entry-level questions intend the first interpretation—writing the BCD groups without spaces. This item tests that understanding. Given Data / Assumptions: BCD groups: 0111 (digit 7), 1001 (digit 9), 0011 (digit 3). “In binary” here means the contiguous bit string formed by these BCD nibbles. No request to compute the single pure-binary value of the decimal number 793. Concept / Approach: In BCD, each decimal digit is independent: 0→0000, 1→0001, …, 7→0111, 8→1000, 9→1001. Therefore, 7→0111, 9→1001, 3→0011. The contiguous BCD representation is just the concatenation of these groups, typically shown with spaces for readability but equivalent without spaces. Step-by-Step Solution: Map 7 to 0111.Map 9 to 1001.Map 3 to 0011.Concatenate: 0111 1001 0011 → 011110010011 (without spaces). Verification / Alternative check: If instead the problem asked for the pure binary of decimal 793, you would compute 793 = 512 + 256 + 16 + 8 + 1, yielding the 10-bit binary 1100011001. That is a different representation than BCD concatenation, which remains digit-wise. Why Other Options Are Wrong: Other choices either scramble the nibble order or use values not matching 7, 9, and 3 in BCD.Some options resemble hex or arbitrary patterns and are not valid BCD concatenations for 793. Common Pitfalls: Confusing BCD concatenation with conversion to a single pure-binary integer; mixing nibble boundaries; or misremembering BCD codes for 8 and 9. Final Answer: 011110010011

Question 18

Terminology clarity: Most computers store data in strings of how many bits, and what is such a unit commonly called?

Accepted Answer

Introduction / Context: Precise terminology matters in digital systems. Although “word” and “byte” are sometimes conflated in casual speech, they have distinct meanings. This question checks whether you know the standard size and name for the fundamental addressable unit on most contemporary computer architectures and interfaces. Given Data / Assumptions: A byte is widely standardized as 8 bits. A word is architecture-dependent (commonly 16, 32, or 64 bits). Many buses, memories, and protocols address or align data on byte boundaries. Concept / Approach: The byte (8 bits) is the de facto universal basic unit for data storage and transfer, especially at the software and ISA levels. While early systems varied, the modern computing ecosystem—from filesystems and networking to instruction sets—assumes an 8-bit byte. The term “word” refers to the natural register size of a CPU and varies by architecture. Step-by-Step Solution: Identify standard: byte = 8 bits.Note variability: word size depends on CPU (e.g., 16, 32, 64 bits).Select the pair that matches conventional usage: “8, byte”. Verification / Alternative check: Inspect any modern ISA or programming language definition: char is typically 8 bits; memory addresses index bytes; network protocols count octets (another name for 8-bit bytes). Why Other Options Are Wrong: “8, word”: word is not universally 8 bits.“16, word/byte”: neither is the ubiquitous base unit; “byte” ≠ 16 bits. Common Pitfalls: Assuming “word” equals “byte” on all systems; ignoring historical exceptions where byte size differed; overlooking the term “octet”. Final Answer: 8, byte

Question 19

Even parity encoding for a given value: Represent decimal 37 using binary along with an even parity bit (assume the parity bit is placed as the most significant bit of an 8-bit pattern). Which option matches?

Accepted Answer

Introduction / Context: Parity bits add minimal redundancy to detect transmission errors. In many examples, a single parity bit is prepended as the most significant bit, forming an 8-bit pattern from a 7-bit data value. This question asks which 8-bit pattern correctly represents decimal 37 with even parity when the parity bit is the MSB. Given Data / Assumptions: Decimal 37 in binary (7-bit data) is 0100101. Parity scheme: even parity, parity bit placed as MSB. Total number of 1s across all 8 bits must be even. Concept / Approach: Compute the binary of the data value, count the number of 1s, and choose the parity bit so that the total count is even. Then confirm the option whose MSB matches the required parity while the lower 7 bits match the data. Step-by-Step Solution: Write 37 in binary: 37 = 32 + 4 + 1 → data bits 0100101 (seven bits).Count 1s in 0100101 → there are 3 ones (odd).Even parity requires total ones even → parity bit must be 1 to make 3 + 1 = 4 (even).Form 8-bit pattern: parity (1) + data (0100101) → 10100101. Verification / Alternative check: Count ones in 10100101: there are 4 ones — even. The lower 7 bits still equal 0100101, confirming the data value is intact and parity is correct. Why Other Options Are Wrong: 00100101: total ones 3 (odd), parity not even.11000100 and 01001011: do not preserve the 7-bit data for 37 and/or parity count is incorrect for the stated scheme. Common Pitfalls: Forgetting whether the parity bit is MSB or LSB in a given protocol; miscounting ones; writing 8-bit binary for 37 (00100101) and treating its MSB as parity when it is actually a data bit in that representation. Final Answer: 10100101

Question 20

Binary basics: The two symbols (digits) used in the binary number system are which pair?

Accepted Answer

Introduction / Context: Binary is the foundation of digital electronics and computing. All logic states, data values, and instructions are ultimately encoded using only two symbols. This question checks your recall of those symbols and dispels common misconceptions arising from related labels like “high/low” or “true/false.” Given Data / Assumptions: Binary is a base-2 numeral system. Many physical encodings (voltage levels, light on/off) map to two abstract symbols. Human-readable labels (H/L, T/F) are convenient but not the digits themselves. Concept / Approach: In a positional numeral system, the set of digits must equal the base. For base-2, there must be two digits. The canonical binary digits are 0 and 1. Other labels simply describe states that are mapped onto these digits in a particular technology or logic convention. Step-by-Step Solution: Recognize base-2 → two unique symbols required.Canonical digits for binary: 0 and 1.Select the option that exactly lists 0, 1 in either order (here 0, 1). Verification / Alternative check: Check any binary representation: counting from zero goes 0, 1, 10, 11, 100, …, demonstrating that only 0 and 1 are used as digits; additional characters do not appear. Why Other Options Are Wrong: “1, 2” represents decimal digits, not binary.“H, L” and “T, F” are labels for logic states, not digits in the numeral system. Common Pitfalls: Confusing logic-level labels with numeral digits; assuming “true/false” are numeric digits; misapplying decimal intuition to binary. Final Answer: 0, 1

Number Systems and Codes Questions