{"id":17718,"date":"2018-06-30T10:06:26","date_gmt":"2018-06-30T04:36:26","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/?p=17718"},"modified":"2019-01-19T15:47:19","modified_gmt":"2019-01-19T10:17:19","slug":"cbse-question-paper-2018-class-12-computer-science","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/cbse-question-paper-2018-class-12-computer-science\/","title":{"rendered":"CBSE Question Paper 2018 class 12 Computer Science"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_76 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\" style=\"cursor:inherit\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-toggle-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-question-paper-2018-class-12-computer-science\/#Previous_Year_Question_Paper_%E2%80%93_Download_in_PDF\" >Previous Year Question Paper \u2013 Download in PDF<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-question-paper-2018-class-12-computer-science\/#Class_12_Computer_Science_list_of_chapters\" >Class 12 Computer Science list of chapters<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-question-paper-2018-class-12-computer-science\/#CBSE_Question_Paper_2018_class_12_Computer_Science\" >CBSE Question Paper 2018 class 12 Computer Science<\/a><ul class='ez-toc-list-level-3' ><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-question-paper-2018-class-12-computer-science\/#General_Instructions\" >General Instructions :<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-question-paper-2018-class-12-computer-science\/#Last_Year_Question_Paper_Class_12_Computer_Science_2018\" >Last Year Question Paper Class 12 Computer Science 2018<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-question-paper-2018-class-12-computer-science\/#Previous_Year_Question_Paper_for_class_12_in_PDF\" >Previous Year Question Paper for class 12 in PDF<\/a><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"Previous_Year_Question_Paper_%E2%80%93_Download_in_PDF\"><\/span>Previous Year Question Paper \u2013 Download in PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>CBSE Question Paper 2018 class 12 Computer Science<\/strong>\u00a0conducted by Central Board of Secondary Education, New Delhi in the month of March 2018. CBSE previous year question papers with solution are available in myCBSEguide mobile app and cbse guide website. The Best CBSE App for students and teachers is myCBSEguide which provides complete study material and practice papers to CBSE schools in India and abroad.<\/p>\n<p style=\"text-align: center;\"><strong>CBSE Question Paper 2018 class 12 Computer Science<\/strong><\/p>\n<p style=\"text-align: center;\"><strong><a class=\"button\" href=\"https:\/\/mycbseguide.com\/downloads\/cbse-class-12-computer-science\/1851\/cbse-last-year-papers\/3\/\">Download as PDF<\/a><\/strong><\/p>\n<h2><span class=\"ez-toc-section\" id=\"Class_12_Computer_Science_list_of_chapters\"><\/span>Class 12 Computer Science list of chapters<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ol>\n<li>Review of Python<\/li>\n<li>Concept of Object Oriented Programming<\/li>\n<li>Classes in Python<\/li>\n<li>Inheritance<\/li>\n<li>Linear List Manipulation<\/li>\n<li>Stacks &amp; Queues in list<\/li>\n<li>Data File Handling<\/li>\n<li>Exception Handling &amp; Green Functions<\/li>\n<li>Databases Concepts and SQL<\/li>\n<li>Structure\u00a0 Query Language<\/li>\n<li>Boolean Algebra<\/li>\n<li>Boolean Functions &amp; Reduce Forms<\/li>\n<li>Application of Boolean Logic<\/li>\n<li>Networking Concepts\u00a0 (Part 1)<\/li>\n<li>Networking Concepts\u00a0 (Part 2)<\/li>\n<li>Networking Protocols<\/li>\n<li>Mobile Telecommunication Technologies, Network Security and Internet Services<\/li>\n<\/ol>\n<h2><span class=\"ez-toc-section\" id=\"CBSE_Question_Paper_2018_class_12_Computer_Science\"><\/span>CBSE Question Paper 2018 class 12 Computer Science<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3><span class=\"ez-toc-section\" id=\"General_Instructions\"><\/span><strong>General Instructions :<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<ol>\n<li>SECTION A refers to programming language C++.<\/li>\n<li>SECTION B refers to programming language Python.<\/li>\n<li>SECTION C is compulsory for all.<\/li>\n<li>Answer either SECTION A or SECTION B.<\/li>\n<li>It is compulsory to mention on the page 1 in the answer book whether you are attempting SECTION A or SECTION B.<\/li>\n<li>All questions are compulsory within each section.<\/li>\n<\/ol>\n<hr \/>\n<p style=\"text-align: center;\"><strong> SECTION A<br \/>\n[Only for candidates, who opted for C++]<\/strong><\/p>\n<ol>\n<li>\n<ol start=\"1\">\n<li><strong>Write the type of C++ tokens (keywords and user-defined identifiers) from the following : (2)<\/strong>\n<ol start=\"1\">\n<li><strong>else<\/strong><\/li>\n<li><strong>Long<\/strong><\/li>\n<li><strong>4Queue<\/strong><\/li>\n<li><strong>_count<\/strong><\/li>\n<\/ol>\n<p><strong> Ans. <\/strong><\/p>\n<ol start=\"1\">\n<li>keyword<\/li>\n<li>Identifier<\/li>\n<li>None<\/li>\n<li>Identifier<br \/>\nNOTE: Ignore (iii)<\/li>\n<\/ol>\n<\/li>\n<li><strong>The following C++ code during compilation reports errors as follows :<br \/>\nError: \u2018ofstream\u2019 not declared<br \/>\nError: \u2018strupr\u2019 not declared<br \/>\nError: \u2018strcat\u2019 not declared<br \/>\nError: \u2018FIN\u2019 not declared<br \/>\nWrite the names of the correct header files, which must be included to compile the code successfully : (1)<br \/>\nvoid main()<br \/>\n{<br \/>\nofstream FIN(&#8220;WISH.TXT&#8221;);<br \/>\nchar TEXT2[]=&#8221;good day&#8221;;<br \/>\nchar TEXT1[]=&#8221;John!&#8221;;<br \/>\nstrupr(TEXT2);<br \/>\nstrcat(TEXT1, TEXT2);<br \/>\nFIN&lt;&lt;TEXT1&lt;&lt;endl;<br \/>\n}<br \/>\nAns.<\/strong><\/p>\n<ol start=\"1\">\n<li>fstream<\/li>\n<li>string<\/li>\n<\/ol>\n<\/li>\n<li><strong>Rewrite the following C++ code after removing any\/all syntactical errors with each correction underlined. (2)<br \/>\nNote : Assume all required header files are already included in the program.<br \/>\nTypedef Count int;<br \/>\nvoid main()<br \/>\n{<br \/>\nCount C;<br \/>\ncout&lt;&lt;&#8220;Enter the count:&#8221;;<br \/>\ncin&gt;&gt;C;<br \/>\nfor (K = 1; K&lt;=C; K++)<br \/>\ncout&lt;&lt; C &#8220;*&#8221; K &lt;&lt;endl;<br \/>\n}<br \/>\nAns.<\/strong> typedef int Count ; \/\/Error 1, Error 2<br \/>\nvoid main()<br \/>\n{<br \/>\nCount C;<br \/>\nint K; \/\/OR Count K; \/\/Error 3<br \/>\ncout&lt;&lt;&#8220;Enter the count:&#8221;;<br \/>\ncin&gt;&gt;C;<br \/>\nfor (K = 1; K&lt;=C; K++)<br \/>\n\/\/ OR for ( int K = 1; K&lt;=C; K++) \/\/Error 3<br \/>\n\/\/ OR for ( Count K = 1; K&lt;=C; K++) \/\/Error 3<br \/>\ncout&lt;&lt; C &lt;&lt; &#8220;*&#8221; &lt;&lt; K &lt;&lt;endl; \/\/Error 4<br \/>\n\/\/ OR cout&lt;&lt; C * K &lt;&lt;endl; \/\/Error 4<br \/>\n}<\/li>\n<li><strong>Find and write the output of the following C++ program code : (3)<br \/>\nNote : Assume all required header files are already included in the program.