Question 1

Transmission reliability insight: Whenever information is transmitted from one digital device to another, what is a realistic possibility that systems must plan for?

Accepted Answer

Introduction / Context: Digital communication channels—wired or wireless—are susceptible to noise, interference, attenuation, and synchronization issues. Practical protocols assume a non-zero probability of corruption and incorporate mechanisms to detect (and often correct) errors. This question asks you to identify the realistic condition that robust systems always consider. Given Data / Assumptions: Physical channels are imperfect; bit flips and frame errors can occur. Noise and interference vary with environment and hardware quality. Protocols typically include parity, checksums, CRCs, and possibly FEC. Concept / Approach: Engineering for reliability means acknowledging that errors may happen and designing layers to detect and manage them. Error *detection* methods (parity, checksums, CRC) flag corrupted data for retransmission. Error *correction* codes (Hamming, Reed–Solomon, LDPC) can repair certain errors without retransmission. Higher layers add sequence control, acknowledgments, and timeouts. Step-by-Step Solution: Assume non-zero bit-error rate on any channel.Add redundancy (e.g., CRC) to detect corruption.Use ARQ (acknowledgment/retry) or FEC to ensure reliable delivery.Monitor performance and adapt (rate control, coding strength) as needed. Verification / Alternative check: Measure packet error rates on real links; even high-quality links show occasional errors. Protocol standards (Ethernet, Wi-Fi, LTE) all include robust error-detection/correction features, proving the engineering consensus. Why Other Options Are Wrong: “One device is off”: Not a design assumption; links are established when devices are on.“Full moon”: A joke option; not an engineering factor.“Always received exactly”: Unrealistic and unsafe assumption; contradicts standard practice. Common Pitfalls: Relying on raw links without integrity checks; assuming testbench perfection translates to field reliability; ignoring synchronization and clock recovery issues. Final Answer: Errors will occur (non-zero probability) and must be detected

Question 2

Basic conversion: What is the correct pure binary representation of the decimal number 42?

Accepted Answer

Introduction / Context: Converting between decimal and binary is a core skill in digital electronics. This item asks for the pure binary form of the decimal number 42, avoiding confusion with hexadecimal, octal, or ASCII encodings that sometimes appear similar. Given Data / Assumptions: Decimal value: 42. Target: binary (base 2), no parity or additional coding. Do not confuse with hex or character encodings. Concept / Approach: Use decomposition into powers of 2 or repeated division by 2. 42 can be written as a sum of powers of 2: 32 + 8 + 2. Setting bits at those positions yields the binary representation. Reading from the highest relevant power downward forms the final bit string without unnecessary leading zeros. Step-by-Step Solution: Find largest power of 2 ≤ 42 → 32 (2^5) → set bit 2^5 = 1, remainder 10.Next power ≤ 10 → 8 (2^3) → set bit 2^3 = 1, remainder 2.Next power ≤ 2 → 2 (2^1) → set bit 2^1 = 1, remainder 0.Bits (from 2^5 to 2^0): 1 0 1 0 1 0 → 101010. Verification / Alternative check: Convert back: 1*32 + 0*16 + 1*8 + 0*4 + 1*2 + 0*1 = 32 + 8 + 2 = 42. The value matches the original decimal. Why Other Options Are Wrong: 01000010: This is 66 in decimal (also the ASCII code for the character “B”).52: That is the octal representation of 42 (base 8), not binary.2A: That is hexadecimal for 42, not binary. Common Pitfalls: Confusing bases due to similar-looking strings; assuming leading zeros are required; forgetting to express the full set of bits from the highest set bit down to 2^0. Final Answer: 101010

Question 3

Signed two's-complement representation (8-bit): In an 8-bit signed two's-complement system, which bit pattern correctly represents the decimal value −128?

