Question 1

Binary-coded decimal (BCD) mapping — The decimal number 3428 corresponds to which 4-bit-per-digit BCD bitstring (concatenate the 4-bit codes for each decimal digit in order)?

Accepted Answer

Introduction: Binary-coded decimal (BCD) encodes each decimal digit individually into a 4-bit binary nibble. Reading and writing BCD correctly is essential when interfacing with displays, keypads, and financial computations where exact decimal representation is preferred over pure binary. Given Data / Assumptions: Decimal value: 3428. BCD uses 4 bits per decimal digit (8421 code). Digits are concatenated in the same left-to-right order as the decimal number. Concept / Approach: Encode each digit independently: 3 → 0011, 4 → 0100, 2 → 0010, 8 → 1000. Concatenate the nibbles without separators to obtain the final BCD bitstring. Leading zeros in the leftmost nibble may be omitted when options show shortened strings; the logical value is unchanged. Step-by-Step Solution: Write digit codes: 3 = 0011; 4 = 0100; 2 = 0010; 8 = 1000.Concatenate: 0011 0100 0010 1000.Remove spaces: 0011010000101000.If leading zeros are dropped, this becomes 11010000101000, which matches the offered option. Verification / Alternative check: Decode the selected string in 4-bit groups: 0011 (3), 0100 (4), 0010 (2), 1000 (8) → confirms 3428. Why Other Options Are Wrong: 11010001001000 / 011010010000010 / 110100001101010: At least one nibble does not correspond to a valid decimal digit or order is incorrect. 0011010000101000: This is actually the full form with leading zeros included; if the platform treats leading zeros as optional, it represents the same value but the keyed answer among options is the shortened form. Common Pitfalls: Confusing pure binary of the whole number with BCD, which encodes per digit. Dropping or misplacing a nibble boundary, leading to invalid digit codes. Final Answer: 11010000101000

Question 2

Hexadecimal to decimal conversion — What is the decimal (base-10) value of the hexadecimal number 777 (base 16)?

Accepted Answer

Introduction: Converting between hexadecimal and decimal is a routine task in digital electronics, debugging, and low-level programming. Understanding positional weights (powers of 16) allows quick mental or written conversion for any hex string, including repeating digits like 777. Given Data / Assumptions: Hex number: 777 (base 16). Digit values: 7 in hex equals 7 in decimal. Weights: 16^0, 16^1, 16^2 for the three positions. Concept / Approach: Use positional expansion: value = d2 * 16^2 + d1 * 16^1 + d0 * 16^0. Each “7” contributes 7 times its place weight. Sum the contributions to obtain the decimal result. Step-by-Step Solution: Compute 16^2 = 256 and 16^1 = 16; 16^0 = 1.Multiply: 7 * 256 = 1792; 7 * 16 = 112; 7 * 1 = 7.Add: 1792 + 112 + 7 = 1911.Therefore, hex 777 equals decimal 1911. Verification / Alternative check: As a quick check, note that 0x3FF (1023) is 10 bits set; 0x777 is smaller but in the same ballpark. Exact arithmetic confirms 1911. Why Other Options Are Wrong: 191 / 19 / 19111: Incorrect sums or extra digits. 777: Merely restates the hex string, not a base-10 value. Common Pitfalls: Treating hex digits as decimal weights (base-10) instead of base-16 powers. Forgetting that A–F represent 10–15 in hex for other problems. Final Answer: 1911

Question 3

Binary fraction to decimal — Convert the binary number 1001.0010 (base 2) into its decimal (base-10) equivalent.

Accepted Answer

Introduction: Binary numbers with fractional parts are common in fixed-point representations, digital filters, and timing calculations. Converting them into decimal clarifies magnitudes and assists in sanity checks when designing or debugging digital systems. Given Data / Assumptions: Binary number: 1001.0010 (base 2). Digits to the left of the point carry weights 2^n, n ≥ 0. Digits to the right carry fractional weights 2^(−n), n ≥ 1. Concept / Approach: Expand the value as a weighted sum of powers of two. The integer portion 1001_2 equals 8 + 1 = 9. The fractional portion .0010_2 has a single 1 in the 2^(−3) place, equal to 1/8 = 0.125. Summing yields the final decimal result. Step-by-Step Solution: Integer side: 12^3 + 02^2 + 02^1 + 12^0 = 8 + 0 + 0 + 1 = 9.Fractional side: 02^(−1) + 02^(−2) + 12^(−3) + 02^(−4) = 0 + 0 + 0.125 + 0 = 0.125.Add: 9 + 0.125 = 9.125.Therefore, 1001.0010_2 = 9.125_10. Verification / Alternative check: Convert to an exact fraction: 9 + 1/8 = 73/8. Dividing 73 by 8 indeed gives 9.125, confirming the calculation without rounding error. Why Other Options Are Wrong: 125 / 12.5 / 90.125 / 9.5: These result from misplacing the binary point or misweighting fractional bits (e.g., treating 2^(−1) as 1/10). Common Pitfalls: Confusing binary fraction weights with decimal fractions; right of the binary point halves with each step. Dropping leading or trailing zeros that actually set bit positions and therefore the correct weights. Final Answer: 9.125

Question 4

Binary place value — In a pure binary number, what is the weight (place value) of the least significant bit (LSB)?

