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Class 11 Computer System: cheat sheet
CBSE Computer Science. The parts of a computer (CPU, memory, input and output devices, buses), the units of memory from a bit to a petabyte, and the software that runs it.
Key ideas
- Input devices take data in, the CPU (ALU, control unit and registers) processes it, and output devices give out the results.
- Memory gets larger, cheaper and slower away from the CPU: registers, cache, RAM, then secondary storage such as an SSD, a hard disk or an optical disc.
- RAM, cache and registers are volatile and lose their contents without power; ROM, disks, SSDs and pen drives are non-volatile.
- The address bus carries memory addresses, the data bus carries data both ways, and the control bus carries signals such as read and write.
- System software (the operating system, device drivers, utilities) runs the computer; application software does a task for the user.
- A compiler translates the whole program before it runs, an interpreter one statement at a time, and an assembler turns assembly language into machine code.
- The operating system manages processes, memory, files and devices, and gives the user a command-line or graphical interface.
Remember
| Small units | 4 bits = 1 nibble; 8 bits = 1 byte; 1 KB = 1024 bytes |
| Each step up | 1024 KB = 1 MB, 1024 MB = 1 GB, 1024 GB = 1 TB, 1024 TB = 1 PB |
| Converting | to a smaller unit × 1024, to a larger unit ÷ 1024: 5 GB = 5,120 MB |
| Bits in a file | 3 KB = 3 × 1024 × 8 = 24,576 bits |
| Address bus of n lines | 2ⁿ locations: 16 lines address 65,536 |
| Data bus of d bits | d ÷ 8 bytes in one transfer: 64 bits move 8 bytes |
| How many files fit | take the whole part: 4 GB of 6 MB photos = 4,096 ÷ 6 → 682 photos |
Watch out
- Using 1000 instead of 1024: 2 GB is 2,048 MB, not 2,000.
- Rounding up the number of files that fit: only whole files count.
- Calling ROM or a pen drive volatile: only RAM, cache and registers lose their data when the power goes off.
- Mixing up the compiler (whole program first) and the interpreter (one statement at a time).
Worked example
A 16 GB pen drive holds 336 photos of 5 MB each and 11 videos of 661 MB each. (a) How many MB are used? (b) How many MB are free? (c) How many more photos of 5 MB can it take?
Working: 336 × 5 + 11 × 661 = 8,951; 16,384 − 8,951 = 7,433; 7,433 ÷ 5
Answer: (a) 8,951 MB; (b) 7,433 MB; (c) 1,486
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