CS-2004
Q1.
Opt(C)
Q2.
Opt(D)
(a) False, Call swap(x, y) does not swap the value of (x, y), because it’s call by value and in call by
value a and b does not share common memory location.
(b) false, its swap the address of x and y.
(c) False,
(d) True, swap(x, y) cannot be used as the parameters are passing by value.
Q3.
Opt(D)
Correct condition for stack full is:
Top2= Top1+1 or Top1= Top2 -1,
Q4.
Opt(B) and (C)
The binary search tree for given order is:
Here are two possibility of height; if we count the height of root from 0 then Height of the tree is 3.
Q5.
Opt(B)
Best Data Structure is stack to check arithmetic expression has balanced parentheses.
Q6.
Q7.
Opt(D)
Opt(C)
Statement i and ii are true.
Q8.
Opt(B)
Operator precedence Grammar does not have two consecutive non terminals and no epsilon
production, so statement ‘i’ and ‘iii’ violate the rule of Grammar.
Q9.
Opt(C)
CS-2004
Q10.
Opt(B)
Because, by doing so, time required for saving and loading E1 as the first operand is saved.
Q11.
Opt(A)
I. False, Context Switching is Faster in user level threads, not in kernel level threading.
II. True
III. True
IV. True
Q12.
Opt(D)
Because, it is single sequential process it doesn’t make any change using FCFS or SSTF.
Q13.
Opt(B)
c(R1)-d(R2)
This query contain the minus operator which means result of query is tuple selected from R1 cannot be the
part of the tuple selected from R2 but in tables there exist the concept of referential integrity, Colum C is the
foreign key of table R1 which is define on the primary key of the relation R2(D), so result of query
necessarily a empty relation.
Q14.
Opt(A)
Student (RollNO, Name, Address)
Enroll (RollNo, CourseNo, CourseName)
Maximum no of tuple in (Student* Enroll): Number of tuple presents in child table, because of
Natural join i.e 8.
Minimum no of tuple in (Student* Enroll): Number of tuple presents in child table, because of
Natural join i.e 8.
Ans is 8, 8
Q15.
Opt(A)
P-1, Q-4, R-3
Q16.
Opt(B)
Bridge use MAC address while routers uses IP address
Q17.
Opt(D)
x’y’+ xy + x’y
Ans: X’+ Y
CS-2004
Q18.
Opt(C)
Indeterminate state means that state can not be determined because it may be toggling. But in SR ff
toggling does not occur. So no indeterminate state is possible.
Q19.
Opt(D)
X=8, Y=11
(73)8=(54)11
7x+3=5y+4
7x-5y=1
Put x=8 and y=11
7*8-5*11=1
Q20.
Q21.
Opt(C)
Opt(A)
Q22.
Opt(B)
Here mode of the transmission is asynchronous; in asynchronous mode we have to transmit
Some extra bits per character, like start bit, end bit and parity bits. Start and end bit are compulsory,
if not given then you have to add 1-bit for both respectively.
So, transmit character will be: 1 start bit + 8 data bit + 2 stop bit +1 parity bit = 12 bits to transmit
per character.
if encoding tech is not given then Bit rate= Baud Rate
So 8 bit character can be transmits in 1 sec= baud rate/ total bits for one character
9600/12 =800 characters
Q23.
None of the option is right.
x [ boy(x)  y [ girl(y) taller(x, y) ] ]
Q24.
Opt(B)
For reflexive (x = y)
And for transitive y>x then correct option:
{(x,y) | y>=x and x,y  {0,1,2,3,…}}
Q25.
Opt(A)
Total favorable Cases: {TTHH, THTH, HTHT, HTTH, THHT, HHTT} =6
Total no of Cases: 24=16
Then probability = 6/16= 3/8
CS-2004
Q26.
Opt(C)
No of symmetric Relations will Be
Q27.
2N(N+1)/2
Opt(B)
ABCD =I
AB
= D-1C-1
(AB)-1 = (D-1C-1)-1
B-1A-1 = CD
B-1
= CDA
Q28.
Opt(A)
Taking first expression:
(113. + -111.) + 7.51
2. + 7.51
9.51
Here all three steps satisfied three digit floating point.
Now taking second expression:
113. + ( -111. + 7.51)
113. + ( -118.51) {but it is wrong because -118.51 is greater than three digit floating point number so
first we have to round up 7.51 to 8 then next step will be:}
113. + ( -111. + 8)
113. + - 103.