Accepted Answer

Introduction: Understanding positional weights is foundational in digital logic, number systems, and data conversion. The least significant bit (LSB) determines the smallest increment that can be represented in an unsigned integer and directly relates to resolution in digital-to-analog and analog-to-digital conversions when the binary number is mapped to a physical quantity. Given Data / Assumptions: Unsigned binary integers, no fractional bits. Positional weighting follows powers of two. LSB is the rightmost bit of the integer portion. Concept / Approach: Binary is a base-2 system. From right to left, the weights are 2^0, 2^1, 2^2, and so on. Therefore, the LSB has weight 2^0 = 1. Each step left doubles the weight; each step right (into fractional bits) halves it. This concept generalizes: in fixed-point schemes, the first fractional bit has weight 2^(−1) = 0.5, but that is not the LSB of an integer. Step-by-Step Solution: List weights for a 4-bit example: positions (from LSB) are 2^0, 2^1, 2^2, 2^3 → 1, 2, 4, 8.By definition, the LSB is at position 2^0.Hence, its weight is 1 for integer binary numbers.This smallest weight defines one “count” or 1 LSB step. Verification / Alternative check: Consider the binary number 0001_2 = 1_10; toggling only the LSB changes the value by exactly 1. Similarly, adding 1 to any integer changes only the LSB (with carries) confirming the unit weight at that position. Why Other Options Are Wrong: 2 / 3 / 4: These correspond to higher-order bit weights (2^1, 2^? not valid for 3, 2^2). 0.5: This is the weight of the first fractional bit 2^(−1), not an integer LSB. Common Pitfalls: Mixing integer and fixed-point fractional interpretations. Assuming LSB means “smallest decimal step”; the base dictates weights. Final Answer: 1

Question 5

Decimal to hexadecimal conversion — Convert the base-10 number 5238 into base 16 (hexadecimal). Provide the correct hexadecimal representation.

Accepted Answer

Introduction: Converting decimal integers to hexadecimal is routine when working with memory addresses, register values, and encoded data. Performing the conversion by repeated division by 16 builds intuition for positional weights and helps verify tool outputs during debugging. Given Data / Assumptions: Decimal value N = 5238. Hexadecimal base B = 16; digits 0–9 and A–F represent 10–15. We assume an integer without fractional part. Concept / Approach: Use successive division by 16 to obtain remainders (least significant digit first). The quotient becomes the new dividend. Stop when the quotient is zero. Then write the remainders in reverse order to form the hexadecimal number. Step-by-Step Solution: Compute 5238 ÷ 16 = 327 remainder 6 → least significant hex digit = 6.Compute 327 ÷ 16 = 20 remainder 7 → next digit = 7.Compute 20 ÷ 16 = 1 remainder 4 → next digit = 4.Compute 1 ÷ 16 = 0 remainder 1 → most significant digit = 1.Write digits in reverse of discovery: 1 4 7 6 → hexadecimal 1476. Verification / Alternative check: Expand to decimal: 116^3 + 416^2 + 7*16 + 6 = 4096 + 1024 + 112 + 6 = 5238, confirming correctness. Why Other Options Are Wrong: 327.375: A decimal/hex mixed form; not a proper integer hex representation. 12166 / 1388: Decimal-like strings that do not correspond to 5238 in hex. 0x1476: Correct value but includes a programming prefix; the expected answer is the bare hexadecimal digits. Common Pitfalls: Writing remainders in the wrong order; always reverse the remainder sequence. Confusing decimal digits with hex digits A–F when remainders exceed 9 (not encountered here but common elsewhere). Final Answer: 1476

Question 6

Digital electronics: Convert the decimal number 151.75 into its binary representation. Be sure to show the correct integer part and the fractional part (to two binary fractional places as appropriate).