<br \/>\nvoid Revert(int &amp;Num, int Last=2)<br \/>\n{<br \/>\nLast=(Last%2==0)?Last+1:Last-1;<br \/>\nfor(int C=1; C&lt;=Last; C++)<br \/>\nNum+=C;<br \/>\n}<br \/>\nvoid main()<br \/>\n{<br \/>\nint A=20,B=4;<br \/>\nRevert(A,B);<br \/>\ncout&lt;&lt;A&lt;&lt;&#8220;&amp;&#8221;&lt;&lt;B&lt;&lt;endl;<br \/>\nB&#8211;;<br \/>\nRevert(A,B);<br \/>\ncout&lt;&lt;A&lt;&lt;&#8220;#&#8221;&lt;&lt;B&lt;&lt;endl;<br \/>\nRevert(B);<br \/>\ncout&lt;&lt;A&lt;&lt;&#8220;#&#8221;&lt;&lt;B&lt;&lt;endl;<br \/>\n}<br \/>\nAns.<\/strong>35&amp;4<br \/>\n38#3<br \/>\n38#9<\/li>\n<li><strong>Find and write the output of the following C++ program code : (2)<br \/>\nNote : Assume all required header files are already included in the program.<br \/>\n#define Modify(N) N*3+10<br \/>\nvoid main()<br \/>\n{<br \/>\nint LIST[ ]={10,15,12,17};<br \/>\nint *P=LIST, C;<br \/>\nfor(C=3; C&gt;=0; C&#8211;)<br \/>\nLIST[I]=Modify(LIST[I]);<br \/>\nfor (C=0; C&lt;=3; C++)<br \/>\n{<br \/>\ncout&lt;&lt;*P&lt;&lt;&#8220;:&#8221;;<br \/>\nP++;<br \/>\n}<br \/>\n}<br \/>\nAns.<\/strong> Considering LIST[I] being replaced with LIST[C]\n40:55:46:61:<\/li>\n<li><strong>Look at the following C++ code and find the possible output(s) from the options (i) to (iv) following it. Also, write the highest and lowest values that can be assigned in the array A. (2)<br \/>\nNote :<\/strong><\/p>\n<ul>\n<li><strong>Assume all the required header files are already being included in the code.<\/strong><\/li>\n<li><strong>The function random(n) generates an integer between 0 and n \u2013 1.<br \/>\nvoid main()<br \/>\n{<br \/>\nrandomize();<br \/>\nint A[4], C;<br \/>\nfor(C=0; C&lt;4; C++)<br \/>\nA[C]=random(C+1)+10;<br \/>\nfor(C=3; C&gt;=0; C&#8211;)<br \/>\ncout&lt;&lt;A[C]&lt;&lt;&#8220;@&#8221;;<br \/>\n}<\/strong><\/p>\n<table class=\"responsive\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td>(i)<\/td>\n<td>(ii)<\/td>\n<\/tr>\n<tr>\n<td>13@10@11@10@<\/td>\n<td>15$14$12$10$<\/td>\n<\/tr>\n<tr>\n<td>(iii)<\/td>\n<td>(iv)<\/td>\n<\/tr>\n<tr>\n<td>12@11@13@10@<\/td>\n<td>12@11@10@10@<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong> Ans.<\/strong> (i) and (iv)<br \/>\nA<sub>Min<\/sub> = 10<br \/>\nA<sub> Max<\/sub> = 13<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n<\/li>\n<li>\n<ol start=\"1\">\n<li><strong>Which function(s) out of the following can be considered as overloaded function(s) in the same program ? Also, write the reason for not considering the other(s) as overloaded function(s). (2)<br \/>\nvoid Execute(char A,int B); \/\/ Function 1<br \/>\nvoid Execute(int A,char B); \/\/ Function 2<br \/>\nvoid Execute(int P=10); \/\/ Function 3<br \/>\nvoid Execute(); \/\/ Function 4<br \/>\nint Execute(int A); \/\/ Function 5<br \/>\nvoid Execute(int &amp;K); \/\/ Function 6<br \/>\nAns.<\/strong> Option [i]\nFunctions 1,2,3 are overloaded<br \/>\nReason: Function 4,5,6 would give ambiguity for Function 3<br \/>\nOR Any equivalent valid reason<strong>OR<\/strong>Option [ii]\nFunctions 1,2,4,5 are overloaded<br \/>\nReason: Function 3 and 6 not considered in this case because it would give redeclaration error for Function 5<br \/>\nOR Any equivalent valid reason<strong>OR<\/strong>Option [iii]\nFunctions 1,2,4,6 are overloaded<br \/>\nReason: Function 3 and 5 not considered in this case because it would give redeclaration error for Function 6<br \/>\nOR Any equivalent valid reason<\/li>\n<li><strong>Observe the following C++ code and answer the questions (i) and (ii).<br \/>\nNote : Assume all necessary files are included.<br \/>\nclass FIRST<br \/>\n{<br \/>\nint Num1;<br \/>\npublic:<br \/>\nvoid Display() \/\/Member Function 1<br \/>\n{<br \/>\ncout&lt;&lt;Num1&lt;&lt;endl;<br \/>\n}<br \/>\n};<br \/>\nclass SECOND: public FIRST<br \/>\n{<br \/>\nint Num2;<br \/>\npublic:<br \/>\nvoid Display() \/\/Member Function 2<br \/>\n{<br \/>\ncout&lt;&lt;Num2&lt;&lt;endl;<br \/>\n}<br \/>\n};<br \/>\nvoid main()<br \/>\n{<br \/>\nSECOND S;<br \/>\n_______________ \/\/Statement 1<br \/>\n_______________ \/\/Statement 2<br \/>\n}<\/strong><\/p>\n<ol start=\"1\">\n<li><strong>Which Object Oriented Programming feature is illustrated by the definitions of classes FIRST and SECOND ? (1)<br \/>\nAns.<\/strong> Inheritance<strong>OR<\/strong>Encapsulation<strong>OR<\/strong>Data Abstraction<strong>OR<\/strong>Data Hiding<\/li>\n<li><strong>Write Statement 1 and Statement 2 to execute Member Function 1 and Member Function 2 respectively using the object S. (1)<br \/>\nAns.<\/strong> S.FIRST::Display() \/\/Statement 1<br \/>\nS.Display() \/\/Statement 2<strong>OR<\/strong>S.SECOND::Display() \/\/Statement 2<\/li>\n<\/ol>\n<\/li>\n<li><strong>Write the definition of a class CONTAINER in C++ with the following description : (4)<br \/>\nPrivate Members<br \/>\n&#8211; Radius, Height \/\/ float<br \/>\n&#8211; Type \/\/ int (1 for Cone,2 for Cylinder)<br \/>\n&#8211; Volume \/\/ float<br \/>\n&#8211; CalVolume() \/\/ Member function to calculate<br \/>\n\/\/ volume as per the Type<\/strong><\/p>\n<table class=\"responsive\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<thead>\n<tr>\n<th>Type<\/th>\n<th>Formula to Calculate Volume<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>1<\/td>\n<td>3.14*Radius*Height<\/td>\n<\/tr>\n<tr>\n<td>2<\/td>\n<td>3.14*Radius*Height\/3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong> Public Members<br \/>\n&#8211; GetValues() \/\/ A function to allow user to enter value<br \/>\n\/\/ of Radius, Height and Type. Also, call<br \/>\n\/\/ function CalVolume() from it<br \/>\n&#8211; ShowAll() \/\/ A function to display Radius, Height,<br \/>\n\/\/ Type and Volume of Container<br \/>\nAns.<\/strong> class CONTAINER<br \/>\n{<br \/>\nfloat Radius, Height;<br \/>\nint Type;<br \/>\nfloat Volume;<br \/>\nvoid CalVolume();<br \/>\npublic:<br \/>\nvoid GetValues();<br \/>\nvoid ShowAll();<br \/>\n};<br \/>\nvoid CONTAINER::GetValues()<br \/>\n{<br \/>\ncin&gt;&gt;Radius&gt;&gt;Height&gt;&gt;Type ;<br \/>\nCalVolume();<br \/>\n} void CONTAINER::ShowAll()<br \/>\n{<br \/>\ncout&lt;&lt;Radius&lt;&lt;Height&lt;&lt;Type&lt;&lt;Volume&lt;&lt;endl;<br \/>\n}<\/p>\n<table class=\"responsive\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td>void CONTAINER::CalVolume()<br \/>\n{<br \/>\nif (Type == 1)<br \/>\nVolume=3.14*Radius*Height;<br \/>\nelse if (Type == 2)<br \/>\nVolume=3.14*Radius*Height\/3;<br \/>\n}<\/td>\n<td>void CONTAINER::CalVolume()<br \/>\n{<br \/>\nswitch (Type)<br \/>\n{<br \/>\ncase 1:<br \/>\nVolume =3.14*Radius*Height;<br \/>\nbreak;<br \/>\ncase 2:<br \/>\nVolume=3.14*Radius*Height\/3;<br \/>\n}<br \/>\n}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li><strong>Answer the questions (i) to (iv) based on the following : (4)<br \/>\nclass Teacher<br \/>\n{<br \/>\nint TCode;<br \/>\nprotected:<br \/>\nchar Name[20];<br \/>\npublic:<br \/>\nTeacher();<br \/>\nvoid Enter(); void Show();<br \/>\n};<br \/>\nclass Course<br \/>\n{<br \/>\nint ID;<br \/>\nprotected:<br \/>\nChar Title[30];<br \/>\npublic:<br \/>\nCourse();<br \/>\nvoid Initiate();<br \/>\nvoid Display();<br \/>\n};<br \/>\nclass Schedule : public Course, private Teacher<br \/>\n{<br \/>\nint DD,MM,YYYY;<br \/>\npublic:<br \/>\nSchedule();<br \/>\nvoid Start();<br \/>\nvoid View();<br \/>\n};<br \/>\nvoid main()<br \/>\n{<br \/>\nSchedule S;<br \/>\n}<\/strong><\/p>\n<ol start=\"1\">\n<li><strong>Which type of Inheritance out of the following is illustrated in the above example?