Question 1

Binary-to-decimal conversion practice Convert the binary number 10011010 (base 2) to its decimal (base 10) equivalent.

Accepted Answer

Introduction / Context: Binary-to-decimal conversion is a fundamental skill in digital electronics and computer science. Each binary digit (bit) represents a power of 2, and the decimal value is the sum of those powers where bits are 1. Given Data / Assumptions: Binary input: 10011010₂. We seek the decimal (base 10) value. No leading zeros change the magnitude. Concept / Approach: For a binary string b7 b6 b5 b4 b3 b2 b1 b0, the decimal value is sum(bi * 2^i) for i from 0 to 7. Only positions with bi = 1 contribute to the sum. Step-by-Step Solution: 1) Write place values: 2^7=128, 2^6=64, 2^5=32, 2^4=16, 2^3=8, 2^2=4, 2^1=2, 2^0=1.2) Map bits (10011010): 1·128 + 0·64 + 0·32 + 1·16 + 1·8 + 0·4 + 1·2 + 0·1.3) Sum contributions: 128 + 16 + 8 + 2 = 154.4) Therefore, decimal value = 154. Verification / Alternative check: Group into nibbles: 1001 1010 → 0x9A in hex. Hex 0x9A = 9*16 + 10 = 144 + 10 = 154, confirming the result. Why Other Options Are Wrong: 153, 155, 157: off by small amounts, typically from misreading one bit's place value (16, 8, or 2). Common Pitfalls: Forgetting that the leftmost bit is the highest power of 2, or mixing up 2^3=8 and 2^4=16. Final Answer: 154

Question 2

Binary subtraction basics Compute the result of 111₂ − 001₂ and express the answer in binary.

Accepted Answer

Introduction / Context: Binary subtraction mirrors decimal subtraction but with base 2. Recognizing basic patterns such as subtracting 1 from a number simplifies mental computation and checking of results. Given Data / Assumptions: Minuend: 111₂ (which equals 7 decimal). Subtrahend: 001₂ (which equals 1 decimal). No borrow issues across more than one bit in this simple case. Concept / Approach: Since 111₂ is all ones, subtracting 001₂ simply clears the LSB and leaves the higher bits unchanged, equivalent to 7 − 1 = 6 decimal. Convert back to binary after subtraction to confirm. Step-by-Step Solution: 1) Convert: 111₂ → 7₁₀, 001₂ → 1₁₀.2) Subtract in decimal: 7 − 1 = 6.3) Convert 6 to binary: 6 = 4 + 2 = 2^2 + 2^1 → 110₂.4) Therefore, 111₂ − 001₂ = 110₂. Verification / Alternative check: Perform column-wise binary subtraction: from right to left, 1−1=0; next bits 1−0=1; next 1−0=1 → 110. Why Other Options Are Wrong: 100: equals 4, not 6. 111: equals the original minuend (no subtraction performed). 001: equals 1, the subtrahend. Common Pitfalls: Attempting unnecessary borrows when subtracting 1 from an odd number ending in 1. Final Answer: 110

Question 3

Decimal-to-hexadecimal conversion Convert the decimal number 5238 (base 10) to base 16 (hexadecimal). Provide the hex digits without any prefix.

Accepted Answer

Introduction / Context: Converting decimal integers to hexadecimal is common in embedded systems, debugging, and memory address representation. Repeated division by 16 yields the hex digits from least significant to most significant. Given Data / Assumptions: Decimal input N = 5238. Target base is 16 (digits 0–9, A–F). No fractional part is present; result will be an integer in hex. Concept / Approach: Use repeated division by 16. Each remainder (0–15) maps to one hex digit (0–9, A–F). Collect remainders bottom-up to form the final hexadecimal string. Step-by-Step Solution: 1) 5238 / 16 = 327 remainder 6 → least significant hex digit = 6.2) 327 / 16 = 20 remainder 7 → next digit = 7.3) 20 / 16 = 1 remainder 4 → next digit = 4.4) 1 / 16 = 0 remainder 1 → most significant digit = 1.5) Read remainders in reverse: 1 4 7 6 → 1476₁₆. Verification / Alternative check: Expand back to decimal: 116^3 + 416^2 + 7*16 + 6 = 4096 + 1024 + 112 + 6 = 5238, confirming correctness. Why Other Options Are Wrong: 1388: common confusion from base-10 division steps; does not re-expand to 5238. 327.375 and 12.166: include decimal points; input was an integer. Common Pitfalls: Listing remainders in the forward order (top-down) instead of reverse order; mis-mapping remainders above 9 to A–F. Final Answer: 1476

Question 4

Decimal fraction to binary conversion Convert the decimal number 151.75 (base 10) to binary (base 2).

