Gate 2005-IT
Question 1 |
A bag contains 10 blue marbles, 20 green marbles and 30 red marbles. A marble is drawn from the bag, its colour recorded and it is put back in the bag. This process is repeated 3 times. The probability that no two of the marbles drawn have the same colour is
1/36 | |
1/6 | |
1/4 | |
1/3 |
Question 1 Explanation:
Question 2 |
If the trapezoidal method is used to evaluate the integral obtained 0∫1x2dx ,then the value obtained
is always > (1/3) | |
is always < (1/3) | |
is always = (1/3) | |
may be greater or lesser than (1/3) |
Question 2 Explanation:
Note: Out of syllabus.
Question 3 |
-1 | |
0 | |
1 | |
2 |
Question 3 Explanation:
determinant = product of diagonal element [upper triangular matrix]
= -1 * 1 * 1 * 1
= -1
Question 4 |
Let L be a regular language and M be a context-free language, both over the alphabet Σ. Let Lc and Mc denote the complements of L and M respectively. Which of the following statements about the language if Lc ∪ Mc is TRUE?
It is necessarily regular but not necessarily context-free. | |
It is necessarily context-free. | |
It is necessarily non-regular. | |
None of the above. |
Question 4 Explanation:
Context-free languages not closed under complementation. So, Lc ∪ Mc is neither regular nor context-free. It might be context sensitive language.
Question 5 |
Which of the following statements is TRUE about the regular expression 01*0?
It represents a finite set of finite strings. | |
It represents an infinite set of finite strings. | |
It represents a finite set of infinite strings. | |
It represents an infinite set of infinite strings.
|
Question 5 Explanation:
The given expression01*0 is regular. So this is a finite string. So options C and D are false and * is placed. So this is infinite set.
So, given regular expression represents an infinite set of finite strings.
So, given regular expression represents an infinite set of finite strings.
Question 6 |
The language {0n 1n 2n | 1 ≤ n ≤ 106} is
regular | |
context-free but not regular | |
context-free but its complement is not context-free | |
not context-free |
Question 6 Explanation:
In this the value of n is finite then we can be able to construct a finite state automata for this language.
So, given language is regular.
So, given language is regular.
Question 7 |
Which of the following expressions is equivalent to (A⊕B)⊕C
 | |
 | |
 | |
None of these |
Question 7 Explanation:
Question 8 |
Using Booth’s Algorithm for multiplication, the multiplier -57 will be recoded as
0 -1 0 0 1 0 0 -1 | |
1 1 0 0 0 1 1 1 | |
0 -1 0 0 1 0 0 0 | |
0 1 0 0 -1 0 0 1 |
Question 8 Explanation:
Consider option A:
Question 9 |
A dynamic RAM has a memory cycle time of 64 nsec. It has to be refreshed 100 times per msec and each refresh takes 100 nsec. What percentage of the memory cycle time is used for refreshing?
10 | |
6.4 | |
1 | |
.64 |
Question 9 Explanation:
We know that 1ms = 106ns
In 106ns refresh 100 times.
Each refresh takes 100ns.
Memory cycle time = 64ns
Refresh time per 1ms i.e., per 106ns = 100 * 100 = 104ns
Refresh time per 1ns = (104)/(106) ns
Refresh time per cycle = (104*64)/(106) = 64ns
Percentage of the memory cycle time is used for refreshing = (64*100)/64 = 1%
In 106ns refresh 100 times.
Each refresh takes 100ns.
Memory cycle time = 64ns
Refresh time per 1ms i.e., per 106ns = 100 * 100 = 104ns
Refresh time per 1ns = (104)/(106) ns
Refresh time per cycle = (104*64)/(106) = 64ns
Percentage of the memory cycle time is used for refreshing = (64*100)/64 = 1%
Question 10 |
 | |
 | |
 | |
 |
Question 10 Explanation:
The bulb will be on when both the switch S1 and S2 are in same state, either OFF (or) ON:
From this we can clearly know that given is EX-NOR operation i.e.,
(S1⊙S2) = (S1⊕S2)'
From this we can clearly know that given is EX-NOR operation i.e.,
(S1⊙S2) = (S1⊕S2)'
Question 11 |
How many pulses are needed to change the contents of a 8-bit up counter from 10101100 to 00100111 (rightmost bit is the LSB)?
134 | |
133 | |
124 | |
123 |
Question 11 Explanation:
The 8 bit counter will be 0-255 to move from 10101100 (172) to 1000111 (39).
→ First counter is move from 172 to 255 = 83 pulses
→ 255 to 0 = 1 pulse
→ 0 to 39 = 39 pulses
Total = 83 + 1 + 39 = 123 pulses
→ First counter is move from 172 to 255 = 83 pulses
→ 255 to 0 = 1 pulse
→ 0 to 39 = 39 pulses
Total = 83 + 1 + 39 = 123 pulses
Question 12 |
The numbers 1, 2, .... n are inserted in a binary search tree in some order. In the resulting tree, the right subtree of the root contains p nodes. The first number to be inserted in the tree must be
p | |
p + 1 | |
n - p | |
n - p + 1 |
Question 12 Explanation:
Total element = n
RST contains elements = p
Root contains = 1 element
1st contains = n - (p + 1) element
Root contains the value is n - p.
RST contains elements = p
Root contains = 1 element
1st contains = n - (p + 1) element
Root contains the value is n - p.
Question 13 |
6 | |
4 | |
3 | |
2 |
Question 13 Explanation:
2, -3, 2, -1, 2
f(Ø)=0 and f(push(S,i) = max(f(S),0) + i;
Initially stack is empty and for empty stack 0 is given.
f(push(0,2)) = max(f(Ø),0) + 2 = max(Ø,0) + 2 = 2
f(push(2,-3)) = max(2,0) + (-3) = 2 - 3 = -1
f(push(-1,2)) = max(-1,0) + 2 = 0 + 2 = 2
f(push(2,-1)) = max(2,0)+ (-1) = 2 - 1 = 1
f(push(1,2)) = max(1,0) + 2 = 1 + 2 = 3
So, 3 will be the answer.
∴ Option C is correct.
f(Ø)=0 and f(push(S,i) = max(f(S),0) + i;
Initially stack is empty and for empty stack 0 is given.
f(push(0,2)) = max(f(Ø),0) + 2 = max(Ø,0) + 2 = 2
f(push(2,-3)) = max(2,0) + (-3) = 2 - 3 = -1
f(push(-1,2)) = max(-1,0) + 2 = 0 + 2 = 2
f(push(2,-1)) = max(2,0)+ (-1) = 2 - 1 = 1
f(push(1,2)) = max(1,0) + 2 = 1 + 2 = 3
So, 3 will be the answer.
∴ Option C is correct.
Question 14 |
In a depth-first traversal of a graph G with n vertices, k edges are marked as tree edges. The number of connected components in G is
k | |
k + 1 | |
n - k - 1 | |
n - k |
Question 14 Explanation:
In a graph G with n vertices and p component then G has n - p edges(k).
In this question, we are going to applying the depth first search traversal on each component of graph where G is conected (or) disconnected which gives minimum spanning tree
i.e., k = n-p
p = n - k
In this question, we are going to applying the depth first search traversal on each component of graph where G is conected (or) disconnected which gives minimum spanning tree
i.e., k = n-p
p = n - k
Question 15 |
1→ C, 2 → A, 3 → B, 4 → D | |
1→ B, 2 → D, 3 → C, 4 → A | |
1→ C, 2 → D, 3 → A, 4 → B | |
1→ B, 2 → A, 3 → C, 4 → D |
Question 15 Explanation:
Bellman-ford algorithm → O(nm)
Krushkal's algorithm → O(m log n)
Floyd-Warshall algorithm → O(n3)
Topological sorting → O(n+m)
Krushkal's algorithm → O(m log n)
Floyd-Warshall algorithm → O(n3)
Topological sorting → O(n+m)
Question 16 |
A hash table contains 10 buckets and uses linear probing to resolve collisions. The key values are integers and the hash function used is key % 10. If the values 43, 165, 62, 123, 142 are inserted in the table, in what location would the key value 142 be inserted?
