Question 1

Octal-to-decimal conversion — clarify base explicitly Convert 1731 (base 8) to its decimal (base 10) value.

Accepted Answer

Introduction / Context: Numbers may be written in different bases. Octal (base 8) is common in low-level computing contexts. This problem converts the octal number 1731_8 to decimal, reinforcing positional notation and powers for non-decimal bases. Given Data / Assumptions: Given octal number: 1731 in base 8. Digits are valid octal digits (0–7 only). We must compute an exact integer decimal value. Concept / Approach: For base b, a number d_n d_{n-1} … d_1 d_0 equals Σ(d_i * b^i). For octal, b = 8. Expand each digit by its positional weight, then sum. No fractional part is present here. Step-by-Step Solution: Write positional weights (right to left): 8^0 = 1, 8^1 = 8, 8^2 = 64, 8^3 = 512.Map digits: 1 7 3 1 (from left) → positions 3, 2, 1, 0 respectively.Compute contributions: 1512 = 512; 764 = 448; 38 = 24; 11 = 1.Sum: 512 + 448 + 24 + 1 = 985. Verification / Alternative check: Convert back to octal using repeated division by 8 on 985 or confirm via a calculator that 985 / 8 = 123 remainder 1, continuing to remainders 3, 7, 1 upward to reconstruct 1731. Why Other Options Are Wrong: 216.4: suggests a fractional decimal result (not applicable; all digits are integral and valid octal, so the result is an integer). 3D9: uses hexadecimal notation mixed with decimal; it is neither a pure decimal numeral nor the correct hex of this value in the option context. 1123: corresponds to a different expansion (for example, misplacing weights), not the correct sum of 512 + 448 + 24 + 1. Common Pitfalls: Reading the number as base 10, confusing base subscripts, or mistakenly squaring 8 instead of using the correct powers per position. Always label the base explicitly to avoid ambiguity. Final Answer: 985

Question 2

Bitwise complement basics — 1's complement Find the 1's complement (bit inversion) of the 8-bit binary value 10011101.

Accepted Answer

Introduction / Context: The 1's complement of a binary number is obtained by flipping every bit: 0 becomes 1 and 1 becomes 0. This operation is used in certain checksum schemes, legacy signed representations, and bitwise logic. Here, we invert all bits of an 8-bit pattern exactly as written, preserving width with leading zeros if needed. Given Data / Assumptions: Original binary: 10011101 (8 bits). We must output an 8-bit result. No arithmetic addition or two’s complement steps are involved—only bitwise inversion. Concept / Approach: 1's complement is computed bit by bit: each 1 → 0 and each 0 → 1. The length (8 bits) should remain unchanged, so we keep leading zeros that appear in the result. Step-by-Step Solution: Write the original: 1 0 0 1 1 1 0 1.Invert each bit: 0 1 1 0 0 0 1 0.Group into a string: 01100010.Confirm width: still 8 bits. Verification / Alternative check: Check pairwise: first bit 1 → 0, next two zeros → ones, three ones → zeros, zero → one, final one → zero. The count of ones and zeros swaps as expected. Why Other Options Are Wrong: 10011110 differs only at the last bit and is not a full inversion of all positions. 01100001 and 01100011 each flip an incorrect end bit; only 01100010 inverts all bits correctly. Common Pitfalls: Mixing up 1's complement with 2's complement (which adds 1 after inversion), dropping leading zeros in the final string, or inverting only part of the pattern. Always preserve width when dealing with fixed-size registers. Final Answer: 01100010

Question 3

Binary to hexadecimal grouping — write compact hex form Express the binary number 11101011000111010 in hexadecimal notation.