Accepted Answer

Introduction / Context: Converting mixed decimal numbers (integer plus fraction) to binary involves two coordinated processes: repeated division by 2 for the integer part and repeated multiplication by 2 for the fractional part. Mastery of both is essential for representing fixed-point values in binary. Given Data / Assumptions: Decimal input: 151.75. We want an exact finite binary since 0.75 is a dyadic fraction. No rounding is required for this case. Concept / Approach: Split the number into 151 (integer) and 0.75 (fraction). Convert 151 using division-by-2; convert 0.75 by multiplying by 2 repeatedly and recording integer bits produced at each step. Step-by-Step Solution (Integer): 1) 151 decimal → binary: 151 = 128 + 16 + 4 + 2 + 1.2) Place bits: 2^7=128 → 1; 2^6=64 → 0; 2^5=32 → 0; 2^4=16 → 1; 2^3=8 → 0; 2^2=4 → 1; 2^1=2 → 1; 2^0=1 → 1.3) Integer part = 10010111. Step-by-Step Solution (Fraction): 1) 0.75 * 2 = 1.5 → record 1, keep 0.5.2) 0.5 * 2 = 1.0 → record 1, fraction ends.3) Fractional part = .11. Verification / Alternative check: Reconstruct: 10010111.11₂ = 151 + 0.75 = 151.75, exact. Why Other Options Are Wrong: 10000111.11: equals 135.75, not 151.75. 11010011.01: equals 211.25 in decimal. 00111100.00: equals 60 exactly. Common Pitfalls: Placing 1s for the wrong powers of 2, or mishandling the fractional multiply-by-2 steps. Final Answer: 10010111.11

Question 5

Decimal-to-binary by repeated division When converting a positive decimal integer to binary using repeated division by 2, the first remainder obtained corresponds to which bit of the final binary result?

Accepted Answer

Introduction / Context: Repeated division by 2 is a standard manual algorithm to convert decimal integers to binary. Understanding which end of the binary number each remainder represents avoids reversed outputs. Given Data / Assumptions: The method divides by 2 repeatedly and records the remainder at each step. Remainders are 0 or 1. The process stops when the quotient becomes 0. Concept / Approach: Every division by 2 extracts the least significant bit of the current quotient. Therefore, the first remainder corresponds to the least significant bit (LSB). Writing remainders from last to first reconstructs the binary number from MSB to LSB. Step-by-Step Solution: 1) Start with N and compute N mod 2 → this is the LSB.2) Update N = floor(N / 2) and repeat.3) Collect remainders in order of computation: LSB first, then next bit, and so on.4) To form the final binary, reverse the remainder sequence so the last remainder becomes the MSB. Verification / Alternative check: Example: N=13. Remainders: 13 mod 2=1 (LSB), 6 mod 2=0, 3 mod 2=1, 1 mod 2=1 → bits (from first to last remainder) 1,0,1,1 ⇒ reversed 1101₂, which is correct. Why Other Options Are Wrong: MSB choices suggest the first remainder is the MSB, which is incorrect. The conditional statements add confusion; the algorithm always yields LSB first. Common Pitfalls: Writing remainders in the order generated without reversing, which results in a mirrored binary number. Final Answer: LSB

Question 6

Detecting invalid BCD digits Which 4-bit pattern below is not a valid Binary-Coded Decimal (BCD) code for a single decimal digit?

Accepted Answer

Introduction / Context: BCD encodes each decimal digit (0–9) as a 4-bit binary value. Recognizing valid versus invalid 4-bit patterns is essential for input validation, display drivers, and error checking in digital systems that use BCD instead of pure binary. Given Data / Assumptions: Valid BCD digit range: 0000 through 1001 (0 to 9). Invalid patterns are any 4-bit values from 1010 to 1111. We evaluate each option against these ranges. Concept / Approach: Check whether the given 4-bit code is less than or equal to 1001. If it exceeds 1001, it cannot represent a decimal digit in BCD and is therefore invalid. Step-by-Step Solution: 1) 0011 → decimal 3: valid.2) 0101 → decimal 5: valid.3) 1001 → decimal 9: valid.4) 1101 → decimal 13: not a valid BCD digit (invalid). Verification / Alternative check: List all invalid BCD codes: 1010, 1011, 1100, 1101, 1110, 1111. 1101 is in this set. Why Other Options Are Wrong: 0011, 0101, 1001: all lie within 0000–1001 and therefore are valid BCD digits. Common Pitfalls: Confusing BCD with hexadecimal, where 1010–1111 would be valid (A–F). In BCD, those patterns are illegal for single digits. Final Answer: 1101

