Binary numbers form the mathematical backbone of computers and digital systems. For NDA Mathematics, this is a focused, mechanical chapter — the questions are almost entirely about conversion between decimal and binary, and performing addition or subtraction in binary. Once the method is practised, this chapter becomes one of the easiest full-marks opportunities in the paper.
📌 What to expect in NDA (based on 2022–2024 papers): (1) Converting a given decimal number (typically 2–3 digits) to binary using repeated division;
(2) Converting a given binary number to decimal using place values;
(3) Adding two binary numbers with carry;
(4) Subtracting two binary numbers with borrow;
(5) Verifying a binary arithmetic result by converting to decimal;
(6) Identifying the decimal equivalent of a given binary number from options.
Topics at a Glance
① Number Systems
Decimal (base 10) vs Binary (base 2)
② Binary Place Values
2⁰, 2¹, 2², 2³, … positional weights
③ Decimal → Binary
Repeated division by 2, read remainders ↑
④ Binary → Decimal
Multiply each bit by its place value, sum
⑤ Binary Addition
0+0=0, 0+1=1, 1+1=10 (carry)
⑥ Binary Subtraction
0−1=1 (borrow from next bit)
1. Number Systems — Decimal vs Binary
1.1
What is a Number System? — Base and Digits
Understanding the base is the key to understanding all conversions
A number system is a way of representing numbers using a specific set of symbols (digits) and a base. The base (or radix) tells you how many distinct digits the system uses and how place values are computed.
Bit position 0 is the rightmost (Least Significant Bit, LSB). Position increases leftward. The leftmost non-zero bit is the Most Significant Bit (MSB).
Place Value Layout — Reading Binary Number 10110101₂
2⁷=128
128
1
2⁶=64
64
0
2⁵=32
32
1
2⁴=16
16
1
2³=8
8
0
2²=4
4
1
2¹=2
2
0
2⁰=1
1
1
Green cells = 1 bits. Value = 128+32+16+4+1 = 181. So (10110101)₂ = 181₁₀.
2. Binary → Decimal Conversion
2.1
Method: Multiply Each Bit by Its Place Value
The easiest direction — always start from the rightmost bit (position 0)
1
Write the binary number and assign position numbers starting from 0 at the rightmost bit, increasing to the left.
2
For each bit that is 1, note the corresponding place value (2^position). Bits that are 0 contribute nothing — skip them.
3
Add all the place values of the 1-bits together. The sum is the decimal equivalent.
Fig 1: Binary to Decimal — only 1-bits contribute their place value. (101101)₂ = 32+8+4+1 = 45.
156÷2 = 78 rem 0; 78÷2 = 39 rem 0; 39÷2 = 19 rem 1; 19÷2 = 9 rem 1;
9÷2 = 4 rem 1; 4÷2 = 2 rem 0; 2÷2 = 1 rem 0; 1÷2 = 0 rem 1.
Read bottom to top: 10011100. So 156₁₀ = (10011100)₂.
Verify: 128+16+8+4 = 156 ✓
📌 Shortcut — Subtraction of Powers of 2 Method (for NDA speed):
Instead of dividing repeatedly, find the largest power of 2 that fits in the number, mark a 1, subtract it, then repeat for the remainder.
Example: 45 → 32 fits (1_ _ _ _ _), 45−32=13 → 8 fits (1 _ 1 _ _ _), 13−8=5 → 4 fits (1 _ 1 1 _ _), 5−4=1 → 1 fits (1 0 1 1 0 1).
Result: (101101)₂. Same answer, faster for mental calculation.
🧩 T1. How many bits are needed to represent the decimal number 100 in binary?
100 in binary = 1100100. Count the bits: 7 bits.
General rule: Number of bits needed = ⌊log₂ N⌋ + 1.
log₂ 100 = log 100/log 2 = 2/0.3010 ≈ 6.64 → ⌊6.64⌋+1 = 7. Trap: Students count digits in the decimal number (100 has 3 digits) and say 3 bits. Always convert to binary first or use the log₂ formula.
🧩 T2. What is the largest decimal number that can be represented with 5 binary bits?
With 5 bits, the maximum binary number is 11111.
Value = 2⁴+2³+2²+2¹+2⁰ = 16+8+4+2+1 = 31.