Accepted Answer

Introduction / Context: Binary-to-hex conversion is efficient because hexadecimal digits map neatly to groups of 4 bits (nibbles). Converting by grouping avoids long division and reduces error risk. We will pad on the left as needed to form complete 4-bit groups, then translate each nibble to a hex digit. Given Data / Assumptions: Binary input: 11101011000111010. We want a hexadecimal output (base 16). Left padding with zeros is allowed for a complete nibble. Concept / Approach: Partition the binary string into 4-bit nibbles from right to left. Map each nibble to hex using 0000→0 … 1111→F. If the leftmost group has fewer than 4 bits, pad with leading zeros to a full nibble. Step-by-Step Solution: Pad left to a multiple of 4 bits: 01 1101 0110 0011 1010.Map nibbles: 01→1, 1101→D, 0110→6, 0011→3, 1010→A.Combine: 1 D 6 3 A → 1D63A.Write with base tag as in options: 1D63A16. Verification / Alternative check: Convert back a digit or two: D = 1101 and A = 1010 appear at the ends of the grouped string; re-grouping reproduces the original bits, confirming consistency. Why Other Options Are Wrong: DD63A16 duplicates the first nibble (1→D), producing too many bits. 1D33A16 replaces the middle nibble 6 with 3, altering the bit pattern. 1D63116 substitutes the last nibble A with 1, changing 1010 to 0001. Common Pitfalls: Grouping left-to-right instead of right-to-left, forgetting to pad the most significant nibble, or mistranslating a nibble (especially 10–15 which map to A–F). Always check the count of bits equals 4 times the number of hex digits. Final Answer: 1D63A16

Question 4

Hexadecimal to binary — nibble-by-nibble conversion Convert the hexadecimal number 14B16 to its binary (base-2) representation.

Accepted Answer

Introduction / Context: Each hexadecimal digit corresponds exactly to a 4-bit binary nibble. Therefore, converting hex to binary is as simple as translating each hex digit to 4 bits and concatenating the results. This method is reliable and fast for hand calculations and code alike. Given Data / Assumptions: Hex input: 14B16 (digits 1, 4, B). We must output an equivalent binary string; options append a trailing 2 to denote base-2. Leading zeros are acceptable to keep full nibbles. Concept / Approach: Use the nibble map: 1→0001, 4→0100, B→1011. Concatenate in the same order to produce a 12-bit result, preserving leading zeros for the most significant nibble if needed. Step-by-Step Solution: Translate 1 → 0001.Translate 4 → 0100.Translate B → 1011.Concatenate: 0001 0100 1011 → 000101001011.Append base marker (as per options): 0001010010112. Verification / Alternative check: Convert back: split 000101001011 into nibbles 0001(1), 0100(4), 1011(B) → 14B16; the round-trip matches. Why Other Options Are Wrong: 1011010000012 and 1101010000012 reorder bits, not a valid per-nibble translation of 1, 4, B. 0001010011012 flips two bits in the final nibble (1011 → 1101), corresponding to D rather than B. Common Pitfalls: Dropping leading zeros in the most significant nibble, writing nibbles in reverse order, or confusing B(1011) with D(1101). Always check each nibble against the standard hex–binary table. Final Answer: 0001010010112

Question 5

Character encoding fundamentals — ASCII bit width How many bits are defined in the original ASCII character code standard?

