Number System and Binary Numbers

What This Topic Covers

The NDA Maths syllabus groups Number System and Binary Numbers under the broader "Sets, Relations, Functions and Number System" chapter. The binary-numbers component specifically tests:

• Converting a decimal integer to its binary equivalent (e.g., 292, 251, 235, 83, 45, 31)
• Converting a binary integer to decimal (e.g., 10101111, 110001, 101011, 11101011, 1011)
• Converting a decimal fraction to binary (e.g., 32.25, 57.375, 127.25, 0.3125, 0.3)
• Converting a binary fraction to decimal (e.g., 11101.001 → 29.125)
• Binary addition (summing two or three binary numbers)
• Binary division including quotient and remainder

Exam Pattern & Weightage

The table below is drawn from the PYQ source material covering the "Sets, Relations, Functions and Number System" chapter from 2010 to 2022. Binary-specific questions are extracted from the combined chapter data.

The pattern is clear: at least one binary question appears in almost every NDA sitting. Integer conversion is the most common type; fractional conversion appears roughly once every two to three papers; binary arithmetic appears less frequently but has been tested in recent years.

Decimal to Binary Conversion (Integers)

The method is called successive division by 2. Divide the decimal number by 2 repeatedly, recording the remainder (0 or 1) at each step. When the quotient reaches 0, stop. Read the remainders from bottom to top — that is your binary number.

Example: Convert 45 to binary (NDA 2013-I).

• \(45 \div 2 = 22\) remainder 1
• \(22 \div 2 = 11\) remainder 0
• \(11 \div 2 = 5\) remainder 1
• \(5 \div 2 = 2\) remainder 1
• \(2 \div 2 = 1\) remainder 0
• \(1 \div 2 = 0\) remainder 1
• Reading remainders from bottom to top: 101101

So 45 in decimal = \((101101)_2\). The NDA 2013-I correct answer confirms this.

Example: Convert 292 to binary (NDA 2012-I).

• \(292 \div 2 = 146\) remainder 0
• \(146 \div 2 = 73\) remainder 0
• \(73 \div 2 = 36\) remainder 1
• \(36 \div 2 = 18\) remainder 0
• \(18 \div 2 = 9\) remainder 0
• \(9 \div 2 = 4\) remainder 1
• \(4 \div 2 = 2\) remainder 0
• \(2 \div 2 = 1\) remainder 0
• \(1 \div 2 = 0\) remainder 1
• Reading bottom to top: 100100100

So \(292 = (100100100)_2\), matching the NDA 2012-I answer.

Example: Convert 31 to binary (NDA 2018-I).

• \(31 \div 2 = 15\) remainder 1
• \(15 \div 2 = 7\) remainder 1
• \(7 \div 2 = 3\) remainder 1
• \(3 \div 2 = 1\) remainder 1
• \(1 \div 2 = 0\) remainder 1
• Reading bottom to top: 11111

So \(31 = (11111)_2\). Notice that \(2^5 - 1 = 31\): any number of the form \(2^n - 1\) gives \(n\) ones in binary — a pattern worth remembering.

Binary to Decimal Conversion

Every binary digit (bit) has a place value that is a power of 2, starting from \(2^0\) on the right. Multiply each bit by its place value and add all the results.

Example: Convert binary \((10101111)_2\) to decimal (NDA 2011-I).

• Write the bits with their place values: $$1\cdot 128 + 0\cdot 64 + 1\cdot 32 + 0\cdot 16 + 1\cdot 8 + 1\cdot 4 + 1\cdot 2 + 1\cdot 1$$
• \(= 128 + 0 + 32 + 0 + 8 + 4 + 2 + 1\)
• = 175

So \((10101111)_2 = 175\). The NDA 2011-I answer is 175.

Example: Convert \((110001)_2\) to decimal (NDA 2013-I / 2013-II).

• $$1\cdot 32 + 1\cdot 16 + 0\cdot 8 + 0\cdot 4 + 0\cdot 2 + 1\cdot 1$$
• \(= 32 + 16 + 0 + 0 + 0 + 1\)
• = 49

So \((110001)_2 = 49\). Option (d) 49 is the correct answer in the PYQ.

