Class 9 Computer Science SEBA Chapter 1 Solution​ SEBA Solution 

  1. Choose the correct option.
    (a) Which of the following is not an output device?
    (1) Speaker
    (2) Mouse
    (3) Printer
    (4) Monitor
    Ans: (2) Mouse.

(b) Which one of the following is the “brain” of a computer?
(1) CPU
(2) ALU
(3) CU
(4) MU
Ans: (1) CPU.

(c) A collection of eight bits is called______.
(1) Byte
(2) Word
(3) Record
(4) Nibble
Ans: (1) Byte.

(d) Second-generation computers were manufactured using which technology?
(1) Vacuum Tubes
(2) Transistors
(3) ICs
(4) None of these
Ans: (2) Transistors.

(e) Storage of 1KB means the following number of bytes:
(1) 1000
(2) 1024
(3) 1064
(4) 1012
Ans: (2) 1024.

(f) Which of the following is not hardware?
(1) Hard disk
(2) Printer
(3) Assembler
(4) CPU
Ans: (3) Assembler.

(g) Fourth generation computers used
(1) Vacuum tubes
(2) Transistors
(3) Microprocessors
(4) ICs
Ans: (3) Microprocessors.

(h) Which one of the following is a special-purpose application software?
(1) word-processing software
(2) Compiler
(3) Operating system software
(4) Payroll system
Ans: (4) Payroll system.

(i) The base of the Octal Number System is _.
(1) 2
(2) 10
(3) 16
(4) 8
Ans: (4) 8.

(j) who is known as the father of the modern computer?
(1) Charles Babbage
(2) Herman Hollerith
(3) Blaise Pascal
(4) John Atanasoff
Ans: (1) Charles Babbage.

  1. Fill in the blanks.
    (a) The first-generation computers used______for storage.
    Ans: magnetic tape.

(b) The third-generation computer replaced______with _.
Ans: transistors, Integrated Circuits (ICs)

(c) ______are high speed temporary storage area.
Ans: Registers.

(d) _____is the short form of binary digit.
Ans: Bit

(e) A group of four bits is called __.
Ans: Nibble.

(f) _____holds data , programs and instructions for computer.
Ans: RAM .

(g) A microphone is ____device.
Ans: input.

(h) A ____is set of instructions to perform a specific task.
Ans: program.

(i) _____software are used to scan the computers for viruses.
Ans: Antivirus.

(j) The decimal system is composed of __ digits.
Ans: 10.

(k) _____was the first fully electronic general-purpose computer.
Ans: ENIAC.

(l) ____was one of the first commercially available computers.
Ans: UNIVAC.

(m) An _____converts the program written in assembly language into machine language.
Ans: assembler.

(n) Processing takes place in the_____.
Ans: CPU.

(o) The ALU performs the ____and ____operations.
Ans: arithmetic, logical.

(p) RAM is or .
Ans: temporary, volatile.

(q) ROM stores basic _____instructions to operate the computer.
Ans: input / output.

(r) _____memory is an electronic non-volatile storage medium that can be electrically erased and reprogrammed.
Ans: EEPROM.

(s) Data and instructions are given to the computer through its ____device.
Ans: input.

(t) Raw data is processed by the computer to produce_____.
Ans: output.

(u) Results are obtained from the computer through its ____device.
Ans: output.

(v) 1 MB = __ KB.
Ans: 1024.

(w) ______is an example of primary memory.
Ans: RAM.

(x) The octal number system uses the digits from to.
Ans: 0,7.

(y) A computer system comprise of both _ and_.
Ans: hardware, software.

(z) _____represents the set of program that directs the computer how to work.
Ans: software.

  1. State true or false.
    (a) Another name for internal memory is primary memory.
    Ans: TRUE.

(b) A hard disk can store less data than CD.
Ans: FALSE.

(c) software represents the physical components of a computer.
Ans: FALSE

(d) A projector is an input device.
Ans: FALSE.

(e) Interpreter is a utility software.
Ans: TRUE.

(f) The first-generation computers used transistors.
Ans: FALSE.

(g) Atanasoff Berry computer was the first electronic digital computer.
Ans: TRUE.

(h) IBM 604 is an example of second-generation computers.
Ans: TRUE.

(i) UNIVAC was one of the first commercially available computers.
Ans: TRUE.

(j) ROM is volatile.
Ans: FALSE.

(k) A group of eight bits is called nibble.
Ans: FALSE.

(l) Computers do not understand any language other than machine language.
Ans: TRUE.

(m) Pen drives use flash memory to store data.
Ans: TRUE.

(n) The monitor is a soft copy output device.
Ans: TRUE.