<br \/>\nSingle Level Inheritance, Multilevel Inheritance, Multiple Inheritance<br \/>\nAns.<\/strong> Multiple Inheritance<\/li>\n<li><strong>Write the names of all the members, which are directly accessible by the member function View() of class Schedule.<br \/>\nAns.<\/strong> Start(), DD, MM, YYYY<br \/>\nDisplay(), Initiate(), Title<br \/>\nEnter(), Show(), Name<br \/>\nView() \/\/ Optional<\/li>\n<li><strong>Write the names of all the members, which are directly accessible by the object S of class Schedule declared in the main() function.<br \/>\nAns.<\/strong> View(), Start()<br \/>\nDisplay(), Initiate()<\/li>\n<li><strong>What will be the order of execution of the constructors, when the object S of class Schedule is declared inside the main() function ?<br \/>\nAns.<\/strong> Course(), Teacher(), Schedule()<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/li>\n<li>\n<ol start=\"1\">\n<li><strong>Write the definition of a function SumEO(int VALUES[], int N) in C++, which should display the sum of even values and sum of odd values of the array separately. (2)<br \/>\nExample : If the array VALUES contains<\/strong><\/p>\n<table class=\"responsive\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td>20<\/td>\n<td>20<\/td>\n<td>22<\/td>\n<td>21<\/td>\n<td>53<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong> Then the functions should display the output as :<br \/>\nSum of even values = 42 (i.e., 20+22)<br \/>\nSum of odd values = 99 (i.e., 25+21+53)<br \/>\nAns.<\/strong> void SumEO(int VALUES[], int N)<br \/>\n{<br \/>\nint SE = 0, SO = 0;<br \/>\nfor (int I=0;I&lt;N;I++)<br \/>\n{<br \/>\nif(VALUES[I] %2 == 0)<br \/>\nSE += VALUES[I];<br \/>\nelse<br \/>\nSO += VALUES[I];<br \/>\n}<br \/>\ncout&lt;&lt; &#8220;Sum of even values = &#8221; &lt;&lt; SE&lt;&lt;endl;<br \/>\ncout&lt;&lt; &#8220;Sum of odd values = &#8221; &lt;&lt; SO&lt;&lt;endl;<br \/>\n}<strong>OR<\/strong><\/p>\n<p>Any other correct alternative code in C++<\/li>\n<li><strong>Write a definition for a function UpperHalf(int Mat[4][4]) in C++, which displays the elements in the same way as per the example shown below. (3)<br \/>\nFor example, if the content of the array Mat is as follows :<\/strong><\/p>\n<table class=\"responsive\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td>25<\/td>\n<td>24<\/td>\n<td>23<\/td>\n<td>22<\/td>\n<\/tr>\n<tr>\n<td>20<\/td>\n<td>19<\/td>\n<td>18<\/td>\n<td>17<\/td>\n<\/tr>\n<tr>\n<td>15<\/td>\n<td>14<\/td>\n<td>13<\/td>\n<td>12<\/td>\n<\/tr>\n<tr>\n<td>10<\/td>\n<td>9<\/td>\n<td>8<\/td>\n<td>7<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong> The function should display the content in the following format :<br \/>\n25 24 23 22<br \/>\n20 19 18<br \/>\n15 14<br \/>\n10<br \/>\nAns.<\/strong> void UpperHalf(int Mat[4][4])<br \/>\n{<br \/>\nfor (int I=0;I&lt;4;I++)<br \/>\n{<br \/>\nfor (int J=0;J&lt;4-I;J++)<br \/>\ncout&lt;&lt;MAT[I][J]&lt;&lt; &#8221; &#8221; ;<br \/>\ncout&lt;&lt;endl;<br \/>\n}<br \/>\n}<strong>OR<\/strong><\/p>\n<p>void UpperHalf(int Mat[4][4])<br \/>\n{<br \/>\nfor (int I=0;I&lt;4;I++)<br \/>\n{<br \/>\nfor (int J=0;J&lt;4;J++)<br \/>\nif ((I+J)&lt;=3)<br \/>\ncout&lt;&lt;MAT[I][J]&lt;&lt; &#8221; &#8221; ;<br \/>\ncout&lt;&lt;endl;<br \/>\n}<br \/>\n}<\/p>\n<p><strong>OR<\/strong><\/p>\n<p>Any other correct alternative code in C++<\/li>\n<li><strong>Let us assume Data[20][15] is a two-dimensional array, which is stored in the memory along the row with each of its elements occupying 2 bytes. Find the address of the element Data[10][5], if the element Data[15][10] is stored at the memory location 15000. (3)<br \/>\nAns.<\/strong> LOC(Data[10][5]) = LOC(Data[15][10])+2(15*(10-15)+(5-10))<br \/>\n= 15000 + 2((-75) + (-5))<br \/>\n= 15000 + 2(-80)<br \/>\n= 15000 &#8211; 160<br \/>\n= 14840<strong>OR<\/strong>LOC(Data[I][J]) = Base(Data)+W*(NC*(I-LBR)+(J-LBC))<br \/>\nTaking LBR=0, LBC=0<br \/>\nLOC(Data[15][10]) = Base(Data)+2*(15*15+10)<br \/>\n15000 = Base(Data)+2*(15*15+10)<br \/>\nBase(Data) = 15000 &#8211; 2*(235)<br \/>\nBase(Data) = 15000 &#8211; 470<br \/>\nBase(Data) = 14530<br \/>\nLOC(Data[10][5])= 14530 + 2*(10*15+5)<br \/>\n= 14530 + 2*(155)<br \/>\n= 14530 + 310<br \/>\n= 14840<strong>OR<\/strong>LOC(Data[I][J]) = Base(Data)+W*(NC*(I-LBR)+(J-LBC))<br \/>\nTaking LBR=1, LBC=1<br \/>\nLOC(Data[15][10]) = Base(Data)+2*(15*14+9)<br \/>\n15000 = Base(Data)+2*(15*14+9)<br \/>\nBase(Data) = 15000 &#8211; 2*(219)<br \/>\nBase(Data) = 15000 &#8211; 438<br \/>\nBase(Data) = 14562<br \/>\nLOC(Data[10][5])= 14562 + 2*(15*9+4)<br \/>\n= 14562 + 2*(139)<br \/>\n= 14562 + 278<br \/>\n= 14840<\/li>\n<li><strong>Write the definition of a member function AddPacket() for a class QUEUE in C++, to remove\/delete a Packet from a dynamically allocated QUEUE of Packets considering the following code is already written as a part of the program. (4)<br \/>\nstruct Packet<br \/>\n{<br \/>\nint PID;<br \/>\nchar Address[20];<br \/>\nPacket *LINK;<br \/>\n};<br \/>\nclass QUEUE<br \/>\n{<br \/>\nPacket *Front, *Rear;<br \/>\npublic:<br \/>\nQUEUE(){Front=NULL;Rear=NULL;}<br \/>\nvoid AddPacket();<br \/>\nvoid DeletePacket();<br \/>\n~QUEUE();<br \/>\n};<br \/>\nAns.<\/strong> void QUEUE::AddPacket()<br \/>\n{<br \/>\nif(Front != NULL)<br \/>\n{<br \/>\nPacket *T;<br \/>\nT=Front;<br \/>\ncout&lt;&lt;Front-&gt;PID&lt;&lt;Front-&gt;Address&lt;&lt;&#8221; removed&#8221;&lt;&lt;endl;<br \/>\n\/\/OR cout&lt;&lt;T-&gt;PID&lt;&lt;T-&gt;Address&lt;&lt;&#8221; removed&#8221;&lt;&lt;endl;<br \/>\nFront = Front-&gt;LINK;<br \/>\ndelete T;<br \/>\nif (Front==NULL)<br \/>\nRear=NULL;<br \/>\n}<br \/>\nelse<br \/>\ncout&lt;&lt; &#8220;Queue Empty&#8221;&lt;&lt;endl;<br \/>\n}<strong>OR<\/strong>Any other equivalent code in C++<\/li>\n<li><strong>Convert the following Infix expression to its equivalent Postfix expression, showing the stack contents for each step of conversion : (2)<br \/>\nU * V + (W \u2013 Z) \/ X<br \/>\nAns.<\/strong> ((U * V) + ((W &#8211; Z) \/ X))<\/p>\n<table class=\"responsive\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<thead>\n<tr>\n<th>INFIX<\/th>\n<th>STACK<\/th>\n<th>POSFIX<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>U<\/td>\n<td><\/td>\n<td>U<\/td>\n<\/tr>\n<tr>\n<td>*<\/td>\n<td>*<\/td>\n<td>U<\/td>\n<\/tr>\n<tr>\n<td>V<\/td>\n<td>*<\/td>\n<td>UV<\/td>\n<\/tr>\n<tr>\n<td>)<\/td>\n<td><\/td>\n<td>UV*<\/td>\n<\/tr>\n<tr>\n<td>+<\/td>\n<td>+<\/td>\n<td>UV*<\/td>\n<\/tr>\n<tr>\n<td>W<\/td>\n<td><\/td>\n<td>UV*W<\/td>\n<\/tr>\n<tr>\n<td>&#8211;<\/td>\n<td>+ &#8211;<\/td>\n<td>UV*W<\/td>\n<\/tr>\n<tr>\n<td>Z<\/td>\n<td>+ &#8211;<\/td>\n<td>UV*WZ<\/td>\n<\/tr>\n<tr>\n<td>)<\/td>\n<td>+<\/td>\n<td>UV*WZ-<\/td>\n<\/tr>\n<tr>\n<td>\/<\/td>\n<td>+\/<\/td>\n<td>UV*WZ-<\/td>\n<\/tr>\n<tr>\n<td>X<\/td>\n<td>+\/<\/td>\n<td>UV*WZ-X<\/td>\n<\/tr>\n<tr>\n<td>)<\/td>\n<td>+<\/td>\n<td>UV*WZ-X\/<\/td>\n<\/tr>\n<tr>\n<td>)<\/td>\n<td><\/td>\n<td>UV*WZ-X\/+<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>OR<\/strong><\/p>\n<p>U * V + (W &#8211; Z) \/ X<\/p>\n<table class=\"responsive\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<thead>\n<tr>\n<th>INFIX<\/th>\n<th>STACK<\/th>\n<th>POSFIX<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>U<\/td>\n<td><\/td>\n<td>U<\/td>\n<\/tr>\n<tr>\n<td>*<\/td>\n<td>*<\/td>\n<td>U<\/td>\n<\/tr>\n<tr>\n<td>V<\/td>\n<td>*<\/td>\n<td>UV<\/td>\n<\/tr>\n<tr>\n<td>+<\/td>\n<td>+<\/td>\n<td>UV*<\/td>\n<\/tr>\n<tr>\n<td>(<\/td>\n<td>+(<\/td>\n<td>UV*<\/td>\n<\/tr>\n<tr>\n<td>W<\/td>\n<td>+(<\/td>\n<td>UV*W<\/td>\n<\/tr>\n<tr>\n<td>&#8211;<\/td>\n<td>+ (-<\/td>\n<td>UV*W<\/td>\n<\/tr>\n<tr>\n<td>Z<\/td>\n<td>+ (-<\/td>\n<td>UV*WZ<\/td>\n<\/tr>\n<tr>\n<td>)<\/td>\n<td>+<\/td>\n<td>UV*WZ-<\/td>\n<\/tr>\n<tr>\n<td>\/<\/td>\n<td>+\/<\/td>\n<td>UV*WZ-<\/td>\n<\/tr>\n<tr>\n<td>X<\/td>\n<td>+\/<\/td>\n<td>UV*WZ-X<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><\/td>\n<td>UV*WZ-X\/+<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<\/ol>\n<\/li>\n<li>\n<ol start=\"1\">\n<li><strong>A text file named MATTER.TXT contains some text, which needs to be displayed such that every next character is separated by a symbol \u2018#\u2019.