Question 7

Binary vs. BCD — conceptual distinction What is the correct statement that distinguishes pure binary coding from Binary-Coded Decimal (BCD)?

Accepted Answer

Introduction / Context: Binary coding represents entire numbers directly in base 2, whereas BCD represents each decimal digit separately using a 4-bit code. This difference affects arithmetic efficiency, storage density, and ease of human-readable display. Given Data / Assumptions: Binary uses powers of 2 for the whole number. BCD encodes digits 0–9 with 0000–1001. Higher 4-bit patterns (1010–1111) are not used for a single BCD digit. Concept / Approach: “Pure binary” means a number like 57 is stored as 111001 in base 2. In BCD, 57 is stored as two nibbles: 0101 (5) and 0111 (7). Thus binary coding is inherently base 2; BCD is a hybrid representation convenient for decimal I/O but less compact and slower for arithmetic without special hardware. Step-by-Step Solution: 1) Identify how binary encodes values: directly as powers of 2.2) Identify how BCD encodes values: per digit, 4 bits each.3) Conclude that “Binary coding is pure binary” is the accurate distinguishing statement.4) Recognize that other statements invert or misstate the relationship. Verification / Alternative check: Compare sizes: the number 99 is 1100011 in binary (7 bits) but 9 9 in BCD requires 8 bits (1001 1001), illustrating BCD’s overhead. Why Other Options Are Wrong: BCD is pure binary: false, it encodes decimal digits in binary. Binary coding has a decimal format: false, it is base 2. BCD has no decimal format: false, it explicitly preserves decimal digits. Common Pitfalls: Assuming BCD arithmetic equals binary arithmetic performance; BCD often needs adjust/pack operations. Final Answer: Binary coding is pure binary.

Question 8

Binary to Gray-code conversion Convert the 4-bit binary number 1100₂ to its equivalent Gray code.

Accepted Answer

Introduction / Context: Gray code changes only one bit between successive values, reducing errors in position encoders and asynchronous signal transfers. Converting between binary and Gray is a common interview and lab exercise. Given Data / Assumptions: Binary input: 1100 (b3 b2 b1 b0). We want standard reflected binary Gray code. No arithmetic beyond XOR operations is required. Concept / Approach: Rules for binary → Gray: g3 = b3; g2 = b3 XOR b2; g1 = b2 XOR b1; g0 = b1 XOR b0. Apply bitwise XORs from the binary input to produce the Gray output. Step-by-Step Solution: 1) b3 b2 b1 b0 = 1 1 0 0.2) g3 = b3 = 1.3) g2 = b3 XOR b2 = 1 XOR 1 = 0.4) g1 = b2 XOR b1 = 1 XOR 0 = 1.5) g0 = b1 XOR b0 = 0 XOR 0 = 0.6) Gray result = g3 g2 g1 g0 = 1010. Verification / Alternative check: Reverse (Gray → Binary) to confirm: b3 = g3 = 1; b2 = b3 XOR g2 = 1 XOR 0 = 1; b1 = b2 XOR g1 = 1 XOR 1 = 0; b0 = b1 XOR g0 = 0 XOR 0 = 0 → 1100, matching the original. Why Other Options Are Wrong: 0011, 1001, 1100: do not satisfy Gray-code conversion for 1100₂. Common Pitfalls: Using XOR with the wrong neighbor bit or copying the binary number unchanged. Final Answer: 1010

Question 9

Octal-to-decimal conversion Interpret 17 (base 8) as an octal number and convert it to decimal (base 10).

