{"id":47367,"date":"2019-08-28T14:54:51","date_gmt":"2019-08-28T09:24:51","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=47367"},"modified":"2020-12-14T11:05:14","modified_gmt":"2020-12-14T05:35:14","slug":"isc-class-12-computer-science-previous-year-question-papers-solved-2013","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/isc-class-12-computer-science-previous-year-question-papers-solved-2013\/","title":{"rendered":"ISC Computer Science Question Paper 2013 Solved for Class 12"},"content":{"rendered":"

ISC Computer Science Previous Year Question Paper 2013 Solved for Class 12<\/h2>\n

Maximum Marks: 70
\nTime allowed: 3 hours<\/p>\n

Part – I<\/strong>
\nAnswer all questions<\/strong><\/p>\n

While answering questions in this Part, indicate briefly your working and reasoning, wherever required.<\/p>\n

Question 1.
\n(a) State the Principle of Duality. Write the dual of: [2]
\n(P + Q’).R.1 = P.R + Q’.R
\n(b) Minimize the expression using Boolean laws: [2]
\nF = (A + B’)(B + CD)’
\n(c) Convert the following cardinal form of expression into its canonical form: [2]
\nF (P, Q, R) = \u03c0 (1, 3)
\n(d) Using a truth table verify: [2]
\n(~p => q) \u2227 p = (p \u2227 ~q) \u2228 (p \u2227 q)
\n(e) If A = 1 and B = 0, then find: [2]
\n(i) (A’ + 1).B
\n(ii) (A + B7
\nAnswer:
\n(a) To every Boolean equation there exists another equation which is dual to the previous equation. This is done by changing AND\u2019s to OR\u2019s and vice-versa, 0\u2019s to Fs and vice-versa, complements remain unchanged.
\nDual: (P.Q’) + R + 0 = (P + R). (Q’+ R)<\/p>\n

(b) F = (A + B’).(B + CD)’
\nF = (A + B’). (B’. (CD)’)
\nF = AB’+B’B’.(C’+D’)
\nF = B’.(C’+D’)<\/p>\n

(c) F(P, Q, R) = \u03c0(1, 3)
\n= 001, 011
\n= (P + Q + R’).(P + Q’ + R’)<\/p>\n

(d) (~p => q) \u2227 p = (p \u2227 ~q) \u2228 (p \u2227 q)
\n\"ISC<\/p>\n

(e) (i) (A’ + 1).B = (0 + 1). 0 = 0
\n(ii) (A+B’)’ = (1 + 1)’ = (1)’ = 0<\/p>\n

Question 2.
\n(a) Differentiate between throw and throws with respect to exception handling. [2]
\n(b) Convert the following infix notation to its postfix form: [2]
\nE*(F\/(G-H)*I) + J
\n(c) Write the algorithm for push operation (to add elements) in an array based stack. [2]
\n(d) Name the File Stream classes to: [2]
\n(i) Write data to a file in binary form.
\n(ii) Read data from a file in text form.
\n(e) A square matrix M [ ] [ ] of size 10 is stored in the memory’ with each element requiring 4 bytes of storage. If the base address at M [0][0] is 1840, determine the address at M [4] [8] when the matrix is stored in Row Major Wise. [2]
\nAnswer:
\n(a) Throw: This clause is used to explicitly raise a exception within the program, the statement would throw new exception.
\nThrows: This clause is used to indicate the exception that are not handled by the method.<\/p>\n

(b) E * (F\/(G-H) * I) +J
\n= E*(F\/GH- *I) + J
\n= E * FGH-\/I * + J
\n= EFGH-\/I**J +<\/p>\n

(c) Step 1: Start
\nStep 2: if top >= capacity then OVERFLOW, Exit
\nStep 3: top = top+1
\nStep 4: Stack [top] = value
\nStep 5: Stop<\/p>\n

(d) (i) FileOutputStream\/DataOutputStream\/FileWriter\/OutputStream
\n(ii) FileReader \/ DatalnputStream\/ InputStream\/ FilelnputStream<\/p>\n

(e) Row Major address formula:
\nM[i] [j] = BA+W [(i – Ir) * column + (j – Ic)]
\nBA: 1840, Ir = 0, Ic = 0, W = 4, rows = 10, column = 10, i = 4, j = 8
\nM[4] [8] = 1840 + 4 [(4 – 0) \u00d7 10+ (8 – 0)]
\n= 1840 + 192
\n= 2032<\/p>\n

Question 3.
\n(a) The following function Recur is a part of some class. What will be the output of the function Recur () when the value of n is equal to 10. Show the dry run \/ working. [5]<\/p>\n