Accepted Answer

Introduction / Context: Two's-complement is the dominant method for representing signed integers in digital systems because addition and subtraction use the same binary adder hardware for both positive and negative numbers. This question asks which 8-bit pattern encodes the decimal value −128 in two's-complement representation. Given Data / Assumptions: Word width is exactly 8 bits. Signed integers use standard two's-complement rules. The value in question is −128, which is the most negative value in 8-bit two's-complement. Concept / Approach: For an N-bit two's-complement integer, the representable range is −2^(N−1) to +2^(N−1) − 1. With N = 8, the range is −128 to +127. The most negative value −128 has a unique representation because there is no positive counterpart of the same magnitude within the range. In two's-complement, −128 is represented by setting only the most significant bit (MSB) to 1 and all other bits to 0, i.e., 1000 0000. Step-by-Step Solution: Establish range for 8-bit two's-complement: −128 to +127.Identify that −128 equals −2^7.The canonical encoding of −2^(N−1) is a 1 in the MSB and 0 in all lower bits.Thus the correct pattern is 1000 0000. Verification / Alternative check: Consider the definition method: to negate a value X, two's-complement uses ~X + 1. The positive counterpart +128 cannot be represented in 8 bits (it would need 1000 0000 as an unsigned value), which is why 1000 0000 is reserved to represent −128 only. Arithmetic and overflow rules are consistent with this reservation. Why Other Options Are Wrong: 1111 1110 equals −2 (since it is ~0000 0001 + 1). 0111 1111 equals +127. 1111 1111 equals −1. None of these are −128. Common Pitfalls: Assuming symmetry and expecting a +128 value; forgetting that the negative range has one extra magnitude. Confusing sign-magnitude or one's-complement with two's-complement encodings. Final Answer: 1000 0000

Question 4

Hexadecimal digit count: How many distinct symbol positions are required to represent all digits in the base-16 (hexadecimal) number system?

Accepted Answer

Introduction / Context: Hexadecimal (base 16) is widely used in digital electronics as a compact way to represent binary numbers. Each hexadecimal digit maps exactly to a 4-bit binary nibble. This question tests basic understanding of the radix and the number of distinct symbols that a base requires. Given Data / Assumptions: The radix is 16. Digits for hex are conventionally written as 0–9 and A–F. Each unique digit symbol represents a distinct value from 0 up to base − 1. Concept / Approach: In any base-R system, there are R unique digit symbols. Therefore, base 16 requires 16 symbols. Hexadecimal uses 10 numeric digits (0–9) for values 0 through 9 and 6 alphabetic symbols (A–F) for values 10 through 15. This mapping aligns perfectly with 4-bit binary groups since 16 = 2^4. Step-by-Step Solution: Recognize base 16 implies 16 distinct digit symbols.List symbols: 0–9 (10 symbols) and A–F (6 symbols).Count total symbols: 10 + 6 = 16.Conclude: sixteen symbols are needed. Verification / Alternative check: Map a binary nibble to a hex symbol: for example, 1111 maps to F (value 15). The existence of F confirms the need for 16 distinct symbols (0 through 15 inclusive). Why Other Options Are Wrong: Six or ten would correspond to bases 6 or 10; twelve is base 12 and does not match 4-bit grouping; none match base 16. Common Pitfalls: Confusing the number of symbols with the largest symbol value; misremembering that A–F correspond to 10–15, not 11–16. Final Answer: Sixteen

Question 5

Hexadecimal bytes to ASCII characters: Interpret the three-byte sequence 4B 52 58 (hex) as ASCII and select the correct 3-character string.

