**Chapter 3: NUMBER SYSTEM AND THEIR CONVERSION**

## Short Answer Questions

**1. Convert 333₁ _{0} Denary numbers into Hexadecimal and back to base two number system.**

First converting, decimal to hexadecimal,

16 | 333 | Remainder
13=D |

16 | 20 | 4 |

16 | 1 | 1 |

0 |

Hence, (333)_{10} = (14D)_{16}

Now, converting back to base two number system (i.e. binary number system)

Arranging in 4-bits,

Hexadecimal | 1 | 4 | D |

Binary | 0001 | 0100 | 1101 |

Hence, (14D) _{16} = (000101001101)_{2}

** 2. Convert the following numbers according to the given instruction**

- 24010 into Octal number

Hence, (240)_{10 }= (360)_{8 }

- ABC
_{16}in to Binary number.

Arranging in 4-bits,

Hexadecimal | A | B | C |

Binary | 1010 | 1011 | 1100 |

Hence, (ABC) _{16} = (101010111100)_{2 }

_{ }

**3. What is octal number system? Convert 35610 int base 8.**

Octal number is a base 8 number system. It uses 8 digits (0, 1, 2, 3, 4, 5 6 and 7). Base of this number system is 8 or 0.

Example: (405)s or (405)o.

For second part of the question,

Hence, (356)10 = (544)8

** **** **

**4. What is binary number system? Convert 52010 into base 16.**

Binary number system is a base 2 number system. It uses 2 digits, 0 and 1. It is the only number system, which is directly understood by the computer. Example: (10101)_{2} or (10101) _{B}

For second part of the question,

Hence, (520)_{10} =(208)_{16}

** **

** 5. What is hexadecimal number system? Convert 11 10 112 into base 16.**

Hexadecimal number system is a base 16 number system. It uses 16 digits, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F.

Example: (A05)_{16} or (A05)_{H}. For second part of the question,

Grouping into 4 bits, 0011 1011

Binary | 0011 | 1011 |

Hexadecimal | 3 | B |

Hence, (111011) 2= (3B)16

42.2074 Set A Q.No. 14 OR Subtract (10011), from (11110), by using 1’s and 2’s complement method.

Subtracting (11110)_{2} – (10011)₂

[Using 1’s complement method]

1’s complement of 10011 = 01100

Adding it with minuend (i.e. 11110) = 11110

+ 01100

101010

Since, there exists’ one additional bit,

Difference= 01010

+1

01011

Hence, 11110 – 10011 = (01011)2.

[Using 2’s complement method]

2’s complement of 10011

= 01100

+1

01101

Adding it with 11110 = 11110

+01101

101011

Since, there exists’ one additional bit, Difference = 01011 (Neglect additional bit i.e. 1)

Hence, 11101 – 10011 = (01011)_{2.}