10.0
Q29.
Opt(C)
The tightest lower Bound on the no of comparison in worst case for comparison Based sorting is
(nlong)
Q30.
Opt(C)
3-SAT- belong to NP- Complete, and 2- SAT –belong to P- class
Q31.
Opt(C)
CS-2004
Output=7
Q32.
Opt(A)
Q33.
Opt(D)
It’s print null string, after the execution of the for loop Array P look like this:
\n
g
n
i
r
t
Because %s – print string up to \n (blank character) and first character of array P is \n, the print
statement print only \n and terminate.
Q34.
Opt(A)
Q35.
Opt(B)
Atleast in order are compulsory to identify a tree uniquely So:
(a) False
(b) True
(c) True
Q36.
Opt(A)
To perform in Constant time in circular linked list P should point to the last node of list (Rear):
If P point to last node Enqueue-O(1);
Dequeue-O(1);
Q37.
Opt(A)
32, 15, 20, 30, 12, 25, 16
CS-2004
Q38.
(A)
+,*,- all are left Associative,
^ is right associative.
Precedence order from L to R ^, *, +, Postfix expression for infix expression:
a+b*c-d^e^f
a+b*c-d^ef^
a+b*c-def^^
a+bc*-def^^
abc*+-def^^
abc*+def^^-
Q39.
Opt(D)
Q40.
Opt(D)
(a) Union O(n2): to merge the list first we have to traverse first list up to last node and add pointer to
first node of second list and again n complexity for suppress delicacy in the list, so total complexity
O(n2)
(b) Intersection O(n2), Membership O(n)
(c) Membership O(n), Cardinality O(n)
(d) Union O(n2), intersection O(n2)
Q41.
Opt(C)
Q42.
Opt(C)
Note: Correction in code: m/m is replace by m/x
Q43.
Opt(D)
By closely analyzing the program, it is seen, the height of the tree is the evaluated by comparing the
height of left subtree and right subtree of the tree using MAX function.
Q44.
Opt(B)
CS-2004
Vertex Set { P, Q, R, U, S, T}
Q45.
Opt(C)
Expression: 2 # 3 & 5 # 6 & 4
E. Value= 160
Q46.
Opt(A)
Gantt chart:
Turnaround time = Completion time – Arrival time
P1=12-0=12
P2= 4-1=3
CS-2004
P3=8-2=6
P4=5-4=1
Avg. turnaround time= (12+3+6+1)/4=5.50
Q47.
None of the option are correct
Q48.
Opt(D)
To avoid deadlock the order of waiting call should be same in all processes, so correct ans would be
option D.
Q49.
Opt(C)
10 direct, 1 single, 1 double, 1 triple indirect pointer.
Address size = 32 bit = 4 Byte
No of address in one block = (Block Size)/ Address Size
10 2
8
= 2 /2 = 2
Then maximum total Size of file = ( 10+ 28 + 28 x 28 + 28 x 28 x 28) x Block size
= ( 10+ 28 + 28 x 28 + 28 x 28 x 28) x 210 Byte
=16.06 GB approx
= 2^34 BYTES.
Q50.
Opt(B)
Candidate Key of relation will be: (RollNo, CourseNo) And (Name, CourseNo)
1st FD: It is in BCNF
2nd FD : BCNF
3rd FD: Violate BCNF property but satisfied 3NF Property.
4th FD: Violate BCNF but in 3NF.
So whole data base system in 3NF.
Q51.
Opt(D)
Query uses the theta join.
Name of the girl student with more than all the boy students.
Q52.
Opt(C)
C= 6 byte, K= 14 byte, B= 512 byte
Order of the internal node:
(N-1)K+ N*C= B
(N-1)14+ N*6= 512
14N-14+ 6N= 512
20N= 526
So N = Floor [526/20] = Floor [26.3]= 26.
Q53.
Opt(D)
CS-2004
the average salary of male employee more than the average salary in company.
Q54.
Opt(B)
A and B are two station on Ethernet A and B attempt to transmit a frame, collide and A win the Race
and transmit its first frame. At the end of this successful transmission of A, Now again both attempt
to transmit and collide, it is a first collision for station A and second collision for station B. so First
station select the random number between 0 to 1 and station B select random number between 0 to 3
using exponential back off Algorithm 2i-1, where I is the collision number.