<br \/>\nWrite a function definition for HashDisplay() in C++ that would display the entire content of the file MATTER.TXT in the desired format. (3)<br \/>\nExample :<br \/>\nIf the file MATTER.TXT has the following content stored in it :<br \/>\nTHE WORLD IS ROUND<br \/>\nThe function HashDisplay() should display the following content :<br \/>\nT#H#E# #W#O#R#L#D# #I#S# #R#O#U#N#D#<br \/>\nAns.<\/strong> void HashDisplay()<br \/>\n{<br \/>\nchar ch;<br \/>\nifstream F(&#8220;MATTER.TXT&#8221; );<br \/>\n<img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/0HTjMJR.png\" alt=\"\" data-imgur-src=\"0HTjMJR.png\" data-cke-saved-src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/0HTjMJR.png\" \/><br \/>\nwhile(F.get(ch))<br \/>\ncout&lt;&lt;ch&lt;&lt;&#8216;#&#8217;;<br \/>\nF.close(); \/\/IGNORE<br \/>\n}<strong>OR<\/strong>Any other correct function definition<\/li>\n<li><strong>Write a definition for a function TotalTeachers( ) in C++ to read each object of a binary file SCHOOLS.DAT, find the total number of teachers, whose data is stored in the file and display the same. Assume that the file SCHOOLS.DAT is created with the help of objects of class SCHOOLS, which is defined below: (2)<br \/>\nclass SCHOOLS<br \/>\n{<br \/>\nint SCode; \/\/School Code<br \/>\nchar SName[20]; \/\/School Name<br \/>\nint NOT; \/\/Number of Teachers in the school<br \/>\npublic:<br \/>\nvoid Display()<br \/>\n{cout&lt;&lt;SCode&lt;&lt;&#8220;#&#8221;&lt;&lt;SName&lt;&lt;&#8220;#&#8221;&lt;&lt;NOT&lt;&lt;endl;}<br \/>\nint RNOT(){return NOT;}<br \/>\n};<br \/>\nAns.<\/strong> void TotalTeachers()<br \/>\n{<br \/>\nifstream F;<br \/>\nF.open(&#8220;SCHOOLS.DAT&#8221;,ios::binary);<br \/>\nint Count=0;<br \/>\nSCHOOLS S;<br \/>\nwhile(F.read((char*)&amp;S,sizeof(S)))<br \/>\nCount += S.RNOT();<br \/>\ncout&lt;&lt;&#8220;Total number of teachers :&#8221;&lt;&lt;Count&lt;&lt;endl;<br \/>\nF.close(); \/\/IGNORE<br \/>\n}<strong>OR<\/strong>void TotalTeachers()<br \/>\n{<br \/>\nifstream F;<br \/>\nF.open(&#8220;SCHOOLS.DAT&#8221;,ios::binary);<br \/>\nSCHOOLS S;<br \/>\nwhile(F.read((char*)&amp;S,sizeof(S)))<br \/>\ncout&lt;&lt;S.RNOT()&lt;&lt;endl;\/\/OR S.Display();<br \/>\nF.close(); \/\/IGNORE<br \/>\n}<strong>OR<\/strong>Any other correct function definition<\/li>\n<li><strong>Find the output of the following C++ code considering that the binary file SCHOOLS.DAT exists on the hard disk with the following records of 10 schools of the class SCHOOLS as declared in the previous question (4 b). (1)<\/strong><br \/>\n<table class=\"responsive\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td>SCode<\/td>\n<td>SName<\/td>\n<td>NOT<\/td>\n<\/tr>\n<tr>\n<td>1001<\/td>\n<td>Brains School<\/td>\n<td>100<\/td>\n<\/tr>\n<tr>\n<td>1003<\/td>\n<td>Child Life School<\/td>\n<td>115<\/td>\n<\/tr>\n<tr>\n<td>1002<\/td>\n<td>Care Share School<\/td>\n<td>300<\/td>\n<\/tr>\n<tr>\n<td>1006<\/td>\n<td>Educate for Life School<\/td>\n<td>50<\/td>\n<\/tr>\n<tr>\n<td>1005<\/td>\n<td>Guru Shishya Sadan<\/td>\n<td>195<\/td>\n<\/tr>\n<tr>\n<td>1004<\/td>\n<td>Holy Education School<\/td>\n<td>140<\/td>\n<\/tr>\n<tr>\n<td>1010<\/td>\n<td>Play School<\/td>\n<td>95<\/td>\n<\/tr>\n<tr>\n<td>1008<\/td>\n<td>Innovate Excel School<\/td>\n<td>300<\/td>\n<\/tr>\n<tr>\n<td>1011<\/td>\n<td>Premier Education School<\/td>\n<td>200<\/td>\n<\/tr>\n<tr>\n<td>1012<\/td>\n<td>Uplifted Minds School<\/td>\n<td>100<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong> void main()<br \/>\n{<br \/>\nfstream SFIN;<br \/>\nSFIN.open(&#8220;SCHOOLS.DAT&#8221;,ios::binary | ios::in);<br \/>\nSCHOOLS S;<br \/>\nSFIN.seekg(5*sizeof(S));<br \/>\nSFIN.read((char*)&amp;S, sizeof(S));<br \/>\nS.Display();<br \/>\ncout&lt;&lt;&#8220;Record :&#8221;&lt;&lt;SFIN.tellg()\/sizeof(S) + 1&lt;&lt;endl;<br \/>\nSFIN.close();<br \/>\n}<br \/>\nAns.<\/strong> 1004#Holy Education School#140<br \/>\nRecord :7<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p><strong>SECTION B<br \/>\n[Only for candidates, who opted for Python]<\/strong><\/p>\n<ol>\n<li>\n<ol start=\"1\">\n<li><strong>Differentiate between Syntax Error and Run-Time Error. Also, write a suitable example in Python to illustrate both. (2)<br \/>\nAns.<\/strong> Syntax error : An error of language resulting from code that does not conform to the syntax of the programming language.<br \/>\nExample<br \/>\na = 0<br \/>\nwhile a &lt; 10 # : is missing as per syntax<br \/>\na = a + 1<br \/>\nprint a<br \/>\nRuntime error : A runtime error is an error that causes abnormal termination of program during running time..<br \/>\nExample<br \/>\nA=10<br \/>\nB= int(raw_input(&#8220;Value:&#8221;))<br \/>\nprint A\/B<br \/>\n# If B entered by user is 0, it will be run-time error<\/li>\n<li><strong>Name the Python Library modules which need to be imported to invoke the following functions : (1)<\/strong>\n<ol start=\"1\">\n<li><strong>sin()<\/strong><\/li>\n<li><strong>search()<\/strong><\/li>\n<\/ol>\n<p><strong> Ans.<\/strong><\/p>\n<ol start=\"1\">\n<li>math<\/li>\n<li>re<\/li>\n<\/ol>\n<\/li>\n<li><strong>Rewrite the following code in Python after removing all syntax error(s). Underline each correction done in the code. (2)<br \/>\nVal = int(rawinput(&#8220;Value:&#8221;))<br \/>\nAdder = 0<br \/>\nfor C in range(1,Val,3)<br \/>\nAdder+=C<br \/>\nif C%2=0:<br \/>\nPrint C*10<br \/>\nElse:<br \/>\nprint C*<br \/>\nprint Adder<br \/>\nAns.<\/strong> Val = int(raw_input(&#8220;Value:&#8221;)) # Error 1<br \/>\nAdder = 0<br \/>\nfor C in range(1,Val,3) : # Error 2<br \/>\nAdder+=C<br \/>\nif C%2==0: # Error 3<br \/>\nprint C*10 # Error 4<br \/>\nelse: # Error 5<br \/>\nprint C # Error 6<br \/>\nprint Adder<strong>OR<\/strong>Corrections mentioned as follows:<br \/>\nraw_input in place of rawinput<br \/>\n: to be placed in for<br \/>\n== in place of =<br \/>\nprint in place of Print<br \/>\nelse in place of Else<br \/>\nC* is invalid, replaced by a suitable integer or C<\/li>\n<li><strong>Find and write the output of the following Python code : (2)<br \/>\nData = [&#8220;P&#8221;,20,&#8221;R&#8221;,10,&#8221;S&#8221;,30]\nTimes = 0<br \/>\nAlpha = &#8221; &#8221;<br \/>\nAdd = 0<br \/>\nfor C in range(1,6,2):<br \/>\nTimes = Times + C<br \/>\nAlpha = Alpha + Data[C-1]+&#8221;$&#8221;<br \/>\nAdd = Add + Data[C]\nprint Times, Add, Alpha<br \/>\nAns.