Example: Convert \((11101011)_2\) to decimal (NDA 2015-I).

• $$1\cdot 128 + 1\cdot 64 + 1\cdot 32 + 0\cdot 16 + 1\cdot 8 + 0\cdot 4 + 1\cdot 2 + 1\cdot 1$$
• \(= 128 + 64 + 32 + 0 + 8 + 0 + 2 + 1\)
• = 235

So \((11101011)_2 = 235\). The NDA 2015-I answer is 235.

Fractional Binary Conversion

For mixed numbers (integer part + fractional part), convert both halves separately, then join them with a binary point.

Example: Convert 32.25 to binary (NDA 2010-II).

• Integer part 32: \(32 = 2^5\), so binary is 100000.
• Fractional part 0.25: \(0.25 \times 2 = 0.50\) → bit = 0; \(0.50 \times 2 = 1.00\) → bit = 1. Fractional part is now 0, stop.
• Fractional binary = .10
• Result: \((100000.10)_2\)

The NDA 2010-II correct answer is \((100000.10)_2\).

Example: Convert 0.3125 to binary (NDA 2016-I).

• \(0.3125 \times 2 = 0.625\) → bit = 0
• \(0.625 \times 2 = 1.25\) → bit = 1
• \(0.25 \times 2 = 0.50\) → bit = 0
• \(0.50 \times 2 = 1.00\) → bit = 1
• Fractional part = 0, stop. Reading top to bottom: .0101
• Result: \((0.0101)_2\)

The NDA 2016-I correct answer is 0.0101.

Example: Convert 57.375 to binary (NDA 2012-II).

• Integer part 57: $$57 = 32+16+8+1 = 2^5+2^4+2^3+2^0 = (111001)_2$$
• Fractional part 0.375: \(0.375 \times 2 = 0.75\) → bit = 0; \(0.75 \times 2 = 1.5\) → bit = 1; \(0.5 \times 2 = 1.0\) → bit = 1. Stop.
• Fractional binary = .011
• Result: \((111001.011)_2\)

The NDA 2012-II correct answer is \((111001.011)_2\).

Converting binary fractions to decimal uses negative powers of 2: \(2^{-1} = 0.5\), \(2^{-2} = 0.25\), \(2^{-3} = 0.125\), and so on.

Example: Convert \((11101.001)_2\) to decimal (NDA 2012-I, Q.32 — the answer is 29.125).

• Integer part: \(1\cdot 16 + 1\cdot 8 + 1\cdot 4 + 0\cdot 2 + 1\cdot 1 = 16+8+4+0+1 = 29\)
• Fractional part: $$0\cdot \tfrac{1}{2} + 0\cdot \tfrac{1}{4} + 1\cdot \tfrac{1}{8} = 0 + 0 + 0.125 = 0.125$$
• Result: 29.125

Binary Arithmetic

NDA occasionally tests binary addition and binary division. The rules for binary addition are simpler than decimal:

Example: What is \((11110)_2 + (1010)_2\)? (NDA 2014-I)

• Align to the right: 11110 and 01010.
• Rightmost column: \(0+0 = 0\)
• Next: \(1+1 = 10\), write 0 carry 1
• Next: \(1+0+1\text{(carry)} = 10\), write 0 carry 1
• Next: \(1+1+1\text{(carry)} = 11\), write 1 carry 1
• Next: \(1+0+1\text{(carry)} = 10\), write 0 carry 1
• Carry: write 1
• Result reading left to right: \((101000)_2\)

The NDA 2014-I correct answer is \((101000)_2\).

Binary division works like long division in decimal but only uses 0 and 1. NDA 2017-I asked for the quotient and remainder when \((101110)_2\) is divided by \((110)_2\). The correct answer from the PYQ is remainder \((100)_2\) and quotient \((111)_2\).

For the NDA 2020-I question: the sum \((1101101)_2 + (1011011)_2\) in decimal — convert each to decimal first (\(109 + 91 = 200\)) or add in binary and then convert. The NDA 2020-I correct answer is \((200)_{10}\).