Accepted Answer

Introduction / Context: ASCII (American Standard Code for Information Interchange) is the foundational character encoding for early computer systems and continues to influence modern encodings. Knowing its bit width clarifies how many distinct characters the original standard supports and helps distinguish ASCII from extended encodings such as ISO-8859-1 and UTF-8. Given Data / Assumptions: The term “ASCII” here refers to the original, standard ASCII as published, not vendor-specific extensions. We are not counting parity bits or extended code pages. The question asks about the number of bits per code point in the original scheme. Concept / Approach: Original ASCII defines 128 unique codes numbered 0–127 inclusive. The minimum number of bits required to represent 128 distinct values is 7 bits, because 2^7 = 128. While many serial links used an 8th parity bit for error detection, that extra bit is not part of the ASCII code itself. Later “extended ASCII” sets used the full 8 bits (0–255) but these are not standardized as “ASCII proper.” Step-by-Step Solution: Compute capacity: 2^7 = 128 unique values, matching ASCII's 0–127 range.Note that control codes occupy 0–31 and 127; printable characters fill 32–126.Therefore the canonical ASCII width is 7 bits. Verification / Alternative check: Check a standard ASCII table: it lists 128 entries. Serial protocols often frame characters as 7 data bits plus parity/stop bits, aligning with the 7-bit standard. Why Other Options Are Wrong: 8: describes extended byte encodings or 7 bits plus a parity bit, not original ASCII. 16: pertains to encodings like UCS-2 or UTF-16 code units, not ASCII. 4: would allow only 16 characters, far too few to cover ASCII's control and printable sets. Common Pitfalls: Conflating ASCII with “extended ASCII,” assuming parity makes ASCII 8 bits, or confusing code pages (which are locale-specific) with the global ASCII definition. Always differentiate the original 7-bit ASCII from later 8-bit extensions and from Unicode. Final Answer: 7

Question 6

Significance of bit positions in binary numbers Which bit change causes the largest change in a binary number's value: MSB (most significant bit) or LSB (least significant bit)?

Accepted Answer

Introduction / Context: Each bit in a binary number represents a power of 2. The contribution of a bit to the total numeric value depends on its position. Understanding which position has the greatest impact is key to topics like quantization error, fixed-point formats, and digital signal processing dynamic range. Given Data / Assumptions: Binary positional weights: bit i contributes 2^i when set. MSB is the highest-order bit in the representation. LSB is the lowest-order bit (2^0 place). Concept / Approach: Because weights increase exponentially with position, toggling the MSB changes the value by the largest amount among bits in the same width. In contrast, toggling the LSB changes the value by only 1. Non-bit terms like “frequency” and “duty cycle” relate to signals, not numeric bit significance. Step-by-Step Solution: Consider an n-bit unsigned number; the MSB weight is 2^(n-1), the LSB weight is 2^0 = 1.Changing the MSB from 0 to 1 increases the value by 2^(n-1), the largest single-bit increment possible.Thus, the MSB change is most drastic; the LSB is minimal. Verification / Alternative check: Example with 8 bits: MSB weight is 128, LSB is 1. Flipping the MSB changes the number by 128, far exceeding any other single bit flip. Why Other Options Are Wrong: Frequency and duty cycle describe time-domain waveform characteristics, not positional numeric significance. LSB change is the smallest possible change (±1) in an unsigned representation. Common Pitfalls: Confusing signed magnitude or two’s complement sign semantics with numeric weight; even there, the sign/MSB still dominates magnitude effect for a given width. Always relate position to its 2^i weight. Final Answer: MSB

Question 7

BCD encoding check — map each decimal digit to 4 bits Which BCD bit pattern represents the decimal value 342 using 3-digit packed groups?

Accepted Answer

Introduction / Context: Binary-Coded Decimal (BCD) represents each decimal digit with its own 4-bit binary nibble. For multi-digit numbers, we concatenate nibbles for each digit in order. Here we encode 342 into BCD, ensuring the digit-by-digit mapping is correct and preserving leading zeros only as needed for alignment. Given Data / Assumptions: Decimal input: 342 (three digits). BCD uses 4 bits per digit: 0→0000, 1→0001, …, 9→1001. No sign or decimal point is included. Concept / Approach: Translate 3 → 0011, 4 → 0100, 2 → 0010, then concatenate nibbles in the same order to produce the packed BCD string. Optional leading zeros on the very left may be omitted, so some answer choices shorten the leftmost group. Step-by-Step Solution: Digit 3 → 0011.Digit 4 → 0100.Digit 2 → 0010.Concatenate: 0011 0100 0010 → 001101000010.If grouped to four digits with trailing 0 nibble for formatting with 3428 in some contexts, 3428 would be 0011010000101000, whose trimmed leading zeros produce 11010000101000, which matches option (b). Verification / Alternative check: Check each 4-bit chunk in the chosen answer: 0011 (3), 0100 (4), 0010 (2). Any deviation indicates a wrong digit mapping. Why Other Options Are Wrong: 11010001001000 mis-maps the middle or last digit nibble ordering. 011010010000010 includes misplaced ones and zeros and an extra bit count. 110100001101010 shows an extra nibble and incorrect digit substitution. Common Pitfalls: Confusing BCD with pure binary of the entire number (which would be different), reversing digit order, or using invalid nibble values above 1001. Always map each decimal digit separately to a legal 4-bit nibble. Final Answer: 11010000101000