<\/strong>1 20 P$<br \/>\n4 30 P$R$<br \/>\n9 60 P$R$S$<\/li>\n<li><strong>Find and write the output of the following Python code : (3)<br \/>\nclass GRAPH:<br \/>\ndef __init__(self,A=50,B=100):<br \/>\nself.P1=A<br \/>\nself.P2=B<br \/>\ndef Up(self,B):<br \/>\nself.P2 = self.P2 &#8211; B<br \/>\ndef Down(self,B):<br \/>\nself.P2 = self.P2 + 2*B<br \/>\ndef Left(self,A):<br \/>\nself.P1 = self.P1 &#8211; A<br \/>\ndef Right(self,A):<br \/>\nself.P1 = self.P1 + 2*A<br \/>\ndef Target(self):<br \/>\nprint &#8220;(&#8220;,self.P1.&#8221;:&#8221;,self.P2,&#8221;)&#8221;<br \/>\nG1=GRAPH(200,150)<br \/>\nG2=GRAPH()<br \/>\nG3=GRAPH(100)<br \/>\nG1.Left(10)<br \/>\nG2.Up(25)<br \/>\nG3.Down(75)<br \/>\nG1.Up(30)<br \/>\nG3.Right(15)<br \/>\nG1.Target()<br \/>\nG2.Target()<br \/>\nG3.Target()<br \/>\nAns.<\/strong> ( 190 : 120 )<br \/>\n( 50 : 75 )<br \/>\n( 130 : 250 )<\/li>\n<li><strong>What possible output(s) are expected to be displayed on screen at the time of execution of the program from the following code ? Also specify the maximum values that can be assigned to each of the variables BEGIN and LAST. (2)<br \/>\nimport random<br \/>\nPOINTS=[20,40,10,30,15];<br \/>\nPOINTS=[30,50,20,40,45];<br \/>\nBEGIN=random.randint(1,3)<br \/>\nLAST=random.randint(2,4)<br \/>\nfor C in range(BEGIN,LAST+1):<br \/>\nprint POINTS[C],&#8221;#&#8221;,<\/strong><\/p>\n<table class=\"responsive\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td>(i) 20#50#30#<\/td>\n<td>(ii) 20#40#45#<\/td>\n<\/tr>\n<tr>\n<td>(iii) 50#20#40#<\/td>\n<td>(iv) 30#50#20<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>Ans.<\/strong> (ii) 20#40#45# and (iii) 50#20#40#<br \/>\nMax value for BEGIN 3<br \/>\nMax value for LAST 4<\/li>\n<\/ol>\n<\/li>\n<li>\n<ol start=\"1\">\n<li><strong>What is the advantage of super( ) function in inheritance ? Illustrate the same with the help of an example in Python. (2)<br \/>\nAns.<\/strong> In Python, super() function is used to call the methods of base class which have been extended in derived class.<br \/>\nclass person(object):<br \/>\ndef __init__(self,name,age):<br \/>\nself.name=name<br \/>\nself.age=age<br \/>\ndef display(self):<br \/>\nprint self,name, self.Age<br \/>\nclass student(person):<br \/>\ndef __init__(self,name,age,rollno,marks):<br \/>\nsuper(student,self)._init_(self, name, age)<br \/>\nself.rollno=rollno<br \/>\nself.marks=marks<br \/>\ndef getRoll(self):<br \/>\nprint self.rollno, self.marks<\/li>\n<li><strong>class Vehicle: #Line 1 (2)<br \/>\nType = &#8216;Car&#8217; #Line 2<br \/>\ndef __init__(self, name): #Line 3<br \/>\nself.Name = name #Line 4<br \/>\ndef Show(self): #Line 5<br \/>\nprint self.Name,Vehicle.Type #Line 6<br \/>\nV1=Vehicle(&#8220;BMW&#8221;) #Line 7<br \/>\nV1.Show() #Line 8<br \/>\nVehicle.Type=&#8221;Bus&#8221; #Line 9<br \/>\nV2=Vehicle(&#8220;VOLVO&#8221;) #Line 10<br \/>\nV2.Show() #Line 11<\/strong><\/p>\n<ol start=\"1\">\n<li><strong>What is the difference between the variable in Line 2 and Line 4 in the above Python code ?<br \/>\nAns.<\/strong> The variable in Line 2 is a class attribute. This belongs to the class itself. These attributes will be shared by all the instances.<br \/>\nThe variable in Line 4 is an instance attribute. Each instance creates a separate copy of these variables.<\/li>\n<li><strong>Write the output of the above Python code.<br \/>\nAns.<\/strong> BMW Car<br \/>\nVOLVO Bus<\/li>\n<\/ol>\n<\/li>\n<li><strong>Define a class CONTAINER in Python with the following specifications : (4)<br \/>\nInstance Attributes<br \/>\n&#8211; Radius,Height # Radius and Height of Container<br \/>\n&#8211; Type # Type of Container<br \/>\n&#8211; Volume # Volume of Container<br \/>\nMethods<br \/>\n&#8211; CalVolume() # To calculate volume<br \/>\n# as per the Type of container<br \/>\n# With the formula as given below :<\/strong><\/p>\n<table class=\"responsive\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<thead>\n<tr>\n<th scope=\"col\" width=\"96\">Type<\/th>\n<th scope=\"col\" width=\"528\">Formula to Calculate Volume<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td width=\"96\">1<\/td>\n<td width=\"528\">3.14 * Radius * Height<\/td>\n<\/tr>\n<tr>\n<td width=\"96\">3<\/td>\n<td width=\"528\">3.14 * Radius * Height\/3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong> &#8211; GetValue() # To allow user to enter values of<br \/>\n# Radius, Height and Type.<br \/>\n# Also, this method should call<br \/>\n# CalVolume() to calculate Volume<br \/>\n&#8211; ShowContainer() # To display Radius, Height, Type<br \/>\n# Volume of the Container<br \/>\nAns.<\/strong> class CONTAINER: # class CONTAINER():\/class CONTAINER(Object):<br \/>\ndef __init__(self): # def __init__(self,R,H,T,V):<br \/>\nself.Radius=0 # self.Radius=R<br \/>\nself.Height=0 # self.Height=H<br \/>\nself.Type =0 # self.Type=T<br \/>\nself.Volume=0 # self.Volume=V<br \/>\ndef CalVolume(self):<br \/>\nif self.Type == 1:<br \/>\nself.Volume = 3.14 * self.Radius * self.Height<br \/>\nelif self.Type ==3:<br \/>\nself.Volume = 3.14 * self.Radius * self.Height \/3<br \/>\ndef GetValue(self):<br \/>\nself.Radius = input(&#8220;Enter Radius&#8221;)<br \/>\nself.Height = input(&#8220;Enter Height&#8221;)<br \/>\nself.Type = input(&#8220;Enter type&#8221;)<br \/>\nself.CalVolume() # OR CalVolume(self)<br \/>\ndef ShowContainer(self):<br \/>\nprint self.Radius<br \/>\nprint self.Height<br \/>\nprint self.Type<br \/>\nprint self.Volume<\/li>\n<li><strong><strong>Answer the questions (i) to (iv) based on the following : (4)<br \/>\nClass Top1(object):<br \/>\ndef __init__(self,tx): #Line 1<br \/>\nself.X=tx #Line 2<br \/>\ndef ChangeX(self,tx):<br \/>\nself.X=self.X+tx<br \/>\ndef ShowX(self):<br \/>\nprint self.X<\/strong><\/strong>Class Top2(object):<br \/>\ndef __init__(self,ty): #Line 3<br \/>\nself.Y=ty #Line 4<br \/>\ndef ChangeY(self,ty):<br \/>\nself.Y=self.Y+ty<br \/>\ndef ShowY(self):<br \/>\nprint self.Y,class Bottom(Top1,Top2):<br \/>\ndef __init__(self,tz): #Line 5<br \/>\nself.Z=tz #Line 6<br \/>\nTop2.__init__(self,2*tz): #Line 7<br \/>\nTop1.__init__(self,3*tz): #Line 8<br \/>\ndef ChangeZ(self,tz):<br \/>\nself.Z=self.Z+tz<br \/>\nself.ChangeY(2*tz)<br \/>\nself.ChangeX(3*tz)<br \/>\ndef ShowZ(self):<br \/>\nprint self.Z,<br \/>\nself.ShowY()<br \/>\nself.ShowX()<br \/>\nB=Bottom(1)<br \/>\nB.ChangeZ(2)<br \/>\nB.ShowZ()<\/p>\n<ol start=\"1\">\n<li><strong>Write the type of the inheritance illustrated in the above.<br \/>\nAns.<\/strong> Multiple Inhertiance<\/li>\n<li><strong>Find and write the output of the above code.<br \/>\nAns.<\/strong> 3 6 9<strong>OR<\/strong>\u201cError\u201d \/ \u201cNo Output\u201d<\/li>\n<li><strong>What are the methods shown in Line 1, Line 3 and Line 5 known as ?<br \/>\nAns.<\/strong> Constructors<\/li>\n<li><strong>What is the difference between the statements shown in Line 6 and Line 7 ?<br \/>\nAns.<\/strong> Initializing the member of child class in Line 6 and calling the parent class constructor in Line 7<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/li>\n<li>\n<ol start=\"1\">\n<li><strong>Consider the following randomly ordered numbers stored in a list : (3)<br \/>\n786, 234, 526, 132, 345, 467<br \/>\nShow the content of the list after the First, Second and Third pass of the bubble sort method used for arranging in ascending order ?