Accepted Answer

Introduction / Context: Converting between hexadecimal byte values and ASCII characters is a routine skill in debugging, embedded development, and protocol analysis. This question asks you to translate three hexadecimal bytes into their ASCII character equivalents to form a readable 3-character string. Given Data / Assumptions: Hex values: 4B, 52, 58. Standard ASCII mapping (7-bit encoded within an 8-bit byte). Uppercase letters A–Z occupy decimal 65–90 (hex 41–5A). Concept / Approach: Each byte represents one ASCII character. The mapping is as follows: hex 41 = 'A', 42 = 'B', …, 4B = 'K'; hex 52 falls within the uppercase range and is 'R'; hex 58 is 'X'. By decoding each byte, the final string is built by concatenation in order. Step-by-Step Solution: Convert 4B (hex) to ASCII: 0x4B = decimal 75 → 'K'.Convert 52 (hex) to ASCII: 0x52 = decimal 82 → 'R'.Convert 58 (hex) to ASCII: 0x58 = decimal 88 → 'X'.Concatenate characters: 'K' + 'R' + 'X' = 'KRX'. Verification / Alternative check: Cross-check with an ASCII table: the range 0x41–0x5A corresponds to uppercase letters. All three values lie in that range and map to the expected letters, confirming the result. Why Other Options Are Wrong: FGM, KLM, and JQW correspond to different hex sequences and do not match 4B 52 58. Therefore, they are plausible-looking but incorrect distractors. Common Pitfalls: Confusing hex values with decimal indices; reading bytes in reverse order; mixing ASCII with extended encodings that change character mappings. Final Answer: KRX

Question 6

Identifying the base-16 system: Which named number system uses a radix of sixteen for its digits and is commonly used to compactly represent binary values?

Accepted Answer

Introduction / Context: Number systems with different bases are used in electronics and computing for convenience and clarity. Hexadecimal is especially useful because each digit maps to exactly four binary bits. This question checks your understanding of which named system corresponds to base sixteen (radix 16). Given Data / Assumptions: Base sixteen implies 16 distinct digit symbols. In hex, symbols are 0–9 and A–F. Binary is base two, octal is base eight. Concept / Approach: Match the given bases to their conventional names: binary → base 2, octal → base 8, hexadecimal → base 16. Johnson is not a number system; it typically refers to a Johnson counter in digital design, unrelated to radix notation. Step-by-Step Solution: Associate base 16 with the correct term: hexadecimal.Confirm digit set: 0–9 and A–F.Validate use case: compact binary representation (4 bits per hex digit).Conclude: hexadecimal is the correct choice. Verification / Alternative check: Consider conversion: 0xFF maps to binary 1111 1111, clearly demonstrating the 4-bits-per-digit relationship. Why Other Options Are Wrong: Johnson is not a radix system; binary is base 2; octal is base 8. Common Pitfalls: Confusing device names or counter types with number systems; mixing up octal and hexadecimal bases. Final Answer: hexadecimal

Question 7

Signed addition overflow rule: In two's-complement arithmetic, overflow can occur when adding two signed numbers under which condition?

Accepted Answer

Introduction / Context: Detecting overflow in signed two's-complement addition is essential for reliable arithmetic units and software that checks bounds. Overflow indicates the true mathematical result is out of range for the chosen word width and cannot be represented exactly. Given Data / Assumptions: Two's-complement representation is used. Word width is fixed (for example, 8, 16, 32 bits). We are adding two signed integers. Concept / Approach: In two's-complement, overflow occurs if and only if adding two positives yields a negative, or adding two negatives yields a positive. This requires that the addends have the same sign, and the sign of the result differs from that sign. If the addends have opposite signs, the result moves toward zero and cannot overflow the representable range. Step-by-Step Solution: Case 1: positive + positive → overflow if result sign bit is 1.Case 2: negative + negative → overflow if result sign bit is 0.Case 3: opposite signs → result in range; no overflow.Therefore, same-sign addition is the necessary condition for overflow. Verification / Alternative check: Hardware rule: overflow flag V = Cout(msb) XOR CarryInto(msb). This condition triggers only when same-sign addition produces a sign flip in the result, confirming the conceptual rule. Why Other Options Are Wrong: Opposite signs cannot overflow in two's-complement; “both” and “none” contradict the established overflow criterion. Common Pitfalls: Confusing carry out of the MSB with signed overflow; relying solely on the carry flag instead of the dedicated signed overflow indicator. Final Answer: they have the same sign

Question 8

Hex-to-binary conversion practice: Convert the hexadecimal value 12 (base 16) into an 8-bit binary representation with leading zeros.