Accepted Answer

Introduction / Context: Binary conversion is a foundational skill in computer engineering and digital electronics. This problem asks for converting a mixed decimal number (an integer plus a fractional part) into binary, which reinforces the two complementary procedures: repeated division by 2 for the integer portion and repeated multiplication by 2 for the fractional portion. Mastering both methods is essential for interpreting numeric data, fixed-point formats, and low-level representations used in microcontrollers and digital signal processing. Given Data / Assumptions: Decimal value: 151.75. Target base: base 2 (binary). Precision: fraction has exactly 0.75, which is exactly representable in two binary fractional bits. No rounding is required because 0.75 is an exact dyadic fraction. Concept / Approach: Convert the integer part using repeated division by 2 and noting remainders. Convert the fractional part using repeated multiplication by 2 and recording integer carries. The final binary number is integer_bits.fraction_bits. Since 0.75 equals 3/4, which is 0.11 in binary, the fractional conversion is straightforward. Step-by-Step Solution: Integer part: 151 / 2 → remainders give bits from LSB to MSB.151 = 128 + 16 + 4 + 2 + 1 → bits at 128, 16, 4, 2, 1 set → 10010111.Fractional part: 0.75 * 2 = 1.5 → take 1, keep 0.5.0.5 * 2 = 1.0 → take 1, fraction becomes 0 → fractional bits = .11.Combine: 151.75 → 10010111.11 (binary). Verification / Alternative check: Convert back: 10010111.11 = 128 + 16 + 4 + 2 + 1 + 1/2 + 1/4 = 151 + 0.75 = 151.75. The round-trip confirms the correctness without rounding error. Why Other Options Are Wrong: 10000111.11: Integer part 135, not 151. 11010011.01: Represents a different integer and a non-matching fractional part. 00111100.00: Equals 60 decimal with zero fraction, not 151.75. Common Pitfalls: Reversing remainder order for the integer conversion, misplacing the binary point, or confusing 0.75 with 0.3 repeating in binary. Remember that fractions with denominators that are powers of 2 convert exactly to a finite number of binary fractional bits. Final Answer: 10010111.11

Question 7

Binary-coded decimal (BCD) storage width: How many bits are used to store a single BCD digit (that is, one decimal digit encoded in BCD)?

Accepted Answer

Introduction / Context: Binary-coded decimal (BCD) is widely used in digital systems that need to display or process human-readable decimal digits, such as clocks, calculators, meters, and financial devices. Knowing the encoding width for one BCD digit is essential for estimating memory, designing buses, and mapping between binary arithmetic and decimal display logic. Given Data / Assumptions: BCD encodes one decimal digit 0–9. We seek the number of bits required per digit in standard straight BCD (8421 code). No packed-nibble vs. unpacked-byte trickery is implied; the question is about the minimal code width. Concept / Approach: Standard straight BCD uses a 4-bit nibble to represent each decimal digit. The 4 bits are often labeled with weights 8-4-2-1 (8421). Ten valid patterns (0000 to 1001) represent digits 0 to 9, while 1010 to 1111 are invalid or used in special variants. Systems may store two BCD digits in one 8-bit byte (packed BCD), but per-digit encoding is still 4 bits. Step-by-Step Solution: Identify the value range: 10 decimal symbols require at least 4 bits because 2^3 = 8 is insufficient, 2^4 = 16 is sufficient.Map digits to 4-bit patterns using 8421 weighting.Conclude that each BCD digit occupies exactly one 4-bit nibble. Verification / Alternative check: Check common devices: seven-segment drivers, RTC chips, and microcontroller libraries all treat each decimal digit as a nibble, frequently packing two digits into a byte for efficient storage. Why Other Options Are Wrong: 8: One byte is used to store two BCD digits when packed; not per digit. 1 or 2: Too few states to represent ten symbols. Common Pitfalls: Confusing packed BCD storage (two digits per 8-bit byte) with the number of bits per digit, and assuming a pure binary representation rather than digit-wise encoding. Final Answer: 4

Question 8

Number systems refresher: The term 'base 10' refers to which positional number system that uses digits 0 through 9?

Accepted Answer

Introduction / Context: Different number systems are used in computing and electronics for convenience: decimal for human interaction, binary for machine-level representation, octal and hexadecimal for compact notation. Understanding the base of a system clarifies how place values work and how to convert between representations. Given Data / Assumptions: Base is the count of unique digit symbols before carry occurs. We compare common systems: binary (base 2), octal (base 8), decimal (base 10), and hexadecimal (base 16). Binary-coded decimal (BCD) is an encoding of decimal digits using binary, not a separate base. Concept / Approach: Base 10 means each place value represents a power of 10. Digits 0–9 are valid symbols. When a column exceeds 9, a carry is propagated to the next higher place value, which is weighted by an additional factor of 10. This positional structure is the familiar decimal system used in everyday arithmetic. Step-by-Step Solution: Identify that base 10 uses digits 0..9 and place weights 10^0, 10^1, 10^2, and so on.Recognize that BCD does not define a new base; it encodes decimal digits with binary nibbles.Match the term 'base 10' to 'decimal' unequivocally. Verification / Alternative check: Try representing a number such as 347 in base 10: 310^2 + 410^1 + 7*10^0. This confirms the positional meaning of base 10 and distinguishes it from other bases which would use different weights. Why Other Options Are Wrong: Binary coded decimal: An encoding scheme for decimal digits, not a different base. Octal: Base 8 uses digits 0–7 and weights 8^n. Hexadecimal: Base 16 uses digits 0–9 and A–F with weights 16^n. Common Pitfalls: Confusing BCD with a number base, and assuming that a different symbol set automatically implies a different base. The base is defined by the count of symbols and the positional weights, not the physical encoding. Final Answer: decimal