Accepted Answer

Introduction / Context: Octal (base 8) is another positional number system. Although less common today than hexadecimal, it appears in legacy systems and permissions in Unix-like operating systems. Converting small octal numbers to decimal is straightforward and reinforces the concept of place value. Given Data / Assumptions: Octal input: 17₈ (digits allowed 0–7). We want the decimal value. No fractional part is present. Concept / Approach: For an octal number d1 d0, the decimal value is d1 * 8^1 + d0 * 8^0. Here, 1 is the “eights” place and 7 is the “ones” place. Ensure digits are within the octal range (0–7); digit 8 would be invalid in octal notation. Step-by-Step Solution: 1) Identify place values: 8^1 = 8, 8^0 = 1.2) Multiply digits: 1 * 8 + 7 * 1 = 8 + 7.3) Sum: 8 + 7 = 15 decimal.4) Therefore, 17₈ = 15₁₀. Verification / Alternative check: Cross-check via binary: 17₈ = 001 111₂ = 15 decimal, confirming the result. Why Other Options Are Wrong: 51, 57, 82: do not match the correct place-value computation for 17₈. Common Pitfalls: Interpreting “17” as base 10 or assuming octal digits can include “8” or “9”. Final Answer: 15

Question 10

Number System Representation — BCD vs. Straight Binary How many BCD code bits and how many straight binary bits are required to represent the decimal number 643? Provide the total bit counts for each scheme.

Accepted Answer

Introduction / Context: This question checks understanding of how many bits are needed to encode a given decimal value using two different schemes: BCD, which codes each decimal digit separately, and straight binary, which represents the entire value in base 2 using the minimum number of bits. Given Data / Assumptions: Target value: decimal 643. BCD uses 4 bits per decimal digit. Straight binary uses the smallest N such that 2^N > 643. Concept / Approach: For BCD, count decimal digits and multiply by 4. For straight binary, locate the smallest power of two strictly greater than the number, then take the exponent as the bit count. If 2^N is equal to the number, N bits suffice; otherwise still N bits when 2^(N-1) <= value < 2^N. Step-by-Step Solution: 1) 643 has three digits, so BCD bits = 3 * 4 = 12.2) Check powers of two: 2^9 = 512, 2^10 = 1024.3) Since 512 <= 643 < 1024, 10 bits are required in straight binary.4) Therefore the pair is 12 BCD, 10 binary. Verification / Alternative check: Encoding 643 in binary gives 1010000011, which indeed fits in 10 bits. BCD digits 6, 4, 3 encode as 0110 0100 0011, which is 12 bits total. Why Other Options Are Wrong: 12 BCD, 12 binary: Overestimates binary bits. 12 BCD, 9 binary: 643 exceeds 2^9 - 1 = 511. 16 BCD, 9 binary: BCD for three digits is 12, not 16, and 9 bits are insufficient. Common Pitfalls: Confusing 2^N thresholds, or assuming BCD packs digits more tightly than 4 bits per digit. Final Answer: 12 BCD, 10 binary

Question 11

Octal Number System — Practical Advantage Which statement best describes the common advantage of using octal notation in digital electronics and computer systems?

Accepted Answer

Introduction / Context: Octal (base 8) representation is an older but still instructive shorthand for binary data. It maps cleanly to binary because 8 equals 2^3, so every octal digit corresponds to exactly three binary bits, making manual conversion fast and reducing transcription errors. Given Data / Assumptions: Octal base is 8. Each octal digit maps to three binary bits. Use in contexts such as permissions, older instruction sets, and compact notations. Concept / Approach: Shorthand bases that align with powers of two compress long binary strings while preserving easy mental conversion. Because each octal digit is three bits, grouping is straightforward and reduces mistakes compared with writing long runs of 0s and 1s. Step-by-Step Solution: 1) Recognize mapping: 0..7 maps to 000..111.2) Conversion method: split a binary string into 3-bit groups from the right and convert each to an octal digit.3) This direct grouping accelerates reading, writing, and checking values.4) Hence octal simplifies tasks and saves time compared to raw binary. Verification / Alternative check: Compare with hexadecimal which uses groups of 4 bits; both are efficient. Octal is historically popular in systems with word sizes that are multiples of 3 bits. Why Other Options Are Wrong: simplifies tasks: Incomplete; ignores time benefit. groups binary numbers in groups of 4: That describes hexadecimal, not octal. saves time: True but incomplete without the simplification rationale. Common Pitfalls: Confusing octal with hex grouping size; forgetting to pad the leftmost group with leading zeros to make three bits. Final Answer: simplifies tasks and saves time

Question 12

Base Conversion — Binary to Hexadecimal Convert the binary number 1100101000110101 to its hexadecimal equivalent. Choose the correct hex representation.

