Home Class 11 Applied Maths Chapter 1: Numbers and Quantification Exercise 1.1: Binary Number System
Chapter 1 • Unit I

Exercise 1.1: Binary Number System

Comprehensive guide and practice questions for CBSE Class 11 Applied Mathematics curriculum (2026-27).

Concept Map & Real-Life Applications

The concept of expressing numbers in different bases comes from the broader study of number systems, which has evolved over thousands of years. Early civilizations such as the Babylonians used a base-60 system, while the Romans used a non-positional numeral system. The most significant advancement came from the Bhartiya number system, which introduced the concept of place value and zero, forming the foundation of modern positional systems.

Real-World Applications of Binary System:
1. Computer Science Application: The binary system is the fundamental language of digital electronics and computers. It uses only two digits, 0 and 1, which makes data processing and storage simple and efficient in digital circuits (OFF/ON states).
2. Storage Devices: Hard disks, SSDs, pen drives, CDs, and flash memory store information as binary patterns of electric charges or magnetic polarities.
3. Logic Gates: Digital electronic devices operate on binary logic gates (AND, OR, NOT, NAND, NOR, XOR, XNOR) using binary inputs.
4. Telecommunications: Signals in optical fibers, Wi-Fi, Bluetooth, and 5G networks are transmitted as binary pulses (0s and 1s).
5. Cryptography: Binary bitwise operations (like XOR) form the backbone of modern encryption standards (AES, RSA).

1.1.1 The Binary System: Structure and Significance

In the binary number system, every number is represented using only two digits: 0 and 1. Each digit in a binary number is called a bit (binary digit). These bits serve as the smallest units of information in digital technology. For example, $(110100)_2$ is an example of a binary number.

In the Decimal Number System, we have 10 digits (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) with base 10. In the Binary Number System, the base is 2. Any decimal number can be expressed as a sum of powers of 2. This representation is called its binary expansion.

Example: Binary Expansion of Decimal Number 23456

$23456 = 2^5(1 + 732) = 2^5 + 2^7 + 2^8 + 2^9 \times 2(1 + 44) = 2^5 + 2^7 + 2^8 + 2^{11} + 2^{12} + 2^{14}$

$23456 = (101101110100000)_2$

Note: Five zeros on the right of the binary expansion indicate that $2^5$ is the exact power of 2 dividing 23456.

1.2.1 Conversion from Decimal to Binary (Base 10 → Base 2)

To convert a decimal number to binary, follow the step-by-step repeated division method:

  1. Divide the given decimal number by 2 and note the quotient and remainder.
  2. Take the quotient from the previous step and divide it again by 2.
  3. Continue dividing the new quotient by 2 and record each remainder.
  4. Stop when the quotient becomes 0 or 1.
  5. Write the remainders from bottom to top (Most Significant Bit to Least Significant Bit).
Example 1: Convert the following numbers to binary: (i) 6, (ii) 15, (iii) 239, (iv) 963.
Show Step-by-Step Solution
(i) Convert 6:
• $6 \div 2 = 3$, Remainder = 0
• $3 \div 2 = 1$, Remainder = 1
• $1 \div 2 = 0$, Remainder = 1
Reading remainders from bottom to top: $(6)_{10} = (110)_2$.

(ii) Convert 15:
• $15 \div 2 = 7$, Remainder = 1
• $7 \div 2 = 3$, Remainder = 1
• $3 \div 2 = 1$, Remainder = 1
Reading from bottom to top: $(15)_{10} = (1111)_2$.

(iii) Convert 239:
• $239 \div 2 = 119$, Remainder = 1
• $119 \div 2 = 59$, Remainder = 1
• $59 \div 2 = 29$, Remainder = 1
• $29 \div 2 = 14$, Remainder = 1
• $14 \div 2 = 7$, Remainder = 0
• $7 \div 2 = 3$, Remainder = 1
• $3 \div 2 = 1$, Remainder = 1
Reading from bottom to top: $(239)_{10} = (11101111)_2$.

(iv) Convert 963:
• Successive division remainders: $963 \to 1, 481 \to 1, 240 \to 0, 120 \to 0, 60 \to 0, 30 \to 0, 15 \to 1, 7 \to 1, 3 \to 1, 1 \to 1$.
Reading bottom to top: $(963)_{10} = (1111000011)_2$.
Example 2: Convert the binary number $(1010101010)_2$ to its decimal equivalent.
Show Step-by-Step Solution
Using the positional formula: $N = \sum_{i=0}^{n} x_i \cdot 2^i$
$(1010101010)_2 = 1\cdot 2^9 + 0\cdot 2^8 + 1\cdot 2^7 + 0\cdot 2^6 + 1\cdot 2^5 + 0\cdot 2^4 + 1\cdot 2^3 + 0\cdot 2^2 + 1\cdot 2^1 + 0\cdot 2^0$
$= 512 + 0 + 128 + 0 + 32 + 0 + 8 + 0 + 2 + 0 = 682$.
Answer: $(1010101010)_2 = (682)_{10}$.
Example 3: Convert the binary number $(101011)_2$ to decimal.
Show Step-by-Step Solution
$(101011)_2 = 1\cdot 2^5 + 0\cdot 2^4 + 1\cdot 2^3 + 0\cdot 2^2 + 1\cdot 2^1 + 1\cdot 2^0$
$= 32 + 0 + 8 + 0 + 2 + 1 = 43$.
Answer: $(101011)_2 = (43)_{10}$.
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