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ISC Class XII Board Exam 2026 : Computer Science

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COMPUTER SCIENCE PAPER 1 (THEORY) Maximum Marks: 70 Time Allotted: Three Hours Reading Time: Additional Fifteen minutes Instructions to Candidates You are allowed additional 15 minutes for only reading the question paper. You must NOT start writing during the reading time. This question paper has 12 printed pages. It is divided into two parts and has 11 questions in all. Part I is compulsory and has two questions. Answer all questions. Part II is divided into three sections: A, B and C. While attempting Multiple Choice Questions in Part I, you are required to write only ONE option as the answer. Each section in Part II has three questions. Any two questions have to be attempted from each section. The intended marks for questions are given in brackets [ ]. Instruction to Supervising Examiner Kindly read aloud the Instructions given above to all the candidates present in the examination hall. --------------------------------------------------------------------------------------------------------------------ISC SPECIMEN QUESTION PAPER 2025 1 PART I 20 MARKS Answer all questions. While answering questions in this Part, indicate briefly your working and reasoning, wherever required. Question 1 (i) (ii) The compliment of the Boolean expression A (B C + B C) (Application) [1] Given below are two statements marked Assertion and Reason. Read the two statements carefully and choose the correct option. Assertion: Recursion utilises more memory as compared to iteration. [1] (a) A (B+C+B +C) (b) A+ (B+C ) (B+C ) (c) A+(B +C) (B+C ) (d) A (B +C +B C) Reason: Time complexity of recursion is higher due to the overhead of maintaining the function call stack. (Analysis) (a) Both Assertion and Reason are true and Reason is the correct explanation for Assertion. (iii) (iv) (v) (b) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion. (c) Assertion is true and Reason is false. (d) Assertion is false and Reason is true. According to the Principle of duality, the Boolean equation (A + B) (1 + B) = A + B will be equivalent to: (a) (A + B ) (0 + B) = A + B (b) (A B) + (0 B) = A B (c) (A B) + (0 B) = A + B (d) (A + B) (0 + B) = A + B Distributive law states that: [1] (Application) (Recall) [1] (d) A + B C = A B + A C The complement of the reduced expression of F(A,B) = (0,1,2,3) is: (Application) (a) 1 [1] (a) A + B C = (A + B) (A +C) (b) A + ( A B) = A (c) A (B + C) = (A B) + (B C) (b) A B (c) 0 (d) A + B ---------------------------------------------------------------------------------------------------------------------------------- ISC SPECIMEN QUESTION PAPER 2025 2 (vi) Study the given propositions and the statements marked Assertion and Reason that follow it. Choose the correct option on the basis of your analysis. p = I am a triangle q = I am a three-sided polygon s1 = p q s2 = q p Assertion: s2 is converse of s1 Reason: Three-sided polygon must be a triangle. (Analysis) (a) Both Assertion and Reason are true and Reason is the correct explanation for Assertion. (b) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion. (c) Assertion is true and Reason is false. (d) Assertion is false and Reason is true. (vii) Given below are two statements marked Assertion and Reason. Read the two statements carefully and choose the correct option. Assertion: In Java, the String class is used to create and manipulate strings, and it is immutable. Reason : Immutability ensures that once a String object is created, its value cannot be changed. (Analysis) (a) Both Assertion and Reason are true and Reason is the correct explanation for Assertion. (b) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion. (c) Assertion is true and Reason is false. (d) Assertion is false and Reason is true. (viii) Consider the following statement written in class Circle where pi is its data member. static final double pi = 3.142; Which of the following statements are valid for pi? I. It contains a common value for all objects class Circle. II. Its value is non-changeable. III. At a time two access modifiers, static and final, cannot be applied to a single data member pi. (Application) (a) I and II (b) II and III (c) I and III (d) Only III [1] [1] [1] (ix) For Big O