ISC Computer Science 2017 Class-12 Previous Year Question Papers
ISC Computer Science 2017 Class-12 Previous Year Question Paper Solved for practice. Step by step Solutions with Questions of Part-1 and Part-2 (Section-A,B and C). By the practice of Computer Science 2017 Class-12 Solved Previous Year Question Paper you can get the idea of solving.
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ISC Computer Science 2017 Class-12 Previous Year Question Paper Solved
-: Select Your Topics :-
Maximum Marks: 70
Time allowed: 3 hours
- Candidates are allowed additional 15 minutes for only reading the paper. They must NOT start writing during this time.
- Answer all questions in Part-I (compulsory) and six questions from Part-II, choosing two questions from Section-A, two from Section-B and two from Section-C.
- All working, including rough work, should be done on the same sheet as the rest of the answer.
- The intended marks for questions or parts of questions are given in brackets [ ].
Part – I (20 Marks)
Answer all questions.
ISC Computer Science 2017 Class-12 Previous Year Question Paper Solved
While answering questions in this Part, indicate briefly your working and reasoning, wherever required.
Question 1.
(a) State the law represented by the following proposition and prove it with the help of a truth table: P ∨ P = P [1]
(b) State the Principle of Duality. [1]
(c) Find the complement of the follow ing Boolean expression using De Morgan’s law: [1]
F(a, b, c) = (b’ + c) + a
(d) Draw the logic diagram and truth table for a 2 input XNOR gate. [1]
(e) If (~P => Q) then write its: [1]
(i) Inverse
(ii) Converse
Answer 1:
(a) The law represented by the given proposition is Idempotent law.
Truth Table:
(b) The duality principle states that every algebraic expression deducible from the postulates of Boolean algebra remains valid if the operators and identity elements are interchanged.
(c) ((b’ + c) + a)’
= (b’+c)’. a’
= ((b’)’ . c’) . a’
= bc’a’
(e) (i) Inverse of the given statement
P => ~Q
(ii) Converse of the given statement Q >= ~P
Question 2.
(a) What is an interface? How is it different from a class? [2]
(b) Convert the following infix expression to postfix form: [2]
P * Q/R + (S + T)
(c) AmatrixP[15] [10] is stored with each element requiring 8 bytes of storage. If the base address at P[0] [0] is 1400, determine the address at P[10] [7] when the matrix is stored in Row Major Wise. [2]
(d) (i) What is the worst-case complexity of the following code segment: [2]
for(int x = 1; x<=a; x++) { statements; } for(int y = 1; y <= b; y++) { for (int z = 1; z <= c; z++) { statements; } }
(ii) How would the complexity change if all the three loops went to N instead of a, b and c?
(e) Differentiate between a constructor and a method of a class. [2]
Answer 2:
(a) An interface in Java is a blueprint of a class. It has static constants and abstract methods. The interface in java is a mechanism to achieve abstraction. There can be only abstract methods in the Java interface, not the method body. It is used to achieve abstraction and multiple inheritances in Java.
It cannot be instantiated just like an abstract class.
A Java Interface is also a virtual construct in the programming world but is contrary to the Java Class, Interface cannot be instantiated or created as an object. An Interface denotes a group of logical entities. It can also act as a contract between two subsystems while communicating with each other.
(b) PQ*R/ST++
(c) Row Major Address Formula: M[i] [j] = B(A) + W[(i – Ir)*column + (j – Ic)]
In the given problem, Address of P[10][[7] is asked.
1400 is base address, column =10, W = 8 bytes, i = 10, j = 7, Ir = 0 and Ic = 0.
Address of P[10] [7]:
= 1400 + 8[(10 – 0)*10 + (7 – 0)]
= 1400 + 8[100 + 7]
= 1400 + 8*107
= 2256
(d) (i) O(a + bc)
(ii) O(n^2)
(e)
Constructor | Method |
1. A constructor is used to initialize the state of an object. | 1. The method is used to expose the behaviour of an object. |
2. The constructor must not have a return type. | 2. The method must have a return type. |
3. Constructor name must be same as the class name. | 3. Method name may or may not be the same as a class name. |
Question 3.
The following function magicfun() is a part of some class. What will the function magicfun() return, when the value of n=7 and n=10, respectively? Show the dry run/working: [5]
int magicfun (int n) { if(n = = 0) return 0; else return magicfim(n/2) * 10 + (n%2); }
Answer 3:
At n = 7 ⇒ 111
At n = 10 ⇒ 1010
Binary equivalent of N
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.
Previous Year Question Paper Solved ISC Computer Science 2017 Class-12
Question 4.
(a) Given the Boolean function F(A, B, C, D) = X (2, 3, 4, 5, 6, 7, 8, 10, 11).
(i) Reduce the above expression by using 4-variable Karnaugh map, showing the various groups (i.eoctal, quads and pairs). [4]
(ii) Draw the logic gate diagram for the reduced expression. Assume that the variables and their complements are available as inputs. [1]
(b) Given the Boolean function F(P, Q, R, S) = π(0, 1, 2, 4, 5, 6, 8, 10).
(i) Reduce the above expression by using the 4-variable Karnaugh map, showing the various groups (i.e., octal, quads and pairs). [4]
(ii) Draw the logic gate diagram for the reduced expression. Assume that the variables and their complements are available as inputs. [1]
Answer 4:
(a)
(i) F(A, B, C, D) = Σ (2, 3, 4, 5, 6, 7, 8, 10, 11)
= CA’ + A’B + B’C + AB’D’
= A’C + A’B + B’C + AB’D’
(ii) Logic gate diagram:
(b)
(i) F(P, Q, R, S) = π (0, 1, 2, 4, 5, 6, 8, 10)
= (A + C)(A + D)(B + D)
(ii) Logic gate diagram:
Question 5.
(a) A school intends to select candidates for an Inter-School Essay Competition as per the criteria are given below: [5]
The student has participated in an earlier competition and is very creative.
OR
The student is very creative and has excellent general awareness, but has not participated in any competition earlier.
OR
The student has excellent general awareness and has won a prize in an inter-house competition.
The inputs are:
Inputs | |
A | participated in a competition earlier |
B | is very creative |
C | won a prize in an inter-house competition |
D | has an excellent general awareness |
(In all the above cases 1 indicates yes and 0 indicates no).
Output: X [1 indicates yes, 0 indicates no for all cases]
Draw the truth table for the inputs and outputs given above and write the POS expression for X(A, B, C, D).
(b) State the application of a Half Adder. Draw the truth table and circuit diagram for a Half Adder. [3]
(c) Convert the following Boolean expression into its canonical POS form: [2]
F(A, B, C) = (B + C’).(A’ + B)
Answer 5:
(a) Truth Table:
The POS expression is,
X(A, B, C, D) = π (0, 1, 2, 4, 6, 8, 9, 10)
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