The probability of A wins the second Back off race is:
Total Favorable Cases = {01, 02, 03, 12, 13}= 5
Total Number of Cases = 8
P (win’s)= 5/8= 0.625
Q55.
Opt(A)
To forward the Packet Routers use the concept of “Longest Matching Prefix”, in this Net ID with
highest number of one’s in sub net mask is used to check the route of packet.
Then next subnet mask:
So 1st packet will be delivered to Eth1.
Similarly for next packet 192.12.17.10 will be forward to Eth2, which is the default Address of
Router.
Q56.
Opt(D)
TCP Header = 20 byte,
IP Header = 20 byte,
Network caring capacity(MTU):
For A = 1000 byte
B = 100 byte
C = 1000 byte
CS-2004
So At the Network C delivered byte including Header will be 260 (80+20+80+20+40+20).
Note: Header of Same layer of the other network will be removing by network and add itself header
to packet.
Q57.
Opt(B)
Taking 512KB as data rate because it is bottleneck :
By unitary method 512/260 = x/180
 x= (512*180)/260 = 354.46
Q58.
Opt(B)
O/p = 1 If the Digit > 5 then corresponding Minterms are F= (5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15)
K-Map for Minterms:
F = BD+ CB+A = B(C+D) + A
So minimum number of Gates required is=3
Q59.
Opt(A)
F( a, b, c) = a’c + ac’ + bc’
Make K-Map of given expression.
ac’ and ac’ are essential prime implicants.
Q60.
Opt(A)
CS-2004
Choose option one by one using option elimination method you would get R to X, 1 to Y, T to Z
Q61.
Opt(D)
F= Q1’Q2+ Q1Q2’
Q62.
Opt(B)
4 bit carry look adder has 4 level of circuitry if EX-OR gate is used, if EX-OR gate is implemented
by AND & OR gate then it became 6 level circuitry.
Q63.
Opt(D)
Return Address Save in Stack will be:
(2+ 1 +1+2+1)* size of the word + Base Address;
7*4+1000=1028
Size of the word = 32 bit = 4 byte
Q64.
Opt(B)
CS-2004
Q65.
Opt(C)
2 –way set associative, 4 blocks then
Number of sets = 4/2 = 2
(Block Address) mod (No of set) = set number
8 mod 2 =
0
miss
12 mod 2 =
0
miss
0 mod 2 =
0
miss
12 mod 2 =
0
hit
8 mod 2 =
0
miss
Total number of miss: = 4
Q66.
Opt(A)
A = 11111010
B = 00001010
A is the negative 2’s compliment number so original number is: - (00000110)2 = - (6)10
B = (00001010)2 = (10)10
So A* B= -(6)*(10) = -(60) = -(00111100)2
Negative number is represent is system in 2’s compliments form so number is: (11000100)2
Q67.
Opt (A)
In MUX there are 8 status lines for that we need 3 selection line so 3 bit for MUX, And control
memory is 26 bit, in this 13- bit for micro operation and 3 bit for MUX and remaining 10-bit for
control add register. So X = 10, Y = 3, 1024 words in control memory.
Q68.
Opt(D)
Transfer Rate = 10Mbyte,
Frequency of processor = 600Mhz,
Instillation Takes = 300 cycle
CS-2004
Termination takes = 900 cycle
Data to transfer = 20Kbyte
So data transfer time =
Total time = Initiate + Data transfer + termination
= 2002 microsec
So % of processor time consumed = (Initialization + Termination)/ Total time;
= (0.5 + 1.5)/ 2002
= 0.09% approx. 0.1%
Q69.
Opt(C)
Clock cycle time of pipeline processor: 160+5 = 165 ns
So first instruction takes = 165*4 ns
And remaining 999 takes = 999*165 ns
So total number of cycle = 165*4+999*165 = 165495 ns = 165.5s
Q70.
Opt(A)
Q71.
(C)
Since there are 3 equations while there are only 2 variables,,,,This system of equations will always
have some unique solution.
Q72.
Opt(D)
E is the identity element of group and by using the associative property and using option elimination
method correct option (d)
a*a-1 = e inverse property
now a*c = e = c*a by using commutative law.
Q73.
Opt(A)
CS-2004
Only {1} is sufficient to make s a complete Poset.
Q74.