Accepted Answer

Introduction / Context: Hexadecimal is a compact way to represent binary values because 16 equals 2^4, allowing a direct 4-bit to 1-hex-digit mapping. This conversion skill is essential for reading memory dumps, opcodes, and register values. Given Data / Assumptions: Binary: 1100101000110101. Group into nibbles of 4 bits from the right. Map each nibble to its hex digit. Concept / Approach: Pad on the left if necessary so the length is a multiple of 4, then translate each 4-bit group using the mapping 0000=0 through 1111=F. Step-by-Step Solution: 1) Group: 1100 1010 0011 0101.2) Convert: 1100=C, 1010=A, 0011=3, 0101=5.3) Combine: CA35.4) Therefore, the hexadecimal result is CA35. Verification / Alternative check: Convert back: C=12=1100, A=10=1010, 3=0011, 5=0101. Concatenating reproduces the original binary, confirming correctness. Why Other Options Are Wrong: 121035: Not a valid consistent hex mapping of the groups. 53AC1 and 530121: Wrong length and ordering relative to 16-bit grouping. Common Pitfalls: Failing to start grouping from the right, forgetting to pad leading bits, or reversing nibble order. Final Answer: CA35

Question 13

Binary Fraction to Decimal — Point Positioning Convert the binary number 1001.0010 to its decimal value. Show the integer and fractional contributions clearly.

Accepted Answer

Introduction / Context: Interpreting binary fractions requires understanding negative powers of two for positions to the right of the binary point. This conversion appears often in fixed point arithmetic and digital signal processing basics. Given Data / Assumptions: Binary value: 1001.0010. Positions left of the point use powers 2^3, 2^2, 2^1, 2^0. Positions right of the point use 2^-1, 2^-2, 2^-3, 2^-4. Concept / Approach: Evaluate each set bit as its weight and sum them. The fractional part adds small contributions determined by negative exponents of two. Step-by-Step Solution: 1) Integer part 1001 = 12^3 + 02^2 + 02^1 + 12^0 = 8 + 0 + 0 + 1 = 9.2) Fractional part .0010 = 02^-1 + 02^-2 + 12^-3 + 02^-4 = 0 + 0 + 0.125 + 0 = 0.125.3) Sum = 9 + 0.125 = 9.125.4) Therefore decimal value is 9.125. Verification / Alternative check: Multiply fractional part by 2 repeatedly to recover bits: 0.1252 = 0.25 gives 0, 0.252 = 0.5 gives 0, 0.52 = 1.0 gives 1, 0.02 gives trailing zeros, matching .0010. Why Other Options Are Wrong: 90.125: Misplaced point by a factor of 10 in decimal. 125: Ignores the base and the point entirely. 12.5: Misinterprets the contribution of bits. Common Pitfalls: Treating binary digits as decimal weights, or reading fractional bits left to right without applying correct negative exponents. Final Answer: 9.125

Question 14

Hexadecimal Digits A to F — Decimal Equivalents In hexadecimal notation, the letters A through F correspond to which consecutive decimal values?

Accepted Answer

Introduction / Context: Hexadecimal extends decimal digits with letters to represent values ten through fifteen. Knowing these mappings is essential for fast mental conversions and interpreting hex dumps or color codes. Given Data / Assumptions: Base 16 uses symbols 0..9 and A..F. Each hex digit represents a 4-bit nibble in binary. Letter mapping is standard across computing. Concept / Approach: Assign A=10, B=11, C=12, D=13, E=14, F=15. This preserves increasing order and provides a compact single digit notation for values beyond 9. Step-by-Step Solution: 1) List the mapping explicitly: A=10, B=11, C=12, D=13, E=14, F=15.2) Observe that these are consecutive decimal numbers.3) Therefore the correct range description is 10 through 15. Verification / Alternative check: Confirm by checking binary correspondence: for example, F=1111 equals 15 decimal, E=1110 equals 14 decimal, and so on. Why Other Options Are Wrong: 1 through 6: Off by a factor of 9 and not consistent with hex usage. 9 through 14: Starts too low and ends too low. 11 through 17: Not aligned with base 16 digits, which stop at 15. Common Pitfalls: Assuming F could mean 16; remember single hex digits only span 0 to 15 inclusive. Final Answer: 10 through 15

Question 15

Binary to Octal Conversion — Grouping by Threes Convert the binary number 001101011 to octal. Choose the correct base 8 representation.

