Question 1

Signal classification — identify the type of information when binary digits are used. An informational signal encoded using binary digits (0 and 1) is best described as:

Accepted Answer

Introduction / Context: Communication and computation systems categorize signals by how information is represented. Binary digits (bits) use two symbols to encode data, forming the foundation of modern digital electronics and computing. Given Data / Assumptions: Encoding set: {0, 1} (binary) Receiver uses thresholds or symbol decision rules No requirement for continuous amplitude fidelity Concept / Approach: A digital signal conveys information using a finite set of discrete symbols. Binary is the simplest case with two levels. Analog signals, by contrast, can (ideally) assume a continuum of amplitudes; small changes matter directly to the information carried. Step-by-Step Reasoning: Identify symbol set: two distinct states ⇒ digital (binary).Determine classification: “digital” describes representation, not device technology.Examples: logic 0/1, ASCII bits on a UART line, Ethernet symbols. Verification / Alternative check: Digital receivers specify decision thresholds (e.g., VIH/VIL). As long as the signal resides in the correct region, it decodes to 0 or 1 regardless of minor analog fluctuations, matching the definition of digital signaling. Why Other Options Are Wrong: Solid state: describes device technology (semiconductors), not information representation. Analog: uses continuous amplitudes, not discrete digits. Non-oscillating: relates to waveform behavior, not encoding type. Stochastic: refers to randomness, not signal representation. Common Pitfalls: Confusing “digital” with “discrete-time” (sampling) versus “discrete-amplitude” (symbol levels); here the emphasis is amplitude symbols. Final Answer: digital

Question 2

Base conversion — convert the decimal value to binary. Task: Express 18₁₀ as a binary number (no prefixes or suffixes).

Accepted Answer

Introduction / Context: Converting decimal to binary is a fundamental skill for understanding memory layouts, bitwise operations, and low-level protocols. The standard approach decomposes a number into sums of powers of two or uses repeated division by 2. Given Data / Assumptions: Decimal value: 18 Unsigned, positive integer Concept / Approach: Find the largest power of two not exceeding 18, subtract it, and continue until the remainder is zero. Alternatively, perform successive dividing by 2 and record remainders from least significant to most significant bit. Step-by-Step Solution: Powers of two: 16, 8, 4, 2, 118 = 16 + 2 ⇒ bits for 16 and 2 are 1; others 0Binary pattern: 1 (16) 0 (8) 0 (4) 1 (2) 0 (1) ⇒ 10010 Verification / Alternative check: Using division by 2: 18/2=9 r0, 9/2=4 r1, 4/2=2 r0, 2/2=1 r0, 1/2=0 r1; remainders upward give 10010. Convert back: 10010₂ = 16 + 2 = 18₁₀, confirming correctness. Why Other Options Are Wrong: 11110: equals 30₁₀. 10001: equals 17₁₀. 1111000: equals 120₁₀. 010010: correct value with a leading zero; while numerically same, the canonical minimal form is 10010. Common Pitfalls: Misplacing weights (confusing the 8 and 4 positions). Dropping necessary zeros between set bits. Final Answer: 10010

Question 3

Hexadecimal to decimal — evaluate 777₁₆. Compute the decimal value of the hex number 0x777 using positional weights.

Accepted Answer

Introduction / Context: Hexadecimal expands by powers of 16. Rewriting multi-digit hex numbers into decimal clarifies positional weighting and strengthens mental math for address offsets and constants in code. Given Data / Assumptions: Hex number: 777₁₆ Each digit 7 represents decimal 7 Concept / Approach: Apply place values: value = d216^2 + d116^1 + d016^0. With d2=d1=d0=7, the computation is straightforward. Step-by-Step Solution: 16^2 = 256 ⇒ 7256 = 179216^1 = 16 ⇒ 716 = 11216^0 = 1 ⇒ 71 = 7Sum: 1792 + 112 + 7 = 1911 Verification / Alternative check: Convert to binary: 7 → 0111, so 777₁₆ = 0111 0111 0111₂. Evaluate in decimal: (4+2+1)16^2 + (4+2+1)16 + (4+2+1) = 7256 + 716 + 7, same result 1911₁₀. Why Other Options Are Wrong: 191 and 19: undercount by ignoring higher place values. 19111: overcount by an extra digit place. 2071: arithmetic mismatch of the 16-based expansion. Common Pitfalls: Treating base-16 digits as base-10 and adding 7+7+7=21 (incorrect). Final Answer: 1911

Question 4

BCD to decimal — decode the packed BCD value. Task: Convert the BCD bitstream 0101 0110 1001 (bcd) into its decimal number.