void Recur (int n)\r\n{\r\nif (n>1)\r\n{\r\nSystem.out.print (n + \" \" );\r\nif(n%2 !=0)\r\n{\r\nn = 3* n + 1;\r\nSystem.out.print(n + \" \");\r\n}\r\nRecur (n\/2);\r\n}\r\n}<\/pre>\n
(b) The following function is a part of some class. Assume \u2018n\u2019 is a positive integer. Answer the given questions along with dry run \/ working,<\/p>\n
int unknown (int n)\r\n{\r\nint i, k;\r\nif (n%2 = = 0)\r\n{\r\ni = n\/2; k=1;\r\n}\r\nelse\r\n{\r\nk=n;\r\nn--;\r\ni=n\/2;\r\n}\r\nwhile (i > 0)\r\n{\r\nk=k*i*n;\r\ni--;\r\nn--;\r\n}\r\nreturn k;\r\n}<\/pre>\n
(i) What will be returned by unknown(5)? [2]
\n(ii) What will be returned by unknown(6)? [2]
\n(iii) What is being computed by unknown (int n)? [1]
\nAnswer:
\n(a) Recur (10)
\n10 Recur (5)
\n5
\n16 Recur (8)
\n8 Recur (4)
\n4 Recur (2)
\n2 Recur (1)
\nOUTPUT: 10 5 16 8 4 2
\n(b) (i) 120
\n(ii) 720
\n(iii) calculate factorial\/ product<\/p>\n
Part – II<\/strong><\/p>\n
Answer seven questions in this part, choosing three questions from Section A, two from Section B and two from Section C.<\/p>\n
Section – A<\/strong>
\nAnswer any three questions<\/strong><\/p>\n
Question 4.
\n(a) Given the Boolean function: F(A, B, C, D) = \u03a3 (0, 2, 4, 5, 8, 9, 10, 12, 13)
\n(i) Reduce the above expression by using 4-variable K-Map, showing the various groups (i.e. octal, quads and pairs). [4]
\n(ii) Draw the logic gate diagram of the reduced expression. Assume that the variables and their complements are available as inputs. [ 1]
\n(b) Given the Boolean function : F(P, Q, R, S) = \u03a0 (0, 1, 3, 5, 7, 8, 9, 10, 11, 14, 15)
\n(i) Reduce the above expression by using 4-variable K-Map, showing the various groups (i.e. octal, quads and pairs). [4]
\n(ii) Draw the logic gate diagram of the reduced expression. Assume that the variables and their complements are available as inputs. [1]
\nAnswer:
\n(a) F(A, B, C, D) = \u03a3 (0, 2, 4, 5, 8, 9, 10, 12, 13)
\n\"ISC
\n\"ISC
\n\"ISC<\/p>\n
Question 5.
\nA Football Association coach analyzes the criteria for a win\/draw of his team depending on the following conditions:
\nIf the Centre and Forward players perform well but Defenders do not perform well.
\nor
\nIf Goalkeeper and Defenders perform well but the Centre players do not perform well.
\nor
\nIf all the players perform well.
\nThe inputs are:<\/p>\n\n\n\n\n\n\n\n
Inputs<\/strong><\/td>\n<\/td>\n<\/tr>\n
C<\/td>\nCentre players perform well.<\/td>\n<\/tr>\n
D<\/td>\nDefenders perform well.<\/td>\n<\/tr>\n
F<\/td>\nForward players perform well.<\/td>\n<\/tr>\n
G<\/td>\nGoalkeeper performs well.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
(In all of the above cases 1 indicates yes and 0 indicates no)
\nOutput: X – Denotes the win\/draw criteria [1 indicates win\/draw and 0 indicates defeat in all cases.]
\n(a) Draw the truth table for the inputs and outputs given above and write the POS expression for X(C, D, F, G). [5]
\n(b) Reduce X(C, D, F, G) using Karnaugh\u2019s Map.
\nDraw the logic gate diagram for the reduced POS expression for X (C, D, F, G ) using AND and OR gate. You may use gates with two or more inputs. Assume that the variable and their complements are available as inputs. [5]
\nAnswer:
\n\"ISC
\n\"ISC<\/p>\n
Question 6.
\n(a) In the following truth table, x and y are inputs and B and D are outputs: [3]
\n\"ISC
\nAnswer the following questions:
\n(i) Write the SOP expression for D.
\n(ii) Write the POS expression for B.
\n(iii) Draw a logic diagram for the SOP expression derived for D, using only NAND gates.
\n(b) Using a truth table, verify if the following proposition is valid or invalid:
\n(a =>b) \u2227 (b =>c) = (a =>c) [3]
\n(c) From the logic circuit diagram given below, name the outputs (1), (2) and (3). Finally, derive the Boolean expression and simplify it to show that it represents a logic gate. Name and draw the logic gate. [4]
\n\"ISC
\nAnswer:
\n\"ISC
\n\"ISC<\/p>\n
Question 7.
\n(a) What are Decoders? How are they different from Encoders? [2]
\n(b) Draw the truth table and a logic gate diagram for a 2 to 4 Decoder and briefly explain its working. [4]
\n(c) A combinational logic circuit with three inputs P, Q, R produces output 1 if and only if an odd number of 0\u2019s are inputs. [4]
\n(i) Draw its truth table.
\n(ii) Derive a canonical SOP expression for the above truth table.
\n(iii) Find the complement of the above-derived expression using De Morgan\u2019s theorem and verify if it is equivalent to its POS expression.
\nAnswer:
\n(a) Decoders are a combinational circuit which inputs \u2018n\u2019 lines and outputs 2n or fewer lines. Encoders convert HLL to LLL i.e. Octal, Decimal and Hexadecimal to binary whereas Decoders convert LLL to HLL i.e. Binary to Octal, Decimal and Hexadecimal.
\n\"ISC
\nWorking: If any number is required as output then the inputs should be the binary equivalent. For example, if the input is 01 (A’.B) then the output is 1 and so on.
\n\"ISC
\n(ii) X (P, Q, R) = P’Q’R’ + P’QR + PQ’R + PQR’
\n(iii) Complement of X (P, Q, R) = (P + Q + R). (P + Q’ + R’). (P’ + Q + R’). (P’ + Q’ + R) which is not equal to POS expression for the above Truth Table.<\/p>\n
Section – B<\/strong>
\nAnswer any two questions<\/strong><\/p>\n