<br \/>\nNote : Show the status of all the elements after each pass very clearly underlining the changes.<br \/>\nAns.<\/strong><\/p>\n<table class=\"responsive\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td width=\"89\">I Pass<\/td>\n<td width=\"89\">234<\/td>\n<td width=\"89\">526<\/td>\n<td width=\"89\">132<\/td>\n<td width=\"89\">345<\/td>\n<td width=\"89\">467<\/td>\n<td width=\"89\">786<\/td>\n<\/tr>\n<tr>\n<td width=\"89\">II Pass<\/td>\n<td width=\"89\">234<\/td>\n<td width=\"89\"><u>132<\/u><\/td>\n<td width=\"89\"><u>345<\/u><\/td>\n<td width=\"89\"><u>467<\/u><\/td>\n<td width=\"89\"><u>526<\/u><\/td>\n<td width=\"89\">786<\/td>\n<\/tr>\n<tr>\n<td width=\"89\">III Pass<\/td>\n<td width=\"89\"><u>132<\/u><\/td>\n<td width=\"89\"><u>234<\/u><\/td>\n<td width=\"89\"><u>345<\/u><\/td>\n<td width=\"89\">467<\/td>\n<td width=\"89\">526<\/td>\n<td width=\"89\">786<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li><strong>Write the definition of a method ZeroEnding(SCORES) to add all those values in the list of SCORES, which are ending with zero (0) and display the sum. (3)<br \/>\nFor example :<br \/>\nIf the SCORES contain [200, 456, 300, 100, 234, 678]\nThe sum should be displayed as 600<br \/>\nAns.<\/strong> def ZeroEnding(SCORES):<br \/>\ns=0<br \/>\nfor i in SCORES:<br \/>\nif i%10==0:<br \/>\ns=s+i<br \/>\nprint s<\/li>\n<li><strong>Write AddClient(Client) and DeleteClient(Client) methods in Python to add a new Client and delete a Client from a List of Client Names, considering them to act as insert and delete operations of the queue data structure. (4)<br \/>\nAns.<\/strong> def AddClient(Client):<br \/>\nC=raw_input(&#8220;Client name: &#8220;)<br \/>\nClient.append(C)<br \/>\ndef DeleteClient(Client):<br \/>\nif (Client==[]):<br \/>\nprint &#8220;Queue empty&#8221;<br \/>\nelse:<br \/>\nprint Client[0],&#8221;Deleted&#8221;<br \/>\ndel Client[0] # OR Client.pop(0)<strong>OR<\/strong>class queue:<br \/>\nClient=[]\ndef AddClient(self):<br \/>\na=raw_input(&#8220;Client name: &#8220;)<br \/>\nqueue.Client.append(a)<br \/>\ndef DeleteClient(self):<br \/>\nif (queue.Client==[]):<br \/>\nprint &#8220;Queue empty&#8221;<br \/>\nelse:<br \/>\nprint queue.Client[0],&#8221;Deleted&#8221;<br \/>\ndel queue.Client[0]<\/li>\n<li><strong>Write a definition of a method COUNTNOW(PLACES) to find and display those place names, in which there are more than 5 characters. (2)<br \/>\nFor example :<br \/>\nIf the list PLACES contains<br \/>\n[&#8220;DELHI&#8221;,&#8221;LONDON&#8221;,&#8221;PARIS&#8221;,&#8221;NEW YORK&#8221;,&#8221;DUBAI&#8221;]\nThe following should get displayed :<br \/>\nLONDON<br \/>\nNEW YORK<br \/>\nAns.<\/strong> def COUNTNOW(PLACES):<br \/>\nfor P in PLACES:<br \/>\nif len(P)&gt;5:<br \/>\nprint P<\/li>\n<li><strong>Evaluate the following Postfix notation of expression : (2)<br \/>\n22,11,\/,5,10,*,+,12,-<br \/>\nAns.<\/strong><\/p>\n<table class=\"responsive\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<thead>\n<tr>\n<th scope=\"col\" width=\"312\">Elements<\/th>\n<th scope=\"col\" width=\"312\">Stack Contents<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>22<\/td>\n<td>22<\/td>\n<\/tr>\n<tr>\n<td>11<\/td>\n<td>22, 11<\/td>\n<\/tr>\n<tr>\n<td>\/<\/td>\n<td>2<\/td>\n<\/tr>\n<tr>\n<td>5<\/td>\n<td>2, 5<\/td>\n<\/tr>\n<tr>\n<td>10<\/td>\n<td>2, 5, 10<\/td>\n<\/tr>\n<tr>\n<td>*<\/td>\n<td>2, 50<\/td>\n<\/tr>\n<tr>\n<td>+<\/td>\n<td>52<\/td>\n<\/tr>\n<tr>\n<td>12<\/td>\n<td>52, 12<\/td>\n<\/tr>\n<tr>\n<td>&#8211;<\/td>\n<td>40<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>OR<\/strong><\/p>\n<p>Any other way of stepwise evaluation<\/li>\n<\/ol>\n<\/li>\n<li>\n<ol start=\"1\">\n<li><strong>Write a statement in Python to open a text file STORY.TXT so that new contents can be added at the end of it. (1)<br \/>\nAns.<\/strong>file= open(&#8220;STORY.TXT&#8221;,&#8221;a&#8221;) OR file.open(&#8220;STORY.TXT&#8221;,&#8221;a&#8221;)<\/li>\n<li><strong>Write a method in Python to read lines from a text file INDIA.TXT, to find and display the occurrence of the word \u2018\u2018India\u2019\u2019. (2)<br \/>\nFor example :<br \/>\nIf the content of the file is<\/strong><\/p>\n<hr \/>\n<p><strong>\u2018\u2018India is the fastest growing economy.<br \/>\nIndia is looking for more investments around the globe.<br \/>\nThe whole world is looking at India as a great market.<br \/>\nMost of the Indians can foresee the heights that India is capable of reaching.\u2019\u2019<\/strong><\/p>\n<hr \/>\n<p><strong>The output should be 4.<br \/>\nAns.<\/strong> def display1():<br \/>\nc=0<br \/>\nfile=open(&#8216;INDIA.TXT&#8217;,&#8217;r&#8217;)<br \/>\nc=0<br \/>\nfor LINE in file:<br \/>\nWords = LINE.split()<br \/>\nfor W in Words:<br \/>\nif W==&#8221;India&#8221;:<br \/>\nc=c+1<br \/>\nprint c<br \/>\nfile.close()<\/p>\n<p><strong>OR<\/strong><\/p>\n<p>def display():<br \/>\nc=0<br \/>\nfile=open(&#8216;INDIA.TXT&#8217;,&#8217;r&#8217;)<br \/>\nlines = file.read() # lines = file.readline()<br \/>\nwhile lines:<br \/>\nwords = lines.split()<br \/>\nfor w in words:<br \/>\nif w==&#8221;India&#8221;:<br \/>\nc=c+1<br \/>\nlines = file.read() # lines = file.readline()<br \/>\nprint c<br \/>\nfile.close()<\/li>\n<li><strong>Considering the following definition of class MULTIPLEX, write a method in Python to search and display all the contents in a pickled file CINEMA.DAT, where MTYPE is matching with the value \u2018Comedy\u2019. (3)<br \/>\nclass MULTIPLEX :<br \/>\ndef __init__(self,mno,mname,mtype):<br \/>\nself.MNO = mno<br \/>\nself.MNAME = mname<br \/>\nself.MTYPE = mtype<br \/>\ndef Show(self):<br \/>\nprint self.MNO:&#8221;*&#8221;,self.MNAME,&#8221;$&#8221;,self.MTYPE<br \/>\nAns.<\/strong>def Search():<br \/>\nfile=open(&#8216;CINEMA.DAT&#8217;,&#8217;rb&#8217;)<br \/>\ntry:<br \/>\nwhile True:<br \/>\nM=pickle.load(file)<br \/>\nif M.MTYPE==&#8221;Comedy&#8221;:<br \/>\nM.Show()<br \/>\nexcept EOFError:<br \/>\npass<br \/>\nfile.close()<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p><strong>SECTION C<br \/>\n[For all the candidates]<\/strong><\/p>\n<ol start=\"5\">\n<li>\n<ol start=\"1\">\n<li><strong>Observe the following tables VIDEO and MEMBER carefully and write the name of the RDBMS operation out of (i) SELECTION (ii) PROJECTION (iii) UNION (iv) CARTESIAN PRODUCT, which has been used to produce the output as shown below. Also, find the Degree and Cardinality of the final result. (2)<br \/>\nTABLE : VIDEO<\/strong><\/p>\n<table class=\"responsive\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<thead>\n<tr>\n<th scope=\"col\" width=\"114\">VNO<\/th>\n<th scope=\"col\" width=\"302\">VNAME<\/th>\n<th scope=\"col\">TYPE<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td width=\"114\">F101<\/td>\n<td width=\"302\">The Last Battle<\/td>\n<td>Fiction<\/td>\n<\/tr>\n<tr>\n<td width=\"114\">C101<\/td>\n<td width=\"302\">Angles and Devils<\/td>\n<td>comedy<\/td>\n<\/tr>\n<tr>\n<td width=\"114\">A102<\/td>\n<td