notation, state the difference between O(n) and O(n2). (Analysis) [1] (x) A full adder needs five gates and those are 3 AND gates, 1 OR gate and 1 XOR gate. When a full adder is constructed using 2 half adders, it also requires 5 gates. State the names along with the quantity those gates. (Analysis) [1] -------------------------------------------------------------------------------------------------------------------------------- ISC SPECIMEN QUESTION PAPER 2025 3 Question 2 (i) Convert the following infix notation to prefix form. (Create) [2] (ii) A matrix M[-6 .10, 4 15] is stored in the memory with each element requiring 4 bytes of storage. If the base address is 1025, find the address of M[4][8] when the matrix is stored in column major wise. (Application) [2] (iii) The following function getIt() is a part of some class. Assume x is a positive integer, f is the lower bound of arr[ ] and l is the upper bound of the arr[ ]. (A B)/C*(D+E) Answer the questions given below along with dry run/working. public int getIt(int x,intarr[],int f,int l) { if(f>l) return -1; int m=(f+l)/2; if(arr[m]<x) return getIt(x,m+1,l); else if(arr[m]>x) return getIt(x,f,m-1); else return m; } (a) What will the function getIt( ) return if arr[ ] = {10,20,30,40,50} and x=40? (Analysis) (b) What is function getIt( ) performing apart from recursion? (Analysis) (iv) [2] [1] The following is a function of class Armstrong. This recursive function calculates and returns the sum of the cubes of all the digits of num, where num is an integer data member of the class Armstrong. [A number is said to be Armstrong if the sum of the cubes of all its digits is equal to the original number]. There are some places in the code marked by ?1?, ?2?,?3? which may be replaced by a statement/expression so, that the function works properly. public int sumOfPowers(int num) { if (num == 0) return ?1?; int digit = ?2?; return (int) Math.pow(digit, 3) + ?3?; } (a) What is the expression or statement at ?1? (Analysis) [1] (b) What is the expression or statement at ?2? (Analysis) [1] (c) What is the expression or statement at ?3? (Analysis) [1] ---------------------------------------------------------------------------------------------------------------------------------- ISC SPECIMEN QUESTION PAPER 2025 4 PART II 50 MARKS Answer six questions in this part, choosing two questions from Section A, two from Section B and two from Section C. SECTION - A Answer any two questions. Question 3 (i) A shopping mall announces a special discount on all its products as a festival offer only to those who satisfy any one of the following conditions. If he/she is an employee of the mall and has a service of more than 10 years. OR A regular customer of the mall whose age is less than 65 years and should not be an employee of the mall. OR If he/she is a senior citizen but not a regular customer of the mall. The inputs are : INPUTS E Employee of the mall [5] R Regular customer of the mall S Service of the employee is more than 10 years C Senior citizen of 65 years or above (In all the above cases, 1 indicates yes and 0 indicates no.) Output: X - Denotes eligible for discount [1 indicates YES and 0 indicates NO in all cases] Draw the truth table for the inputs and outputs given above and write the SOP expression for X ( E, R, S, C ). (Application) (ii) Reduce the above expression X ( E, R, S, C ) by using 4-variable Karnaugh map, showing the various groups (i.e. octal, quads and pairs). Draw the logic gate diagram for the reduced expression. Assume that the variables and their complements are available as inputs. (Create) [5] Question 4 (i) (a) Reduce the Boolean function F(P,Q,R,S) = (P+Q+R+S) (P+Q+R+S ) (P+Q+R +S) (P+Q +R+S) (P+Q +R+S ) (P+Q +R +S) (P+Q +R +S ) (P +Q+R+S) (P +Q+R+S ) by using 4-variable Karnaugh map, showing the various groups (i.e. octal, quads and pairs). (Application) [4] (b) Draw the logic gate diagram for the reduced expression. Assume that the variables and their complements are available as inputs. (Create) [1] -------------------------------------------------------------------------------------------------------------------------------- ISC SPECIMEN QUESTION PAPER 2025 5 (ii) From the given logic diagram : (a) Derive Boolean expression and draw the truth table for the derived expression. (Application) [4] (b) If A=1, B=0 