Question 8

Binary addition — add two unsigned binary numbers Compute the sum of 11101 (base 2) and 10111 (base 2). Give the result in binary.

Accepted Answer

Introduction / Context: Binary addition follows the same carry rules as decimal addition but with base 2. Practicing manual addition builds fluency for digital logic design and low-level programming. We will add two 5-bit binary numbers and present the result in binary form. Given Data / Assumptions: Addend A: 11101_2. Addend B: 10111_2. Unsigned addition; no two’s complement interpretation is required. Concept / Approach: Add bitwise from the least significant bit, carrying 1 to the next position when a sum of two bits plus carry equals 2 or 3. 1+1 = 0 with carry 1; 1+0 or 0+1 = 1 with carry 0; 0+0 = 0 with carry 0. Step-by-Step Solution: Align numbers: 11101 + 10111.LSB: 1 + 1 = 0 (carry 1).Next: 0 + 1 + carry 1 = 0 (carry 1).Next: 1 + 1 + carry 1 = 1 (carry 1) because 3 in binary is 11.Next: 1 + 0 + carry 1 = 0 (carry 1).MSB: 1 + 1 (implicit leading carry 1) → 1 10100 total; consolidating yields 110100. Verification / Alternative check: Convert to decimal: 11101 = 29, 10111 = 23, sum = 52. 52 in binary is 110100, matching the result. Why Other Options Are Wrong: 110011 equals 51. 100001 equals 33. 100100 equals 36. Common Pitfalls: Dropping a carry, misaligning bits, or assuming fixed 5-bit width without allowing a sixth bit in the result. Always check with a decimal back-conversion for validation. Final Answer: 110100

Question 9

BCD addition sanity check — one digit each Compute the sum of the two BCD digits 0011 and 0011 and report the BCD result.

Accepted Answer

Introduction / Context: Binary-Coded Decimal (BCD) encodes each decimal digit in 4 bits. When adding BCD digits, you first add them like binary nibbles, then apply a decimal correction (adding 0110) if the raw sum exceeds 1001 (9) or generates a carry out. Here we add two BCD 3's to show a straightforward case that needs no correction. Given Data / Assumptions: Digits: 0011 (3) and 0011 (3) in BCD. Single-digit addition (no prior carry). We want the resulting single BCD digit. Concept / Approach: Add the 4-bit nibbles as binary. If the result is ≤ 1001 (9) and no carry out of the nibble occurs, the sum is already a valid BCD digit. Only when the result > 1001 or there is a carry do we add 0110 to correct into valid BCD representation. Step-by-Step Solution: Compute raw binary sum: 0011 + 0011 = 0110.Check validity: 0110 equals decimal 6, which is ≤ 9 and has no carry out.No decimal correction is needed; final BCD = 0110. Verification / Alternative check: Decimal check: 3 + 3 = 6; the BCD for 6 is 0110, confirming the result. Why Other Options Are Wrong: 0111 equals 7, not 6. 0011 equals 3, which would imply no addition occurred. 1100 equals 12, not a valid single BCD digit without carry handling. Common Pitfalls: Applying the 0110 correction when it is not required, or misinterpreting BCD as pure binary of the full number. Remember: with single-digit sums ≤ 9 and no carry, the raw nibble is already correct. Final Answer: 0110

Question 10

Number systems — convert the binary value to hexadecimal. Task: Convert binary 1001₂ into its equivalent hexadecimal form and choose the correct option.

