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Solved Notes – Unit: 1 – Computing Devices, Software & Operating System | BBA 1st Semester | Computer & IT Application

Unit: 1 – Computing Devices, Software and Operating System Level: BBA 1st Semester

Subject: Computer & It Application

(For Pokhara University)

Very Short answer questions

1. Computer is a diligence machine. Justify.

Ans: Diligence machine means industrious and tirelessness machine. It can be used form complex calculation research or industrial application. Being non-living thing, it performs the allocated task without any objection. Once it is programmed, it keeps doing things that are assigned to it. Today special designed computer aided machines and robot perform difficult task in hazardous area also. Because of these characteristics, we can say computers are diligence machine.

2. Compare between primary and secondary memory with examples.

Ans: Primary memories are directly accessed by CPU whereas secondary memory cannot be directly accessed by CPU. Primary memories are required for interaction and executions of instructions whereas secondary memories are used for backup or permanent storage of data for future use. Examples of primary memories are RAM, ROM, cache etc. Hard Disk, CD-ROM, DVD, Magnetic tapes are secondary memory.

Primary MemorySecondary Memory
1. Primary memory is a semiconductor memory.1. Secondary memory is a magnetic and optical memory.
2. It’s storing capacity is small.2. It’s storing capacity is very large.
3. For example: RAM, ROM3. For example: Hard Disk, DVD.

3. What is application software? What are the benefits of using different types of software in an organization?

Ans: Application software refers to the computers software designed to perform a specific function directly for user or in some cases for another application program. Some general kinds of application software include productive software, presentation software, etc.

Depending upon the task, software is of different types and each one has its own benefits.

System software: This is the software type that helps in the basic functioning of computers inclusive of the hardware and the rest of the system.

Application software: This software type is meant for catering to precise requirements of a user and is highly made use of for commercial purposes.

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4. Differentiation between SRAM and DRAM.

Ans: The difference between SRAM and DRAM are listed as follows:

1. It doesn’t need periodic refresh rate like DRAM.1. It needs periodic refresh.
2. SRAM is much faster than DRAM.2. It is slower than SRAM.
3. SRAM is lower in density.3. DRAM is denser than SRAM.
4. More expensive.4. Less expensive.
5. Transistors are used in memory cell.5. Tiny capacitors are used as memory cell.
6. Transistors don’t loss their charge.The charge slowly leaks from capacitors (cell) so needs refreshing.

5. Find the value of x when (155.84)10=(x)8

Ans: First converting integral part into octal

155/8=19 remainder is 3

19/8=2 remainder is 3

2/8=0 remainder is 2

Now for fractional part

Fractional partFractional part * 8Integral part
0.84 * 8
0.92 * 8
0.36 * 8
0.88 * 8
= 6.92
= 7.36
= 2.88
= 7.04

∴ (233.672)8

Hence, the K is 233.6727

6. List any three differences between GUI and CUI.

Ans: Differentiation between CUI and GUI.

The differences between CUI and GUI are showcased as follows:

1. It is based on command user interface. Instructions and commands are given through keyboards using syntax (character).1. It is based on graphical user interface. Instructions and command are given through interaction with graphical components like icon, menu etc.
2. Only keyboard as input device.2. Keyboard, mouse, joysticks and many other input devices can be used.
3. Single user and single tasking.3. Mostly multi-user and multi-tasking.

7. Convert (B2C)16 into base 8 system.

Ans: (B2C)16 → Octal First let convert it to decimal.

(B2C)16 = B * 162 + 2 * 161 + C * 160

= 11 * 256 + 2 * 16 + 12 * 160

= 2816 +32 + 12

= 2860

Now, converting (2860) decimal into Octal.

2860/8 = 357 remainder 4

357/8 = 44 remainder 5

44/8 = 5 remainder 4

5/8 = 0 remainder 5

⇒ (5454)8

Hence, (B2C)16⇒ (5454)8

8. Explain the term cache memory and buffer.

Ans: Cache memory: Cache memory is small memory that resides between the CPU and RAM whose access time is closer to the processing speed of the CPU.

Buffer: Buffer is a region of physical memory storage used to temporarily store data while it is being moved from one place to another.

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