and C=1 then find the value of X. (Application) [1] Draw the logic circuit to decode the following binary number (0001, 0101, 0111, 1000, 1010, 1100, 1110,1111) to its hexadecimal equivalents. Also state the Hexadecimal equivalents of the given binary numbers. (Application & Analyse) Verify if the following proposition is valid using the truth table: (X Y) =>Z = (Y =>Z) (X =>Y) (Application) [5] Question 5 (i) (ii) [3] (iii) Answer the following questions related to the below image: (a) What is the output of the above gate if input A=0, B=1? (b) What are the values of the inputs if output =1? (Analysis) [1] (Application) [1] ---------------------------------------------------------------------------------------------------------------------------------- ISC SPECIMEN QUESTION PAPER 2025 6 SECTION B Answer any two questions. Each program should be written in such a way that it clearly depicts the logic of the problem. This can be achieved by using mnemonic names and comments in the program. (Flowcharts and Algorithms are not required.) The programs must be written in Java. Question 6 Given are two strings, input string and a mask string that remove all the characters of the mask string from the original string. Example: INPUT: ORIGINALSTRING: communication MASK STRING: mont OUTPUT: cuicai [10] A class StringOp is defined as follows to perform above operation. Some of the members of the class are given below: Class name : StringOp : to store the original string msk : to store the mask string nstr : to store the resultant string StringOp() : default constructor to initialize the data member with legal initial value void accept( ) : void form() : to accept the original string str and the mask string msk in lower case to form the new string nstr after removal of characters present in mask from the original string void display( ) : Data members/instance variables: nstr Methods / Member functions: to display the original string and the newly formed string nstr Specify the class StringOp giving details of the constructor( ), void accept( ), void form() and void display( ). Define a main( ) function to create an object and call all the functions accordingly to enable the task. (Create) -------------------------------------------------------------------------------------------------------------------------------- ISC SPECIMEN QUESTION PAPER 2025 7 Question 7 [10] A class Mixarray contains an array of integer elements along with its capacity (More than or equal to 3). Using the following description, form a new array of integers which will contain only the first 3 elements of the two different arrays one after another. Example: Array1: { 78, 90, 100, 45, 67 } Array2: {10, 67, 200, 90 } Resultant Array: { 78, 90, 100, 10, 67, 200} The details of the members of the class are given below: Class name : Mixarray Data members/instance variables: arr[] : integer array cap : integer to store the capacity of the array Member functions/methods: Mixarray (int mm ) : to initialize the capacity of the array cap=mm void input( ) : to accept the elements of the array Mixarray mix(Mixarray P, Mixarray Q) : returns the resultant array having the first 3 elements of the array of objects P and Q void display( ) : to display the array with an appropriate message. Specify the class Mixarraygiving details of the constructor(int), void input( ), Mixarray mix(Mixarray,Mixarray) and void display( ). Define a main( ) function to create objects and call all the functions accordingly to enable the task. (Create) ---------------------------------------------------------------------------------------------------------------------------------- ISC SPECIMEN QUESTION PAPER 2025 8 Question 8 [10] A class LCM has been defined to find the Lowest Common Multiple of two integers. Some of the data members and member functions are given below: Class name : LCM Data members/instance variables: n1 : to store an integer number n2 : to store an integer number large : integer to store the largest from n1,n2 sm : integer to store the smallest from n1,n2 l : to store lcm of two numbers LCM( ) : default constructor to initialize data members with legal initial values void accept( ) : to accept n1 and n2 int getLCM( ) : returns the lcm of n1 and n2 using the recursive technique void display( ) : to print the numbers n1, n2 and lcm