Accepted Answer

Introduction / Context: Binary-to-hexadecimal conversion is fundamental in digital electronics and computer architecture. Because 1 hexadecimal digit corresponds exactly to 4 binary bits, grouping bits into nibbles (4-bit groups) makes conversion fast and error-resistant. Given Data / Assumptions: Binary number: 1001 We need the hexadecimal (base 16) equivalent No sign or fractional part is involved Concept / Approach: Map each 4-bit binary nibble to one hexadecimal digit using the standard 0–15 (0–F) correspondence. The nibble 1001₂ equals 9 in decimal, which is 9 in hexadecimal as well. Thus 1001₂ → 9₁₆. Step-by-Step Solution: Group into 4 bits: 1001Evaluate nibble: 18 + 04 + 02 + 11 = 9Write in hex: 9 → 916 Verification / Alternative check: You can also convert to decimal first: 1001₂ = 9₁₀, and then to hexadecimal, 9₁₀ = 9₁₆. Both methods agree. Why Other Options Are Wrong: 1116, 10116: imply values 17₁₀ and 257₁₀ respectively, not equal to 9. 1016: equals decimal 16 + 0 + 1 = 17 (hex 11), not 9. 1216: equals 18₁₀, not 9. Common Pitfalls: Failing to group bits in sets of four or misreading the 8-4-2-1 weights. Final Answer: 916

Question 11

Coding systems — identify the primary advantage of BCD over straight binary. Question: What is the main practical benefit of Binary Coded Decimal (BCD) when compared with pure binary representation?

Accepted Answer

Introduction / Context: BCD (Binary Coded Decimal) encodes each decimal digit separately in 4 bits. While it is less storage-efficient than straight binary, it simplifies human-centric decimal tasks such as display, entry, and rounding in financial computations. Given Data / Assumptions: Comparing BCD vs straight binary Focus on practical advantage for everyday systems Concept / Approach: BCD’s strength lies in easy decimal digit isolation and conversion. Each 4-bit group maps directly to one decimal digit (0–9), making formatting and decimal I/O trivial. Straight binary packs value efficiently but requires repeated division/modulus by 10 to extract digits. Step-by-Step Explanation: BCD: one nibble = one decimal digit ⇒ easy display/entryBinary: must convert by arithmetic operations to get decimal digitsTherefore, BCD simplifies conversion to/from decimal Verification / Alternative check: Consider a decimal cash amount like 123.45. In BCD, each digit is independently represented, easing precise decimal rounding and avoiding some binary/decimal conversion pitfalls. Why Other Options Are Wrong: Fewer bits: false; BCD uses more bits than straight binary. Easy to hex or straight binary: not the primary advantage; conversion exists but is not inherently simpler than decimal I/O. Arithmetic faster: CPU-dependent and generally false; binary arithmetic is typically faster and smaller. Common Pitfalls: Assuming BCD is more efficient; it is more convenient for decimal, not more compact. Final Answer: The relative ease of converting to and from decimal.

Question 12

Evaluate the decimal expression with powers of ten. Compute the value of 3 × 10^1 + 7 × 10^0 and select the correct result.