2 | |
3 | |
4 | |
6 |
Question 16 Explanation:
43%10 = 3 [occupy 3]
165%10 = 5 [occupy 5]
62%10 = 2 [occupy 2]
123%10 = 3 [3 already occupied, so occupies 4]
142%10 = 2 [2, 3, 4, 5 are occupied, so it occupies 6]
165%10 = 5 [occupy 5]
62%10 = 2 [occupy 2]
123%10 = 3 [3 already occupied, so occupies 4]
142%10 = 2 [2, 3, 4, 5 are occupied, so it occupies 6]
Question 17 |
(I) and (II) only | |
(II) and (III) only | |
(II) only | |
(I) and (III) only |
Question 17 Explanation:
Note: Out of syllabus.
Question 18 |
ls passwd | |
cat passwd | |
grep name passwd | |
grep print passwd
|
Question 18 Explanation:
Note: Out of syllabus.
Question 19 |
A user level process in Unix traps the signal sent on a Ctrl-C input, and has a signal handling routine that saves appropriate files before terminating the process. When a Ctrl-C input is given to this process, what is the mode in which the signal handling routine executes?
kernel mode | |
superuser mode | |
privileged mode | |
user mode |
Question 19 Explanation:
Note: Out of syllabus.
Question 20 |
The Function Point (FP) calculated for a software project are often used to obtain an estimate of Lines of Code (LOC) required for that project. Which of the following statements is FALSE in this context.
The relationship between FP and LOC depends on the programming language used to implement the software. | |
LOC requirement for an assembly language implementation will be more for a given FP value, than LOC for implementation in COBOL | |
On an average, one LOC of C++ provides approximately 1.6 times the functionality of a single LOC of FORTRAN | |
FP and LOC are not related to each other
|
Question 20 Explanation:
Note: Out of syllabus.
Question 21 |
Person | |
Hotel Room | |
Lodging | |
None of these |
Question 21 Explanation:
The information should appear as an attribute of lodging, because this is the only common attribute that relating to the hotel room and person.
Question 22 |
1 NF | |
2 NF | |
3 NF | |
None |
Question 22 Explanation:
F1 → F3 ......(i)
F2 → F4 ......(ii)
(F1⋅F2) → F5 .....(iii)
F1F2 is the candidate key.
F1 and F2 are the prime key.
In (i) and (ii) we can observe that the relation from P → NP which is partial dependency. So this is in 1NF.
F2 → F4 ......(ii)
(F1⋅F2) → F5 .....(iii)
F1F2 is the candidate key.
F1 and F2 are the prime key.
In (i) and (ii) we can observe that the relation from P → NP which is partial dependency. So this is in 1NF.
Question 23 |
A B-Tree used as an index for a large database table has four levels including the root node. If a new key is inserted in this index, then the maximum number of nodes that could be newly created in the process are:
5 | |
4 | |
3 | |
2 |
Question 23 Explanation:
No. of nodes in a children of a node = no. of keys in parent node + 1
Here, the tree has 4 levels, then 4+1=5 nodes to be present in the newly created process.
Here, the tree has 4 levels, then 4+1=5 nodes to be present in the newly created process.
Question 24 |
Amongst the ACID properties of a transaction, the 'Durability' property requires that the changes made to the database by a successful transaction persist
Except in case of an Operating System crash | |
Except in case of a Disk crash | |
Except in case of a power failure | |
Always, even if there is a failure of any kind |
Question 24 Explanation:
Durability gurantees that once a transition has been committed, it will remain committed even in the case of a system failure.
Question 25 |
HTTP, SMTP, FTP | |
FTP, HTTP, SMTP | |
HTTP, FTP, SMTP | |
SMTP, HTTP, FTP |
Question 25 Explanation:
Note: Out of syllabus.
Question 26 |
Traceroute reports a possible route that is taken by packets moving from some host A to some other host B. Which of the following options represents the technique used by traceroute to identify these hosts
By progressively querying routers about the next router on the path to B using ICMP packets, starting with the first router | |
By requiring each router to append the address to the ICMP packet as it is forwarded to B. The list of all routers en-route to B is returned by B in an ICMP reply packet | |
By ensuring that an ICMP reply packet is returned to A by each router en-route to B, in the ascending order of their hop distance from A | |
By locally computing the shortest path from A to B |
Question 26 Explanation:
Traceroute works by sending packets with gradually increasing TTL value, starting with TTL value of 1. The first router receives the packet, decrements the TTL value and drops the packet because it then has TTL value zero. The router sends an ICMP time exceeded message back to the source. The next set of packets are given a TTL value of 2.
So the first router forwards the packets, but the second router drops them and replies with ICMP time exceeded. Proceeding in this way, traceroute uses the returned ICMP time exceeded messages to build a list of routers that packets traverse, until the destination is reached and returns an ICMP echo reply message.
So the first router forwards the packets, but the second router drops them and replies with ICMP time exceeded. Proceeding in this way, traceroute uses the returned ICMP time exceeded messages to build a list of routers that packets traverse, until the destination is reached and returns an ICMP echo reply message.
Question 27 |
Which of the following statements is TRUE about CSMA/CD
IEEE 802.11 wireless LAN runs CSMA/CD protocol | |
Ethernet is not based on CSMA/CD protocol | |
CSMA/CD is not suitable for a high propagation delay network like satellite network | |
There is no contention in a CSMA/CD network |
Question 27 Explanation:
For CSMA/CD requires that sender is to be transmitting atleast till the first bit reaches the receiver. So the collision will be eliminated in case if it is present.
For networks with high propagation delay this time becomes too long hence the minimum packet size required becomes too big to be feasible.
For networks with high propagation delay this time becomes too long hence the minimum packet size required becomes too big to be feasible.
Question 28 |
Which of the following statements is FALSE regarding a bridge?
Bridge is a layer 2 device | |
Bridge reduces collision domain | |
Bridge is used to connect two or more LAN segments | |
Bridge reduces broadcast domain |
Question 28 Explanation:
Bridge devices works at the data link layer of the open system interconnected (OSI) model, connecting two different networks together and providing communication between them. So, option A, C are true.
The bridge acts as a interface between two networks and speed the traffic between them and there by reduces the collision domain.
So, option B also True.
The bridge acts as a interface between two networks and speed the traffic between them and there by reduces the collision domain.
So, option B also True.
Question 29 |
Count to infinity is a problem associated with
link state routing protocol. | |
distance vector routing protocol. | |
DNS while resolving host name. | |
TCP for congestion control. |
Question 29 Explanation:
Distance vector routing protocol uses the Bellman-Ford algorithm and Ford-Fulkerson algorithm.
The Bellman-Ford algorithm does not prevent routing loops from happening and suffers from the count-to-infinity problem.
The Bellman-Ford algorithm does not prevent routing loops from happening and suffers from the count-to-infinity problem.
Question 30 |
A HTML form is to be designed to enable purchase of office stationery. Required items are to be selected (checked). Credit card details are to be entered and then the submit button is to be pressed. Which one of the following options would be appropriate for sending the data to the server. Assume that security is handled in a way that is transparent to the form design.
Only GET | |
Only POST | |
Either of GET or POST | |
Neither GET nor POST |
Question 30 Explanation:
Note: Out of syllabus.