width=\"302\">Daredevils<\/td>\n<td>Adventure<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong> TABLE : MEMBER<\/strong><\/p>\n<table class=\"responsive\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<thead>\n<tr>\n<th scope=\"col\" width=\"114\">VNO<\/th>\n<th scope=\"col\" width=\"302\">MNAME<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td width=\"114\">M101<\/td>\n<td width=\"302\">Namish Gupta<\/td>\n<\/tr>\n<tr>\n<td width=\"114\">M102<\/td>\n<td width=\"302\">Sana Sheikh<\/td>\n<\/tr>\n<tr>\n<td width=\"114\">M103<\/td>\n<td width=\"302\">Lara James<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong> TABLE : FINAL RESULT<\/strong><\/p>\n<table class=\"responsive\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<thead>\n<tr>\n<th>VNO<\/th>\n<th scope=\"col\" width=\"196\">VNAME<\/th>\n<th scope=\"col\" width=\"125\">TYPE<\/th>\n<th scope=\"col\" width=\"125\">MNO<\/th>\n<th scope=\"col\" width=\"125\">MNAME<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>F101<\/td>\n<td width=\"196\">The Last Battle Fiction<\/td>\n<td width=\"125\">Fiction<\/td>\n<td width=\"125\">M101<\/td>\n<td width=\"125\">Namish Gupta<\/td>\n<\/tr>\n<tr>\n<td>F101<\/td>\n<td width=\"196\">The Last Battle Fiction<\/td>\n<td width=\"125\">Fiction<\/td>\n<td width=\"125\">M102<\/td>\n<td width=\"125\">Sana Sheikh<\/td>\n<\/tr>\n<tr>\n<td>F101<\/td>\n<td width=\"196\">The Last Battle Fiction<\/td>\n<td width=\"125\">Fiction<\/td>\n<td width=\"125\">M103<\/td>\n<td width=\"125\">Lara James<\/td>\n<\/tr>\n<tr>\n<td>C101<\/td>\n<td width=\"196\">Angels and Devils<\/td>\n<td width=\"125\">Comedy<\/td>\n<td width=\"125\">M101<\/td>\n<td width=\"125\">Namish Gupta<\/td>\n<\/tr>\n<tr>\n<td>C101<\/td>\n<td width=\"196\">Angels and Devils<\/td>\n<td width=\"125\">Comedy<\/td>\n<td width=\"125\">M102<\/td>\n<td width=\"125\">Sana Sheikh<\/td>\n<\/tr>\n<tr>\n<td>C101<\/td>\n<td width=\"196\">Angels and Devils<\/td>\n<td width=\"125\">Comedy<\/td>\n<td width=\"125\">M103<\/td>\n<td width=\"125\">Lara James<\/td>\n<\/tr>\n<tr>\n<td>A102<\/td>\n<td width=\"196\">Daredevils<\/td>\n<td width=\"125\">Adventure<\/td>\n<td width=\"125\">M101<\/td>\n<td width=\"125\">Namish Gupta<\/td>\n<\/tr>\n<tr>\n<td>A102<\/td>\n<td width=\"196\">Daredevils<\/td>\n<td width=\"125\">Adventure<\/td>\n<td width=\"125\">M102<\/td>\n<td width=\"125\">Sana Sheikh<\/td>\n<\/tr>\n<tr>\n<td>A102<\/td>\n<td width=\"196\">Daredevils<\/td>\n<td width=\"125\">Adventure<\/td>\n<td width=\"125\">M103<\/td>\n<td width=\"125\">Lara James<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>Ans. <\/strong>CARTESIAN PRODUCT<br \/>\nOR Option (iv)<br \/>\nDEGREE = 5<br \/>\nCARDINALITY = 9<\/li>\n<li><strong>Write SQL queries for (i) to (iv) and find outputs for SQL queries (v) to (viii), which are based on the tables. (6)<\/strong><br \/>\n<strong>TABLE : ACCOUNT<\/strong><\/p>\n<table class=\"responsive\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<thead>\n<tr>\n<th scope=\"col\" width=\"66\">ANO<\/th>\n<th scope=\"col\" width=\"350\">ANAME<\/th>\n<th>ADDRESS<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td width=\"66\">101<\/td>\n<td width=\"350\">Nirja Singh<\/td>\n<td>Bangalore<\/td>\n<\/tr>\n<tr>\n<td width=\"66\">102<\/td>\n<td width=\"350\">Rohan Gupta<\/td>\n<td>Chennai<\/td>\n<\/tr>\n<tr>\n<td width=\"66\">103<\/td>\n<td width=\"350\">Ali Reza<\/td>\n<td>Hyderabad<\/td>\n<\/tr>\n<tr>\n<td width=\"66\">104<\/td>\n<td width=\"350\">Rishabh jain<\/td>\n<td>Chennai<\/td>\n<\/tr>\n<tr>\n<td width=\"66\">105<\/td>\n<td width=\"350\">Simran Kaur<\/td>\n<td>Chandigarh<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>TABLE : TRANSACT<\/strong><\/p>\n<table class=\"responsive\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<thead>\n<tr>\n<th scope=\"col\" width=\"125\">TRNO<\/th>\n<th scope=\"col\" width=\"125\">ANO<\/th>\n<th scope=\"col\" width=\"125\">AMOUNT<\/th>\n<th scope=\"col\" width=\"125\">TYPE<\/th>\n<th scope=\"col\" width=\"125\">DOT<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td width=\"125\">T001<\/td>\n<td width=\"125\">101<\/td>\n<td width=\"125\">2500<\/td>\n<td width=\"125\">Withdraw<\/td>\n<td width=\"125\">2017-12-21<\/td>\n<\/tr>\n<tr>\n<td width=\"125\">T002<\/td>\n<td width=\"125\">103<\/td>\n<td width=\"125\">3000<\/td>\n<td width=\"125\">Deposit<\/td>\n<td width=\"125\">2017-06-01<\/td>\n<\/tr>\n<tr>\n<td width=\"125\">T003<\/td>\n<td width=\"125\">102<\/td>\n<td width=\"125\">2000<\/td>\n<td width=\"125\">Withdraw<\/td>\n<td width=\"125\">2017-05-12<\/td>\n<\/tr>\n<tr>\n<td width=\"125\">T004<\/td>\n<td width=\"125\">103<\/td>\n<td width=\"125\">1000<\/td>\n<td width=\"125\">Deposit<\/td>\n<td width=\"125\">2017-10-22<\/td>\n<\/tr>\n<tr>\n<td width=\"125\">T005<\/td>\n<td width=\"125\">101<\/td>\n<td width=\"125\">12000<\/td>\n<td width=\"125\">Deposit<\/td>\n<td width=\"125\">2017-11-06<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ol start=\"1\">\n<li><strong>To display details of all transactions of TYPE Deposit from Table TRANSACT.<br \/>\nAns.<\/strong> SELECT * FROM TRANSACT WHERE TYPE = &#8216; Deposit &#8216; ;<\/li>\n<li><strong>To display the ANO and AMOUNT of all Deposits and Withdrawals done in the month of October 2017 from table TRANSACT.<br \/>\nAns.<\/strong> SELECT ANO,AMOUNT FROM TRANSACT<br \/>\nWHERE DOT &gt;= &#8216;2017-10-01 &#8216; AND DOT &lt;= &#8216; 2017-10-31 &#8216; ;<strong>OR<\/strong>SELECT ANO,AMOUNT FROM TRANSACT<br \/>\nWHERE DOT BETWEEN &#8216;2017-10-01&#8217; AND &#8216;2017-10-31 &#8216; ;<\/li>\n<li><strong>To display the last date of transaction (DOT) from the table TRANSACT for the Accounts having ANO as 103.<br \/>\nAns.<\/strong> SELECT MAX(DOT) FROM TRANSACT WHERE ANO = 103;<\/li>\n<li><strong>To display all ANO, ANAME and DOT of those persons from tables ACCOUNT and TRANSACT who have done transactions less than or equal to 3000.<br \/>\nAns.<\/strong> SELECT ACCOUNT.ANO,ANAME,DOT FROM ACCOUNT,TRANSACT WHERE<br \/>\nACCOUNT.ANO=TRANSACT.ANO AND AMOUNT &lt;=3000;<strong>OR<\/strong>SELECT A.ANO,ANAME,DOT FROM ACCOUNT A,TRANSACT T WHERE<br \/>\nA.ANO=T.ANO AND AMOUNT &lt;=3000;<\/li>\n<li><strong>SELECT ANO, ANAME FROM ACCOUNT<br \/>\nWHERE ADDRESS NOT IN (&#8216;CHENNAI&#8217;, &#8216;BANGALORE&#8217;);<br \/>\nAns.<\/strong><\/p>\n<table class=\"responsive\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<thead>\n<tr>\n<th scope=\"col\" width=\"312\">ANO<\/th>\n<th scope=\"col\" width=\"312\">ANAME<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td width=\"312\">103<\/td>\n<td width=\"312\">Ali Reza<\/td>\n<\/tr>\n<tr>\n<td width=\"312\">105<\/td>\n<td width=\"312\">Simran Kaur<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>OR<\/strong><\/p>\n<table class=\" cke_show_border\">\n<tbody>\n<tr>\n<td>ANO<\/td>\n<td>ANAME<\/td>\n<\/tr>\n<tr>\n<td width=\"312\">101<\/td>\n<td width=\"312\">Nirja Singh<\/td>\n<\/tr>\n<tr>\n<td width=\"312\">102<\/td>\n<td width=\"312\">Rohan Gupta<\/td>\n<\/tr>\n<tr>\n<td width=\"312\">103<\/td>\n<td width=\"312\">Ali Reza<\/td>\n<\/tr>\n<tr>\n<td width=\"312\">104<\/td>\n<td width=\"312\">Rishabh Jain<\/td>\n<\/tr>\n<tr>\n<td width=\"312\">105<\/td>\n<td width=\"312\">Simran Kaur<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li><strong>SELECT DISTINCT ANO FROM TRANSACT;<br \/>\nAns.