+1] 1. Fundamentals of Computer - Previous Questions chapter wise

               PLUS ONE COMPUTER APPLICATION

          First Year Computer Application(Commerce) Previous Questions Chapter wise ..

                                         Chapter  1. Fundamentals of Computer

1. Meaningful and processed form of data is known as ................... 

Ans. Information

2. which of the following is known as the brain of the computer.

 a). Central Processing unit 

 b).Control Unit 

 c). Arithmetic Logic Unit 

 d). Monitor    

Ans. a

3.  Name the character representation coding scheme developed in India       and approved by  the             Bureau of Indian Standards(BIS).

     Ans. ISCII stands for Indian Standard Code for Information Interchange or Indian Script Code for Information Interchange.

4. .Find the smallest number in the list.

   (a) (1101)2

   b) (A)16

   c)  (13)8

    d)   (15)10

      Ans. b

5. .Fill the series .       

(151)8 ,(153)8,(155)8,................ ,.................

Ans. (157)8,(160)8

6.  Convert the Hexadecimal (A2D) into octal equivalent

   Ans. (A2D)16= (1010 0010 1101)2

            101,000,101,101 = (5055)8

7. Represent -83 in 1’s Complement form


                             

8. Write a short note on Unicode.

            Using 8-bit ASCII we can represent only 256 characters. This cannot represent all characters of written languages of the world and other symbols. Unicode is developed to resolve this problem. It aims to provide a standard character encoding scheme, which is universal and efficient. It provides a unique number for every character, no matter what the language and platform be. Unicode originally used 16 bits which can represent up to 65,536 characters.

9.  Fill in the blanks
    a).  (DA)16 = (…. …….)2       

     b).  (25)10 =  (…….. ..)8

Ans.a. (11011010)2

          b.  (29)8

10.  Represent -35 in the following forms:
        (Hint: Use 8 bit form of representation)

         a). Sign and magnitude.       
         b). One’s  complement.       
          c). Two’s complement.




                             

Binary value of 35 =100011

8 bit form of 35= (00100011)2

then (-35)10 =( 11011100)2

a) Sign and magnitude = 1 and 1011100

On's complement = 11011100

Two's complement  = 11011100+1= 11011101

11.  If the binary equivalent of 56 is (111000)2 find the 1's complement form,  and Sign & Magnitude form of –56 (negative 56) in 8 bits.

    If (56)=(111000)2

 1's complement form = (000111)2

Binary value of  56 in 8 bit form= (00111000)2

Binary value of -56 in 8 bit form= (11000111)2

sign bit is MSB 1 and remaining bit is magnitude that is 1000111

To get the negative integer number representation on 8 bits, signed binary one's complement, replace all the bits on 0 with 1s and all the bits set on 1 with 0s (reversing the digits):

!(0011 1000) = 1100 0111

12.  Despite of the high speed and accuracy, computers are said to be the  slaves of human beings. why? 

    computers can execute millions of instructions in a second. The results produced after processing the data are very accurate, but computers do not have adequate knowledge or intelligence to interpret the results. They only carry out instructions like an obedient servant. The computer gives correct results only if the data and instructions given are correct. The term Garbage In Garbage Out (GIGO) is used to mean this feature. That is, if a wrong input is given to the computer, it will give a wrong output.

13.  a)Write the two’s complement form of the decimal number -119 

       b)State the benefit of using two’s complement representation as compared  to one’s  complement            form. 

   Ans. a). Binary of 119 in 8-bit form = (01110111)2

                -119 in 1’s complement form = (10001000)2

                2's complement form= 10001000+1= (10001001)2

             b).  The primary advantage of two's complement over one's complement is that two's                                 complement only has one value for zero. One's complement has a "positive" zero and a                        "negative" zero. Next, to add numbers using one's complement you have to first do                              binary   addition, then add in an end-around carry value.

14.  Data processing refers to the activities performed on data to generate  information. List the                   stages of data processing.

               (a) Capturing data (b) Input of data (c) Storage of data (d) Processing / manipulating data  (e)    Output of information (f) Distribution of information

   Capturing data

    When we apply for admission to the higher secondary course, we usually provide details through a prescribed application form. The authority is actually collecting the required data for the admission process through the proforma. This is the first stage in data processing. The proforma, also known as the source document, is so designed that all relevant data to be recorded in proper order and format. Thus, preparation of hard copy of source document and data collection are the activities that take place in this stage.

    Input

       In the case of seeking admission, we submit the filled up application form to the school. There the data is extracted and fed into the computer. Sometimes, we may enter these details directly into the computer. Feeding data to the computer for processing is known as input. The input data is usually stored in computers before it is processed.

     Storage

         In many cases, the amount of data given to the computers will be large. Besides, the data entry may not be completed in a single session or a day. In the case of admissions, the data of lakhs of applicants is input to the computer. It usually takes a few weeks to complete the data entry. So the data input at different times should be stored then and there. The processing will start only after the entire data is stored. The information obtained as a result of processing is also stored in the computer. This stored data and information can be used in future for various purposes.

    Process 

    The data stored in computers is retrieved for processing. Various operations like calculation, classification, comparison, sorting, filtering, summarising etc. are carried out as part of processing. In the case of admission to the higher secondary course, Weighted Grade Point Average (WGPA) of each applicant is calculated. Then the applicants are listed under various categories based on the descending order of WGPA. Here, school of choice, course, and performance in various co-curricular activities are considered. Finally, allotment lists for schools and allotment slips for applicants are prepared.

       Output 

      The information obtained after processing will be available in this stage. Output stage should provide the information in such a form that the beneficiary should be able to take decision or solve the problem. In the case of admission to the higher secondary course, allotment slip for the applicant and allotment list for the school are generated in the desired format as outputs.

       Distribution of information

             The information obtained in the output stage is distributed to the beneficiaries. They take decisions or solve problems according to the information. For example in higher secondary admission, the allotment slips are distributed to applicants for joining the school allotted and allotment lists are issued to the schools for admitting the eligible applicants. The allotment slips may be used to prepare admission register or roll list of classes. The allotment lists may be used to prepare nominal roll for registering the students for public examination.

15.  a). Convert (1010.11)2 to decimal.          

     b). Find the missing terms in the following series.      

     (18)16 , (1A)16 , (1C)16 , ……. , ……….  

 Ans.   (1010.11)2

         decimal part 1010=0*2^0+1*2^1+0*2^2+1*2^3

                                           =0+2+0+8

                                           = 10

fractional part 11=1*2^-1+1*2^-2

          

                            = 1/2+1/4

                   

                              =.5+.25

                              

                              =.75

so the answer is (10.75)10

b). (1E)16,(20)16

16.  If (M)8 = (96)10 = (N)2, find M and N. 

          (M)8=(96)10

         M=140

      (96)10= (1100000)2

       N=1100000

17. .If (11011)2=(A)8=(B)16=(C)10 .Find the value of A,B and C. 

       (11011)2

For octal conversion grouping 3bits format

then one 0 add in MSB 

that is the binary bit 011011=011,111=(37)8

           For hexadecimal conversion grouping 4 bits format

then  three 0s add in MSB

that is our bit becomes 00011011=0001,1011

                                                       (1B)16

(11011)2= 1*2^0+1*2^1+0*2^2+1*2^3+1*2^4

                 = 1+2+0+8+16

                     =(27)10

Then A=37 ,B=1B and C=27

(37)8,(1B)16 & (27)10