Question 31 |
Let f be a function from a set A to a set B, g a function from B to C, and h a function from A to C, such that h(a) = g(f(a)) for all a ∈ A. Which of the following statements is always true for all such functions f and g?
g is onto ⇒ h is onto | |
h is onto ⇒ f is onto | |
h is onto ⇒ g is onto | |
h is onto ⇒ f and g are onto |
Question 31 Explanation:
g(f(a)) is a composition function which is A→B→C.
If h: A→C is a onto function, the composition must be onto, but the first function in the composition need to be onto.
So, B→C is must be onto.
If h: A→C is a onto function, the composition must be onto, but the first function in the composition need to be onto.
So, B→C is must be onto.
Question 32 |
Let A be a set with n elements. Let C be a collection of distinct subsets of A such that for any two subsets S1 and S2 in C, either S1 ⊂ S2 or S2 ⊂ S1. What is the maximum cardinality of C?
n | |
n + 1 | |
2(n-1) + 1 | |
n! |
Question 32 Explanation:
Consider an example:
Let A = {a, b, c}, here n = 3
Now, P(A) = {Ø, {a}, {b}, {c}, {a,b}, {b,c}, {{a}, {a,b,c}}
Now C will be contain Ø (empty set) and {a,b,c} (set itself) as Ø is the subset of every set. And every other subset is the subset of {a,b,c}.
Now taking the subset of cardinality, we an take any 1 of {a}, {b}, {c} as none of the set is subset of other.
Let's take {2}
→ Now taking the sets of cardinality 2 -{a,b}, {b,c}
→ {b} ⊂ {a,b} and {b,c} but we can't take both as none of the 2 is subset of the other.
→ So let's take {c,a}.
So, C = {Ø, {b}, {b,c}, {a,b,c}}
→ So, if we observe carefully, we can see that we can select only 1 set from the subsets of each cardinality 1 to n
i.e., total n subsets + Ø = n + 1 subsets of A can be there in C.
→ So, even though we can have different combinations of subsets in C but maximum cardinality of C will be n+1 only.
Let A = {a, b, c}, here n = 3
Now, P(A) = {Ø, {a}, {b}, {c}, {a,b}, {b,c}, {{a}, {a,b,c}}
Now C will be contain Ø (empty set) and {a,b,c} (set itself) as Ø is the subset of every set. And every other subset is the subset of {a,b,c}.
Now taking the subset of cardinality, we an take any 1 of {a}, {b}, {c} as none of the set is subset of other.
Let's take {2}
→ Now taking the sets of cardinality 2 -{a,b}, {b,c}
→ {b} ⊂ {a,b} and {b,c} but we can't take both as none of the 2 is subset of the other.
→ So let's take {c,a}.
So, C = {Ø, {b}, {b,c}, {a,b,c}}
→ So, if we observe carefully, we can see that we can select only 1 set from the subsets of each cardinality 1 to n
i.e., total n subsets + Ø = n + 1 subsets of A can be there in C.
→ So, even though we can have different combinations of subsets in C but maximum cardinality of C will be n+1 only.
Question 33 |
An unbiased coin is tossed repeatedly until the outcome of two successive tosses is the same. Assuming that the trials are independent, the expected number of tosses is
3 | |
4 | |
5 | |
6 |
Question 34 |
Let n = p2q, where p and q are distinct prime numbers. How many numbers m satisfy 1 ≤ m ≤ n and gcd (m, n) = 1? Note that gcd (m, n) is the greatest common divisor of m and n.
p(p - 1) | |
pq | |
(p2 - 1)(q - 1) | |
p(p - 1)(q - 1) |
Question 34 Explanation:
n = p2q, [p, q are prime numbers]
→ No. of multiples of p in n = pq [n = p⋅p⋅q]
→ No. of multiples of q in n = p2 [n = p2q]
→ Prime factorization of n contains only p & q.
→ gcd(m,n) is to be multiple of p and (or) 1.
→ So, no. of possible m such that gcd(m,n) is 1 will be
n - number of multiples of either p (or) q
= n - p2 - pq + p
= p2q - p2 - pq + p
= p(pq - p - q + 1)
= p(p - 1)(q - 1)
→ No. of multiples of p in n = pq [n = p⋅p⋅q]
→ No. of multiples of q in n = p2 [n = p2q]
→ Prime factorization of n contains only p & q.
→ gcd(m,n) is to be multiple of p and (or) 1.
→ So, no. of possible m such that gcd(m,n) is 1 will be
n - number of multiples of either p (or) q
= n - p2 - pq + p
= p2q - p2 - pq + p
= p(pq - p - q + 1)
= p(p - 1)(q - 1)
Question 35 |
-1 | |
1 | |
0 | |
π |
Question 35 Explanation:
In the limits are be -π to π, one is odd and another is even product of even and odd is odd function and integrating function from the same negative value to positive value gives 0.
Question 36 |
Let P(x) and Q(x) be arbitrary predicates. Which of the following statements is always TRUE?
((∀x(P(x)∨Q(x))))⟹((∀xP(x))∨(∀xQ(x))) | |
(∀x(P(x)⟹Q(x)))⟹((∀xP(x))⟹(∀xQ(x))) | |
(∀x(P(x))⟹∀x(Q(x)))⟹(∀x(P(x)⟹Q(x))) | |
(∀x(P(x))⇔(∀x(Q(x))))⟹(∀x(P(x)⇔Q(x)))
|
Question 36 Explanation:
LHS: (P(x) ⟹Q(x)) which is True for every value thus ∀x(P(x) ⟹ Q(x)) becomes True.
RHS: (∀xP(x)) and (∀xQ(x)) both becomes False for assumed values which implies F→F and result will be True.
∴ LHS = RHS
RHS: (∀xP(x)) and (∀xQ(x)) both becomes False for assumed values which implies F→F and result will be True.
∴ LHS = RHS
Question 37 |
L1 = L2 | |
L1 ⊂ L2 | |
L2 ⊂ L1 | |
None of the above |
Question 37 Explanation:
Based on Arden's theorem write the Y and Z in terms of incoming arrows,
Y = X0 + Y0 + Z1
Z = X0 + Y0 + Z;
⇒ X = Z;
⇒ L1 = L2
Y = X0 + Y0 + Z1
Z = X0 + Y0 + Z;
⇒ X = Z;
⇒ L1 = L2
Question 38 |
Let P be a non-deterministic push-down automaton (NPDA) with exactly one state, q, and exactly one symbol, Z, in its stack alphabet. State q is both the starting as well as the accepting state of the PDA. The stack is initialized with one Z before the start of the operation of the PDA. Let the input alphabet of the PDA be Σ. Let L(P) be the language accepted by the PDA by reading a string and reaching its accepting state. Let N(P) be the language accepted by the PDA by reading a string and emptying its stack. Which of the following statements is TRUE?
L(P) is necessarily Σ* but N(P) is not necessarily Σ* | |
N(P) is necessarily Σ* but L(P) is not necessarily Σ* | |
Both L(P) and N(P) are necessarily Σ* | |
Neither L(P) nor N(P) are necessarily Σ*
|
Question 38 Explanation:
Since, it is NPDA, so we might not have any transitions over any alphabet. So option (D) is correct.
Question 39 |
2 | |
3 | |
4 | |
5 |
Question 39 Explanation:
L = {aa, aaa, aaaaa, ...}
The minimum string length is 2 [aa], so we require 3 states to construct DFA.
The minimum string length is 2 [aa], so we require 3 states to construct DFA.
Question 40 |
A language L satisfies the Pumping Lemma for regular languages, and also the Pumping Lemma for context-free languages. Which of the following statements about L is TRUE?
L is necessarily a regular language | |
L is necessarily a context-free language, but not necessarily a regular language | |
L is necessarily a non-regular language | |
None of the above |
Question 40 Explanation:
As we know that pumping lemma is a negative test, which can be use to disprove the given language is not regular. But reverse is not True.
