--- title: "Number conversions (903)10 to BCD etc." subject: "Basic Electronics Engineering" chapter: 6 key: "semester_2/basic_electronics_engineering:number-conversion-2068-magh" source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/number-conversion-2068-magh site: BCT Engineering --- # Number conversions (903)10 to BCD etc. Basic Electronics Engineering · Chapter 6: Digital Electronics - Asked in: 2068 Magh, 1.5×4 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/number-conversion-2068-magh - Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-6 ## Question Perform the following: a) (903)10 = (?)BCD b) (624.03)8 = (?)16 c) (101101)2 + (10111)2 = ? d) Subtract using 2's complement: (14)10 − (11)10 ## Short answer a) 903 BCD: 1001 0000 0011. b) Octal 624.03 to hex via binary: 194.0C. c) 101101+10111 = 1000100 (68). d) 2’s complement of 11 is 0101; 14+0101=0011 (3). ## Answer #### a) (903)₁₀ to BCD 9 = 1001, 0 = 0000, 3 = 0011 **(903)₁₀ = (1001 0000 0011)BCD** #### b) (624.03)₈ to hexadecimal ```text 624 = 110 010 100 .03 = 000 011 Integer: 1 1001 0100 -> 1 9 4 Fraction: 0000 1100 -> 0 C ``` **(624.03)₈ = (194.0C)₁₆** #### c) (101101)₂ + (10111)₂ ```text 101101 (45) + 010111 (23) -------- 1000100 (68) ``` Carries are propagated from the LSB: 1+1 = 10, etc. **Answer: 1000100₂ (= 68₁₀)** #### d) (14)₁₀ − (11)₁₀ using 2's complement (4 bits) - 14 = 1110; 11 = 1011 - 2's complement of 1011 = 0100 + 1 = 0101 ```text 1110 + 0101 ------ 1 0011 discard the end carry ``` End carry present, so the result is positive: 0011. **Answer: 14 − 11 = (0011)₂ = 3** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. (903)10 in BCD is: A. 1001 0000 0011 B. 1001 0000 0101 C. 1001 0011 D. 1110000111 Answer: A. The answer states this: 1001 0000 0011. (Key point: BCD of 903) 2. (624.03)8 equals: A. 194.03 B. 194.0C C. 1A4.0C D. 293.0C Answer: B. The answer states this: 194.0C. (Key point: (624.03) octal to hex) 3. (101101)2 + (10111)2 equals: A. 1000010 B. 110100 C. 1000100 D. 1001100 Answer: C. The answer states this: 1000100. (Key point: Binary addition) 4. Using 2's complement, 1110 + 0101 = 1 0011 and the end carry is discarded, so 14 - 11 equals: A. 1101 (-3) B. 0101 (5) C. 10011 (19) D. 0011 (3) Answer: D. The answer states this: 0011 (3). (Key point: 14 - 11) ## Close variants set in other exams - [Number conversions (375.37)8 etc.](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Anumber-conversion-2071-magh) — Basic Electronics Engineering, chapter 6, asked 3× - [Number conversions (122)10 to BCD etc.](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Anumber-conversion-2069-poush) — Basic Electronics Engineering, chapter 6, asked 3× ## Related in Basic Electronics Engineering - [Number conversions (10111.101)2 etc.](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Anumber-conversion-2067-mangsir) — Basic Electronics Engineering, chapter 6 - [Number conversions (25.5)10, (EAB)16](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Anumber-conversion-2073-magh) — Basic Electronics Engineering, chapter 6 - [Number conversions (10101.101)2 etc.](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Anumber-conversion-2071-bhadra) — Basic Electronics Engineering, chapter 6 - [Number conversions 33 to binary etc.](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Anumber-conversion-2067-chaitra) — Basic Electronics Engineering, chapter 6 ## Related in other subjects - [(10.001)₂ − (11.101)₂ by 2's complement](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Atwos-complement-binary-subtraction) — Digital Logic, chapter 5 - [FFH AND C0H; (16)₁₀ − (26)₁₀ by 2's complement](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Ahex-and-twos-complement-subtract) — Digital Logic, chapter 5 - [K-map Σ(1,2,3,8,9,10,11,14)+d(0,4,12), NAND](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Akmap-nand-1238-dc-0412) — Digital Logic, chapter 3, asked 5× - [K-map Σ(1,2,3,8,9,10,11,13,14)+d, POS](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Akmap-pos-form-12389) — Digital Logic, chapter 3