Accepted Answer

Introduction / Context: Octal conversion is convenient because each octal digit corresponds to three binary bits. Grouping from the right and mapping each triplet eliminates arithmetic and reduces errors during manual conversion. Given Data / Assumptions: Binary: 001101011. Octal mapping uses triplets: 000..111 maps to 0..7. Pad on the left with zeros to complete a triplet if needed. Concept / Approach: Partition the binary string into groups of three bits starting at the least significant end, translate each group to its octal digit, and then concatenate in the same order from most significant to least significant group. Step-by-Step Solution: 1) Grouping: 001 101 011.2) Convert each: 001=1, 101=5, 011=3.3) Combine digits: 1 5 3 → 153 base 8.4) Therefore the correct octal form is 153. Verification / Alternative check: Check by expanding to decimal: 153_8 = 164 + 58 + 3 = 64 + 40 + 3 = 107. Binary 001101011 equals 107 decimal as well, confirming correctness. Why Other Options Are Wrong: 3518, 2538, 3528: These reflect different groupings or digit orderings that do not match the correct triplet mapping. Common Pitfalls: Grouping from the left instead of the right, or forgetting to pad the leading zeros to complete the most significant triplet. Final Answer: 1538

Question 16

Data Size Terminology — Relative Length Considering the common terms bit, byte, nibble, and word, which is typically the longest unit of data width in general computer architecture usage?

Accepted Answer

Introduction / Context: Understanding standard data width terminology prevents confusion when reading documentation. The terms bit, nibble, byte, and word describe increasing sizes often aligned to hardware registers, buses, and memory operations. Given Data / Assumptions: Bit is a single binary digit. Nibble is 4 bits. Byte is 8 bits. Word is the native register width of an architecture, commonly 16, 32, or 64 bits. Concept / Approach: Rank units by typical width. In most contexts, a word represents the processor natural data size and exceeds a byte. While specific sizes vary by architecture, word is widely the largest among the listed standard terms. Step-by-Step Solution: 1) Order: bit < nibble (4) < byte (8) < word (16 or more).2) Compare magnitudes: word is at least 16 bits in classic usage and can be larger.3) Therefore, word is typically the longest unit listed. Verification / Alternative check: Check examples: x86 historically uses 16-bit word, 32-bit double word, 64-bit quad word; ARM and many others use 32-bit word in some documentation. Why Other Options Are Wrong: Bit: smallest possible unit. Nibble: only 4 bits. Byte: 8 bits, smaller than typical word sizes. Common Pitfalls: Assuming byte equals word; forgetting that the exact size of word can vary but remains larger than a byte for most systems. Final Answer: Word

Question 17

Word Size Identification — Valid Widths Which of the following is not commonly used as a processor word size in mainstream computing contexts?

Accepted Answer

Introduction / Context: Word size refers to the typical width of the CPU internal registers and the size the processor handles efficiently. Recognizing common widths helps when reading ISA specifications and ABI documentation. Given Data / Assumptions: Common word sizes historically include 16, 32, and 64 bits. Bytes are 8 bits but usually not called a word in mainstream usage. Exotic or niche machines may vary but standards dominate practice. Concept / Approach: Identify which of the listed values is not a typical standard word width across common architectures. While 8, 16, and 64 all have roles in terminology or extended word concepts, 28 is unusual and not standardized as a word size in mainstream ISAs. Step-by-Step Solution: 1) 16 and 64 are widely recognized as word and double word sizes in many ISAs.2) 8 is the byte size; though not a word, it is a standard width term.3) 28 is not a common register or bus width for word definitions.4) Therefore 28 is not a typical word size. Verification / Alternative check: Survey mainstream ISAs such as x86, ARM, RISC-V, MIPS: none define 28-bit word sizes as a norm. Why Other Options Are Wrong: 64: Common in 64-bit architectures. 16: Canonical word size in many legacy systems. 8: Standard byte size; although not called a word, it is a recognized fundamental size. Common Pitfalls: Equating byte and word terminology; overlooking that architectures may define word differently but still not use uncommon values like 28. Final Answer: 28

Question 18

Octal to Decimal — Positional Evaluation Convert the octal number 35 (base 8) to its decimal equivalent. Show the weighted sum explicitly.