Question 41 |
Both 1 and 2 are true | |
1 is true but 2 is false | |
2 is true but 1 is false | |
Both 1 and 2 are false |
Question 41 Explanation:
P1 can be starves on P, then P2 loops forever.
P2 can be starves on P, then P1 loops forever.
Both statements (i) and (ii) are True.
P2 can be starves on P, then P1 loops forever.
Both statements (i) and (ii) are True.
Question 42 |
1 | |
2 | |
3 | |
4 |
Question 42 Explanation:
It needs two semaphores such as X=0; Y=0
Question 43 |
Which of the following input sequences will always generate a 1 at the output Z at the end of the third cycle?
000 101 111 | |
101 110 111 | |
011 101 111 | |
001 110 111 | |
None of these |
Question 43 Explanation:
While filling done in reverse order, all operations are not satisfied.
Question 44 |
We have two designs D1 and D2 for a synchronous pipeline processor. D1 has 5 pipeline stages with execution times of 3 nsec, 2 nsec, 4 nsec, 2 nsec and 3 nsec while the design D2 has 8 pipeline stages each with 2 nsec execution time How much time can be saved using design D2 over design D1 for executing 100 instructions?
214 nsec | |
202 nsec | |
86 nsec | |
-200 nsec |
Question 44 Explanation:
k = total no. of stages
n = no. of instructions
Total execution time = (k+n-1) * maximum clock cycle
In case of D1:
k = 5
n = 100
Maximum clock cycle = 4 ns
Total execution time = (5+100-1) * 4 = 416
In case of D2:
k = 8
n = 100
Maximum clock cycle = 2 ns
Total execution time = (8+100-1) * 2 = 214
Starved time D2 over D1 = 416 - 214 = 202
n = no. of instructions
Total execution time = (k+n-1) * maximum clock cycle
In case of D1:
k = 5
n = 100
Maximum clock cycle = 4 ns
Total execution time = (5+100-1) * 4 = 416
In case of D2:
k = 8
n = 100
Maximum clock cycle = 2 ns
Total execution time = (8+100-1) * 2 = 214
Starved time D2 over D1 = 416 - 214 = 202
Question 45 |
S5=T1+I2⋅T3 and S10=(I1+I3)⋅T4+(I2+I4)⋅T5 | |
S5=T1+(I2+I4)⋅T3 and S10=(I1+I3)⋅T4+(I2+I4)⋅T5 | |
S5=T1+(I2+I4)⋅T3 and S10=(I2+I3+I4)⋅T2+(I1+I3)⋅T4+(I2+I4)⋅T5 | |
S5=T1+(I2+I4)⋅T3 and S10=(I2+I3)⋅T2+I4⋅T3+(I1+I3)⋅T4+(I2+I4)⋅T5 |
Question 45 Explanation:
It is an example of Hardwired CU programming used in RISC processors.
→ If we look at the table, we need to find those timestamps which are using these control signals.
→ For example consider S5 = T1 has been used for control signal for all the instructions, or we can say that irrespective of the instruction.
→ Also S5 is used by instructions I2 and I4 for the timestamp T3 so that
S5 = T1 + I2⋅T3 + I4⋅T3 = T1 + (I2+I4)⋅T3
→ Like that
S10 = (I2+I3)⋅T2 + I4⋅T3 +(I1+I3)⋅T4 + (I2⋅I4)⋅T5
→ If we look at the table, we need to find those timestamps which are using these control signals.
→ For example consider S5 = T1 has been used for control signal for all the instructions, or we can say that irrespective of the instruction.
→ Also S5 is used by instructions I2 and I4 for the timestamp T3 so that
S5 = T1 + I2⋅T3 + I4⋅T3 = T1 + (I2+I4)⋅T3
→ Like that
S10 = (I2+I3)⋅T2 + I4⋅T3 +(I1+I3)⋅T4 + (I2⋅I4)⋅T5
Question 46 |
A line L in a circuit is said to have a stuck-at-0 fault if the line permanently has a logic value 0. Similarly a line L in a circuit is said to have a stuck-at-1 fault if the line permanently has a logic value 1. A circuit is said to have a multiple stuck-at fault if one or more lines have stuck at faults. The total number of distinct multiple stuck-at faults possible in a circuit with N lines is
3N | |
3N - 1 | |
2N - 1 | |
2 |
Question 46 Explanation:
Answer should be 3N-1.
This is because the total possible combinations (i.e., a line may either be at fault (in 2 ways i.e., stuck at 0 or 1) or it may not be, so there are only 3 possibilities for a line) is 3N. In only one combination the circuit will have all lines to be correct (i.e., not a fault). Hence, total combinations in which distinct multiple stuck-at-faults possible in a circuit with N lines is 3N - 1.
This is because the total possible combinations (i.e., a line may either be at fault (in 2 ways i.e., stuck at 0 or 1) or it may not be, so there are only 3 possibilities for a line) is 3N. In only one combination the circuit will have all lines to be correct (i.e., not a fault). Hence, total combinations in which distinct multiple stuck-at-faults possible in a circuit with N lines is 3N - 1.
Question 47 |
(34.4)8 × (23.4)8 evaluates to
(1053.6)8 | |
(1053.2)8 | |
(1024.2)8 | |
None of these |
Question 47 Explanation:
First convert (34.4)8 and (23.4)8 to decimal.
(34.4)8 = 3×81 + 4×80 + 4×8-1
= 24 + 4 + 0.5
= (28.5)10
(23.4)8 = 2×81 + 3×80 + 4×8-1
= 16 + 3 + 0.5
= (19.5)10
Now,
(28.5)10 × (19.5)01 = (555.75)10
Now,
(555.75)10 = ( ? )8
To convert the integer part,
We get, 1053.
To convert the fractional part, keep multiplying by 8 till decimal part becomes 0,
∴ (555.75)10 = (1053.6)8
(34.4)8 = 3×81 + 4×80 + 4×8-1
= 24 + 4 + 0.5
= (28.5)10
(23.4)8 = 2×81 + 3×80 + 4×8-1
= 16 + 3 + 0.5
= (19.5)10
Now,
(28.5)10 × (19.5)01 = (555.75)10
Now,
(555.75)10 = ( ? )8
To convert the integer part,
We get, 1053.
To convert the fractional part, keep multiplying by 8 till decimal part becomes 0,
∴ (555.75)10 = (1053.6)8
Question 48 |
1, 0, B | |
1, 0, A | |
0, 1, B | |
0, 1, A |
Question 48 Explanation:
In MUX1, the equation is
g = Ax + Bz'
In MUX2, the equation is
f = xg + yg'
= x(Az+Bz') + y(Az+Bz')'
Function f should be equal to (A+B)'.
Just try to put the values of option (D), i.e., x=0, y=1, z=A,
f = 0(AA+BA') +1(AA+BA')'
= (A+B)'
∴ Option (D) is correct.
g = Ax + Bz'
In MUX2, the equation is
f = xg + yg'
= x(Az+Bz') + y(Az+Bz')'
Function f should be equal to (A+B)'.
Just try to put the values of option (D), i.e., x=0, y=1, z=A,
f = 0(AA+BA') +1(AA+BA')'
= (A+B)'
∴ Option (D) is correct.