Accepted Answer

Introduction / Context: Converting between hexadecimal and binary is fundamental in digital design, debugging, and low-level programming. Hexadecimal compresses binary by grouping bits in units of four, making the conversion a straightforward nibble-by-nibble mapping. Given Data / Assumptions: Hex value is 12 (base 16). We seek an 8-bit binary form with leading zeros. Mapping: each hex digit corresponds to 4 binary bits. Concept / Approach: Split the hex number into digits: '1' and '2'. Convert each digit to a 4-bit binary nibble and concatenate left to right. Hex 1 maps to 0001; hex 2 maps to 0010. Concatenating yields 0001 0010, which as a continuous 8-bit string is 00010010. Step-by-Step Solution: Take hex 12 → digits: 1 and 2.Map 1 → 0001; map 2 → 0010.Concatenate: 0001 0010.Remove space: 00010010 (8 bits). Verification / Alternative check: Convert 00010010 back to hex by grouping into nibbles: 0001 = 1, 0010 = 2 → 0x12, confirming correctness. Why Other Options Are Wrong: 00010111 equals 0x17; 00010100 equals 0x14; 00100001 equals 0x21. None match 0x12. Common Pitfalls: Reversing nibble order; forgetting leading zeros; confusing decimal 12 with hex 12 (which equals decimal 18). Final Answer: 00010010

Question 9

Binary-to-hexadecimal translation: What is the hexadecimal equivalent of the 8-bit binary pattern 1011 1111?

Accepted Answer

Introduction / Context: Translating binary to hexadecimal is a quick way to express long bit strings more compactly. The conversion is exact and relies on 4-bit groupings, because each hex digit corresponds to a binary nibble. This is commonly used in datasheets, debuggers, and memory dumps. Given Data / Assumptions: Binary pattern: 1011 1111. We assume big-end nibble grouping from the left as shown. Hexadecimal uses digits 0–9 and A–F. Concept / Approach: Group the bits into nibbles: 1011 and 1111. Convert each nibble to its hex digit: 1011 = B (decimal 11), 1111 = F (decimal 15). Combine digits to form the hex value BF. The provided option includes a base annotation “16” as BF16, which is acceptable as a notation for base 16. Step-by-Step Solution: Split: 1011 1111 → 1011 and 1111.Map: 1011 → B; 1111 → F.Concatenate hex digits: BF.Select option indicating BF in base 16: BF16. Verification / Alternative check: Convert BF16 back to binary: B = 1011, F = 1111, which reconstructs 1011 1111 exactly. Why Other Options Are Wrong: FB16 reverses nibble order; 27716 expresses an octal-like form and is unrelated; 10111111 is simply the original binary, not a hex conversion. Common Pitfalls: Swapping nibble order; misreading 1111 as E instead of F; ignoring the space that separates nibbles for readability only. Final Answer: BF16

Question 10

Typical CD-ROM capacity: Select the closest standard capacity specification for a typical CD-ROM in terms of stored digital data.

Accepted Answer

Introduction / Context: Understanding storage media capacities is part of foundational knowledge in computer hardware. CD-ROMs were a standard distribution medium for software and data for many years. This question asks for the typical capacity specification for a CD-ROM in digital data terms. Given Data / Assumptions: The question refers to standard CD-ROM, not DVD or Blu-ray. We want capacity expressed as a digital storage figure. Common formats include 650 MB and 700 MB discs. Concept / Approach: Standard CD-ROMs commonly store approximately 650 MB (earlier standard) or 700 MB (later common variant). The option set includes 650 megabytes, which is a widely recognized specification for a typical CD-ROM. Time-based measures like 74 or 80 minutes refer to audio CD capacity, but the question explicitly requests digital data capacity. Therefore, 650 megabytes is the best answer among the options provided. Step-by-Step Solution: Identify that CD-ROM capacity is typically 650 MB or 700 MB.Match against provided choices.Select the canonical figure present: 650 megabytes.Confirm that other options do not express valid digital data capacity. Verification / Alternative check: Historical media specs show 74-minute audio CDs correspond roughly to 650 MB of data capacity, while 80-minute discs map to about 700 MB. Either validates that 650 MB is a standard figure. Why Other Options Are Wrong: 90 minutes is an audio time length and not the common CD-ROM data rating; “analog portions” is irrelevant because CD-ROM stores digital data; “octal bytes” is not a recognized capacity unit. Common Pitfalls: Confusing audio minutes with data megabytes; mixing up CD-ROM (650/700 MB) with DVD (4.7 GB) or Blu-ray (25 GB or more). Final Answer: 650 megabytes

Question 11

Hexadecimal value to decimal and the role of parity: Compute the decimal value of 9E (hex). The phrase “odd parity” is incidental and does not alter the numeric value.