Question 9

Analog-to-digital converters (ADCs): What is the primary purpose of the sample-and-hold (S/H) circuit inside or ahead of an ADC during the conversion interval?

Accepted Answer

Introduction / Context: Many ADC architectures require a constant input during conversion to avoid conversion errors. Real-world signals may change rapidly or contain noise. The sample-and-hold (S/H) circuit captures (samples) the instantaneous analog value and maintains (holds) it stable while the converter performs quantization. This is crucial for achieving accurate, glitch-free results, especially at high resolution or when input bandwidth is significant. Given Data / Assumptions: An ADC needs a quasi-static input during its internal decision process. The S/H circuit consists of a switch and a hold capacitor with low droop and fast acquisition. We focus on the role with respect to the analog input, not internal digital nodes. Concept / Approach: During the brief sample phase, the hold capacitor charges to the input level. The switch then opens for the hold phase, isolating the capacitor so the input presented to the ADC core remains constant. By eliminating input movement while comparisons are made, the S/H prevents code ambiguity, reduces distortion, and improves effective number of bits (ENOB) for dynamic signals. Step-by-Step Solution: Identify the problem: moving inputs during conversion cause decision errors.Solution: capture the input level at a precise time (sampling aperture).Hold: maintain that level with minimal droop and feed it to the ADC core until conversion completes. Verification / Alternative check: Datasheets specify acquisition time, aperture time, aperture jitter, hold step, and droop rate. Meeting these ensures the input is effectively stable, which directly correlates with dynamic performance metrics such as SINAD and THD for fast-changing signals. Why Other Options Are Wrong: Binary counter output: Relevant to successive-approximation timing in some ADCs, not what S/H stores. ADC threshold voltage: Stabilized by internal references and regulators, not the S/H. ADC staircase waveform: A DAC inside an SAR can generate a staircase, but the S/H does not hold that waveform. Common Pitfalls: Confusing S/H storage target (the analog input) with internal DAC or comparator nodes; overlooking the impact of aperture jitter on high-frequency inputs. Final Answer: stabilize the input analog signal during the conversion process

Question 10

Base conversions: Convert the binary number 1011010₂ to its hexadecimal representation (use standard uppercase hex digits).

Accepted Answer

Introduction / Context: Converting between binary and hexadecimal is a routine task because hex compactly represents binary in groups of four bits (nibbles). This saves space and reduces errors when reading or writing long binary values in firmware, datasheets, and debugging logs. Given Data / Assumptions: Given binary: 1011010₂. Goal: hexadecimal (base 16) using digits 0–9 and A–F. Leading zeros may be added to complete a 4-bit group. Concept / Approach: Group bits from the binary point outward into sets of four, padding the most significant group with leading zeros if needed. Translate each nibble to its corresponding hex digit using the 8421 weighting (8-4-2-1). Step-by-Step Solution: Pad: 1011010 → 0101 1010 (groups of four).Left nibble 0101 = 5 (since 08 + 14 + 02 + 11 = 5).Right nibble 1010 = A (since 18 + 04 + 12 + 01 = 10 → A).Combine: 5A. Verification / Alternative check: Convert back: 5A_hex = 5*16 + 10 = 80 + 10 = 90. Binary 1011010₂ = 90 decimal. Round-trip confirms correctness. Why Other Options Are Wrong: 5B: Corresponds to 0101 1011, which has an extra 1 in the least significant bit. 5F: Corresponds to 0101 1111, different lower nibble. 5C: Corresponds to 0101 1100, not the given pattern. Common Pitfalls: Grouping from the wrong end or forgetting to pad with a leading zero. Always ensure four-bit alignment when mapping to hex. Final Answer: 5A

Question 11

Convert the 8-bit binary number 11001001₂ into its decimal (base 10) value.