General: n bits can represent 0 to 2ⁿ−1. For n=5: 0 to 31. Trap: Students say 2⁵ = 32. The max representable number is 2ⁿ−1 = 31, not 2ⁿ = 32.
🧩 T3. If (x)₁₀ = (10010)₂, find x + 5.
(10010)₂ = 2⁴+2¹ = 16+2 = 18.
x = 18. x + 5 = 23.
As a bonus: 23 in binary = (10111)₂. NDA often asks: convert to decimal, perform arithmetic, convert back. Always complete the binary-to-decimal step before doing any arithmetic.
4. Binary Addition
4.1
Rules of Binary Addition
Only four cases — memorise the carry rule for 1+1
Binary addition follows the same column-by-column method as decimal addition, but with only 0s and 1s. The key rule is: 1 + 1 = 10 in binary (write 0, carry 1 to the next column).
Addition Truth Table
A
B
Sum (S)
Carry (C)
0
0
0
0
0
1
1
0
1
0
1
0
1
1
0
1
1
1+carry(1)
1
1
Last row: 1+1+1 (with carry-in) = 3 = 11 in binary → sum=1, carry=1.
⚡ Binary Addition Summary Rules
0 + 0 = 0 (write 0, carry 0)
0 + 1 = 1 (write 1, carry 0)
1 + 0 = 1 (write 1, carry 0)
1 + 1 = 10 (write 0, carry 1) ← most important rule
1 + 1 + 1 = 11 (write 1, carry 1) ← when there is a carry-in
The rules mirror decimal: in decimal, 9+1=10 (write 0, carry 1). In binary, 1+1=10 (write 0, carry 1). The process is identical — only the threshold differs.
The only difficult case: 0 − 1 requires borrowing from the next column
⚡ Binary Subtraction Rules
0 − 0 = 0 (no borrow)
1 − 0 = 1 (no borrow)
1 − 1 = 0 (no borrow)
0 − 1 = 1 (borrow 1 from the next column → that column loses 1)
When you borrow:
The current column gets +2 (since you borrow 1 from the 2× column)
The next column to the left has its value reduced by 1 (the borrow)
Borrow works just like in decimal: 10 − 1 = 9 in decimal → borrow 1 from tens → 10+0−1 = 9. In binary: 10 − 1 = 1 → borrow 1 from the 2s column → current column becomes 2, subtract 1 = 1.
Worked Example — Binary Subtraction
Subtract (1011)₂ − (0110)₂
Borrow: 0 1 1 0
1 0 1 1
− 0 1 1 0
0 1 0 1
Col 0: 1−0 = 1, no borrow
Col 1: 1−1 = 0, no borrow
Col 2: 0−1 → borrow! becomes 10−1 = 1, borrow from col 3
Col 3: 1−0−1(borrow) = 0
Result: (0101)₂ = 5₁₀. Verify: 11−6 = 5 ✓
16 − 5 = 11. (01011)₂ = 8+2+1 = 11 ✓.
Borrow chains across multiple zeros — this is the most common error trap. Always work carefully column by column.
⚠ Successive Borrow Trap — Most Common Error:
When subtracting 0 − 1 in a column where the next column is also 0 (like 10000 − 1), the borrow must cascade leftward until a 1 is found. Each 0 that is borrowed from becomes a 1 (it borrowed from the next column and gave 1 to the current). Students often stop borrowing at the first 0 and write an incorrect result. Always chain the borrow all the way to the nearest 1.
📌 Verification Shortcut — Always Use Decimal:
After any binary arithmetic result, convert both numbers to decimal, perform the operation in decimal, and verify the decimal matches your binary answer. This takes 10 seconds and catches all errors. NDA allocates marks only for correct final answers — verification saves marks.
🧩 T4. Add (11111)₂ + (00001)₂ and express the result in decimal.
(11111)₂ = 31. (00001)₂ = 1. Sum = 32.
Binary: 11111 + 00001 = 100000 (all 1s become 0 with a chain of carries, and a new leading 1 appears). (100000)₂ = 2⁵ = 32. Insight: Adding 1 to a binary number that is all 1s always produces a 1 followed by all zeros — equivalent to the next power of 2. This is analogous to 999+1 = 1000 in decimal.
🧩 T5. Find the value of (11010)₂ − (1011)₂ + (101)₂ in decimal.