Accepted Answer

Introduction / Context: Hexadecimal numbers are frequently converted to decimal for readability and to verify computations. Sometimes protocol fields mention parity for error checking, but parity does not change the numeric interpretation of a hexadecimal number itself. This question asks for the decimal value of 9E16 and clarifies that “odd parity” is not part of the numeric conversion. Given Data / Assumptions: Hex value: 9E16 (i.e., 0x9E). Standard base conversion rules apply. Parity, when mentioned, is a separate error-checking bit and not a component of the hexadecimal magnitude. Concept / Approach: Convert each hex digit to its decimal value and apply positional weights: the left digit is multiplied by 16^1 and the right digit by 16^0. The letter E corresponds to decimal 14. The sum gives the final decimal value. Step-by-Step Solution: Write 9E16 = 9 * 16^1 + E * 16^0.Compute 16^1 = 16 and 16^0 = 1.Substitute digits: 9 * 16 + 14 * 1 = 144 + 14.Add results: 144 + 14 = 158. Verification / Alternative check: Convert 158 back to hex: 158 / 16 = 9 remainder 14 → 9E16, confirming the conversion is correct. Why Other Options Are Wrong: 30 and 78 are unrelated decimal values; saying “Nothing. Parity does not check.” confuses an integrity check with numeric conversion and is not a decimal value. Common Pitfalls: Treating E as 15 instead of 14; misinterpreting the parity comment as modifying the numeric value; reversing digit weights. Final Answer: 158

Question 12

Positional weight in hexadecimal: For the hexadecimal number 1AB16, what is the column weight (positional value) of the leftmost digit “1”?

Accepted Answer

Introduction / Context: Understanding positional weights in different bases is crucial for converting and analyzing numbers in digital systems. Hexadecimal uses base 16, so each position moving left multiplies the weight by 16. This question asks for the positional value (column weight) of the leftmost digit “1” in the hex number 1AB16. Given Data / Assumptions: Number is 1AB16 (three hex digits). Positions from rightmost: 16^0, 16^1, 16^2. We need the weight of the leftmost “1”. Concept / Approach: In base 16, the rightmost digit has weight 16^0 = 1. The next digit to the left has weight 16^1 = 16. The next one (leftmost in this 3-digit number) has weight 16^2 = 256. Therefore, the column weight of the leftmost “1” is 256, independent of the actual digit values in the other positions. Step-by-Step Solution: Identify positions: B → 16^0, A → 16^1, 1 → 16^2.Compute 16^2 = 256.State the positional weight: 256.Conclude that this is the correct column weight for the leftmost digit. Verification / Alternative check: Expand 1AB16 to decimal: 1 * 256 + A * 16 + B * 1 = 256 + 10 * 16 + 11 = 256 + 160 + 11 = 427. The component for the leftmost digit is indeed 256. Why Other Options Are Wrong: 64 is 16^1 for a different position count; 512 and 1024 correspond to 16^2 * 2 and 16^2 * 4 and are not the direct column weight for the leftmost digit in a 3-digit hex number. Common Pitfalls: Confusing value contribution (digit * weight) with weight itself; miscounting positions from the rightmost digit; assuming decimal place values. Final Answer: 256

Question 13

Binary length estimation — how many bits does a decimal need? In base-2 representation, determine how many binary digits (bits) are required to represent the decimal number 93 without leading zeros.