Accepted Answer

Introduction / Context: Binary-to-decimal conversion is fundamental for interpreting register contents, configuration fields, and data payloads. Here we translate an 8-bit binary number into its base-10 value using positional weights, a method that scales to larger bit widths as well. Given Data / Assumptions: Binary input: 11001001₂. Unsigned interpretation (no sign or two's-complement concerns). Standard positional weights: 2^7 down to 2^0 for 8 bits. Concept / Approach: Each bit contributes its weight if it is 1. Sum the weights to obtain the decimal value. For an 8-bit number, the weights are 128, 64, 32, 16, 8, 4, 2, 1 from MSB to LSB. Step-by-Step Solution: Write weights under bits: 1 1 0 0 1 0 0 1 → 128, 64, 0, 0, 8, 0, 0, 1.Add contributions: 128 + 64 + 8 + 1 = 201.Hence, 11001001₂ = 201₁₀. Verification / Alternative check: Break into nibbles: 1100 1001 = C9_hex. Convert C9_hex to decimal: 12*16 + 9 = 192 + 9 = 201. Both methods agree. Why Other Options Are Wrong: 2001: Digit concatenation error, not a base conversion. 20: Misses high-order bits 128 and 64. 210: Close but incorrect; likely a mis-sum of 128 + 64 + 16 + 2. Common Pitfalls: Reading bits right-to-left without mapping correct weights, or confusing hex and decimal during intermediate steps. Keep a clear table of weights to avoid mistakes. Final Answer: 201

Question 12

Concept check: What is the key difference between generic binary coding and binary-coded decimal (BCD) when representing numeric data?

Accepted Answer

Introduction / Context: There are multiple ways to represent decimal numbers inside digital systems. Pure binary coding represents the entire numeric value as a single base-2 quantity. Binary-coded decimal (BCD), on the other hand, encodes each decimal digit separately using a 4-bit nibble. Understanding this distinction affects arithmetic design, storage efficiency, and user-interface formatting choices. Given Data / Assumptions: Pure binary uses powers of 2 for the entire number. BCD uses 4-bit nibbles for digits 0 through 9. We consider standard straight BCD (8421). Concept / Approach: In pure binary, a value like 59 is stored as 00111011₂ (one quantity). In BCD, 59 is stored as two nibbles: 0101 (five) and 1001 (nine), often packed into one byte as 0x59. BCD eases decimal digit extraction and display, while pure binary is more storage efficient and supports native binary arithmetic without digit-wise constraints. Step-by-Step Solution: Define pure binary: one multi-bit field encodes the entire magnitude with weights 2^n.Define BCD: each decimal digit 0–9 mapped to a 4-bit code; ten valid codes per nibble.Contrast: efficiency vs. ease of decimal I/O and rounding. Verification / Alternative check: Consider the number 99: pure binary uses 7 bits (1100011₂), while BCD uses two nibbles 1001 1001 (8 bits). This illustrates BCD's overhead and its convenience for decimal display and financial calculations. Why Other Options Are Wrong: BCD is pure binary: False; BCD is a decimal-digit-oriented code using binary symbols but not a pure binary representation of the whole number. Binary coding has a decimal format: Incorrect; pure binary is base 2. BCD has no decimal format: Incorrect; BCD is explicitly a decimal digit format. Common Pitfalls: Confusing BCD with a different base, assuming arithmetic works the same as binary (BCD arithmetic needs digit corrections like adding 6 in some adders), and forgetting invalid BCD codes 1010–1111. Final Answer: Binary coding is pure binary.

Question 13

Binary result of a decimal addition: Compute 49 + 1 in decimal and give the 8-bit binary representation of the result.

Accepted Answer

Introduction / Context: Moving between decimal arithmetic and fixed-width binary encodings is a common task in programming and embedded systems. This exercise reinforces how to add decimal numbers and then correctly encode the result as an 8-bit binary value, taking care with leading zeros for fixed-width formats used in registers and memory. Given Data / Assumptions: Addends: 49 and 1 (decimal). Output format: 8-bit unsigned binary. No overflow concerns because the sum is small compared to 255. Concept / Approach: First, perform the base-10 addition. Next, convert the decimal result to binary using positional weights or successive division by 2. Finally, represent the binary value using eight bits by adding leading zeros if the natural binary length is less than 8 bits. Step-by-Step Solution: Compute sum: 49 + 1 = 50 decimal.Convert 50 to binary: 50 = 32 + 16 + 2 = 2^5 + 2^4 + 2^1 → bits set at positions 5, 4, and 1.Natural binary: 110010₂.Pad to 8 bits: 00110010. Verification / Alternative check: Convert back: 00110010₂ = 32 + 16 + 2 = 50, matching the decimal sum. Hex check: 0x32 equals 50 decimal, further confirming the result. Why Other Options Are Wrong: 01010101: Equals 85 decimal. 00110101: Equals 53 decimal. 00110001: Equals 49 decimal (the original, not the sum). Common Pitfalls: Forgetting to pad to 8 bits, misplacing the binary point, or mis-summing powers of two. Always double-check by reconverting to decimal or cross-checking with hexadecimal. Final Answer: 00110010

