ISC Computer Science 2014 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 2014 Class-12 Solved Previous Year Question Paper you can get the idea of solving.

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## ISC Computer Science 2014 Class-12 Previous Year Question Paper Solved

Part-I

Section-A of Part-II

Section-B of Part-II

Section-C of Part-II

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)

### ISC Computer Science 2014 Class-12 Previous Year Question Paper Solved

While answering questions in this Part, indicate briefly your working and reasoning, wherever required.

Question 1.
(a) From the logic circuit diagram given below, find the output ‘F’ and simplify it.  Also, state the law represented by the logic diagram. (b) Write the truth table for a 2-input conjunction and disjunction in a proposition.  (c) Find the complement of XY’Z + XY + YZ’  (d) Convert the following expression into its canonical POS form:  F(A, B) = (A + B).A’
(e) Minimize the following Boolean expression using the Karnaugh map: $\mathrm{F}(\mathrm{A}, \mathrm{B}, \mathrm{C})=\overline{\mathrm{A}} \mathrm{B} \overline{\mathrm{C}}+\overline{\mathrm{A}} \mathrm{BC}+\mathrm{AB} \overline{\mathrm{C}}+\mathrm{ABC}$  Question 2.
(a) State two advantages of using the concept of inheritance in Java.  (b) An array AR [-4…. 6, -2 ….. 12 ], stores elements in Row Major Wise, with the address AR  as 4142 . If each element requires 2 bytes of storage, find the Base address.  (c) State the sequence of traversing a binary tree in: 
(i) preorder
(ii) postorder
(d) Convert the following infix expression into its postfix form: 
(A/B + C) * (D/(E – F))
(e) State the difference between the functions int nextlnt() and boolean hasNextInt().  Answer 2:
(a)

(i) Existing code can be reused and functionality can be extended.
(ii) New members to the derived class can be added.
(iii) Implementation of existing methods can be replaced by overriding a method that already exists in the base class.

(b) AR [-4,… 6, -2 … 12] Row- wise order,
AR  = B + W(n(i-1)+(j-1))
Given i = 2, j = 3, W = 2 bytes
B = ?
n = Uc – Lc + 1
= 12 – (-2) + 1
= 12 + 2 + 1
= 15
Now, 414 = B + 2 [15 (2 -(-2)) + (3-(-4))] B + 2[60 + 7] = 4142
B + 134 = 4142 or B = 4008

(c)

(i) Preorder is Root, Left, Right
(ii) Postorder is Left, Right, Root

(d) (A/B + C) * (D/(E -F))
= (A/B + C) (D/(E-F))*
= (AB/+C) (D/(EF-))*
= (AB/C+) (DEF-/)*
= AB/C + DEF-/*

(e) nextlnt() returns an integer present in a Scanner class object whereas hasNextInt() checks whether Scanner class object contains an integer or not and returns true if it contains an integer, otherwise false.

Question 3.
(a) The following functions are part of some class:

void fun 1 (char s[ ],int x)
{
System.out.println(s);
char temp;
if(x<s.length/2)
{
temp=s[x];
s[x]=s[s.length-x-1];
s[s.length-x-1 ]=temp;
fun1(s, x+1);
}
}
void fun2(String n)
{
char c[ ]=new char[n.length()];
for(int i=0;i<c.length; i++)
c[i]=n.charAt(i);
fun1(c,0);
}

(i) What will be the output of fun1() when the value of s[ ]={‘J’,‘U’,‘N’,‘E’} and x = 1?  (ii) What will be the output of fun2( ) when the value of n = ‘SCROLL”?
(iii) State in one line what does the function fim1() do apart from recursion.  (b) The following is a function of some class which sorts an integer array a[ ] in ascending order using selection sort technique. There are some places in the code marked by ?1?, ?2?, ?3?, ?4?, ?5? which may be replaced by a statement/expression so that the function works properly:

void selectsort(int [ ]a)
{
int i, j, t, min, minpos;
for(i=0;i {
min=a[i];
minpos = i;
for(j=?2?;y<a.length;j++) { if(min>a[j])
{
?3?=j;
min = ?4?;
}
}
t=a[minpos];
a[minpos]=a[i];
a[i]= ?5?;
}
for(int k=0;k<a.length;k++)
System. out.println(a[k]);
}

(i) What is the expression or statement at ?1?  (ii) What is the expression or statement at ?2?  (in) What is the expression or statement at ?3?  (iv) What is the expression or statement at ?4?  (v) What is the expression or statement at ?5?  Answer 3:
(a)

(i) JUNE
JNUE
(ii) SCROLL
LCROLS
LLROCS
LLORCS
(iii) Reverses the part of a string from a specified location.
(b)

(i) a. length
(ii) i + 1
(iii) minpos=j
(iv) min = a [j] (v) t

Part- II (50 Marks)

Answer seven questions in this part, choosing two questions from Section A, two from Section B and two from Section C.

Section – A

### Previous Year Question Paper Solved ISC Computer Science 2014 Class-12

Question 4.
(a) Given the Boolean function F(A, B, C, D) = Σ (0, 1, 2, 3, 5, 6, 7, 10, 13, 14, 15)
(i) Reduce the above expression by using, 4-variable Karnaugh map, showing the various groups (i.e., octal, quads and pairs).  (ii) Draw the logic gate diagram for the reduced expression. Assume that the variables and their complements are available as inputs.  (b) Given the Boolean function P(A, B, C, D) = Π (0, 1, 2, 3, 5, 6, 7, 10, 13, 14, 15)
(i) Reduce the above expression by using the 4-variable Karnaugh map, showing the various groups (i.e., octal, quads and pairs).  (ii) Draw the logic gate diagram for the reduced expression. Assume that the variables and their complements are available as inputs.  Answer 4:
(a)

(i) F(A, B, C, D) = Σ (0, 1, 2, 3, 5, 6, 7, 10, 13, 14, 15)    Question 5.
A school intends to select candidates for the Inter-School Athletic Meet, as per the criteria are given below:
The candidate is from the Senior School and has participated in an Inter-School Athletic Meet earlier.
OR
The candidate is not from the Senior School, but the height is between 5 ft. and 6 ft. and weight is between 50 kg. and 60 kg.
OR
The candidate is from the Senior School and has a height between 5 ft. and 6 ft., but the weight is not between 50 kg. and 60 kg.
The inputs are:

 Inputs S A student is from the Senior School W Weight is between 50 kg. and 60 kg. H Height is between 5 ft. and 6 ft. A Taken part in Inter-School Athletic Meet earlier

(In all of the above cases, 1 indicates yes and 0 indicates no)
Output: X – Denotes the selection criteria [1 indicates selected and 0 indicates rejected in all cases] (a) Draw the truth table for the inputs and outputs given above and write the SOP expression for X (S, W, H, A).  (b) Reduce X(S, W, H, A) using Karnaugh map.  Draw the logic gate diagram for the reduced SOP expression for X(S, W, H, A) 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.   