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Number systems look like the easiest marks in Class 11 computer science, and they are, until you read the remainders in the wrong order or write 12 where you meant C. This post walks through every conversion the syllabus asks for with one number, shows the shortcut that saves most of the work, and ends with a short practice routine.
What does the CBSE syllabus ask about number systems?
The CBSE Computer Science curriculum for 2026-27 lists "Number System: Binary, Octal, Decimal and Hexadecimal number system; conversion between number systems" in Unit 1, Computer Systems and Organisation, which carries 10 of the 70 theory marks in Class 11. The same unit covers encoding schemes: ASCII, ISCII and Unicode (UTF-8, UTF-32).
The matching chapter is Chapter 2, Encoding Schemes and Number System, in the NCERT Class 11 Computer Science textbook. It also covers numbers with a fractional part, so practise those too. Check the curriculum for your own year before the exam.
Why do computers need octal and hexadecimal?
Every number system is a set of digits plus a base. Decimal uses ten digits, binary uses two (0 and 1), octal uses eight (0 to 7) and hexadecimal uses sixteen (0 to 9, then A to F for 10 to 15). The value of a digit depends on its place: in base b, the places are worth 1, b, b², b³ and so on, from the right.
Computers store everything in binary, but long strings of bits are hard for people to read. Octal and hex are shorthand. Since 2³ = 8, one octal digit stands for exactly three bits. Since 2⁴ = 16, one hex digit stands for exactly four bits. The NCERT chapter gives a 16-bit memory address, 1100000011110001, which is C0F1 in hex and 140361 in octal. Web colours work the same way: pure red is 11111111 00000000 00000000 in binary, written FF0000 in hex.
How do you convert a decimal number to binary, octal and hex?
Divide repeatedly by the new base, write down each remainder, and read the remainders from the last one to the first.
Worked example: convert (156)₁₀.
To binary, divide by 2:
156 ÷ 2 = 78, remainder 0
78 ÷ 2 = 39, remainder 0
39 ÷ 2 = 19, remainder 1
19 ÷ 2 = 9, remainder 1
9 ÷ 2 = 4, remainder 1
4 ÷ 2 = 2, remainder 0
2 ÷ 2 = 1, remainder 0
1 ÷ 2 = 0, remainder 1
Reading from the bottom up gives (10011100)₂.
To octal, divide by 8: 156 ÷ 8 = 19 remainder 4, 19 ÷ 8 = 2 remainder 3, 2 ÷ 8 = 0 remainder 2. So (156)₁₀ = (234)₈.
To hexadecimal, divide by 16: 156 ÷ 16 = 9 remainder 12, and 12 is C. Then 9 ÷ 16 = 0 remainder 9. So (156)₁₀ = (9C)₁₆.
Check by converting back. Multiply each digit by its place value and add:
(10011100)₂ = 128 + 16 + 8 + 4 = 156
(234)₈ = 2 × 64 + 3 × 8 + 4 = 128 + 24 + 4 = 156
(9C)₁₆ = 9 × 16 + 12 = 156
What is the grouping shortcut?
You never need to pass through decimal to go between binary and octal or hex. Take (11010110)₂.
To octal: group bits in threes from the right, adding zeros on the left if needed: 011 010 110. Each group is one digit: 3, 2, 6. So the answer is (326)₈.
To hex: group in fours from the right: 1101 0110. That is 13 and 6, so (D6)₁₆.
Both are 214 in decimal: 3 × 64 + 2 × 8 + 6 = 214, and 13 × 16 + 6 = 214. To go the other way, replace each octal digit with three bits, or each hex digit with four.
Fractions. For the part after the point, multiply by the base and collect the whole-number parts in order. For (0.375)₁₀: 0.375 × 2 = 0.75, digit 0; 0.75 × 2 = 1.5, digit 1; 0.5 × 2 = 1.0, digit 1, and the fraction is now zero. So (0.375)₁₀ = (0.011)₂. Check: 0.25 + 0.125 = 0.375.
Where do ASCII and Unicode fit in?
Encoding schemes turn characters into numbers, which the computer then stores in binary. In ASCII, the letter A is 65, which is 1000001 in 7-bit binary. Unicode extends this to every script. NCERT's table shows the Devanagari letter अ at hexadecimal 0905. According to the Unicode FAQ, UTF-8 stores a character in one to four 8-bit bytes, while UTF-32 always uses a single 32-bit unit. So A takes one byte in UTF-8 but four in UTF-32, which answers a question NCERT asks in its margin.
Which mistakes cost marks?
Reading remainders top-down. The first remainder is the last digit, not the first.
Writing 10 to 15 as two digits in hex. Write A to F, or the answer cannot be read back.
Padding on the wrong side when grouping, especially with fractions.
Dropping the base subscript. 101 could be binary, octal, decimal or hex. Write (101)₂.
Using a digit that does not exist in the base. There is no 8 in octal and no 2 in binary. If one appears, recheck your division.
How can you check your answers with Python?
Python's built-in functions do every conversion, so you can mark your own practice:
print(bin(156), oct(156), hex(156)) # 0b10011100 0o234 0x9c
print(int("9C", 16), int("10011100", 2)) # 156 156
print(ord("A"), chr(97)) # 65 aUse this for checking only. In the exam you must show the working. If the idea of place value itself feels shaky, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, so you can revise bases and powers before the conversions. Number systems lead straight into Boolean logic for CBSE Class 11, and the same `ord()` and `chr()` functions come back in strings in Python.
A 20-minute practice routine
Pick three random numbers between 50 and 500.
Convert each to binary, octal and hex by division.
Convert the binary answers to octal and hex by grouping, and check that they match.
Convert one fraction, such as 0.625 or 0.8125, to binary.
Check everything with Python and log each error by type: order, digit, padding or arithmetic.
Do this three times this week, then once a week until the exam. Start today with (200)₁₀ and check your answers against Python.