Accepted Answer

Introduction / Context: Knowing how many bits are required to represent a given unsigned decimal number in binary is a foundational skill in digital electronics, impacting memory sizing, register widths, and bus design. Here, we determine the minimum number of binary digits needed to represent the decimal value 93 without using leading zeros. Given Data / Assumptions: Unsigned representation (no sign bit). Value: 93 in base-10. No leading zeros; we want the minimal bit length. Concept / Approach: The number of bits n required to represent a positive integer N satisfies 2^(n-1) ≤ N ≤ 2^n - 1. Equivalently, n = floor(log2(N)) + 1. Alternatively, we can convert 93 to binary and count the digits directly, or bound it between nearby powers of two. Step-by-Step Solution: Compute powers of two near 93: 2^6 = 64 and 2^7 = 128.Since 64 ≤ 93 ≤ 127, any binary form of 93 must fit within 7 bits (range for 7 bits is 0 to 127).Optional explicit conversion: 93 = 64 + 16 + 8 + 4 + 1 → binary 1011101.Count digits in 1011101 → 7 bits. Verification / Alternative check: Using n = floor(log2(93)) + 1 → log2(93) is between 6 and 7, so floor(·) = 6 → n = 7. This matches the direct conversion count. Why Other Options Are Wrong: eight: 8 bits can represent 93 but are not required; the question asks for the minimal number. 11: Far too many for such a small value; 11 bits cover up to 2047. five: 5 bits cover only up to 31; insufficient for 93. Common Pitfalls: Confusing the maximum representable value with the minimum needed bits, or accidentally adding a sign bit when the question specifies an unsigned value. Final Answer: seven

Question 14

Hex conversion with a parity red herring: Treat the bit pattern 01101000 as an 8-bit quantity and convert it to hexadecimal. Ignore the mention of even parity for the purpose of base conversion.

Accepted Answer

Introduction / Context: Parity bits are used for error detection, but they have nothing to do with converting a given bit pattern between bases. This question checks whether you can convert an 8-bit binary value to hexadecimal while recognizing that parity information is irrelevant to the numerical translation itself. Given Data / Assumptions: Binary pattern: 01101000 (eight bits). Unsigned interpretation for conversion. Parity mention is a distractor; no extra parity bit is provided. Concept / Approach: Hexadecimal maps cleanly to binary using 4-bit groups (nibbles). Split the 8-bit value into two nibbles from the left: the high nibble and the low nibble. Convert each nibble to a hex digit and concatenate. Step-by-Step Solution: Group: 0110 1000.Convert 0110₂ → 6₁₆.Convert 1000₂ → 8₁₆.Combine → 0x68 (written as 6816). Verification / Alternative check: Decimal cross-check: 01101000₂ = 64 + 32 + 8 = 104; 104₁₀ = 0x68, confirming the result. Why Other Options Are Wrong: C816: 0xC8 = 200₁₀, not 104₁₀. D016: 0xD0 = 208₁₀, not 104₁₀. Nothing. Parity does not check.: Parity checking is unrelated to numeric base conversion of the given 8 data bits. Common Pitfalls: Letting the parity note mislead you into thinking the value is invalid; mixing up nibble boundaries; or converting using decimal steps instead of direct nibble mapping which is faster and less error-prone. Final Answer: 6816

Question 15

Two’s-complement edge case: Find the 2’s complement of the 4-bit binary number 1000, assuming fixed 4-bit arithmetic.

Accepted Answer

Introduction / Context: Two’s-complement arithmetic is the dominant scheme for signed integers in digital systems. An important corner case is the most negative number in a fixed-width representation, which exhibits a unique self-inverse property under two’s-complement. This question spotlights that behavior in 4-bit arithmetic. Given Data / Assumptions: Fixed width: 4 bits (values wrap modulo 2^4). Input: 1000₂ (which represents -8 in 4-bit two’s-complement). Two’s-complement operation is defined as bitwise invert then add 1, within the same width. Concept / Approach: In n-bit two’s-complement, the range is -2^(n-1) to +2^(n-1)-1. For n=4, the most negative number is -8, encoded as 1000₂. Taking two’s-complement of a number gives its negation modulo 2^n. The negation of -8 is +8, which cannot be represented in 4 bits; modulo arithmetic brings it back to 1000. Hence, 1000 is its own two’s complement in 4-bit width. Step-by-Step Solution: Invert 1000 → 0111.Add 1 → 0111 + 0001 = 1000 (carry beyond 4 bits is discarded).Result remains 1000. Verification / Alternative check: Interpretation: 1000₂ = -8. Negate: -(-8) = +8. But +8 is not representable in 4-bit two’s-complement (max is +7). Modular wrap yields the same pattern 1000, confirming the self-inverse behavior. Why Other Options Are Wrong: 111 and 0110 and 1110 are unrelated to the defined 4-bit two’s-complement of 1000 and do not reflect the fixed-width arithmetic rule. Common Pitfalls: Forgetting fixed width and accidentally using wider arithmetic; assuming every number’s two’s-complement is different from itself without considering the most negative value special case. Final Answer: 1000

