<p> KENDRIYA VIDYALAYA SANGATHAN (AGRA REGION) PRE BOARD EXAMINATION 2017-18 MARKING SCHEME</p><p>COMPUTER SCIENCE SET SERIES: KVS (AGRA), CODE: 083/1</p><p>General Instructions:</p><p>1. The answers given in the marking scheme are SUGGESTIVE; Examiners are requested to award marks for all alternative correct Solutions / Answers conveying the similar meaning.</p><p>2. All programming questions have to be answered with respect to C++ Language only.</p><p>3. In C++ ignore case sensitivity for identifiers (Variable/ Functions/ Structures/ Class Names).</p><p>4. In SQL related questions - both ways of text/character entries should be acceptable for Example: “AMAR” and ‘amar’ both are correct. </p><p>5. In SQL related questions, ignore case sensitivity.</p><p>6. In SQL related questions – semicolon should be ignored for terminating the SQL statements 7. In SQL related questions – all date entries should be acceptable for Example: ‘ YYYY-MM-DD’, ‘YY-MM-DD’, ‘DD-Mon-YY’, “DD/MM/YY”, ‘DD/MM/YY’,“MM/DD/YY”, ‘MM/DD/YY’ and {MM/DD/YY} are correct.</p><p>1 Question Suggested Answer/ Description Marks No. Distribution 1. (a) Assuming that target variable is res of type int 1 marks for correct statement & 1 int res= (int) x1+ (int) x2; mark for correct data type conversion: explicit type (b) iostream.h or iomanip.h or fstream.h ½ mark for string.h each correct stdio.h (½ Mark each for writing any two correct expansion header files) ½ +½ NOTE: Ignore additional header file(s) (c) #include <iostream.h> ½ mark for void Awazdo(int Arg1, int Arg2=20); void main( ) each error { identified int Mukki=10, kukki=20; Awazdo(Mukki,kukki); answer Awazdo(kukki); ½ *4=2 } void Awazdo( int Arg1, int Arg2=20) { Arg1=Arg1+Arg2 ; cout<<Arg1 << Arg2; } (d) The structure s1 comprises of 2 pointers; one a char 2 marks for pointer and another a pointer to structure s1. An array of correct answer structures of type s1 named as arr [ ] is declared and initialized. Next, an array of pointers to structure, p [3] is declared. Each element pointer of p [3] will hold the address of a structure of type s1. Now look at the following figure before understanding the reasons for producing the given output: arr [ ]</p><p> str ptr</p><p>3000 X0 3004 3004 X1 3008 3008 X2 3000</p><p> p[ ] 0 3004 5018 1 3008 5020 2 3000 5022</p><p>X0 Teeya \0 X2 Mukesh\02 X1 Paki \0 Address where from strings start</p>