---
title: "Gray→BCD, decimal→Excess-3 and 2's complement addition"
subject: "Digital Logic"
chapter: 1
key: "semester_3/digital_logic:gray-bcd-excess3-twos-addition"
source: https://www.bctengineer.com/q/semester_3/digital_logic/gray-bcd-excess3-twos-addition
site: BCT Engineering
---

# Gray→BCD, decimal→Excess-3 and 2's complement addition

Digital Logic · Chapter 1: Introduction

- Asked in: 2080 Baisakh, 2+2+4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/digital_logic/gray-bcd-excess3-twos-addition
- Whole chapter: https://www.bctengineer.com/md/semester_3/digital_logic/chapter-1

## Question

Perform the following: (i) (1110)gray = ( )BCD (ii) (1430)₁₀ = ( )excess-3 (iii) Use 2's complement method to perform the addition (-28 +17).

## Short answer

Gray 1110 → binary 1011 (11₁₀) → BCD 0001 0001. (1430)₁₀ = 0100 0111 0110 0011 in Excess‑3 (add 3 to each digit). In 8‑bit 2’s complement, –28 (1110 0100) + 17 (0001 0001) = 1111 0101, which is –11₁₀.

## Answer

#### (i) $(1110)_{Gray}$ to BCD
Gray to binary: MSB is copied; each next binary bit = previous binary bit XOR current Gray bit.

| Step | Gray bit | Binary bit |
|---|---|---|
| MSB | 1 | 1 |
| 2 | 1 | 1 ⊕ 1 = 0 |
| 3 | 1 | 0 ⊕ 1 = 1 |
| LSB | 0 | 1 ⊕ 0 = 1 |

$(1110)_{Gray} = (1011)_2 = (11)_{10}$. Each decimal digit in BCD: 1 → 0001, 1 → 0001.

**Answer: $(1110)_{Gray} = (0001\ 0001)_{BCD}$**

#### (ii) $(1430)_{10}$ to Excess-3
Add 3 to each digit and write in 4 bits:

| Digit | +3 | XS-3 |
|---|---|---|
| 1 | 4 | 0100 |
| 4 | 7 | 0111 |
| 3 | 6 | 0110 |
| 0 | 3 | 0011 |

**Answer: $(1430)_{10} = (0100\ 0111\ 0110\ 0011)_{XS-3}$**

#### (iii) (−28) + 17 using 2's complement (8 bits)
- $+28 = 0001\ 1100$
- 1's complement $= 1110\ 0011$; add 1: $-28 = 1110\ 0100$
- $+17 = 0001\ 0001$

```text
    1110 0100   (-28)
  + 0001 0001   (+17)
  -----------
    1111 0101   (no end carry)
```

The MSB is 1 and there is no end carry, so the result is negative in 2's complement form. Magnitude: 2's complement of $1111\ 0101$ = $0000\ 1010 + 1 = 0000\ 1011 = 11$.

Check: $-28 + 17 = -11$.

**Answer: $1111\ 0101_2 = (-11)_{10}$**

## Check yourself

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

1. What is (1110) Gray in binary?
   A. 1110
   B. 1101
   C. 1011
   D. 0111
   Answer: C. MSB 1; 1^1=0; 0^1=1; 1^0=1 gives 1011. (Key point: Gray to binary)

2. (1110) Gray equals decimal 11. What is it in BCD?
   A. 1011
   B. 0001 1011
   C. 0010 0001
   D. 0001 0001
   Answer: D. Each decimal digit 1 and 1 is coded separately as 0001 0001. (Key point: Gray to BCD)

3. What is (1430) decimal in Excess-3?
   A. 0100 0111 0110 0011
   B. 0001 0100 0011 0000
   C. 0100 0111 0110 0000
   D. 0101 1000 0111 0100
   Answer: A. Add 3 to each digit: 4,7,6,3 gives 0100 0111 0110 0011. (Key point: Excess-3 of 1430)

4. For -28 + 17 in 8 bits, what is the final answer?
   A. 1111 0101 = +245
   B. 1111 0101 = -11
   C. 0000 1011 = +11
   D. 1110 0100 = -28
   Answer: B. The sum is 1111 0101; its 2's complement is 0000 1011 = 11, so the answer is -11. (Key point: 2's complement addition)

## Related in Digital Logic

- [(1110)gray→BCD and (1430)₁₀→Excess-3](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Agray-to-bcd-decimal-to-excess3) — Digital Logic, chapter 1
- [39 to Gray/Excess-3 and 2's complement addition](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Aconvert-39-gray-excess3-addition) — Digital Logic, chapter 1
- [Convert (17)excess-3 to BCD](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Aconvert-17-excess3-bcd) — Digital Logic, chapter 1
- [39 to binary/hex and 2's complement addition](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Aconvert-39-binary-hex-addition) — Digital Logic, chapter 1

## Related in other subjects

- [2's complement: (1111)2 − (1110)2](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Atwo-complement-1111-1110) — Basic Electronics Engineering, chapter 6