Question 49 |
0 | |
103 | |
22 | |
55 |
Question 49 Explanation:
In horizontal microprogramming we need 1 bit for every control word, therefore total bits in horizontal microprogramming
= 20 + 70 + 2 + 10 + 23
= 125
Now lets consider vertical microprogramming. In vertical microprogramming no. of bits required to activate 1 signal in group of N signals, is ⌈log2 N⌉. And in the question 5 groups contains mutually exclusive signals,
group 1 = ⌈log2 20⌉ = 5
group 2 = ⌈log2 70⌉ = 7
group 3 = ⌈log2 2⌉ = 1
group 4 = ⌈log2 10⌉ = 4
group 5 = ⌈log2 23⌉ = 5
Total bits required in vertical microprogramming
= 5 + 7 + 1 + 4 + 5
= 22
So, number of bits saved is
= 125 - 22
= 103
= 20 + 70 + 2 + 10 + 23
= 125
Now lets consider vertical microprogramming. In vertical microprogramming no. of bits required to activate 1 signal in group of N signals, is ⌈log2 N⌉. And in the question 5 groups contains mutually exclusive signals,
group 1 = ⌈log2 20⌉ = 5
group 2 = ⌈log2 70⌉ = 7
group 3 = ⌈log2 2⌉ = 1
group 4 = ⌈log2 10⌉ = 4
group 5 = ⌈log2 23⌉ = 5
Total bits required in vertical microprogramming
= 5 + 7 + 1 + 4 + 5
= 22
So, number of bits saved is
= 125 - 22
= 103
Question 50 |
In a binary tree, for every node the difference between the number of nodes in the left and right subtrees is at most 2. If the height of the tree is h > 0, then the minimum number of nodes in the tree is:
2h-1 | |
2h-1 + 1 | |
2h - 1 | |
2h |
Question 50 Explanation:
Let's take an example,
Above tree satisfies the property given in the question. Hence, only option (B) satisfies it.
Above tree satisfies the property given in the question. Hence, only option (B) satisfies it.
Question 51 |
T(n) = θ(log n) | |
T(n) = θ(√n) | |
T(n) = θ(n) | |
T(n) = θ(n log n) |
Question 51 Explanation:
Apply Master's theorem.
Question 52 |
Let G be a weighted undirected graph and e be an edge with maximum weight in G. Suppose there is a minimum weight spanning tree in G containing the edge e. Which of the following statements is always TRUE?
There exists a cutset in G having all edges of maximum weight | |
There exists a cycle in G having all edges of maximum weight | |
Edge e cannot be contained in a cycle | |
All edges in G have the same weight
|
Question 52 Explanation:
(A) True, because if there is heaviest edge in MST, then there exist a cut with all edges with weight equal to heaviest edge.
(B) False, because the cutset of heaviest edge may contain only one edge.
(C) False. The cutset may form cycle with other edge.
(D) False. Not always true.
(B) False, because the cutset of heaviest edge may contain only one edge.
(C) False. The cutset may form cycle with other edge.
(D) False. Not always true.
Question 53 |
A : count [a[j]]++ and B : count[b[j]]– | |
A : count [a[j]]++ and B : count[b[j]]++ | |
A : count [a[j++]]++ and B : count[b[j]]– | |
A : count [a[j]]++and B : count[b[j++]]– |
Question 53 Explanation:
A: Increments the count by 1 at each index that is equal to the ASCII value of the alphabet, it is pointing at.
B: Decrements the count by 1 at each index that is equal to the ASCII value of the alphabet it is pointing at. Also it increments the loop counter for next iteration.
If one string is permutation of other, there would have been equal increments and decrements at each index of array, and so count should contain zero at each index, that is what the loop checks at last and if any non-zero elements is found, it returns 0 indicating that strings are not anagram to each other.
B: Decrements the count by 1 at each index that is equal to the ASCII value of the alphabet it is pointing at. Also it increments the loop counter for next iteration.
If one string is permutation of other, there would have been equal increments and decrements at each index of array, and so count should contain zero at each index, that is what the loop checks at last and if any non-zero elements is found, it returns 0 indicating that strings are not anagram to each other.
Question 54 |
1, 2, 3, 4, 5, 6, 7 | |
2, 1, 4, 3, 6, 5, 7 | |
1, 3, 2, 5, 4, 7, 6 | |
2, 3, 4, 5, 6, 7, 1 |
Question 54 Explanation:
It is nothing but a pairwise swapping of the linked list.
Question 55 |
A binary search tree contains the numbers 1, 2, 3, 4, 5, 6, 7, 8. When the tree is traversed in pre-order and the values in each node printed out, the sequence of values obtained is 5, 3, 1, 2, 4, 6, 8, 7. If the tree is traversed in post-order, the sequence obtained would be
8, 7, 6, 5, 4, 3, 2, 1 | |
1, 2, 3, 4, 8, 7, 6, 5 | |
2, 1, 4, 3, 6, 7, 8, 5 | |
2, 1, 4, 3, 7, 8, 6, 5 |
Question 55 Explanation:
Pre-order is given as
5, 3, 1, 2, 4, 6, 8, 7
In-order is the sorted sequence, i.e.,
1, 2, 3, 4, 5, 6, 7, 8
And we know that with given inorder and preorder, we can draw a unique BST.
So, the BST will be,
Hence, postorder traversasl sequence is,
2, 1, 4, 3, 7, 8, 6, 5
5, 3, 1, 2, 4, 6, 8, 7
In-order is the sorted sequence, i.e.,
1, 2, 3, 4, 5, 6, 7, 8
And we know that with given inorder and preorder, we can draw a unique BST.
So, the BST will be,
Hence, postorder traversasl sequence is,
2, 1, 4, 3, 7, 8, 6, 5
Question 56 |
Let G be a directed graph whose vertex set is the set of numbers from 1 to 100. There is an edge from a vertex i to a vertex j iff either j = i + 1 or j = 3i. The minimum number of edges in a path in G from vertex 1 to vertex 100 is
4 | |
7 | |
23 | |
99 |
Question 56 Explanation:
Edge set consists of edges from i to j, using either
j = i +1
(or)
j = 3i
Second option will help us reach from 1 to 100 rapidly. The trick to solve this question is to think in reverse way. Instead of finding a path from 1 to 100, try to find a path from 100 to 1.
So, the edge sequence with minimum number of edges is
1 → 3 → 9 → 10 → 11 → 33 → 99 → 100
which consists of 7 edges.
j = i +1
(or)
j = 3i
Second option will help us reach from 1 to 100 rapidly. The trick to solve this question is to think in reverse way. Instead of finding a path from 1 to 100, try to find a path from 100 to 1.
So, the edge sequence with minimum number of edges is
1 → 3 → 9 → 10 → 11 → 33 → 99 → 100
which consists of 7 edges.
Question 58 |
a[j] – a[i] > S | |
a[j] – a[i] < S | |
a[i] – a[j] < S | |
a[i] – a[j] > S
|
Question 58 Explanation:
For some 'i' if we find that difference of (A[j] - A[i] < S) we increment 'j' to make this difference wider so that it becomes equal to S.
If at times difference becomes greater than S we know that it won't reduce further for same 'i' and so we increment the 'i'.
If at times difference becomes greater than S we know that it won't reduce further for same 'i' and so we increment the 'i'.
Question 59 |
only I and II | |
only I and IV | |
only II and III | |
only III and IV |
Question 59 Explanation:
a[i] ≥ b[i]
Since both 'a' and 'b' are sorted in the beginning, there are 'i' elements than or equal to a[i] and similarly 'i' elements smaller than or equal to b[i]. So, a[i] ≥ b[i] means there are 2i elements smaller than or equal to a[i] and hence in the merged array, a[i] will come after these 2i elements. So, c[2i] ≤ a[i].
Similarly, a[i] ≥ b[i] says for b that, there are not more than 2i elements smaller than b[i] in the sorted array. So, b[i] ≤ c[2i].
So, option (C) is correct.
Since both 'a' and 'b' are sorted in the beginning, there are 'i' elements than or equal to a[i] and similarly 'i' elements smaller than or equal to b[i]. So, a[i] ≥ b[i] means there are 2i elements smaller than or equal to a[i] and hence in the merged array, a[i] will come after these 2i elements. So, c[2i] ≤ a[i].