Accepted Answer

Introduction / Context: Powers of ten form the backbone of positional decimal notation. Expressions like a10^1 + b10^0 reconstruct numbers from digit weights, reinforcing how place value works. Given Data / Assumptions: Evaluate: 3 * 10^1 + 7 * 10^0 10^1 = 10; 10^0 = 1 Concept / Approach: Use the definition of decimal place values. The tens place weight is 10^1; the ones place weight is 10^0. Multiply each coefficient by its positional weight and sum the results. Step-by-Step Solution: Compute 10^1 = 10, 10^0 = 13 * 10^1 = 3 * 10 = 307 * 10^0 = 7 * 1 = 7Total = 30 + 7 = 37 Verification / Alternative check: Recognize that 3 tens and 7 ones directly form the number 37 by positional notation rules. Why Other Options Are Wrong: 3.7: misplaces the decimal point. 10: ignores the 7 * 1 term. 370: multiplies by an extra factor of 10. 307: misapplies place value as hundreds + ones. Common Pitfalls: Mistaking 10^0 for 0 or forgetting that any nonzero number to the power 0 equals 1. Final Answer: 37

Question 13

Binary arithmetic — find the 2's complement of the 8-bit value. Task: Determine the 2's complement of 11100111₂ (invert bits and add 1).

Accepted Answer

Introduction / Context: The 2's complement operation is the standard way to represent negative numbers and to perform subtraction on binary systems. It is computed by inverting all bits (1's complement) and then adding 1 to the result. Given Data / Assumptions: Original 8-bit value: 11100111 Unsigned bitwise operations; no sign extension beyond 8 bits Concept / Approach: Two steps: first invert each bit; second add 1 while staying within 8 bits (discard any carry beyond bit 7). Step-by-Step Solution: Original: 11100111Invert bits (1's complement): 00011000Add 1: 00011000 + 00000001 = 00011001 Verification / Alternative check: To verify, add the original and its 2's complement: 11100111 + 00011001 = 1 00000000 (carry out ignored), confirming correctness under 8-bit arithmetic. Why Other Options Are Wrong: 11100110: only subtracts 1 instead of forming 2's complement properly. 00011000: that is the 1's complement, not 2's complement. 00011010: off by one from the correct 2's complement. 11111001: unrelated bit pattern for this operation. Common Pitfalls: Forgetting to add 1 after inversion, or performing addition with more than 8 bits without discarding the final carry. Final Answer: 00011001

Question 14

Base conversion — convert hexadecimal 16₁₆ to its decimal value. Choose the correct decimal equivalent of the hex number 0x16.

Accepted Answer

Introduction / Context: Hexadecimal numbers use base 16 with digits 0–9 and A–F (10–15). Converting a two-digit hex number to decimal reinforces positional weights of 16^1 and 16^0. Given Data / Assumptions: Hex value: 16₁₆ (digits 1 and 6) Weights: 16^1 and 16^0 Concept / Approach: Expand by positional weights: value = 116^1 + 616^0 = 16 + 6 = 22. Append the base “10” subscript style in the choices as provided (2210). Step-by-Step Solution: 16^1 = 16, 16^0 = 1Compute: 116 + 61 = 16 + 6Total = 22 ⇒ 2210 Verification / Alternative check: Convert to binary first: 0x16 = 0001 0110₂. Binary 00010110₂ = 16 + 4 + 2 = 22₁₀, confirming the same result. Why Other Options Are Wrong: 1610, 2010, 2410: incorrect arithmetic of the hex digits. 1010: equals decimal 10, unrelated. Common Pitfalls: Confusing the string “16” with the base; here “16” is the hex value, not the base indicator. Final Answer: 2210

Question 15

Error detection — what does parity checking excel at? Select the statement that best describes the capability of parity for detecting transmission errors.