Question 14

Character encoding — assess the statement: “ASCII codes are used strictly for representing the letters in the alphabet.” Is this statement accurate, considering that ASCII also includes digits, punctuation, whitespace, and control codes (e.g., line …

Accepted Answer

Introduction / Context: ASCII (American Standard Code for Information Interchange) is a foundational character-encoding scheme in computer systems. Many newcomers assume it covers “letters only,” but ASCII was designed to standardize not just alphabetic characters, but also numerals, punctuation, whitespace, and non-printing control functions. This question asks whether it is correct to say that ASCII is used strictly for letters of the alphabet. Given Data / Assumptions: ASCII is a 7-bit code in its original form, providing 128 distinct values (0–127). It predates Unicode and was intended for telecommunication and computing interoperability. The statement to evaluate: “ASCII codes are used strictly for representing the letters in the alphabet.” Concept / Approach: To judge the statement, recall the ASCII table’s structure. The lower ranges include control characters (0–31 and 127), used historically for device control (e.g., NUL, BEL, CR, LF). Printable characters include space, punctuation, digits 0–9, uppercase and lowercase letters, and a few symbols. Therefore, the scope of ASCII plainly exceeds letters-only usage. Step-by-Step Solution: Identify ASCII's domain → 128 code points, including control and printable characters.List categories → control codes, whitespace, punctuation, digits, uppercase A–Z, lowercase a–z, symbols.Compare with claim → “strictly letters” is contradicted by these additional categories.Conclude → the statement is inaccurate. Verification / Alternative check: Consult any standard ASCII table: positions 48–57 are digits, 32 is space, 33–47 and 58–64 include punctuation, 65–90 uppercase letters, 97–122 lowercase letters, and 0–31 plus 127 are control characters. The presence of these non-letter codes disproves the claim. Why Other Options Are Wrong: Correct: Fails because ASCII is not limited to letters. Applies only to uppercase letters / lowercase letters: Both are narrower subsets and ignore digits, punctuation, and control codes. Common Pitfalls: Confusing ASCII with letter-focused encodings; conflating “text” with “letters,” forgetting that text requires spaces, punctuation, and formatting/control signals. Also, mixing ASCII with ASCII-extended code pages (which are 8-bit and vendor-specific) can cause additional confusion. Final Answer: Incorrect — ASCII includes letters, digits, punctuation, whitespace, and control characters.

Question 15

Gray code concept — evaluate the statement: “In Gray code, each number is 3 greater than the binary representation of that number.” Consider the defining property of Gray code (single-bit transitions) versus ordinary binary counting.

Accepted Answer

Introduction / Context: Gray code (reflected binary code) is a binary numeral system where successive values differ by exactly one bit. It is used in position encoders, error reduction in A/D conversion, and state machines to minimize transition ambiguity. This question asks whether Gray code values are simply “3 greater than” ordinary binary numbers, which would imply a fixed arithmetic offset relationship between the two systems. Given Data / Assumptions: Gray code is defined by single-bit adjacency, not arithmetic displacement. Binary numbers follow standard base-2 positional weighting (1, 2, 4, 8, ...). The statement proposes a fixed +3 mapping from binary to Gray. Concept / Approach: Converting binary to Gray involves bitwise operations: Gray MSB equals binary MSB; each subsequent Gray bit equals the XOR of adjacent binary bits (G_i = B_i XOR B_{i+1}). There is no constant arithmetic offset that globally maps binary to Gray or vice versa. The relationship is positional and bitwise, not additive. Step-by-Step Solution: Recall rule → G = B XOR (B >> 1).Test small cases: binary 0 (000) → Gray 000; binary 1 (001) → Gray 001; binary 2 (010) → Gray 011; binary 3 (011) → Gray 010.Observe → no constant “+3” relation holds across those mappings.Therefore → the claim is incorrect. Verification / Alternative check: An inverse mapping exists but is cumulative XOR, not subtraction of a constant. Numerous tables show that Gray and binary sequences interleave, but do not differ by a uniform numeric offset. Why Other Options Are Wrong: Correct / Only for even numbers / Only for 3-bit words: None fit because the mapping is never a constant +3 for any general subset. Common Pitfalls: Assuming a simple arithmetic formula connects all numeral systems; misunderstanding that Gray code is constructed for transition properties, not arithmetic convenience. Final Answer: Incorrect — Gray code is not obtained by adding 3 to binary; it uses bitwise rules to ensure single-bit transitions.

Question 16

Excess-3 (XS-3) coding — determine the correctness of the statement: “When using the excess-3 code, a value of 3 is added to each decimal digit before converting that digit to a 4-bit binary code.”