(D)
Exp(marks per question) = 0.25*1 + 0.75*(-0.25)=0.0625
Exp(marks obtained per student) = 150*0.0625 = 9.375
Exp(marks obtained by 1000 student) =9.375*1000=9375
Q75.
(C)
let X colors will be needed..then X self pairs of letters will have these X colors and remaining
letters will have C(X,2) colors..such that X + C(X,2)>=26 pair of letters
The minimum number of colors which is sufficient to satisfy this inequality is 7.
Q76.
Opt(D)
Q77.
Q78.
Opt(C)
(A)
CS-2004
Suppose the bit Pattern is 1101 then 1 bit differ pattern are: {0101, 1001, 1111, and 1100} = 4 cases
So total favorable cases corresponding to bit pattern 1101 = 24 * 4.
Q79.
(D)
Total possible edges = C(n,2) = n(n-1)/2
so number of graphs possible with n(n-3)/2 edges are=
= C( n(n-1)/2 , n(n-3)/2) + C( n(n-1)/2 , n(n-3)/2 +1) + C( n(n-1)/2 , n(n-3)/2 + 2) +….. + C( n(n1)/2 , n(n-1)/2 -2)+ C( n(n-1)/2 , n(n-1)/2 -1 ) + C( n(n-1)/2 , n(n-1)/2)……….(1)
Note: since C(n,r) = C(n,n-r)
Hence C( n(n-1)/2 , n(n-3)/2) = C( n(n-1)/2 , n(n-1)/2 -n(n-3)/2)
C( n(n-1)/2 , n(n-3)/2) = C( n(n-1)/2 ,n) …………………(2)
similarly C( n(n-1)/2 , n(n-3)/2 + 1) =C(n(n-1)/2 , n-1)
And also C( n(n-1)/2 , n(n-1)/2) = C( n(n-1)/2 , 0) =1
Thus we can rewrite the expression (1) as follows:
= C(n(n-1)/2 , n) + C(n(n-1)/2 , n-1) + C(n(n-1)/2 , n-2) + …… C(n(n-1)/2 , 2) + C(n(n-1)/2 , 1) +
C(n(n-1)/2 , 0)
or
= C(n(n-1)/2 , 0) + C(n(n-1)/2 , 1) + C(n(n-1)/2 , 2) + …… C(n(n-1)/2 ,n- 2) + C(n(n-1)/2 , n-1) +
C(n(n-1)/2 , n)
or
=
SUM { C ( n(n-1)/2 , k )}
k=0 to n
Hence option (d) is correct.
CS-2004
Q80.
(D)
Q81.
Opt(B)
(a) False, G1U G2 has a cut vertex.
(b) True
(C) False, Because G1U G2 does not have cut-edge.
(D) False, G1 is 3 colorable, G2 is 2 colorable but G1 U G2 is also 3 colorable.
Q82.
Opt(C)
Loop’s run for n- time so complexity of loop run is (n) and for function (m)
So hole complexity =(n) + (m) i.e. (n)
Q83.
Opt(D)
T(n) = 2T(n-1) + 1
Solve the recursion using Iterative method, you will get the Complexity = (2n)
CS-2004
Q84.
Opt(A)
T(1) = 1,
T(n) = 2T(n-1) + n
n >= 2
Check though option elimination method, Put n= 2 in recursion relation.
T(n) = 2 T(2-1) + 2
2 T(1) + 2
2*1 + 2 = 4
(a) 4, True
(b) 2, False
(c) 6, False
(d) 2, False
Q85.
Opt(A)
The program has linear worst case complexity hence complexity is (n);
Q86.
Opt(A)
Number of 1’s is Divisible of 3 and number of 0’s is divisible of 2.
Q87.
Opt(B)
We can design the PDA for that given language hence it is CFG but not Regular because Number of
c’s is the sum of number of a’s and b’s , so here counting is involve and DFA does not count, it is
memory less Automata.
Q88.
Q89.
Opt(C)
Opt(C)
L1: Set of strings which can be enumerated by a TM…thus there can be strings with infinite length
too…but possible to be enumerated effectively.
L2 : {W1#W2 ,W1#W3,….W1#W_infinity……….W2#W3,,,,,,W2#W_infinity…}
It means there is no TM which can enumerate this kind of set of strings.
Hence L2 is the reduced form (i.e. enhanced version) of L1,,which means if L2 is solvable then L1 is
definitely solvable..
Thus only S2 is true.
Q90.
Opt(D)