Accepted Answer

Introduction / Context: Converting from octal to decimal uses positional weights that are powers of eight. This reinforces the general positional numeral system concept used across all bases. Given Data / Assumptions: Octal number: 35_8. Digit positions from right: 8^0, 8^1, 8^2, and so on. Digits are valid octal symbols 0..7. Concept / Approach: Multiply each digit by its positional weight and add. The rightmost digit multiplies by 8^0 = 1, the next by 8^1 = 8, and so forth. Step-by-Step Solution: 1) Write 35_8 as 38^1 + 58^0.2) Compute weights: 8^1 = 8, 8^0 = 1.3) Evaluate: 38 = 24; 51 = 5.4) Sum: 24 + 5 = 29 decimal. Verification / Alternative check: Convert 29 back to octal by repeated division by 8: 29 / 8 = 3 remainder 5, reconstructing 35_8. Why Other Options Are Wrong: 71 and 92: Values inconsistent with the small range of two octal digits. 17: Result of misplacing digit weights or treating the number as base 10. Common Pitfalls: Confusing base when evaluating weights, or reading 35 as thirty five decimal instead of base 8 representation. Final Answer: 29

Question 19

Binary Coded Decimal (BCD) — Digit Representation In BCD coding, each decimal digit 0 through 9 is represented by which type of binary code width?

Accepted Answer

Introduction / Context: BCD is used when decimal digit integrity matters, such as in financial displays and calculators. Each digit is encoded independently, simplifying decimal correction and display driving at the expense of storage efficiency compared to pure binary. Given Data / Assumptions: BCD encodes digits 0..9 individually. Standard BCD uses four bits per digit, producing values 0000..1001 for 0..9. Invalid combinations 1010..1111 do not represent decimal digits in straight BCD. Concept / Approach: The width per digit equals the number of bits required to represent the set of ten symbols. Four bits provide 16 combinations, enough to hold 10 valid codes and 6 invalid ones in straight BCD. Step-by-Step Solution: 1) Determine required combinations: at least 10.2) Evaluate 2^3 = 8 (insufficient), 2^4 = 16 (sufficient).3) Adopt 4-bit codes for each decimal digit.4) Result: each digit in BCD is a 4-bit binary code. Verification / Alternative check: Examine common BCD tables: 0=0000, 1=0001, 2=0010, ..., 9=1001; codes 1010..1111 are unused in straight BCD. Why Other Options Are Wrong: 8-bit and 16-bit: Overkill for a single digit and not standard for BCD. ASCII code: ASCII is a 7 or 8 bit character encoding, not a numeric digit code in BCD sense. Common Pitfalls: Confusing packed BCD storage (two digits per byte) with the width per digit; the per digit code remains 4 bits. Final Answer: 4-bit binary code

Question 20

Binary conversion practice — base-10 to base-2 Convert the decimal number 64 (base 10) into its binary representation (base 2).

Accepted Answer

Introduction / Context: Converting between decimal (base 10) and binary (base 2) is a foundational digital electronics and computer science skill. The goal is to represent the same quantity using powers of 2 rather than powers of 10. Here we convert 64 in decimal to its binary form, reinforcing the idea that each binary digit (bit) corresponds to a power of 2 position. Given Data / Assumptions: Input value: 64 in base 10. We need an unsigned binary result. No leading sign bit is required; leading zeros may be included to match option widths. Concept / Approach: Binary expansion expresses a number as a sum of powers of 2. Identify the highest power of 2 less than or equal to the number, set that bit to 1, and set all lower bits to 0 if no remainder remains. Alternatively, repeated division by 2 provides the bit pattern from least significant bit to most significant bit. Step-by-Step Solution: Note that 64 = 2^6 = 64 exactly.Therefore, bit 6 (counting from 0) is 1 and bits 5..0 are 0.A minimal binary is 1 followed by six zeros: 1000000 (7 bits).With an 8-bit presentation (as in the options), this becomes 01000000. Verification / Alternative check: Convert 01000000 back to decimal using positional weights: 02^7 + 12^6 + 02^5 + … + 02^0 = 64. This confirms correctness. Why Other Options Are Wrong: 01010010 represents 2^6 + 2^4 + 2^1 = 64 + 16 + 2 = 82, not 64. 00110110 equals 32 + 16 + 4 + 2 = 54. 01001000 equals 64 + 8 = 72. Common Pitfalls: Confusing bit positions, miscounting the number of leading zeros, or assuming every answer must be a power of 2 pattern. Always confirm by summing powers of 2 with bit positions set to 1. Final Answer: 01000000

Number Systems and Codes Questions