{"id":47427,"date":"2024-02-09T10:07:25","date_gmt":"2024-02-09T04:37:25","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=47427"},"modified":"2024-02-09T13:16:10","modified_gmt":"2024-02-09T07:46:10","slug":"isc-class-12-computer-science-previous-year-question-papers-solved-2011","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/isc-class-12-computer-science-previous-year-question-papers-solved-2011\/","title":{"rendered":"ISC Computer Science Question Paper 2011 Solved for Class 12"},"content":{"rendered":"

ISC Computer Science Previous Year Question Paper 2011 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 two Absorption laws. Verify any one of them using the truth table. [2]
\n(b) Reduce the following expression: [2]
\nF(A, B, C) = \u03a3 (0, 1, 2, 3, 4, 5, 6, 7)
\nAlso, find the complement of the reduced expression.
\n(c) Name the logic gate for the following circuit diagram and write its truth table. [2]
\n\"ISC
\n(d) Using truth table, verify whether the following is true or false: [2]
\n\\((p \\Rightarrow q)=(\\overline{q} \\Rightarrow \\overline{p})\\)
\n(e) If A = 1, B = 0, C= 1 and D = 1 find its: [2]
\n(i) Maxterm
\n(ii) Minterm
\nAnswer:
\n\"ISC
\n\"ISC
\n\"ISC<\/p>\n

Question 2.
\n(a) How can we override a method in inheritance? [2]
\n(b) A square matrix A[m*m] is stored in the memory with each element requiring 2 bytes of storage.
\nIf the base address A[1] [1] is 1098 and the address at A [4] [5] is 1144, determine the order of the matrix A[m \u00d7 m] when the matrix is stored Column Major wise. [2]
\n(c) What is Big O notation? [2]
\n(d) What is an exception? [2]
\n(e) Convert the following infix expression to its postfix form: [2]
\na + b * c – d\/e
\nAnswer:
\n(a) When we extend a class, you can change the behaviour of a method in the parent class. This is called method overriding. This happens when we write in a subclass a method that has the same signature as a method in the parent class.
\n(b) B = 1098, W = 2, n = m
\n1144 = 1098 + 2 [m(5 – 1) + (4 – 1)]
\n\u21d2 1144 = 1098 + 8m + 6
\n\u21d2 8m = 40
\n\u21d2 m = 5
\nThe order of the matrix is [5 \u00d7 5]
\n(c) Big O is the function with parameter N, where N is usually the size of the input to the algorithm. More the input size, more impact it can have on the growth rate of the algorithm.
\n(d) An unexpected situation or unexpected error, during program execution, is known as an exception.
\n(e) (a + b * c – d\/e)
\n\"ISC<\/p>\n

