---
title: "Number conversions (122)10 to BCD etc."
subject: "Basic Electronics Engineering"
chapter: 6
key: "semester_2/basic_electronics_engineering:number-conversion-2069-poush"
source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/number-conversion-2069-poush
site: BCT Engineering
---

# Number conversions (122)10 to BCD etc.

Basic Electronics Engineering · Chapter 6: Digital Electronics

- Asked in: 2069 Poush, 1×4 marks
- Video: [Decimal to Binary and Binary to Decimal Conversion](https://www.youtube.com/watch?v=AHQwhdz3tlg) — Neso Academy
- 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-2069-poush
- Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-6

## Question

Perform the following: a) (122)10 = (?)BCD  b) (423.25)8 = (?)2  c) (179.03125)10 = (?)2  d) Subtract (25)10 from (49)10 using 2's complement method.

## Short answer

a) 122 BCD: 0001 0010 0010. b) 423.25 octal to binary: 100010011.010101. c) 179=10110011, .03125=2⁻⁵ → 10110011.00001. d) 49-25 by 2’s complement = 24.

## Answer

#### a) (122)₁₀ to BCD

1 = 0001, 2 = 0010, 2 = 0010

**(122)₁₀ = (0001 0010 0010)BCD**

#### b) (423.25)₈ to binary

4 = 100, 2 = 010, 3 = 011 . 2 = 010, 5 = 101

**(423.25)₈ = (100010011.010101)₂**

#### c) (179.03125)₁₀ to binary

- 179 = 128 + 32 + 16 + 2 + 1 = 10110011
- 0.03125 = 2⁻⁵ = .00001

**(179.03125)₁₀ = (10110011.00001)₂**

#### d) 49 − 25 by 2's complement (8 bits)

- 25 = 00011001; 2's complement = 11100111
- 00110001 + 11100111 = 1 00011000
- End carry is discarded, so the result is positive: 00011000

**Answer: 49 − 25 = 24 = (00011000)₂**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. (122)10 in BCD is:
   A. 0001 0010 0010
   B. 0111 1010
   C. 0001 0010 0001
   D. 1111010
   Answer: A. The answer states this: 0001 0010 0010. (Key point: BCD of 122)

2. (423.25)8 equals:
   A. 100010011.101010
   B. 100010011.010101
   C. 100011010.010101
   D. 010100011.010101
   Answer: B. The answer states this: 100010011.010101. (Key point: Octal to binary)

3. 179 = 128 + 32 + 16 + 2 + 1 gives:
   A. 10110101
   B. 10111011
   C. 10110011
   D. 10100111
   Answer: C. The answer states this: 10110011. (Key point: 179 to binary)

4. In 2's complement subtraction, an end carry means:
   A. The result is negative
   B. Overflow occurred
   C. The answer is zero
   D. The result is positive and the carry is discarded
   Answer: D. The answer states this: The result is positive and the carry is discarded. (Key point: Subtraction rule)

## 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 (903)10 to BCD etc.](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Anumber-conversion-2068-magh) — Basic Electronics Engineering, chapter 6, asked 3×

## Related in Basic Electronics Engineering

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- [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
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- [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

- [Gray→BCD, decimal→Excess-3 and 2's complement addition](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Agray-bcd-excess3-twos-addition) — Digital Logic, chapter 1
- [Conversions: (21)₈, (671)₁₀ to BCD, (110101.011)₂](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Anumber-conversions-21-671) — Digital Logic, chapter 1