Methods / Member functions: Specify the class LCM giving details of the constructor( ), void accept( ), int getLCM() and void display( ). Define a main ( ) function to create an object and call the member functions accordingly to enable the task. (Create) -------------------------------------------------------------------------------------------------------------------------------- ISC SPECIMEN QUESTION PAPER 2025 9 SECTION C Answer any two questions. Each program should be written in such a way that it clearly depicts the logic of the problem stepwise. This can be achieved by using comments in the program and mnemonic names or pseudo codes for algorithms. The programs must be written in Java and the algorithms must be written in general / standard form, wherever required / specified. (Flowcharts are not required.) Question 9 Recycle is an entity which can hold at the most 100 integers. The chain enables the user to add and remove integers from both the ends i.e. front and rear. Define a class ReCycle with the following details: Class name : ReCycle Data members/instance variables: ele[ ] : the array to hold the integer elements cap : stores the maximum capacity of the array front : to point the index of the front rear : to point the index of the rear Methods / Member functions: (i) ReCycle (int max) : constructor to initialize the data cap = max, front = rear = 0 and to create the integer array. void pushfront(int v) : to add integers from the front index if possible else display the message( full from front ). int popfront( ) : to remove the return elements from front. If array is empty then return-999. void pushrear(int v) : to add integers from the front index if possible else display the message( full from rear ). int poprear( ) : to remove and return elements from rear. If the array is empty then return-999. Specify the class ReCycle giving details of the functions void pushfront(int) and int poprear( ). Assume that the other functions have been defined. The main( ) function and algorithm need NOT be written. (ii) Name the entity described above and state its principle. [4] (Create) (Understanding) [1] ---------------------------------------------------------------------------------------------------------------------------------- ISC SPECIMEN QUESTION PAPER 2025 10 Question 10 [5] A library issues books on rental basis at a 2% charge on the cost price of the book per day. As per the rules of the library, a book can be retained for 7 days without any fine. If the book is returned after 7 days, a fine will also be charged for the excess days as per the chart given below: Number of excess days Fine per day (Rs.) 1 to 5 2.00 6 to 10 3.00 Above 10 5.00 A super class Library has been defined. Define a sub class Compute to calculate the fine and the total amount. The details of the members of both the classes are given below: Class name : Library name : to store the name of the book author : to store the author of the book p : to store the price of the book (in decimals) Library( ... ) : void show( ) : parameterized constructor to assign values to the data members displays the book details Data members/instance variables: Methods / Member functions: Class name Compute Data members/instance variables: d : number of days taken in returning the book f : to store the fine (in decimals) Methods / Member functions: Compute( ... ) : parameterized constructor to assign values to the data members of both the classes void fine( ) : calculates the fine for the excess days as given in the table above void show( ) : displays the book details along with the number of days, fine and the total amount to be paid. Total amount is (2% of price of book * total no of days) + fine Assume that the super class Library has been defined. Using the concepts of Inheritance, specify the class Compute giving the details of constructor, void fine ( ) and void show ( ) functions. The super class, main function and algorithm need NOT be written. (Create) -------------------------------------------------------------------------------------------------------------------------------- ISC SPECIMEN QUESTION PAPER 2025 11 Question 11 (i) (ii) A linked list is formed from the objects of the class Node. The class structure of the Node is given below: class Node { int n; Node link; } Write an Algorithm OR a Method