Digital Logic Circuits and Discrete Mathematical Structures Topic Wise Pyq

Correct Answer: B

Explanation (EN): Write both in equal length: 1101 0110 Now compare bit by bit: 1 and 0 → different 1 and 1 → same 0 and 1 → different 1 and 0 → different Total differing positions = 3 So Hamming distance = 3

Explanation (HI): दोनों को समान लंबाई में लिखते हैं: 1101 0110 अब bit by bit तुलना करें: 1 और 0 → अलग 1 और 1 → समान 0 और 1 → अलग 1 और 0 → अलग कुल अलग स्थान = 3 अतः Hamming distance = 3

Correct Answer: C

Explanation (EN): Normalized binary scientific notation requires one non-zero digit before the binary point. (0.10101)₂ can be written as 1.0101 × 2⁻¹ because the binary point moves one place to the right, so exponent is −1.

Explanation (HI): Normalized binary scientific notation में binary point के पहले एक non-zero digit होना चाहिए। (0.10101)₂ को 1.0101 × 2⁻¹ लिखा जाता है क्योंकि binary point एक स्थान दाईं ओर गया है, इसलिए exponent −1 होगा।

Correct Answer: D

Explanation (EN): Convert 213 to base 4 by repeated division: 213 ÷ 4 = 53 remainder 1 53 ÷ 4 = 13 remainder 1 13 ÷ 4 = 3 remainder 1 3 ÷ 4 = 0 remainder 3 Reading remainders from bottom to top gives 3111₄

Explanation (HI): 213 को 4 से बार-बार भाग दें: 213 ÷ 4 = 53 शेष 1 53 ÷ 4 = 13 शेष 1 13 ÷ 4 = 3 शेष 1 3 ÷ 4 = 0 शेष 3 शेषों को नीचे से ऊपर पढ़ने पर उत्तर 3111₄ मिलेगा।

Correct Answer: B

Explanation (EN): 15's complement in hexadecimal means subtract each digit from F: F − C = 3 F − 0 = F F − 0 = F F − D = 2 F − 0 = F So answer = (3FF2F)₁₆

Explanation (HI): Hexadecimal में 15's complement निकालने के लिए हर digit को F में से घटाते हैं: F − C = 3 F − 0 = F F − 0 = F F − D = 2 F − 0 = F अतः उत्तर = (3FF2F)₁₆

Correct Answer: C

Explanation (EN): First find 9's complement: 99999 − 20100 = 79899 Now add 1 to get 10's complement: 79899 + 1 = 79900

Explanation (HI): पहले 9's complement निकालते हैं: 99999 − 20100 = 79899 अब 1 जोड़कर 10's complement मिलेगा: 79899 + 1 = 79900

Correct Answer: C

Explanation (EN): Convert (241)₁₆ to decimal: = 2×16² + 4×16 + 1 = 512 + 64 + 1 = 577 Now, (1101)x = x³ + x² + 1 So, x³ + x² + 1 = 577 x³ + x² = 576 Try x = 8: 8³ + 8² = 512 + 64 = 576 So, x = 8

Explanation (HI): (241)₁₆ को decimal में बदलें: = 577 अब, (1101)x = x³ + x² + 1 अतः, x³ + x² + 1 = 577 x³ + x² = 576 x = 8 रखने पर: 8³ + 8² = 512 + 64 = 576 इसलिए x = 8

Correct Answer: B

Explanation (EN): Convert the binary number to decimal: (1010110000111)₂ = 5511 Now write octal expression in decimal: (12?07)₈ = 1×8⁴ + 2×8³ + ?×8² + 0×8 + 7 = 4096 + 1024 + 64? + 7 = 5127 + 64? Equate: 5127 + 64? = 5511 64? = 384 ? = 6

Explanation (HI): दिए गए binary को decimal में बदलने पर मान 5511 मिलता है। अब (12?07)₈ को decimal में लिखें: = 1×8⁴ + 2×8³ + ?×8² + 0×8 + 7 = 5127 + 64? अब बराबर रखें: 5127 + 64? = 5511 64? = 384 ? = 6 अतः missing digit 6 है।

Correct Answer: D

Explanation (EN): In 8-bit 2's complement, range is −128 to +127. The pattern 10000000 represents the smallest value, which is −128.

Explanation (HI): 8-bit 2's complement में range −128 से +127 तक होती है। Pattern 10000000 सबसे छोटी संख्या को दर्शाता है, जो −128 है।

Correct Answer: B

Explanation (EN): Convert each term to decimal: (1010x)₂ = 1×2⁴ + 0×2³ + 1×2² + 0×2 + x = 16 + 4 + x = 20 + x (y7)₈ = 8y + 7 (24)₁₆ = 2×16 + 4 = 36 Now, (20 + x) + (8y + 7) = 36 x + 8y = 9 Try options: x = 1, y = 1 gives 1 + 8 = 9 So missing digits are 1, 1

Explanation (HI): हर संख्या को decimal में बदलते हैं: (1010x)₂ = 20 + x (y7)₈ = 8y + 7 (24)₁₆ = 36 अब, (20 + x) + (8y + 7) = 36 x + 8y = 9 विकल्प जाँचने पर x = 1, y = 1 सही बैठता है। अतः missing digits 1, 1 हैं।

Correct Answer: C

Explanation (EN): First find 1's complement of 101100: 010011 Now add 1: 010011 + 1 = 010100

Explanation (HI): पहले 101100 का 1's complement लें: 010011 अब 1 जोड़ें: 010011 + 1 = 010100