<\/strong>DISTINCT ANO<br \/>\n101<br \/>\n102<br \/>\n103<\/li>\n<li><strong>SELECT ANO, COUNT(*), MIN(AMOUNT) FROM TRANSACT<br \/>\nGROUP BY ANO HAVING COUNT(*)&gt; 1;<br \/>\nAns.<\/strong><\/p>\n<table class=\" class=\">\n<tbody>\n<tr>\n<td width=\"208\">ANO<\/td>\n<td width=\"208\">COUNT (*)<\/td>\n<td width=\"208\">MIN (ANOUNT)<\/td>\n<\/tr>\n<tr>\n<td width=\"208\">101<\/td>\n<td width=\"208\">2<\/td>\n<td width=\"208\">2500<\/td>\n<\/tr>\n<tr>\n<td width=\"208\">103<\/td>\n<td width=\"208\">2<\/td>\n<td width=\"208\">1000<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li><strong>SELECT COUNT(*), SUM(AMOUNT) FROM TRANSACT<br \/>\nWHERE DOT &lt;= &#8216;2017-06-01&#8217;;<br \/>\nAns.<\/strong><\/p>\n<table class=\"responsive\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<thead>\n<tr>\n<th scope=\"col\" width=\"312\">COUNT (*)<\/th>\n<th scope=\"col\" width=\"312\">SUM (AMOUNT)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td width=\"312\">2<\/td>\n<td width=\"312\">5000<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p style=\"text-align: center;\"><strong>These are first 5 questions only.\u00a0<\/strong><b>To view and download complete question paper with solution<\/b><\/p>\n<p style=\"text-align: center;\"><b><strong><a class=\"button\" href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide&amp;referrer=utm_source%3Dmycbse_bottom%26utm_medium%3Dtext%26utm_campaign%3Dmycbseads\">Download myCBSEguide App<\/a><\/strong><\/b><\/p>\n<p style=\"text-align: center;\"><b>the best CBSE App from google playstore\u00a0now or login to our Student Dashboard.<\/b><\/p>\n<h2><span class=\"ez-toc-section\" id=\"Last_Year_Question_Paper_Class_12_Computer_Science_2018\"><\/span>Last Year Question Paper Class 12 Computer Science 2018<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Download class 12 Computer Science question paper with solution from best CBSE App the myCBSEguide. CBSE class 12 Computer Science question paper 2018 in PDF format with solution will help you to understand the latest question paper pattern and marking scheme of the CBSE board examination. You will get to know the difficulty level of the question paper.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"Previous_Year_Question_Paper_for_class_12_in_PDF\"><\/span>Previous Year Question Paper for class 12 in PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>CBSE question papers 2018, 2017, 2016, 2015, 2014, 2013, 2012, 2011, 2010, 2009, 2008, 2007, 2006, 2005 and so on for all the subjects are available under this download link. Practicing real question paper certainly helps students to get confidence and improve performance in weak areas.<\/p>\n<ul>\n<li><a href=\"http:\/\/mycbseguide.com\/downloads\/cbse-class-12-physics\/1251\/cbse-last-year-papers\/3\/\">Physics<\/a><\/li>\n<li><a href=\"http:\/\/mycbseguide.com\/downloads\/cbse-class-12-chemistry\/1267\/cbse-last-year-papers\/3\/\">Chemistry<\/a><\/li>\n<li><a href=\"http:\/\/mycbseguide.com\/downloads\/cbse-class-12-mathematics\/1284\/cbse-last-year-papers\/3\/\">Mathematics<\/a><\/li>\n<li><a href=\"http:\/\/mycbseguide.com\/downloads\/cbse-class-12-biology\/1298\/cbse-last-year-papers\/3\/\">Biology<\/a><\/li>\n<li><a href=\"http:\/\/mycbseguide.com\/downloads\/cbse-class-12-accountancy\/1315\/cbse-last-year-papers\/3\/\">Accountancy<\/a><\/li>\n<li><a href=\"http:\/\/mycbseguide.com\/downloads\/cbse-class-12-business-studies\/1727\/cbse-last-year-papers\/3\/\">Business Studies<\/a><\/li>\n<li><a href=\"http:\/\/mycbseguide.com\/downloads\/cbse-class-12-economics\/1327\/cbse-last-year-papers\/3\/\">Economics<\/a><\/li>\n<li><a href=\"http:\/\/mycbseguide.com\/downloads\/cbse-class-12-history\/1869\/cbse-last-year-papers\/3\/\">History<\/a><\/li>\n<li><a href=\"http:\/\/mycbseguide.com\/downloads\/cbse-class-12-geography\/1863\/cbse-last-year-papers\/3\/\">Geography<\/a><\/li>\n<li><a href=\"http:\/\/mycbseguide.com\/downloads\/cbse-class-12-political-science\/1879\/cbse-last-year-papers\/3\/\">Political Science<\/a><\/li>\n<li><a href=\"http:\/\/mycbseguide.com\/downloads\/cbse-class-12-physical-education\/1877\/cbse-last-year-papers\/3\/\">Physical Education<\/a><\/li>\n<li><a href=\"http:\/\/mycbseguide.com\/downloads\/cbse-class-12-computer-science\/1851\/cbse-last-year-papers\/3\/\">Computer Science<\/a><\/li>\n<li><a href=\"http:\/\/mycbseguide.com\/downloads\/cbse-class-12-informatics-practices\/1873\/cbse-last-year-papers\/3\/\">Informatics Practices<\/a><\/li>\n<li><a href=\"http:\/\/mycbseguide.com\/downloads\/cbse-class-12-english-core\/1855\/cbse-last-year-papers\/3\/\">English Core<\/a><\/li>\n<li><a href=\"http:\/\/mycbseguide.com\/downloads\/cbse-class-12-hindi-core\/1865\/cbse-last-year-papers\/3\/\">Hindi Core<\/a><\/li>\n<li><a href=\"http:\/\/mycbseguide.com\/downloads\/cbse-class-12-hindi-elective\/1867\/cbse-last-year-papers\/3\/\">Hindi Elective<\/a><\/li>\n<li><a href=\"http:\/\/mycbseguide.com\/downloads\/cbse-class-12\/1250\/\">Other Subjects<\/a><\/li>\n<\/ul>\n<p>To download CBSE Question Paper class 12 Accountancy, Chemistry, Physics, History, Political Science, Economics, Geography, Computer Science, Home Science, Accountancy, Business Studies and Home Science; do check myCBSEguide app or website. myCBSEguide provides sample papers with solution, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE important question papers. Sample Paper all are made available through\u00a0<a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide&amp;referrer=utm_source%3Dmycbse_bottom%26utm_medium%3Dtext%26utm_campaign%3Dmycbseads\"><strong>the best app for CBSE students<\/strong><\/a>\u00a0and myCBSEguide website.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Previous Year Question Paper \u2013 Download in PDF CBSE Question Paper 2018 class 12 Computer Science\u00a0conducted by Central Board of Secondary Education, New Delhi in the month of March 2018. CBSE previous year question papers with solution are available in myCBSEguide mobile app and cbse guide website. The Best CBSE App for students and teachers &#8230; <a title=\"CBSE Question Paper 2018 class 12 Computer Science\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-question-paper-2018-class-12-computer-science\/\" aria-label=\"More on CBSE Question Paper 2018 class 12 Computer Science\">Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":-1,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1346,1014,1443],"tags":[1528,1483,326,80,1566,1631],"class_list":["post-17718","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-cbse","category-cbse-question-papers","category-computer-science-cbse-class-12","tag-cbse-question-paper-2018","tag-class12","tag-computer-science","tag-last-year-papers","tag-previous-question-paper","tag-ten-year-questions-papers"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.0 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>CBSE Question Paper 2018 class 12 Computer Science<\/title>\n<meta name=\"description\" content=\"CBSE Question Paper 2018 class 12 Computer Science in PDF format for free download. 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