In NDA 2018-II, the sum of \((11011)_2 + (10110110)_2 + (10011x0y)_2 = (101101101)_2\) was given and the values of \(x\) and \(y\) were asked. The correct answer is \(x = 1\), \(y = 0\).

Worked Examples from PYQs

Below are additional solved examples pulled directly from the PYQ source, with the year of appearance noted.

NDA 2013-II (Q.73) — Binary 83:

• \(83 \div 2 = 41\) r 1
• \(41 \div 2 = 20\) r 1
• \(20 \div 2 = 10\) r 0
• \(10 \div 2 = 5\) r 0
• \(5 \div 2 = 2\) r 1
• \(2 \div 2 = 1\) r 0
• \(1 \div 2 = 0\) r 1
• Reading bottom to top: \((1010011)_2\)

NDA 2015-I — \((1000000001)_2 - (0.0101)_2\): The question asked what \((1000000001)_2 - (0.0101)_2\) equals. \((1000000001)_2 = 512 + 1 = 513\). \((0.0101)_2 = 0 + 0.25 + 0 + 0.0625 = 0.3125\). The result is 512.6875. The NDA 2015-I correct answer is \((512.6875)_{10}\).

NDA 2019-II — Binary number \((cdcdcdcdcdcd)_2\) where \(c > d\): Since binary digits are only 0 or 1 and \(c > d\), we must have \(c = 1\) and \(d = 0\). The number is \((101010101010)_2\). Computing: $$2^{11} + 2^9 + 2^7 + 2^5 + 2^3 + 2^1 = 2048 + 512 + 128 + 32 + 8 + 2 = 2730$$ Wait — the PYQ option closest to this is 2730 but the listed options in the PYQ source are 1848, 2048, 2842, 2872. If the pattern is 12 digits with \(c=1\), \(d=0\) alternating starting from position 11, the sum is indeed 2730 — but the question as printed in the PYQ source lists 2048 as option (b). Use the actual PYQ options when checking your answer in exam conditions, and note that the question is testing positional value reasoning for a 12-bit alternating pattern.

Drill — Convert decimal 25 to binary (powers-of-2 method): The biggest powers of 2 that fit into 25 are 16, 8, and 1: $$25 = 16 + 8 + 1 = 2^4 + 2^3 + 2^0$$

• Place values 16, 8, 4, 2, 1 → bits 1, 1, 0, 0, 1
• Result: \((11001)_2\)
• Cross-check by successive division: \(25 \to 12\) r1, \(12 \to 6\) r0, \(6 \to 3\) r0, \(3 \to 1\) r1, \(1 \to 0\) r1 → reading bottom-up gives 11001. Confirmed.

Drill — Convert decimal 0.625 to binary (multiply-by-2 method): Note that $$0.625 = \tfrac{1}{2} + \tfrac{1}{8} = 2^{-1} + 2^{-3}$$

• \(0.625 \times 2 = 1.25\) → bit = 1
• \(0.25 \times 2 = 0.50\) → bit = 0
• \(0.50 \times 2 = 1.00\) → bit = 1. Fractional part = 0, stop.
• Reading top to bottom: \((0.101)_2\)

Drill — Binary addition \((101)_2 + (11)_2\):

• Rightmost column: \(1 + 1 = 10\) → write 0, carry 1
• Middle column: \(0 + 1 + 1\text{(carry)} = 10\) → write 0, carry 1
• Left column: \(1 + 1\text{(carry)} = 10\) → write 0, carry 1
• Final carry: write 1
• Result: \((1000)_2\) (\(= 5 + 3 = 8\) in decimal — confirmed.)

Exam Shortcuts (Pro-Tips)

Binary questions are among the fastest marks in the NDA paper if you stop doing successive division on the exam sheet. The three hacks below cut a 45-second problem to under 10 seconds and have all appeared in past papers.

Shortcut 1 — The Powers-of-2 Decomposition

Memorise the powers of 2 up to 512: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512. To convert any decimal integer \(N\) to binary, find the largest power of 2 \(\le N\), subtract, repeat. Then mark a 1 under the powers you used and a 0 under the rest.