Accepted Answer

Introduction / Context: Excess-3 (XS-3) is a self-complementing binary-coded decimal (BCD) scheme commonly introduced in number systems and digital logic. Understanding its construction helps with error detection properties and complements in arithmetic operations. This question asks whether the coding rule indeed adds 3 to each decimal digit before encoding in binary. Given Data / Assumptions: Each decimal digit (0–9) is encoded independently. Excess-3 is a 4-bit code for each decimal digit. “Add 3 then convert to 4-bit binary” is the proposed rule. Concept / Approach: In XS-3, every decimal digit D is mapped to D + 3, and this sum is then encoded as a 4-bit binary number. For example, 0 → 3 (0011), 5 → 8 (1000), 9 → 12 (1100). The “excess” of 3 yields a self-complementing property: the 9’s complement of a decimal digit corresponds to the bitwise complement of its XS-3 code, simplifying certain arithmetic and error checks. Step-by-Step Solution: Take digit D.Compute D_excess = D + 3.Encode D_excess as a 4-bit binary value.Verify with a few digits: 2 → 5 (0101); 7 → 10 (1010); 9 → 12 (1100). Verification / Alternative check: Look-up tables in textbooks match these results; decoding reverses the process by subtracting 3 from the 4-bit value to recover the decimal digit. Why Other Options Are Wrong: Incorrect / Only true for odd digits / Only true for digit 9: The rule applies uniformly to all digits 0–9. Common Pitfalls: Confusing XS-3 with straight BCD (8421) where no “+3” offset exists; forgetting that each digit is encoded independently and must remain in 4 bits. Final Answer: Correct — Excess-3 encodes each decimal digit as (digit + 3) in 4-bit binary.

Question 17

Terminology — evaluate the statement: “The process of converting a decimal number to its binary equivalent is called binary conversion.” Does this naming align with standard number-system terminology taught in digital electronics?

Accepted Answer

Introduction / Context: Number-system conversions are core skills in digital electronics. Common tasks include converting between decimal, binary, octal, and hexadecimal. This question checks whether the phrase “binary conversion” appropriately describes converting a decimal value to its binary representation, an operation used in manual computation and software/hardware systems. Given Data / Assumptions: We are naming the process of changing representations: decimal → binary. Standard curricula use generic phrases like “convert to binary,” “binary conversion,” or “base-2 encoding.” No special algorithm is mandated by the wording. Concept / Approach: “Binary conversion” is a straightforward, widely accepted term to describe producing a base-2 representation from a number in another base (commonly decimal). The most taught algorithm divides the decimal number by 2 repeatedly and collects remainders, but other methods exist (e.g., subtracting powers of two). The naming focuses on the destination base. Step-by-Step Solution: Identify the operation → representing a decimal value in base 2.Associate standard wording → “binary conversion,” “convert to binary.”Acknowledge multiple algorithms → division by 2 with remainders, positional weighting checks, or iterative subtraction of powers.Conclusion → the term is acceptable and correct. Verification / Alternative check: Textbooks and tutorials regularly title sections as “Decimal-to-Binary Conversion,” which is colloquially shortened to “binary conversion” when the direction is clear from context. Why Other Options Are Wrong: Incorrect / Only acceptable in software engineering / Only acceptable in mathematics: Artificial restrictions; the term is used across computing, electronics, and math education. Common Pitfalls: Overthinking naming conventions; confusing the term with “binary operations” (AND, OR, XOR) which are unrelated to number base conversion. Final Answer: Correct — converting decimal to base-2 is commonly called binary conversion.

Question 18

Place value — assess the statement: “The most significant digit (MSD) is the rightmost, largest-weight digit in a number.” Use standard positional notation where place value increases to the left.

Accepted Answer

Introduction / Context: Understanding “most significant digit” (MSD) and “least significant digit” (LSD) is essential in number systems, digital design, and data formatting. MSD determines the largest contribution to the overall value due to its positional weight. This question evaluates whether the MSD is the rightmost digit, as the statement claims, or the leftmost digit in standard notation. Given Data / Assumptions: Standard positional system: place value increases to the left for integers. Rightmost integer digit has the smallest weight (units place). We are not discussing bit-endianness in memory layout but numeric place value. Concept / Approach: For integers, the leftmost digit is the most significant because it multiplies the highest power of the base. For example, in base 10, 7 in 7,000 (thousands place) contributes more than any digit to its right. Therefore, the MSD is leftmost; the rightmost is the LSD. For fixed-point fractions to the right of the radix point, significance decreases further to the right (tenths, hundredths), but the integer part's MSD remains on the left side. Step-by-Step Solution: Define significance → contribution to magnitude via place weight.Integer notation → weights increase leftward; hence MSD is leftmost.Rightmost integer digit → smallest weight; that is the LSD.Therefore → the statement given is incorrect. Verification / Alternative check: Consider 9,876: MSD = 9 (thousands place), LSD = 6 (units place). The same principle applies in binary, octal, hex, and BCD representations. Why Other Options Are Wrong: Correct: Conflicts with positional notation. “Correct only for fixed-point fractions” / “Correct only in BCD”: Misinterpret significance; BCD still follows positional weight, and fraction positions to the right of the radix have smaller weights. Common Pitfalls: Confusing endianness (byte order in memory) with digit significance; mixing visual position with storage order in computing systems. Final Answer: Incorrect — the MSD is the leftmost highest-weight digit; the rightmost is the LSD.