to search for a number from an existing linked list. [2] [2] The method declaration is as follows: void FindNode( Node str, int b ) (Analysis) Answer the following questions from the diagram of a Binary Tree given below: A B D F C G E H (a) Name the root of the left sub tree and its siblings. (Understanding) [1] (b) State the size and depth of the right sub tree. (Understanding) [1] (c) Write the in-order traversal of the above tree structure. (Understanding) [1] ---------------------------------------------------------------------------------------------------------------------------------- ISC SPECIMEN QUESTION PAPER 2025 12 COMPUTER SCIENCE PAPER 1 (THEORY) ANSWER KEY PART I 20 MARKS Question 1 (i) (c) A + (B + C) (B + C ) [1] (ii) [1] (iii) (b) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion. (b) (A B) + (0 B) = A B (iv) (a) A + B C = (A + B) (A +C) [1] (v) (c) 0 [1] (b) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion. (vii) (a) Both Assertion and Reason are true and Reason is the correct explanation for Assertion. (viii) (a) I and II (vi) (ix) O(n) Single loop O(n2) Nested loop (x) 2 AND gates, 1 OR gate and 2 XOR gates [1] [1] [1] [1] [1] [1] Question 2 (i) */-ABC+DE [2] (ii) Address of M[4][8] = B + W * ((J LC) * M + (I LR)) = 1025 + 4 ((8-4) 17+4 - - 6) = 1025 + 4(68 +10) = 1025+4 *78 = 1025 + 312 = 1337 (a) 3 [2] (b) Function getIt( ) is performing binary search on arr [ ] [1] (a) 0 [1] (b) num%10 [1] (c) sumOfPowers(num/10) [1] (iii) (iv) [2] --------------------------------------------------------------------------------------------------------------------ISC SPECIMEN ANSWER KEY 2025 1 PART II 50 MARKS SECTION A Question 3 (i) [5] E Employee of the mall 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 R Regular customer of the mall 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 S Service of the employee is more than 10 years 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 C Senior citizen of 65 years or above 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 X OUTPUT 0 1 0 1 1 0 1 0 0 1 1 1 0 0 1 1 X (E,R,S,C) = (1,3,4,6,9,10,11,14,15) = E R S C + E R SC + E RS C + E RSC + ER S C+ ER SC + ER SC + ERSC + ERSC ---------------------------------------------------------------------------------------------------------------------------------- ISC SPECIMEN ANSWER KEY 2025 2 (ii) [5] S C 0 E R 4 3 1 12 ER 7 0 6 15 0 1 14 1 9 11 0 0 0 13 0 8 2 1 5 1 S C SC 1 0 E R E R S C 1 1 10 1 1 There are two quads and a pair: Quad 1 ( m1 + m3 + m9 + m11) = R C Quad 2 ( m10 + m11 + m14 + m15 ) = E S Pair ( m4 + m6 ) = E R C Hence F(E,R,S,C) = R C + E S + E R C -------------------------------------------------------------------------------------------------------------------------------- ISC SPECIMEN ANSWER KEY 2025 3 Question 4 (i) (a) [4] R+S R+S 0 P+Q 1 0 0 15 1 14 1 9 0 6 0 13 8 P +Q 0 7 0 1 2 1 5 12 P +Q R +S 3 0 4 0 P+Q R +S 1 11 0 10 1 1 There are three quads : Quad 1: ( M0 M1 M8 M9) = Q + R Quad 2: ( M4 M5 M6 M7) = P + Q Quad 3: ( M0 M2 M4 M6) = P + S Hence F(P,Q,R,S) = ( Q + R ) .( P + Q ) . ( P + S ) (b) (ii) (a) (b) [1] [(A.B) + B .C] [4] A B C B B .C (A.B) 0 0 0 1 0 1 1 0 0 0 1 1 1 1 1 0 0 1 0 0 0 1 1 0 0 1 1 0 0 1 1 0 1 0 0 1 0 1 1 0 1 0 1 1 1 1 1 0 1 1 0 0 0 0 0 1 1 1 1 0 0 0 0 1 0 (A.B) + B .C [ (A.B) + B .C ] [1] ---------------------------------------------------------------------------------------------------------------------------------- ISC SPECIMEN ANSWER KEY 2025 4 Question 5 (i) [5] (0001) (0101) (0111) (1000) (1010) (1100) (1110) (1111) =1 =5 =7 =8 =A =C =E =F -------------------------------------------------------------------------------------------------------------------------------- ISC SPECIMEN ANSWER KEY 2025 5 (ii) [3] X Y Z X Y (X Y)=>Z Y=>Z 0 0 0 0 1 1 1 1 0 0 1 0 1 1 1 1 0 1 0 0 1 0 1 0 0 1 1 0 1 1 1 1 1 0 0 0 1 1 0 0 1 0 1 0 1 1 0 0 1 1 0 1 0 0 1 0 1 1 1 1 1 1 1 1 X=>Y (Y=>Z ) (X=>Y) INVALID (iii) (a) 0 [1] (b) A=0, B=0 [1] ---------------------------------------------------------------------------------------------------------------------------------- ISC SPECIMEN ANSWER KEY 2025 6 SECTION B Question 6 [10] import java.util.