Accepted Answer

Introduction / Context: Binary Coded Decimal (BCD) represents each decimal digit independently using a 4-bit nibble. Decoding BCD is common in digital clocks, calculators, and interfaces where human-readable decimal is required without complex conversions. Given Data / Assumptions: BCD stream: 0101 0110 1001 Each nibble encodes one decimal digit 0–9 No invalid BCD digits appear (1010–1111 would be invalid) Concept / Approach: Split the bitstream into 4-bit groups and translate each nibble to its decimal digit using 8-4-2-1 weights. Concatenate the resulting digits to form the final decimal number (not binary interpretation). Step-by-Step Solution: 0101₂ = 50110₂ = 61001₂ = 9Combine digits: 569 Verification / Alternative check: Cross-check weights: 0101 = 4+1=5; 0110 = 4+2=6; 1001 = 8+1=9. None of the nibbles exceed 1001, so all are valid BCD digits. Therefore, the decoded decimal is 569. Why Other Options Are Wrong: 539 / 596: one or more nibbles misread. 2551 / 1552: treat the stream as binary groups wrongly or mix decimal positions. Common Pitfalls: Attempting to interpret the entire 12-bit word as a binary integer; BCD is digit-wise, not value-packed like pure binary. Final Answer: 569

Question 5

Base conversion — convert the binary number to decimal. Task: Evaluate the decimal value of binary 1110₂ using positional weights.

Accepted Answer

Introduction / Context: Binary-to-decimal conversion reinforces understanding of bit weights (1, 2, 4, 8, …). This is essential when reading register dumps, flags, and immediate constants in low-level code. Given Data / Assumptions: Binary number: 1110₂ No sign bit or fraction; simple unsigned conversion Concept / Approach: Sum the weights corresponding to the 1 bits. For 4-bit numbers, the weights from most to least significant are 8, 4, 2, 1. Step-by-Step Solution: Pattern: 1 1 1 0Weights: 8 + 4 + 2 + 0Sum: 8 + 4 + 2 = 14 Verification / Alternative check: Using polynomial expansion: 12^3 + 12^2 + 12^1 + 02^0 = 8 + 4 + 2 + 0 = 14. Converting 14 back to binary gives 1110₂, confirming the round-trip. Why Other Options Are Wrong: 3, 1, 7, 12: each omits one or more active bit weights. Common Pitfalls: Reading bits right-to-left and misassigning weights (e.g., adding the 1′s place when it is 0). Final Answer: 14

Question 6

Decimal to 8-bit binary — encode the value with leading zeros if needed. Task: Convert 187₁₀ to an 8-bit binary representation.

Accepted Answer

Introduction / Context: Encoding a decimal integer as an 8-bit binary pattern is common when writing to registers, configuring peripherals, or defining bit masks. The 8-bit form explicitly shows each bit from 2^7 to 2^0. Given Data / Assumptions: Decimal value: 187 Target width: 8 bits (MSB first) Concept / Approach: Decompose 187 into powers of two ≤ 128. Mark a 1 for each included power and 0 otherwise to build the 8-bit string. Alternatively, perform repeated division by 2 and pad to 8 bits. Step-by-Step Solution: Largest power ≤ 187 is 128 (2^7): remainder 59Next 64 (2^6): not used (0); remainder stays 59Use 32 (2^5): remainder 27Use 16 (2^4): remainder 11Use 8 (2^3): remainder 3Use 4 (2^2): not used (0); remainder 3Use 2 (2^1): remainder 1Use 1 (2^0): remainder 0Bits: 1 0 1 1 1 0 1 1 ⇒ 10111011 Verification / Alternative check: Sum the active weights: 128 + 32 + 16 + 8 + 2 + 1 = 187. Also, a calculator check in binary mode will show 187 → 10111011₂. Why Other Options Are Wrong: 11011101, 10111101, 10111100, 10111010: each flips or misses one or more required bit positions. Common Pitfalls: Forgetting the 64’s place is 0 here, or misplacing the 4’s place (which is 0). Final Answer: 10111011

Question 7

Analog signals — counting possibilities over a continuous range. If an analog signal can vary anywhere between 0 V and 5 V, how many distinct analog values are possible in theory?

Accepted Answer

Introduction / Context: Analog signals, in theory, can assume any real-valued amplitude within a specified range. Unlike digital signals that restrict values to a finite set of levels, analog amplitudes are continuous, leading to an uncountably infinite set of possibilities in mathematics (within noise and bandwidth limits in practice). Given Data / Assumptions: Amplitude limits: 0 V to 5 V inclusive. Idealized measurement (no quantization, infinite resolution). No discretization via ADC is applied. Concept / Approach: A continuous interval on the real number line contains infinitely many values. Any subinterval, no matter how small, can be subdivided indefinitely. Therefore, the theoretical count of distinct analog levels is infinite. Finite counts only appear after quantization (e.g., an N-bit ADC yields 2^N discrete codes). Step-by-Step Reasoning: Model the range as the real interval [0, 5].Between any two amplitudes, another amplitude exists (density of reals).Therefore, the set of possible values is infinite. Verification / Alternative check: Compare with a 10-bit ADC: 2^10 = 1024 discrete codes approximate 0–5 V but do not make it truly continuous. Removing quantization restores the infinite set. Laboratory instruments with higher resolution approach but never reach perfect continuity; the theoretical model remains infinite. Why Other Options Are Wrong: 5, 50, 250, 1024: each presumes discretization (step counts). These apply only when a quantizer (ADC, DAC step size) is specified. Common Pitfalls: Confusing device resolution (finite) with the mathematical nature of analog signals (continuous). Final Answer: infinite

Question 8

Binary to decimal conversion Convert the binary number 11001001₂ to its decimal (base 10) value.