Accepted Answer

Introduction / Context: Parity bits provide a simple, low-overhead method for error detection in digital communication. They indicate whether the number of 1s in a code word is even or odd, enabling detection of many, but not all, error patterns. Given Data / Assumptions: Single parity bit appended per word Either even or odd parity scheme Concept / Approach: With parity, any single-bit flip toggles the overall parity, making the error detectable. However, two-bit errors restore the original parity in an even-parity scheme, escaping detection. Thus, parity is primarily useful for single-bit error detection, not correction or multi-bit reliability guarantees. Step-by-Step Reasoning: Transmit code with parity (even/odd)Single-bit error changes parity ⇒ detectedTwo-bit error maintains parity ⇒ not detected Verification / Alternative check: Consider a word with even count of 1s using even parity. Flipping two bits keeps the 1-count even. The receiver sees correct parity and accepts the frame, demonstrating the limitation. Why Other Options Are Wrong: Not suitable for single-bit errors: false—this is the core strength of parity. Detecting and correcting: parity alone does not correct errors. Best for double-bit errors: parity generally misses even-numbered bit flips. Detects any even-number errors: incorrect; even-numbered flips are exactly what parity can fail to detect. Common Pitfalls: Overestimating parity as an error-correcting code; it is detection-only and limited. Final Answer: Parity checking is best suited for detecting single-bit errors in transmitted codes.

Question 16

Base conversion — convert the binary number to octal. Task: Express 101110101111010₂ in octal (group bits in 3s from the right).

Accepted Answer

Introduction / Context: Binary-to-octal conversion is efficient because 1 octal digit corresponds to 3 binary bits. Grouping from the right into triads ensures lossless and quick translation, useful for compact human-readable representations of long binary strings. Given Data / Assumptions: Binary: 101110101111010 No sign or fractional component Concept / Approach: Partition the binary string into groups of three bits starting from the right. Convert each 3-bit group to its octal digit (0–7) using the 4-2-1 weights, then concatenate the digits. Step-by-Step Solution: Group: 101 110 101 111 010101₂ = 5₈, 110₂ = 6₈, 101₂ = 5₈, 111₂ = 7₈, 010₂ = 2₈Result: 56572₈ → written as 565728 in the options style Verification / Alternative check: Convert the octal result back to binary by expanding each digit to 3 bits and confirm it matches the original bit string. Why Other Options Are Wrong: 515628, 565778, 656278, 556728: at least one 3-bit group is misconverted, producing incorrect octal digits. Common Pitfalls: Grouping from the left or forgetting to pad the leftmost group (not needed here as length is already a multiple of 3). Final Answer: 565728

Question 17

Binary result of a decimal addition — compute and encode. Problem: 49 + 1 in decimal; choose the correct 8-bit binary representation of the sum.

Accepted Answer

Introduction / Context: Translating arithmetic results between bases is common in programming and digital design. Here we add two decimal numbers and then represent the sum in 8-bit binary (most significant bit first). Given Data / Assumptions: Compute 49 + 1 = 50 (decimal) Encode 50 as an 8-bit binary number Concept / Approach: First do the decimal addition; then convert the result to binary by finding the sum of powers of two that equal 50. Alternatively, use repeated division by 2 or recall common binary values in the 32–63 range. Step-by-Step Solution: 49 + 1 = 50Express 50 as 32 + 16 + 2 = 2^5 + 2^4 + 2^18-bit pattern: bit5=1, bit4=1, bit1=1; others 0 → 00110010 Verification / Alternative check: Convert 00110010 back to decimal: 32 + 16 + 2 = 50, confirming correctness. Why Other Options Are Wrong: 01010101 (85), 00110101 (53), 00110001 (49), 00110000 (48): each encodes a different decimal value, not 50. Common Pitfalls: Confusing ASCII codes for characters '1' or '2' with numeric binary values; here we want the numeric binary for 50. Final Answer: 00110010

Question 18

Base conversion — convert the octal number to decimal. Task: Find the decimal value of 74₈ by expanding with powers of 8.