*; class StringOp { String str; String nstr; String msk; Scanner sc=new Scanner(System.in); StringOp() { str=""; nstr=""; msk=""; } void accept()//to accept the original and mask string { System.out.println("Enter the original word"); str=sc.next()+sc.nextLine(); System.out.println("enter the mask string"); msk=sc.next(); } void form() //formation of a new string according to the requirement { int l1=str.length();int l2=msk.length(); for (int i=0;i<l1;i++) { int fr=0;char c1=str.charAt(i); if (msk.indexOf(c1)==-1) nstr=nstr+c1; } } void display() //display original and newly formed string { System.out.println("original string: "+str); System.out.println("changed string: "+nstr); } public static void main() { StringOp ob=new StringOp(); ob.accept(); ob.form(); ob.display(); } } -------------------------------------------------------------------------------------------------------------------------------- ISC SPECIMEN ANSWER KEY 2025 7 Question 7 [10] import java.util.*; class Mixarray { int arr[]; int cap; static Scanner sc=new Scanner(System.in); Mixarray(int mm) { cap=mm; arr=new int[cap]; } void input() { System.out.println("Enter the content of the array"); for (int i=0;i<cap;i++) arr[i]=sc.nextInt(); } void display() { for (int i=0;i<cap;i++) System.out.print(arr[i]+" "); System.out.println(); } Mixarray mix(Mixarray P,Mixarray Q) { Mixarray res=new Mixarray(6); int k=0; for (int i=0;i<3;i++) res.arr[k++]=P.arr[i]; for (int i=0;i<3;i++) res.arr[k++]=Q.arr[i]; return res; } public static void main() { System.out.println("enter the capacity of both the array"); int c1=sc.nextInt(); int c2=sc.nextInt(); Mixarray ob1=new Mixarray(c1); Mixarray ob2=new Mixarray(c2); System.out.println("Enter the content of Ist array"); ob1.input(); System.out.println("Enter the content of 2nd array"); ob2.input(); Mixarray r=new Mixarray(c1+c2); Mixarray res=r.mix(ob1,ob2); System.out.println("content of the combined array"); res.display(); } } ---------------------------------------------------------------------------------------------------------------------------------- ISC SPECIMEN ANSWER KEY 2025 8 Question 8 [10] import java.util.*; class LCM { int n1,n2; int large,sm; int l; static Scanner sc=new Scanner(System.in); void accept() { System.out.println("enter 2 different integers:"); n1=sc.nextInt(); n2=sc.nextInt(); if (n1>n2) { large=n1; sm=n2; } elseif (n2>n1) { large=n2; sm=n1; } } int getLCM() { if(large!=sm) { if (large>sm) large=large-sm; else if (large<sm) sm=sm-large; return getLCM(); } else return (n1*n2)/large; } void display() { l=getLCM(); System.out.println("LCM of "+n1 +"and "+n2+"="+l); } public static void main() { LCM ob=new LCM(); ob.accept(); ob.display(); } } -------------------------------------------------------------------------------------------------------------------------------- ISC SPECIMEN ANSWER KEY 2025 9 SECTION C Question 9 (i) [4] (ii) [1] class ReCycle { void pushfront(int v) { if(front !=0) q[front--]=v; else System.out.println("FULL FROM FRONT"); } int poprear() { if(front !=rear) return(q[rear--]); else return -999; } } Entity is dequeue and works on the principle of FIFO Question 10 [5] class Compute extends Library { private int d; private double f; public Compute(String name, String author, double p, int d) { super(name, author, p); this.d=d; f=0.0; } public void fine( ) { int d1=d-7; if(d1>=1 && d1<=5) f = d1*2; else if(d1>=6 && d1<=10) f = d1*3; else f = d1*5; } public void show( ) { super.show( ); System.out.println("Fine =" +f); System.out.println("Total amount ="+ ((0.02*p*d)+f)); } } ---------------------------------------------------------------------------------------------------------------------------------- ISC SPECIMEN ANSWER KEY 2025 10 Question 11 (i) ALGORITHM: Step 1. Start Step 2. Set temporary pointer to the first node Step 3. Repeat steps 4 and 5 until the pointer reaches null. Display number not found Step 4. check for number, if found display, exit Step 5. Move pointer to the next node Step 6. End algorithm [2] METHOD: void FindNode(Node str, int b) { Node temp=str; while(temp.link!=null) { if (temp.n == b) {System.out.prinln(b+ is found ); break; } temp=temp.link; } if (temp.link= =null) System.out.prinln(b+ is not found ); } (ii) (a) Root: B Sibling: F [1] (b) Size: 4 Depth: 2 [1] (c) EDBACFGH [1] -------------------------------------------------------------------------------------------------------------------------------- ISC SPECIMEN ANSWER KEY 2025 11

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