(o) A compiler translates a program written in machine language into a program written high level language.
Ans: TRUE.

(p) Bit is the smallest unit of data in a computer.
Ans: TRUE.

  1. Short answer questions.

 

 

  1. Long answer questions.

 

 

  • Differentiate between the following pairs:

I. Input unit and Output unit

II. RAM and ROM

III. Soft copy and hard copy

IV. Primary memory and secondary memory

V. Impact-printer and non-impact printer

VI. Assembler and Compiler

VII. Hardware and software

VIII. System software and application software

6. Carry out the following conversions:

a. Convert 135(10) to binary:
135 ÷ 2 = 67 r 1
67 ÷ 2 = 33 r 1
33 ÷ 2 = 16 r 1
16 ÷ 2 = 8 r 0
8 ÷ 2 = 4 r 0
4 ÷ 2 = 2 r 0
2 ÷ 2 = 1 r 0
1 ÷ 2 = 0 r 1
Answer: 135(10) = 10000111(2)


b. Convert 1100110(2) to decimal:
= 1×64 + 1×32 + 0×16 + 0×8 + 1×4 + 1×2 + 0×1 = 102
Answer: 1100110(2) = 102(10)


c. Convert 110.110(2) to decimal:
Integer: 1×4 + 1×2 + 0×1 = 6
Fraction: 1×0.5 + 1×0.25 + 0×0.125 = 0.75
Answer: 110.110(2) = 6.75(10)


d. Convert 35.375(10) to binary:
Integer part:
35 ÷ 2 = 17 r 1
17 ÷ 2 = 8 r 1
8 ÷ 2 = 4 r 0
4 ÷ 2 = 2 r 0
2 ÷ 2 = 1 r 0
1 ÷ 2 = 0 r 1 -> 100011
Fraction:
0.375 × 2 = 0.75 -> 0
0.75 × 2 = 1.5 -> 1
0.5 × 2 = 1.0 -> 1 -> .011
Answer: 35.375(10) = 100011.011(2)


e. Convert 265(10) to octal:
265 ÷ 8 = 33 r 1
33 ÷ 8 = 4 r 1
4 ÷ 8 = 0 r 4
Answer: 265(10) = 411(8)


f. Convert 250.375(10) to octal:
Integer part:
250 ÷ 8 = 31 r 2
31 ÷ 8 = 3 r 7
3 ÷ 8 = 0 r 3 -> 372
Fraction:
0.375 × 8 = 3.0 -> 3 -> .3
Answer: 250.375(10) = 372.3(8)


g. Convert 375(8) to decimal:
= 3×64 + 7×8 + 5×1 = 253
Answer: 375(8) = 253(10)


h. Convert 375(16) to decimal:
= 3×256 + 7×16 + 5 = 885
Answer: 375(16) = 885(10)


i. Convert 25.6(8) to decimal:
Integer: 2×8 + 5 = 21
Fraction: 6×(1/8) = 0.75
Answer: 25.6(8) = 21.75(10)


j. Convert 125(16) to decimal:
= 1×256 + 2×16 + 5 = 293
Answer: 125(16) = 293(10)


k. Convert 436(10) to hexadecimal:
436 ÷ 16 = 27 r 4
27 ÷ 16 = 1 r 11 -> B
1 ÷ 16 = 0 r 1
Answer: 436(10) = 1B4(16)


l. Convert 213.03125(10) to hexadecimal:
213 ÷ 16 = 13 r 5 -> D5
13 ÷ 16 = 0 r 13 -> D
Fraction:
0.03125 × 16 = 0.5 -> 0
0.5 × 16 = 8.0 -> 8 -> .08
Answer: 213.03125(10) = D5.08(16)


m. Convert 55.08(16) to decimal:
= 5×16 + 5 = 85
+ 8/256 = 0.03125
Answer: 55.08(16) = 85.03125(10)


n. Convert 11001101111(2) to hexadecimal:
0001 1001 1011 1111 = 1 9 B F
Answer: 11001101111(2) = 19BF(16)


o. Convert 10011001.10111001(2) to hexadecimal:
Integer: 1001 1001 = 9 9
Fraction: 1011 1001 = B 9
Answer: 10011001.10111001(2) = 99.B9(16)


p. Convert B59E(16) to binary:
B = 1011, 5 = 0101, 9 = 1001, E = 1110
Answer: B59E(16) = 1011010110011110(2)


q. Convert 10011001111(2) to octal:
001 001 100 111 = 1 1 4 7
Answer: 10011001111(2) = 1147(8)


r. Convert 6017(8) to binary:
6 = 110, 0 = 000, 1 = 001, 7 = 111
Answer: 6017(8) = 110000001111(2)

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