Convert each to decimal first:
(11010)₂ = 16+8+2 = 26.
(1011)₂ = 8+2+1 = 11.
(101)₂ = 4+1 = 5.
Result = 26 − 11 + 5 = 20.
Binary: 20 = 16+4 = (10100)₂. Key strategy: For multi-step binary arithmetic, convert all to decimal, compute in decimal, then convert back if binary form is required.
🧩 T6. If (1xy1)₂ = 13₁₀, find x and y.
(1xy1)₂ = 1×8 + x×4 + y×2 + 1×1 = 9 + 4x + 2y = 13.
4x + 2y = 4. Since x, y ∈ {0,1}:
Try x=1: 4+2y=4 → 2y=0 → y=0. Check: (1101)₂ = 8+4+1 = 13 ✓. x = 1, y = 0. The number is (1101)₂. This tests reverse engineering — a type of question that appears in NDA where you must find missing binary digits.
📋 Master Formula Sheet — MN08 Binary Numbers
All critical facts and methods for rapid pre-exam revision.
🔢 Place Values (Powers of 2)
2⁰=1, 2¹=2, 2²=4, 2³=8
2⁴=16, 2⁵=32, 2⁶=64, 2⁷=128
2⁸=256, 2⁹=512, 2¹⁰=1024
n bits → max value = 2ⁿ−1
Bits needed for N: ⌊log₂ N⌋ + 1
🔄 Binary → Decimal
Assign position 0 to rightmost bit
Multiply each 1-bit by 2^(position)
Sum all the values
Zero-bits contribute nothing — skip
Check: result must be < 2^(total bits)
🔃 Decimal → Binary
Divide by 2 repeatedly
Record remainder (0 or 1) each time
Continue until quotient = 0
Read remainders BOTTOM to TOP
Shortcut: subtract largest fitting power of 2
➕ Binary Addition Rules
0+0 = 0 (no carry)
0+1 = 1 (no carry)
1+0 = 1 (no carry)
1+1 = 10 → write 0, carry 1
1+1+1 = 11 → write 1, carry 1
➖ Binary Subtraction Rules
0−0 = 0 (no borrow)
1−0 = 1 (no borrow)
1−1 = 0 (no borrow)
0−1 = 1 (borrow 1 from next column)
Borrow cascades through zeros leftward
✅ Verification Method
Always convert binary result back to decimal
Verify: decimal A ± decimal B = decimal result
Saves marks in NDA — never skip verification
Common check: 11111+1 = 100000 = 2⁵
Common check: 1000−1 = 0111
⚡ Quick Revision Booster — MN08 Binary Numbers
🔢 Must-Know Powers of 2
1, 2, 4, 8, 16, 32, 64, 128
2⁰ through 2⁷ — memorise all
256 = 2⁸, 512 = 2⁹, 1024 = 2¹⁰
n bits: values 0 to 2ⁿ−1
All 1s (n bits) = 2ⁿ−1
🔄 Conversion Checklist
Binary→Dec: identify 1-bit positions, sum their 2^k values
Dec→Binary: divide by 2, read remainders bottom-up
Shortcut: subtract successive powers of 2
Verify by converting result back
Leading zeros don't change value
➕ Addition at a Glance
1+1 = 10 (write 0, carry 1)
1+1+1 = 11 (write 1, carry 1)
Carry propagates leftward
11111+1 = 100000 (all carries)
Always verify in decimal
➖ Subtraction at a Glance
0−1 → borrow 1 from next column
Borrowed column loses 1; current gains 2
Borrow cascades through 0s
10000−1 = 01111 (borrow chain)
Always verify in decimal
📋 Quick Conversions
10 = 2, 11 = 3, 100 = 4, 101 = 5
110 = 6, 111 = 7, 1000 = 8
1010 = 10, 1100 = 12, 1111 = 15
10000 = 16, 11111 = 31, 100000 = 32
These arise most often in NDA questions
🚨 Critical Exam Traps
Read remainders bottom-to-top (NOT top-to-bottom)
Max n-bit value = 2ⁿ−1 (not 2ⁿ)
1+1 = 10 in binary (NOT 2 or 11)
Borrow cascades through all zeros
Always check: is the answer sensible in decimal?
Subscript ₂ means binary — don't confuse with decimal
This material is for personal NDA exam preparation only.
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