Question 16

Storage sizing — bytes needed for a value: Determine the minimum whole bytes required to store the unsigned decimal value 2875 in binary (no sign bit, no leading zeros).

Accepted Answer

Introduction / Context: Memory and register sizing problems are routine in digital design. The key steps are finding the minimum number of bits required for a given unsigned value, then converting that bit count to whole bytes (8-bit groups). Here we size storage for the decimal value 2875. Given Data / Assumptions: Unsigned value: 2875. No sign bit; fixed binary storage. 1 byte = 8 bits; we must use a whole number of bytes. Concept / Approach: Compute n = floor(log2(N)) + 1 to get the minimal bit length, or bound N between powers of two. Then compute bytes = ceil(n / 8). This ensures there is enough capacity without wasted fractional bytes. Step-by-Step Solution: Find surrounding powers of two: 2^11 = 2048, 2^12 = 4096.Since 2048 ≤ 2875 ≤ 4095, the value requires 12 bits.Translate to bytes: ceil(12 / 8) = ceil(1.5) = 2 bytes.Therefore, two bytes suffice to store 2875 in binary. Verification / Alternative check: Direct conversion to binary (optional) will produce a 12-bit pattern, reaffirming that 16-bit (2-byte) storage is the nearest whole-byte capacity. Why Other Options Are Wrong: 3, 4, 5: These provide more space than needed. 3 bytes would cover up to 24 bits (overkill for a 12-bit value). Common Pitfalls: Using 8 bits per byte incorrectly or adding an unnecessary sign bit; forgetting to round up to the next whole byte after computing the bit length. Final Answer: 2

Question 17

Direct base conversion — map a small decimal to hex: Convert the decimal number 12 into its standard hexadecimal notation.

Accepted Answer

Introduction / Context: Hexadecimal is widely used in digital electronics because each hex digit corresponds cleanly to four binary bits. Converting small decimals to hex helps build fluency for reading register maps, memory dumps, and microcontroller documentation. Given Data / Assumptions: Decimal input: 12. Standard hex digit set: 0–9 and A–F (A=10, B=11, C=12, …). We express the answer with the “16” subscript marker (as in the options) to clarify the base. Concept / Approach: Translate decimal values 10–15 to single hex digits A–F. Therefore, 12 corresponds to the hex symbol C. The correct notation is C16, indicating base-16. Step-by-Step Solution: Recognize 12 lies in 10–15 range.Map 12 → C (hex digit with value 12).Write final form as C16. Verification / Alternative check: Binary cross-check: 12₁₀ = 1100₂; grouping by 4 bits gives hex C, confirming the mapping. Why Other Options Are Wrong: 1216 suggests a base misinterpretation; 12 in hex would be 0x0C if two digits were required. A216 and A16 correspond to decimal 162 and 10 respectively, not 12. Common Pitfalls: Confusing the hex literal “0xC” with the base-annotated form “C16”; assuming two hex digits are always required even when a single digit suffices. Final Answer: C16

Question 18

Nibble grouping practice — convert a 16-bit binary value to hex: Translate the binary value 0010 1111 0111 1110 into its hexadecimal representation.

