---
title: "Number conversions (375.37)8 etc."
subject: "Basic Electronics Engineering"
chapter: 6
key: "semester_2/basic_electronics_engineering:number-conversion-2071-magh"
source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/number-conversion-2071-magh
site: BCT Engineering
---

# Number conversions (375.37)8 etc.

Basic Electronics Engineering · Chapter 6: Digital Electronics

- Asked in: 2071 Magh, 4×1 marks
- Video: [Decimal To Binary, Octal, & Hexadecimal Number System conversion in Hindi | CSE GYAN](https://www.youtube.com/watch?v=-gyDAMNyu_Y) — CS Engineering Gyan
- 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-2071-magh
- Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-6

## Question

Perform the following: a) (375.37)8 = (?)16  b) (169.03125)10 = (?)2  c) (905)10 = (?)BCD  d) Subtract (25)10 from (49)10 using 2's complement method.

## Short answer

a) Octal 375.37 to hex via binary: FD.7C. b) 169=10101001, fraction 1/32=.00001 → 10101001.00001. c) 905 BCD: 1001 0000 0101. d) 2’s complement of 25 is 11100111; 49+(-25)=24 with end carry discarded.

## Answer

#### a) (375.37)₈ to hexadecimal

Octal to binary (3 bits each), then regroup in 4 bits.

```text
375  = 011 111 101
.37  = 011 111
Integer: 0 1111 1101 -> 0 F D
Fraction: 0111 1100  (padded with zeros) -> 7 C
```

**(375.37)₈ = (FD.7C)₁₆**

#### b) (169.03125)₁₀ to binary

- 169: 128 + 32 + 8 + 1 = 10101001
- 0.03125 = 1/32 = 2⁻⁵ = .00001

**(169.03125)₁₀ = (10101001.00001)₂**

#### c) (905)₁₀ to BCD

9 = 1001, 0 = 0000, 5 = 0101

**(905)₁₀ = (1001 0000 0101)BCD**

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

- 49 = 00110001, 25 = 00011001
- 2's complement of 25 = 11100110 + 1 = 11100111

```text
   00110001
 + 11100111
 ----------
 1 00011000   discard the end carry
```

End carry present, so the result is positive: 00011000 = 24.

**Answer: 49 − 25 = 24**

## Check yourself

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

1. Octal to hex conversion is done by:
   A. Writing octal digits as 3 bits and regrouping in 4 bits
   B. Dividing by 16
   C. Reading the digits directly as hex
   D. Converting each digit to BCD
   Answer: A. The answer states this: Writing octal digits as 3 bits and regrouping in 4 bits. (Key point: Octal to hex)

2. (375.37)8 equals:
   A. FD.37
   B. FD.7C
   C. DF.7C
   D. F5.7C
   Answer: B. The answer states this: FD.7C. (Key point: (375.37) octal)

3. 0.03125 equals 2^-5, so in binary it is:
   A. .0001
   B. .00011
   C. .00001
   D. .1
   Answer: C. The answer states this: .00001. (Key point: 0.03125 in binary)

4. With an end carry in 2's complement subtraction, 49 - 25 gives:
   A. -24, negative
   B. 25
   C. -25
   D. 24, positive (end carry discarded)
   Answer: D. The answer states this: 24, positive (end carry discarded). (Key point: 49 - 25)

## Close variants set in other exams

- [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×
- [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

- [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 (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 (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

- [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
- [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