Similarly, a[i] ≥ b[i] says for b that, there are not more than 2i elements smaller than b[i] in the sorted array. So, b[i] ≤ c[2i].
So, option (C) is correct.
Question 60 |
30 sec, 30 sec | |
30 sec, 10 sec | |
42 sec, 42 sec | |
30 sec, 42 sec
|
Question 60 Explanation:
For preemtive,
TAT of P2 = Completion time of P2 - Arrival time of P2
= 33 - 3
= 30 sec
For non-preemptive,
TAT of P2 = Completion time of P2 - Arrival time of P2
= 45 - 3
= 42 sec
TAT of P2 = Completion time of P2 - Arrival time of P2
= 33 - 3
= 30 sec
For non-preemptive,
TAT of P2 = Completion time of P2 - Arrival time of P2
= 45 - 3
= 42 sec
Question 61 |
0 3 5 7 16 55 | |
0 3 5 7 9 16 55 | |
0 5 7 9 16 55 | |
3 5 7 9 16 55 |
Question 61 Explanation:
The cache is 2-way associative, so in a set, there can be 2 block present at a time.
So,
Since, each set has only 2 places, 3 will be thrown out as its the least recently used block. So final content of cache will be
0 5 7 9 16 55
Hence, answer is (C).
So,
Since, each set has only 2 places, 3 will be thrown out as its the least recently used block. So final content of cache will be
0 5 7 9 16 55
Hence, answer is (C).
Question 62 |
(I) and (IV) | |
(II) and (III) | |
(I) and (II) | |
None of the above |
Question 62 Explanation:
If all resources are allocated to one process then deadlock will never occur. So, if we allocate both R1 and R2 to process P1 or both R1 and R2 to process P2 then deadlock will never occur. So when one process will complete its execution then both the resources are allocated to the other process. So, either condition (I) and (II) or (III) and (IV) ensures that deadlock will never occur.
Question 63 |
In a computer system, four files of size 11050 bytes, 4990 bytes, 5170 bytes and 12640 bytes need to be stored. For storing these files on disk, we can use either 100 byte disk blocks or 200 byte disk blocks (but can’t mix block sizes). For each block used to store a file, 4 bytes of bookkeeping information also needs to be stored on the disk. Thus, the total space used to store a file is the sum of the space taken to store the file and the space taken to store the book keeping information for the blocks allocated for storing the file. A disk block can store either bookkeeping information for a file or data from a file, but not both.
What is the total space required for storing the files using 100 byte disk blocks and 200 byte disk blocks respectively?
35400 and 35800 bytes | |
35800 and 35400 bytes | |
35600 and 35400 bytes | |
35400 and 35600 bytes |
Question 63 Explanation:
For 100 bytes block,
11050 = 111 blocks requiring 111×4=444B of bookkeeping into which requires another 5 disk blocks. So, totally 111+5=116 disk blocks. Similarly,
4990 = 50 + ⌈(50×4)/100⌉ = 52
5170 = 52 + ⌈(52×4)/100⌉ = 55
12640 = 127 + ⌈(127×4)/100⌉ = 133
∴ Total no. of blocks required
= 52 + 55 + 133 + 116
= 356
So, total space required
= 35600 Bytes
For 200 Bytes block:
56 + ⌈(56×4)/200⌉ = 58
25 + ⌈(25×4)/200⌉ = 26
26 + ⌈(26×4)/200⌉ = 27
64 + ⌈(64×4)/200⌉ = 66
∴ Total space required,
(58+26+27+66) × 200
= 177 × 200
= 35400 Bytes
11050 = 111 blocks requiring 111×4=444B of bookkeeping into which requires another 5 disk blocks. So, totally 111+5=116 disk blocks. Similarly,
4990 = 50 + ⌈(50×4)/100⌉ = 52
5170 = 52 + ⌈(52×4)/100⌉ = 55
12640 = 127 + ⌈(127×4)/100⌉ = 133
∴ Total no. of blocks required
= 52 + 55 + 133 + 116
= 356
So, total space required
= 35600 Bytes
For 200 Bytes block:
56 + ⌈(56×4)/200⌉ = 58
25 + ⌈(25×4)/200⌉ = 26
26 + ⌈(26×4)/200⌉ = 27
64 + ⌈(64×4)/200⌉ = 66
∴ Total space required,
(58+26+27+66) × 200
= 177 × 200
= 35400 Bytes
Question 64 |
The availability of a complex software is 90%. Its Mean Time Between Failure (MTBF) is 200 days. Because of the critical nature of the usage, the organization deploying the software further enhanced it to obtain an availability of 95%. In the process, the Mean Time To Repair (MTTR) increased by 5 days.
What is the MTBF of the enhanced software?205 days | |
300 days | |
500 days | |
700 days |
Question 64 Explanation:
Note: Out of syllabus.
Question 66 |
p1 invokes p2, p2 invokes either p3, or p4, or p5 | |
p2 invokes p1, and then invokes p3, or p4, or p5 | |
A new module Tc is defined to control the transaction flow. This module Tc first invokes p1 and then invokes p2, p2 in turn invokes p3, or p4, or p5 | |
A new module Tc is defined to control the transaction flow. This module Tc invokes p2, p2 invokes p1, and then invokes p3, or p4, or p5 |
Question 66 Explanation:
Note: Out of syllabus.
Question 67 |
Execute T1 followed by T2 followed by T3 | |
Execute T2, followed by T3; T1 running concurrently throughout | |
Execute T3 followed by T2; T1 running concurrently throughout | |
Execute T3 followed by T2 followed by T1
|
Question 67 Explanation:
T3 followed by T2 followed by T1 will be the correct execution sequence.
In other cases some people will get two times increment, for example,
if we have T1 followed by T2 and if initial commission is 49500, then he is belonging to <50000.
Hence, 49500 * 1.02 = 50490.
Now again he is eligible for second category. So, he will get again increment as,
50490 * 1.04 = 52509.6
So he will get increment two times, but he is eligible for only one slab of commission.
In other cases some people will get two times increment, for example,
if we have T1 followed by T2 and if initial commission is 49500, then he is belonging to <50000.
Hence, 49500 * 1.02 = 50490.
Now again he is eligible for second category. So, he will get again increment as,
50490 * 1.04 = 52509.6
So he will get increment two times, but he is eligible for only one slab of commission.
Question 68 |
6 | |
4 | |
2 | |
0 |
Question 68 Explanation:
SQL query will return:
Total 7 rows are selected.
Where in relational algebra only distinct values of hostels are selected,i.e., 5, 6, 7 (3 rows).
∴ Answer is 7 - 3 =4
Total 7 rows are selected.
Where in relational algebra only distinct values of hostels are selected,i.e., 5, 6, 7 (3 rows).
∴ Answer is 7 - 3 =4
Question 69 |
do not supply any item | |
supply exactly one item | |
supply one or more items | |
supply two or more items |
Question 69 Explanation:
Here (not unique) in nested query ensures that only for those suppliers it return True which supplies more than 1 item in which case supplier id in inner query will be repeated for that supplier. Hence, the answer is (D) which supply two or more items.
Question 70 |
CD → AC | |
BD → CD | |
BC → CD | |
AC → BC |
Question 70 Explanation:
Apply membership test for all the functional dependencies.
Option (B):
BD → CD
BD+ = BD
i.e., BD cannot derive CD and hence is not implied.
Option (B):
BD → CD
BD+ = BD
i.e., BD cannot derive CD and hence is not implied.
Question 71 |
A network with CSMA/CD protocol in the MAC layer is running at 1 Gbps over a 1 km cable with no repeaters. The signal speed in the cable is 2 × 108 m/sec. The minimum frame size for this network should be
10000 bits | |
10000 bytes | |
5000 bits | |
5000 bytes |
Question 71 Explanation:
For CSMA/CD protocol we know that minimum frame size required is,
L ≤ 2×Tp×B
L ≤ 2×(d/v)×B
d = 1Km = 1000m
v = 2×103 m/s
B = 109 bps
By solving the above equation we will set the value of L as,
10000 bits.