Question 19

Binary subtraction rule — evaluate the statement: “If you borrow from a position that contains a 0, you must borrow from the next more significant 1; all 0s up to that point become 1s, and the digit last borrowed from becomes 0.”

Accepted Answer

Introduction / Context: Manual binary subtraction mirrors decimal subtraction with borrowing, but carries and borrows are base dependent. In binary, a 1 can be borrowed from a more significant position to add 2 to a lower position. This question confirms whether the described cascading-borrow rule is accurate when the immediate higher-order bit is 0 and further searching is needed for a 1. Given Data / Assumptions: Base-2 subtraction of unsigned numbers. Borrow operation when the minuend bit is smaller than the subtrahend bit. Possible runs of zeros among higher-order bits before encountering a 1. Concept / Approach: When borrowing in binary, if the next higher bit is 0, you must continue to propagate the borrow until a 1 is found. The found 1 becomes 0, the intermediate zeros turn into 1s (because each 0 receives a borrow making it 2 in base 2, then one is used to settle the lower position, leaving 1 to pass down), and the original position gains 2 to complete the subtraction. This is the classic cascaded-borrow mechanism. Step-by-Step Solution: Encounter need to borrow at bit k; immediate higher bit k+1 is 0.Search upward until bit m is 1.Set bit m → 0 after borrowing; set all bits between m and k+1 → 1 due to passed borrow.Complete subtraction at bit k with added 2 from the borrow chain. Verification / Alternative check: Work an example: subtract 0001 from 1000. You must borrow across the run of zeros: 1000 − 0001 = 0111. The higher 1 becomes 0; the intervening zeros become ones, consistent with the rule. Why Other Options Are Wrong: Incorrect: Contradicts the standard borrow algorithm. “Only for hexadecimal” / “Only for signed magnitude”: Borrowing logic applies to binary subtraction irrespective of signed representation or other bases. Common Pitfalls: Losing track of bit positions during multi-step borrows; forgetting that intermediate zeros become ones as the borrow propagates. Final Answer: Correct — the described cascaded-borrow rule accurately reflects binary subtraction mechanics.

Question 20

Decimal → binary method — evaluate the statement: “Decimal numbers can be converted into binary by dividing by 2 repeatedly and recording the remainders (LSB first).” Is this standard algorithm description accurate?

Accepted Answer

Introduction / Context: One of the most taught methods for decimal-to-binary conversion in digital fundamentals is the repeated division-by-2 technique. It yields binary digits as remainders, naturally generating the least significant bit first. This question checks whether that description accurately captures the process. Given Data / Assumptions: Unsigned integer conversion is considered. Remainders from successive divisions by 2 are either 0 or 1. The final binary number is read from the last remainder to the first (reverse order). Concept / Approach: Division by the base (2) produces digits in reverse order. Each division step produces a quotient and a remainder; stacking remainders from last to first reconstructs the binary representation. This method is efficient for manual conversions and is analogous to division by 8 for octal and by 16 for hexadecimal. Step-by-Step Solution: Start with decimal N.Divide N by 2 → record remainder (0 or 1).Replace N with the integer quotient → repeat until quotient is 0.Read remainders in reverse order → obtain binary digits from MSB to LSB. Verification / Alternative check: Example: 13 ÷ 2 → r1; 6 ÷ 2 → r0; 3 ÷ 2 → r1; 1 ÷ 2 → r1; reading remainders backward gives 1101 (binary for 13). The procedure works for any nonnegative integer. Why Other Options Are Wrong: Incorrect: Conflicts with a universally taught algorithm. “Applies only to even numbers” / “only up to 8 bits”: The method is general; the number of bits expands as needed. Common Pitfalls: Forgetting to reverse the order of remainders; mishandling 0 and 1 edge cases; confusing integer division with floating-point division for fractional parts (which require a separate multiply-by-2 method). Final Answer: Correct — repeated division by 2 with recorded remainders is the standard decimal-to-binary conversion method.

Number Systems and Codes Questions