No long division on the exam sheet. For NDA 2013-II (decimal 83): \(83 = 64 + 16 + 2 + 1\) → bits at 64, 32, 16, 8, 4, 2, 1 = 1, 0, 1, 0, 0, 1, 1 → \((1010011)_2\). Solved in under 15 seconds.

Shortcut 2 — Number of Binary Digits Formula

If a question asks how many bits the binary form of \(N\) has, do not actually convert. Use:

For \(N = 251\): $$\log_2 251 \approx 7.97$$ so the answer is \(7 + 1 = 8\) bits. Confirms \((11111011)_2\) has 8 digits without dividing.

Shortcut 3 — Spot \(2^n\) and \(2^n - 1\) Instantly

Any power of 2 in binary is a single 1 followed by zeros. Any number of the form \(2^n - 1\) is \(n\) ones in a row.

NDA 2018-I asked for \((31)_{10}\) in binary; the \(2^n - 1\) pattern gives \((11111)_2\) in two seconds. The same logic flags option traps: if an option for "convert 32 to binary" shows anything other than a single 1 followed by zeros, it is wrong by inspection.

Shortcut 4 — Reverse Check Without Reconverting

When you finish a decimal-to-binary conversion, do a 5-second sanity check by adding the place values of the 1-bits back. For \((100100100)_2\) from decimal 292: \(256 + 32 + 4 = 292\). If your sum does not match the original, you miscounted a remainder — go back at once.

Shortcut 5 — Fractional Multiplication Stops at Power of 1/2

A decimal fraction terminates in binary only if it can be written as a sum of negative powers of 2 (i.e., it is a multiple of \(1/2^n\) for some \(n\)). 0.25, 0.125, 0.0625, 0.3125, 0.375 all terminate; 0.1, 0.2, 0.3 do not. If you see a decimal like 0.3 in the question, expect a recurring binary fraction and pick the option with "…" after the first 4–5 bits — do not keep multiplying past 5 iterations.

How NDA Tests This Topic

One additional sub-type that has appeared: a binary equation with missing digits (2017-I question on p, q, r). Treat this as binary addition in reverse — fill in the known bits, propagate carries, and deduce the unknowns.

Preparation Strategy

This is one of the few NDA Maths chapters where pure practice — not conceptual depth — is the differentiator. Follow this four-step plan:

1. Day 1 — Integer conversion both ways: Practise converting 10 decimal integers to binary (try 45, 83, 235, 292, 251, 31) and 10 binary numbers to decimal. Verify every answer against the PYQ source. Speed target: under 60 seconds per conversion.
2. Day 2 — Fractional conversion: Practise the multiply-by-2 method on 0.25, 0.375, 0.3125, 0.125, and the full mixed numbers 32.25 and 57.375. Also practise the reverse (binary fraction to decimal) using powers of 2.
3. Day 3 — Binary arithmetic: Practice three binary additions and one binary long division. Confirm carries carefully — a single missed carry shifts every digit left of it.
4. Day 4 — Timed PYQ drill: Time yourself on the actual NDA questions from the PYQ source. Target 45 seconds per question. If you exceed 90 seconds on any question, restart from the procedure rather than trying to rescue a partial answer.

For exam-day strategy: scan the question in 5 seconds to identify the template (T1, T2, T3, or T4). Start the procedure immediately. Do not re-read the question mid-procedure — the numbers are all you need once you know the template.

Binary arithmetic questions (T4) are the riskiest. If you have not practised binary addition under time pressure, you may make carry errors. If the T4 question looks complex, skip it and return after finishing the other Maths questions — a T1 or T2 question should never take more than a minute.

Revision note: the full chapter — all four templates — can be re-covered in 15 minutes the day before the exam. Keep a single A4 sheet with the place value table (128, 64, 32, 16, 8, 4, 2, 1 and 0.5, 0.25, 0.125, 0.0625) and the binary addition rules. That sheet is all you need for revision.

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