L ≤ 2×Tp×B
L ≤ 2×(d/v)×B
d = 1Km = 1000m
v = 2×103 m/s
B = 109 bps
By solving the above equation we will set the value of L as,
10000 bits.
Question 72 |
A channel has a bit rate of 4 kbps and one-way propagation delay of 20 ms. The channel uses stop and wait protocol. The transmission time of the acknowledgement frame is negligible. To get a channel efficiency of at least 50%, the minimum frame size should be
80 bytes | |
80 bits | |
160 bytes | |
160 bits |
Question 72 Explanation:
Question 73 |
On a TCP connection, current congestion window size is Congestion Window=4 KB. The window size advertised by the receiver is Advertise Window=6 KB. The last byte sent by the sender is LastByteSent=10240 and the last byte acknowledged by the receiver is LastByteAcked=8192. The current window size at the sender is
2048 bytes | |
4096 bytes | |
6144 bytes | |
8192 bytes |
Question 73 Explanation:
Corrent sender window
= min (4KB, 6KB)
= 4KB
= min (4KB, 6KB)
= 4KB
Question 74 |
In a communication network, a packet of length L bits takes link L1 with a probability of p1 or link L2 with a probability of p2. Link L1 and L2 have bit error probability of b1and b2 respectively. The probability that the packet will be received without error via either L1 or L2 is
(1 – b1)L p1 + (1 – b2)Lp2 | |
[1 – (b1 + b2)L]p1p2 | |
(1 – b1)L (1 – b2)Lp1p2 | |
1 – (b1 Lp1 + b2 Lp2) |
Question 74 Explanation:
Probability of choosing link L1 = p1
Probability for no bit error for any single bit = (1 - b1)
Similarly, for link L2,
Probability of no bit error = (1 - b2)
Packet can go either through link L1 or L2, they are mutually exclusive events.
Probability packet will be received without any error = Probability of L1 being chosen and no errors in any of L bits + Probability of L2 being chosen and no error in any of the L bits
= (1 - b1)L p1 + (1 - b2)L p2
Hence, answer is option A.
Probability for no bit error for any single bit = (1 - b1)
Similarly, for link L2,
Probability of no bit error = (1 - b2)
Packet can go either through link L1 or L2, they are mutually exclusive events.
Probability packet will be received without any error = Probability of L1 being chosen and no errors in any of L bits + Probability of L2 being chosen and no error in any of the L bits
= (1 - b1)L p1 + (1 - b2)L p2
Hence, answer is option A.
Question 75 |
In a TDM medium access control bus LAN, each station is assigned one time slot per cycle for transmission. Assume that the length of each time slot is the time to transmit 100 bits plus the end-to-end propagation delay. Assume a propagation speed of 2 × 108 m/sec. The length of the LAN is 1 km with a bandwidth of 10 Mbps. The maximum number of stations that can be allowed in the LAN so that the throughput of each station can be 2/3 Mbps is
3 | |
5 | |
10 | |
20 |
Question 75 Explanation:
Question 76 |
A company has a class C network address of 204.204.204.0. It wishes to have three subnets, one with 100 hosts and two with 50 hosts each. Which one of the following options represents a feasible set of subnet address/subnet mask pairs?
204.204.204.128/255.255.255.192 204.204.204.0/255.255.255.128 204.204.204.64/255.255.255.128 | |
204.204.204.0/255.255.255.192 204.204.204.192/255.255.255.128 204.204.204.64/255.255.255.128 | |
204.204.204.128/255.255.255.128 204.204.204.192/255.255.255.192 204.204.204.224/255.255.255.192 | |
204.204.204.128/255.255.255.128 204.204.204.64/255.255.255.192 204.204.204.0/255.255.255.192 |
Question 76 Explanation:
MSB in last 8 bits is in use to get subnet since it is class C IP.
10000000/128 (mask) - subnet id bit (1) (subnet 1)
01000000/192 (mask) - subnet id bit (01) (subnet 2)
0000000/192 (mask) - subnet id bit (00) (subnet 3)
10000000/128 (mask) - subnet id bit (1) (subnet 1)
01000000/192 (mask) - subnet id bit (01) (subnet 2)
0000000/192 (mask) - subnet id bit (00) (subnet 3)
Question 77 |
Assume that “host1.mydomain.dom” has an IP address of 145.128.16.8. Which of the following options would be most appropriate as a subsequence of steps in performing the reverse lookup of 145.128.16.8? In the following options “NS” is an abbreviation of “nameserver”.
Query a NS for the root domain and then NS for the “dom” domains | |
Directly query a NS for “dom” and then a NS for “mydomain.dom” domains | |
Query a NS for in-addr.arpa and then a NS for 128.145.in-addr.arpa domains | |
Directly query a NS for 145.in-addr.arpa and then a NS for 128.145.in-addr.arpa domains |
Question 77 Explanation:
We are performing reverse lookup of IP address to its hostname.
First we need to locate in-addr.apra, then perform reverse lookup of 8.16.128.145.in-addr.arpa which will point to host1.mydomain.com.
First we need to locate in-addr.apra, then perform reverse lookup of 8.16.128.145.in-addr.arpa which will point to host1.mydomain.com.
Question 78 |
Consider the following message M = 1010001101. The cyclic redundancy check (CRC) for this message using the divisor polynomial x5 + x4 + x2 + 1 is :
01110 | |
01011 | |
10101 | |
10110 |
Question 78 Explanation:
Degree of generator polynomial is 5. Hence, 5 zeroes is appended before division.
M = 1010001101
append 5 zeroes = M = 101000110100000
∴ CRC = 01110
M = 1010001101
append 5 zeroes = M = 101000110100000
∴ CRC = 01110
Question 79 |
Suppose that two parties A and B wish to setup a common secret key (D-H key) between themselves using the Diffie-Hellman key exchange technique. They agree on 7 as the modulus and 3 as the primitive root. Party A chooses 2 and party B chooses 5 as their respective secrets. Their D-H key is
3 | |
4 | |
5 | |
6 |
Question 79 Explanation:
For Diffe-Hellaman the secret key is (pab)modn,
where p is the primitive root and n is the modulus and 'a' and 'b' are the secret values selected by parity A & B.
So answer is,
32×5 mod 7 = 310 mod 7 = 4
where p is the primitive root and n is the modulus and 'a' and 'b' are the secret values selected by parity A & B.
So answer is,
32×5 mod 7 = 310 mod 7 = 4
Question 81 |
(i) 80 MB (ii) 2040 MB | |
(i) 2040 MB (ii) 80 MB | |
(i) 80 MB (ii) 360 MB | |
(i) 80 MB (ii) 360 MB |
Question 81 Explanation:
Constant linear velocity:
Diameter of inner track = d = 1 cm
Circumference of inner track
= 2 * 3.14 * d/2
= 3.14 cm
Storage capacity = 10 MB (given)
Circumference of all equidistant tracks
= 2 * 3.14 * (0.5+1+1.5+2+2.5+3+3.5+4)
= 113.14 cm
Here, 3.14 cm holds 10 MB
Therefore, 1 cm holds 3.18 MB.
So, 113.14 cm holds
113.14 * 3.18 = 360 MB
So, total amount of data that can be hold on the disk = 360 MB.
For constant angular velocity:
In case of CAV, the disk rotates at a constant angular speed. Same rotation time is taken by all the tracks.