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

void mymethod (int counter)\r\n{\r\nif (counter == 0)\r\nSystem.out. println(\u201d \");\r\nelse\r\n{\r\nSystem.out.println (\"Hello\" +counter);\r\nmymethod (--counter);\r\nSystem.out.println (\" \" +counter);\r\n}\r\n}<\/pre>\n
(b) The following function is a part of some class which computes and returns the greatest common divisor of any two numbers. There are some places in the code marked by ?1?, ?2?, ?3?, ?4? and ?5? which must be replaced by statement\/expression so that the function works correctly0<\/p>\n
int gcd(int a, int b)\r\n{\r\nint r.\r\nwhile(?1?)\r\n{\r\nr = ?2?;\r\nb = ?3?;\r\na = ?4?\r\n}\r\nif (a ==0)\r\nreturn ?5?;\r\nelse\r\nreturn-1;\r\n}<\/pre>\n
(i) What is the expression or statement at ?1? [1]
\n(ii) What is the expression or statement at ?2? [1]
\n(iii) What is the expression or statement at ?3? [1]
\n(iv) What is the expression or statement at ?4? [1]
\n(v) What is the expression or statement at ?5? [1]
\nAnswer:
\n\"ISC
\n(b) (i) a * b! = 0
\n(ii) b
\n(iii) a
\n(iv) a%b
\n(v) r<\/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) State the principle of Duality. Give the dual of the following: [3]
\n(A’.B) + (C. 1) = (A’ + C).(B + C)
\n(b) Reduce the Boolean expressions to their simplest forms: [4]
\n(i) {(C.D)’ +A} + A + C.D + A.B
\n(ii) A. {B + C(A.B + A. C)’}
\n(c) Verily using a truth table if: [3]
\n\\((\\mathrm{A} \\odot \\mathrm{B} \\odot \\mathrm{C})^{\\prime}=\\mathrm{A} \\oplus \\mathrm{B} \\oplus \\mathrm{C}\\)
\nAnswer:
\n(a) According to the principle of Duality, “Dual of one expression is obtained by replacing AND (.) with OR (+) and OR with AND togather with replacement of 1 with 0 and 0 with 1.”
\nDual of (A’ . B) + (C.1) is given by (A’ + B). (C + 0) = (A’ + B). C
\nDual of (A’ + C). (B + C) is given by (A’.C) + (B.C)
\nThen Dual of (A’ . B) + (C . 1) = (A’ + C) . (B + C) is (A’ + B). (C + 0) = (A’.C) + (B.C)<\/p>\n
(b) (i) {(C.D)’ + A) + A + C. D + A.B
\n= {(C.D)’ + A} + A + AB + C.D
\n= {(C.D)’ + A) + A + C.D [Absorption Law]
\n= (C’ + D’) + A + A + C.D [De Morgan\u2019s Theorem]
\n= C’ + D’ + A + C.D
\n= C’ + C”.D + D’ + A
\n= C’ + D + D’ + A
\n= C’ + A
\n(ii) A. {B + C (A . B + A. C)’}
\n= A {B + C ((A.B)’ . (A.C)’)} [De Morgan\u2019s theorem]
\n= A. {B + C ((A’ + B’). (A\u2019 + C’))} [De Morgan\u2019s theorem]
\n= A. {B + C(A’+B’C’)} [Distributive law]
\n= A. {B + C.A’ + B’.C’.C.}
\n= A. (B + C.A’ +0) [Complement property]
\n= AB + C.A’.A [Complement property]
\n= AB + 0
\n= AB
\n\"ISC<\/p>\n
Question 5.
\n(a) Given F(P, Q, R, S) = \u03a0 (2, 3, 6, 7, 9, 11, 12, 13, 14, 15) [5]
\nReduce the above expression by using four variable Karnaugh’s Map. Draw the logic gate diagram of the reduced expression using NOR gate only.
\n\\(\\text { (b) Given } \\mathbf{F}(\\mathbf{A}, \\mathbf{B}, \\mathbf{C}, \\mathbf{D})=\\overline{\\mathbf{A}} \\overline{\\mathbf{B}} \\overline{\\mathbf{C}} \\overline{\\mathbf{D}}+\\overline{\\mathbf{A}} \\overline{\\mathbf{B}} \\overline{\\mathbf{C}} \\mathbf{D}+\\mathbf{A} \\overline{\\mathbf{B}} \\overline{\\mathbf{C}} \\overline{\\mathbf{D}}+\\mathbf{A} \\overline{\\mathbf{B}} \\overline{\\mathbf{C}} \\mathbf{D}+\\overline{\\mathbf{A}} \\mathbf{B} \\overline{\\mathbf{C}} \\overline{\\mathbf{D}}+\\overline{\\mathbf{A}} \\mathbf{B} \\mathbf{C} \\overline{\\mathbf{D}}\\)
\nReduce the above expression by using four variable Karnaugh\u2019s Map. Draw the logic gate diagram of the reduced expression using NAND gate only. [5]
\nAnswer:
\n(a) F (P, Q, R, S) = \u03a0 (2, 3, 6, 7, 9, 11, 12, 13, 14, 15)
\n\"ISC
\n\"ISC
\n\"ISC<\/p>\n
Question 6.
\n(a) Show with the help of a logic diagram how a NAND gate is equivalent to an OR gate. [3]
\n(b) Verify if the following is valid: [3]
\n(a => b)^(a => c) = a => (b ^ c)
\n(c) What is a Decoder? Draw the truth table and logic circuit diagram for a 2 to 4 Decoder. [4]
\nAnswer:
\n\"ISC
\n\"ISC<\/p>\n
Question 7.
\n(a) What is Full Adder? Draw the truth table for a Full adder. Also, derive SOP expression for the Full Adder and draw its logic circuit. [4]
\n(b) State how a Decoder is different from a Multiplexer. Also, state one use of each. [3]
\n(c) Convert the following cardinal expression into its canonical form and reduce it using Boolean laws: [3]
\nF(L, M, O, P) = \u03a0 (0, 2, 8, 10)
\nAnswer:
\n(a) A full adder is a logic circuit that can add three bits at a time producing two outputs one of which is the Sum bit and the other is Carry bit.
\n\"ISC
\n(b) A Multiplexer is a circuit that selects one of many input channels and connects it to the output channel. Whereas a decoder is a circuit tan converts binary numbers to denary numbers
\nA multiplexer is used as a common bus system. Whereas a decoder is used in converting binary to denary.
\n\"ISC<\/p>\n
Section – B<\/strong>
\nAnswer any two questions<\/strong><\/p>\n