Accepted Answer

Introduction / Context: Converting a binary number to decimal reinforces the idea that each bit position represents a power of 2. Understanding positional weights lets you evaluate any base-2 number quickly and accurately. Given Data / Assumptions: Binary number: 11001001₂. Bit weights from left (MSB) to right (LSB) for 8 bits: 128, 64, 32, 16, 8, 4, 2, 1. Only positions with a 1 contribute to the sum. Concept / Approach: Write the binary digits aligned with their powers of two and add the weights where the bit is 1. This is the standard positional notation method for base conversion. Step-by-Step Solution: List bits: 1 1 0 0 1 0 0 1.Associate weights: 128, 64, 32, 16, 8, 4, 2, 1.Select weights with bit = 1: 128 + 64 + 8 + 1.Compute sum: 128 + 64 = 192; 192 + 8 = 200; 200 + 1 = 201. Verification / Alternative check: Group into hexadecimal nibbles: 1100 1001₂ = C9₁₆. Convert hex to decimal: C(12)*16 + 9 = 192 + 9 = 201, confirming the result. Why Other Options Are Wrong: 2001: Misplaces the base-10 positional idea; binary 11001001₂ cannot reach thousands. 20: Under-counts by ignoring higher-order bits. 210: Off by 9 due to arithmetic or bit-weight mistake. Common Pitfalls: Reading from the wrong end (mixing up MSB and LSB), forgetting a weight, or confusing binary with decimal digit concatenation. Final Answer: 201

Question 9

Hexadecimal to decimal conversion Convert the hexadecimal number 731₁₆ to its decimal (base 10) value.

Accepted Answer

Introduction / Context: Hexadecimal (base 16) is compact for representing binary values because each hex digit corresponds to exactly four binary bits. Converting hex to decimal uses the same positional method as any base conversion: multiply each digit by its power of 16 and sum the results. Given Data / Assumptions: Hex number: 731₁₆ (digits 7, 3, 1). Powers of 16: 16^0 = 1, 16^1 = 16, 16^2 = 256. Digit values: 7 = 7, 3 = 3, 1 = 1 (A–F would map to 10–15). Concept / Approach: Apply positional notation: value = d216^2 + d116^1 + d016^0. This yields an exact integer result for any hex numeral. Step-by-Step Solution: Write expression: 731₁₆ = 716^2 + 316 + 1.Compute powers: 16^2 = 256.Multiply and add: 7256 = 1792; 316 = 48; 11 = 1.Sum totals: 1792 + 48 + 1 = 1841. Verification / Alternative check: Convert to binary nibbles first: 7→0111, 3→0011, 1→0001 → 0111 0011 0001₂, then evaluate as binary to decimal to confirm 1841. Why Other Options Are Wrong: 216.4: Decimal conversion of an integer in hex must be an integer; fractional result is invalid. 985: Equals 0x3D9 decimal, not 0x731. 3D9: This is a hex string, not a decimal value; 0x3D9 = 985. Common Pitfalls: Reversing the order of remainders, confusing hex digits with decimal digits, or miscomputing 16^2. Final Answer: 1841

Question 10

Decimal to binary conversion The decimal number 188 is equal to which binary (base 2) representation?

Accepted Answer

Introduction / Context: Converting decimal to binary is fundamental for digital design. The standard technique uses successive division by 2 or decomposition into powers of two that sum to the target value. Given Data / Assumptions: Target decimal value: 188. Powers of two near 188: 128, 64, 32, 16, 8, 4, 2, 1. Concept / Approach: Express 188 as a sum of powers of two, then place 1s in those bit positions and 0s elsewhere. Alternatively, use repeated division by 2 and collect remainders from last to first. Step-by-Step Solution: Pick largest power ≤ 188: 128 → remainder 60.Next 64 is too big; bit for 64 = 0; remainder remains 60.Use 32 → remainder 28; use 16 → remainder 12; use 8 → remainder 4; use 4 → remainder 0; bits for 2 and 1 are 0.Bits from 128..1: 1 0 1 1 1 1 0 0 → 10111100. Verification / Alternative check: Compute back: 128 + 32 + 16 + 8 + 4 = 188, matching the pattern 10111100₂. Why Other Options Are Wrong: 0111000 / 1100011 / 1111000: Different bit sums that evaluate to other decimal values, not 188. Common Pitfalls: Missing a needed power (for example, forgetting 4) or misplacing zeros when writing the final 8-bit form. Final Answer: 10111100

Question 11

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 12

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 13

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 14

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 15

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.

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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 16

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 17

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.

Question 18

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

Accepted Answer

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

Question 19

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

Accepted Answer

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

Question 20

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

Accepted Answer

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

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