Accepted Answer

Introduction / Context: Hexadecimal is a compact way to express binary values. Every hex digit corresponds to a 4-bit nibble, making conversion between binary and hex a straightforward grouping exercise. This is essential when reading datasheets or debugging at the register level. Given Data / Assumptions: Binary input: 0010111101111110. We will group from the left as 4-bit nibbles. No sign; pure bit-pattern conversion. Concept / Approach: Partition the 16-bit binary into four nibbles. Convert each nibble independently to a hex digit and concatenate the digits in order. Leading zeros in the most significant nibble are preserved when a fixed width is implied (16 bits here). Step-by-Step Solution: Group: 0010 1111 0111 1110.0010 → 2.1111 → F.0111 → 7.1110 → E.Combine → 2F7E16. Verification / Alternative check: Decimal spot-check: Convert 2F7E16 back to binary nibbles 0010 1111 0111 1110; the round-trip confirms correctness. Why Other Options Are Wrong: 77F216, 4EEE16, and 2F7716 each mismatch at least one nibble compared to the given binary, so their hex digits do not map to the provided bit pattern. Common Pitfalls: Accidentally regrouping bits from the right without maintaining the original fixed width; misreading 1110 (E) and 1111 (F). Final Answer: 2F7E16

Question 19

Hex to binary — direct digit mapping: Convert the hexadecimal value F2 to its binary equivalent using nibble expansion.

Accepted Answer

Introduction / Context: Hexadecimal to binary conversion is quick because each hex digit corresponds exactly to one 4-bit nibble. Converting F2 demonstrates this one-to-one mapping, an everyday skill for interpreting register contents and instruction encodings. Given Data / Assumptions: Hex input: F2. Hex digits: F = 15, 2 = 2. Unsigned bit-pattern conversion. Concept / Approach: Expand each hex digit to its 4-bit binary code and concatenate in order. F → 1111, 2 → 0010, yielding an 8-bit result for the two-digit hex number. Step-by-Step Solution: F → 1111.2 → 0010.Combine → 1111 0010 → 11110010. Verification / Alternative check: Decimal check: F2₁₆ = (15 × 16) + 2 = 240 + 2 = 242; 11110010₂ also equals 242, confirming correctness. Why Other Options Are Wrong: 11100011, 10100001, and 11111100 correspond to different hex values (E3, A1, and FC respectively), not F2. Common Pitfalls: Mixing digit order or mistyping the low nibble 0010 as 0100; dropping leading zeros in the low nibble unintentionally. Final Answer: 11110010

Question 20

Positional weights in decimal — identify the contribution of a digit: In the decimal number 481, what is the powers-of-10 weight contributed by the digit 4 (hundreds place)?

Accepted Answer

Introduction / Context: Place-value systems assign weights to digit positions based on powers of the base. In decimal (base-10), the hundreds, tens, and ones places carry weights 10^2, 10^1, and 10^0, respectively. This question reinforces recognition of positional contribution for a specific digit in a three-digit number. Given Data / Assumptions: Number: 481 = 4×10^2 + 8×10^1 + 1×10^0. We are asked only for the contribution of the leftmost digit (4). Standard decimal notation, no scientific notation needed here. Concept / Approach: The value contributed by a digit equals digit_value * base^position. For the hundreds place, the weight is 10^2 = 100. Multiplying by the digit 4 gives 4 * 100 = 400. Step-by-Step Solution: Identify position: leftmost digit 4 is in the hundreds position.Compute weight: 10^2 = 100.Multiply by digit: 4 * 100 = 400.Therefore, the contribution of the digit 4 is 400. Verification / Alternative check: Reconstruct 481 from place values: 4×100 + 8×10 + 1×1 = 400 + 80 + 1 = 481; the 4 contributes 400, confirming the result. Why Other Options Are Wrong: 102 and 104 are miswritten forms that resemble powers of ten but are not numerical contributions; the correct contribution is a value (400), not an exponent string. 100 is only the positional weight, not the digit’s contribution (needs multiplication by 4). Common Pitfalls: Confusing the positional weight (100) with the digit’s total contribution (400); overlooking that each digit’s effect equals digit × weight. Final Answer: 400

Number Systems and Codes Questions