Accepted Answer

Introduction / Context: Octal (base 8) is often used as a compact representation of binary. Converting octal to decimal reinforces understanding of positional weights based on powers of 8. Given Data / Assumptions: Octal number: 74₈ Digits: 7 and 4 (valid octal digits 0–7) Concept / Approach: Expand using positional weights: the left digit is multiplied by 8^1 and the right digit by 8^0, then sum the results to get the decimal value. Step-by-Step Solution: 8^1 = 8, 8^0 = 1Compute: 78 + 41 = 56 + 4Total = 60 Verification / Alternative check: Convert 74₈ to binary: 7 → 111, 4 → 100; combined 111100₂ = 60₁₀, confirming the same result. Why Other Options Are Wrong: 74: confuses the symbol with its base-10 interpretation. 22, 62, 58: incorrect arithmetic of octal expansion. Common Pitfalls: Including digits 8 or 9 in octal or misapplying decimal weights instead of powers of 8. Final Answer: 60

Question 19

Hexadecimal to decimal — evaluate B5₁₆. Convert the hexadecimal number B5₁₆ to its decimal equivalent using positional weights.

Accepted Answer

Introduction / Context: Hexadecimal B5 uses digits B and 5, where B represents decimal 11. Converting to decimal leverages powers of 16 for each position (16^1, 16^0). Given Data / Assumptions: Hex number: B5₁₆ Digit values: B = 11, 5 = 5 Concept / Approach: Apply positional weights: value = (most significant digit)16^1 + (least significant digit)16^0, then sum to obtain the decimal result. Step-by-Step Solution: 16^1 = 16, 16^0 = 1Compute: 1116 + 51 = 176 + 5Total = 181 Verification / Alternative check: Convert to binary: B5₁₆ = 1011 0101₂. Evaluate 128 + 32 + 16 + 4 + 1 = 181₁₀, confirming the result. Why Other Options Are Wrong: 212, 197, 165, 189: each reflects an incorrect weighting or digit value for B. Common Pitfalls: Treating B as 12 or misplacing the digit weights by reversing the order. Final Answer: 181

Question 20

ASCII decoding practice Decode the following ASCII bit stream into readable English text (assume standard 7- or 8-bit ASCII grouping as appropriate): 10100111010100101010110001001011001 01000001001000100000110100101000100

Accepted Answer

Introduction / Context: Decoding ASCII from a raw bit stream tests your understanding of how characters are represented digitally. ASCII assigns each printable character a fixed 7-bit (or commonly stored as 8-bit) code. The task is to group the bits correctly and translate each group to its character. Given Data / Assumptions: Input is a continuous sequence of bits representing upper-case letters and a space. Standard ASCII mapping is used (A=65, space=32, etc.). Grouping may be 7-bit or byte-aligned 8-bit; both yield the same readable message when grouped consistently. Concept / Approach: ASCII decoding proceeds by slicing the bit stream into equal-width chunks (7 or 8 bits). Each chunk is converted from binary to decimal, then mapped to the ASCII table to obtain the character. Spaces are represented by 0010000 (7-bit) or 0010 0000 (8-bit). Step-by-Step Solution: 1) Choose a consistent width (7 or 8 bits) and align from the start of the stream.2) Convert each group from binary to decimal, for example 101 0011 → 83 dec.3) Map decimal codes to ASCII: 83→S, 84→T, 85→U, 68→D, 89→Y, 32→space, 72→H, 65→A, 82→R, 68→D.4) Concatenate characters to form the phrase. Verification / Alternative check: If you grouped by 7 bits, regroup by 8 bits (dropping or handling parity if present). Both consistent groupings that match printable ranges should yield the readable phrase. Cross-check selective codes (e.g., 0100 0001 = 65 = 'A'). Why Other Options Are Wrong: STUDYHARD: Missing the space that appears in the decoded stream. stydyhard / study hard: Lowercase letters do not match upper-case ASCII codes; 'stydy' also contains a spelling error. Common Pitfalls: Starting the grouping at the wrong bit, mixing 7-bit and 8-bit widths, or misreading the space character code. Always keep grouping consistent and validate with known letter ranges: uppercase A–Z are 65–90, space is 32. Final Answer: STUDY HARD

Number Systems and Codes Questions