Total amount of data that can be stored on the disk
= 8 * 10 = 80 MB
Diameter of inner track = d = 1 cm
Circumference of inner track
= 2 * 3.14 * d/2
= 3.14 cm
Storage capacity = 10 MB (given)
Circumference of all equidistant tracks
= 2 * 3.14 * (0.5+1+1.5+2+2.5+3+3.5+4)
= 113.14 cm
Here, 3.14 cm holds 10 MB
Therefore, 1 cm holds 3.18 MB.
So, 113.14 cm holds
113.14 * 3.18 = 360 MB
So, total amount of data that can be hold on the disk = 360 MB.
For constant angular velocity:
In case of CAV, the disk rotates at a constant angular speed. Same rotation time is taken by all the tracks.
Total amount of data that can be stored on the disk
= 8 * 10 = 80 MB
Question 82 |
13.5 ms | |
10 ms | |
9.5 ms | |
20 ms |
Question 82 Explanation:
Radius of inner track is 0.5cm (where the head is standing) and the radius of outermost track is 4cm.
So, the header has to seek (4 - 0.5) = 3.5cm.
For 10m ------- 1s
1m ------- 1/10 s
100cm ------- 1/(10×100) s
3.5cm ------- 3.5/1000 s = 3.5ms
So, the header will take 3.5ms.
Now, angulur velocity is constant and header is now at end of 5th sector. To start from front of 4th sector it must rotate upto 18 sector.
6000 rotation in 60000ms.
1 rotation in 10ms (time to traverse 20 sectors).
So, to traverse 18 sectors, it takes 9ms.
In 10ms, 10MB data is read.
So, 1MB data can be read in 1ms.
∴ Total time = 1+9+3.5 = 13.5ms
So, the header has to seek (4 - 0.5) = 3.5cm.
For 10m ------- 1s
1m ------- 1/10 s
100cm ------- 1/(10×100) s
3.5cm ------- 3.5/1000 s = 3.5ms
So, the header will take 3.5ms.
Now, angulur velocity is constant and header is now at end of 5th sector. To start from front of 4th sector it must rotate upto 18 sector.
6000 rotation in 60000ms.
1 rotation in 10ms (time to traverse 20 sectors).
So, to traverse 18 sectors, it takes 9ms.
In 10ms, 10MB data is read.
So, 1MB data can be read in 1ms.
∴ Total time = 1+9+3.5 = 13.5ms
Question 83 |
800000 | |
40080 | |
32020 | |
100 |
Question 83 Explanation:
To minimize block accesses, we have to put that table outside having less records because for each outer record one block accesses inside will be required.
Therefore,
putting 2nd table outside,
for every 400 records {
80 block accesses in first table
}
= 32000 blocks
and also 20 blocks accesses of outer table.
So, answer comes out to be,
32000 + 20 = 32020
Therefore,
putting 2nd table outside,
for every 400 records {
80 block accesses in first table
}
= 32000 blocks
and also 20 blocks accesses of outer table.
So, answer comes out to be,
32000 + 20 = 32020
Question 84 |
0 | |
30400 | |
38400 | |
798400 |
Question 84 Explanation:
In Nested loop join the no. of block access will be,
400×80+20
= 32020 (calculated in previous question)
In block Nested loop join, the no. of block access will be
20×80+20 = 1620
∴ The difference is = 32020 - 1620 = 30400
400×80+20
= 32020 (calculated in previous question)
In block Nested loop join, the no. of block access will be
20×80+20 = 1620
∴ The difference is = 32020 - 1620 = 30400
Question 85 |
id + id + id + id | |
id + (id* (id * id)) | |
(id* (id * id)) + id | |
((id * id + id) * id) |
Question 85 Explanation:
Let's draw more than one possible tree for id + id + id + id.
Question 86 |
5 | |
4 | |
3 | |
2 |
Question 86 Explanation:
Total 5 parse trees possible (solved in previous question).
Question 87 |
E1 : A[i][j] and E2 : i = j; | |
E1 : !A[i][j] and E2 : i = j + 1; | |
E1: !A[i][j] and E2 : i = j; | |
E1 : A[i][j] and E2 : i = j + 1;
|
Question 87 Explanation:
If there is a sink in the graph, the adjacency matrix will contain all 1's (except diagonal) in one column and all 0's (except diagonal) in the corresponding row of that vertex. The given algorithm is a smart way of doing this as it finds the sink in O(n) time complexity.
The first part of the code, is finding if there is any vertex which doesn't have any outgoing edge to any vertex coming after it in adjacency matrix. The smart part of the code is E2, which makes rows slip when there is no edge from i to it, making it impossible for them to form a sink. This is done through
E1: A[i][j]
and
E2: i = j
E1 makes sure that there is no edge from i to j and i is a potential sink till A[i][j] becomes 1. If A[i][j] becomes 1, i can no longer be a sink. Similarly, all previous j can also not be a sink. Now, the next potential candidate for sink is j. So, in E2, we must make i = j.
So, answer is (C).
The first part of the code, is finding if there is any vertex which doesn't have any outgoing edge to any vertex coming after it in adjacency matrix. The smart part of the code is E2, which makes rows slip when there is no edge from i to it, making it impossible for them to form a sink. This is done through
E1: A[i][j]
and
E2: i = j
E1 makes sure that there is no edge from i to j and i is a potential sink till A[i][j] becomes 1. If A[i][j] becomes 1, i can no longer be a sink. Similarly, all previous j can also not be a sink. Now, the next potential candidate for sink is j. So, in E2, we must make i = j.
So, answer is (C).
Question 88 |
(A[i][j] && !A[j][i]) | |
(!A[i][j] && A[j][i]) | |
(!A[i][j] | | A[j][i]) | |
(A[i][j] | | !A[j][i]) |
Question 88 Explanation:
First go through the explanation of previous question.
Now, the loop breaks when we found a potential sink-that is a vertex which does not have any outgoing edge to any coming after it in adjacency matrix.
So, if the column in which this vertex comes is all 1's and the row is all 0's (except diagonal), this is the sink. Otherwise there is no sink in the graph. So, E3 is checking this condition. But in the code flag is used for storing the state that sink is present or not. And as per the usage of flag in code, by default sink is considered present.
So, the condition is E3 must make flag = 0, if the found i is not a sink. So, the condition should be
A[i][j] | | !A[j][i]
So, option (D) is the answer.
Now, the loop breaks when we found a potential sink-that is a vertex which does not have any outgoing edge to any coming after it in adjacency matrix.
So, if the column in which this vertex comes is all 1's and the row is all 0's (except diagonal), this is the sink. Otherwise there is no sink in the graph. So, E3 is checking this condition. But in the code flag is used for storing the state that sink is present or not. And as per the usage of flag in code, by default sink is considered present.
So, the condition is E3 must make flag = 0, if the found i is not a sink. So, the condition should be
A[i][j] | | !A[j][i]
So, option (D) is the answer.
Question 90 |
>100 but finite | |
∞ | |
3 | |
>3 and ≤100 |
Question 90 Explanation:
We consider ABDF at t, they are:
The distance between A and the nodes B, D, F respectively are:
t: 1 2 3
t + 1: 3 2 3
t + 2: 3 4 3
t + 3: 5 4 5
t + 4: 5 6 5
t + 5: 7 6 7
t + 6: 7 8 7
t + 7: 9 8 9
t + 8: 9 to 10
and this continues.
So, in every two steps they get incremented by 2.
So,
at t + 99, F is 101
at t + 100, F is 101.
The distance between A and the nodes B, D, F respectively are:
t: 1 2 3
t + 1: 3 2 3
t + 2: 3 4 3
t + 3: 5 4 5
t + 4: 5 6 5
t + 5: 7 6 7
t + 6: 7 8 7
t + 7: 9 8 9
t + 8: 9 to 10
and this continues.
So, in every two steps they get incremented by 2.
So,
at t + 99, F is 101
at t + 100, F is 101.
There are 90 questions to complete.