UGC NET CS 2009-June-Paper-2
Question 1 |
If x and y are independent Gaussian random variables with average value 0 and with same variance, their joint probability density function is given by :
p(x, y)=p(x).p(y) | |
p(x, y)=p(x)+p(y) | |
p(x, y)=p(x+y) | |
p(x, y)=p(x).p(y)+p(x) |
Question 1 Explanation:
As they are independent, answer could be Option-A, though it need to be checked with zero average too.
Note: Original key given option-C is correct answer
Note: Original key given option-C is correct answer
Question 2 |
In order that a code is ‘t’ error correcting, the minimum Hamming distance should be :
t | |
2t-1 | |
2t | |
2t+1 |
Question 2 Explanation:

Question 3 |
The boolean expression x'y'z+yz+xz is equivalent to:
x | |
y | |
z | |
x+y+z |
Question 3 Explanation:

Question 4 |
The characteristic equation of a JK flip flop is :
Qn+1 = J.Qn +K.Qn | |
Qn+1 = J.Q'n +K'Qn | |
Qn+1= QnJ.K | |
Qn+1= (J+K)Qn |
Question 4 Explanation:
The characteristic equation of a JK flip flop is Qn+1 = J.Q'n +K'Qn
Characteristic Table:


Question 5 |
In order to implement a n variable switching function, a MUX must have :
2n inputs | |
2n+1 inputs | |
2n-1 inputs | |
2n-1 inputs |
Question 5 Explanation:
In order to implement a n variable switching function, a MUX must have 2n inputs.
Question 6 |
The throughput of pure ALOHA is given by :
S=G | |
S=e2G | |
S=Ge2G | |
S=Ge-2G |
Question 6 Explanation:
→ The throughput of pure ALOHA is given by S=Ge-2G
→ The throughput of pure ALOHA is given by S=Ge-G
→ The throughput of pure ALOHA is given by S=Ge-G
Question 7 |
The Fiber Distributed Data Interface(FDDI) uses :
single mode fibers and LEDs | |
multimode fibers and LEDs | |
single mode fibers and ILDs | |
multimode fibers and ILDs |
Question 7 Explanation:
→ FDDI uses a multimode fiber because the cost of single mode fiber is not justified for networks running at only 100 Mbps.
→ It also uses LEDs instead of Lasers not only because of the lower cost but also because FDDI may sometimes be used to connect directly to user workstations, and safety against exposure to LASER radiation is difficult to maintain in that case.
→ It also uses LEDs instead of Lasers not only because of the lower cost but also because FDDI may sometimes be used to connect directly to user workstations, and safety against exposure to LASER radiation is difficult to maintain in that case.
Question 8 |
To employ multi-access in GSM, users are given different :
time slots | |
bandpass filters | |
handsets | |
frequency bands |
Question 8 Explanation:
To employ multi-access in GSM, users are given different frequency bands.
Question 9 |
With a four programs in memory and with 80% average I/O wait, the CPU utilization is ?
60% | |
70% | |
90% | |
100% |
Question 9 Explanation:
Given data
-- Average I/O wait=80%
-- Total programs in memory=4
-- CPU Utilization=?
Step-1: To find CPU utilization, we have standard formula is 1-pn
Where,
-- ‘n’ is number of process
-- ‘P’ is waiting time for I/O.
Step-2: CPU Utilization=1-0.84
= 1-0.4096
= 0.5904 (approximation is 60%)
-- Average I/O wait=80%
-- Total programs in memory=4
-- CPU Utilization=?
Step-1: To find CPU utilization, we have standard formula is 1-pn
Where,
-- ‘n’ is number of process
-- ‘P’ is waiting time for I/O.
Step-2: CPU Utilization=1-0.84
= 1-0.4096
= 0.5904 (approximation is 60%)
Question 10 |
Assume N segments in memory and a page size of P bytes. The wastage on account of internal fragmentation is :
NP/2 bytes | |
P/2 Bytes | |
N/2 Bytes | |
NP Bytes |
Question 10 Explanation:
→ The wastage on account of internal fragmentation is NP/2 bytes.
where,
Segments in memory=N
Page size= P bytes.
where,
Segments in memory=N
Page size= P bytes.
Question 11 |
Assertion (A) : Bit maps are not often used in memory management.
Reason (R) :Searching a bitmap for a run of given length is a slow operation.
Reason (R) :Searching a bitmap for a run of given length is a slow operation.
Both (A) and (R) are true and (R) is correct explanation for (A) | |
Both (A) and (R) are true but (R) is not correct explanation for (A) | |
(A) is true (R) is false | |
(A) is false (R) is true |
Question 11 Explanation:
→ Bit maps are not often used in memory management because searching a bitmap for a run of given length is a slow operation.
Question 12 |
The complete graph with four vertices has k edges where k is :
3 | |
4 | |
5 | |
6 |
Question 12 Explanation:
To find number of edges in complete graph using standard formula is n(n-1)/2

n=4
= 4(4-3)/2
= (4*3)/2
= 12/2
= 6

n=4
= 4(4-3)/2
= (4*3)/2
= 12/2
= 6
Question 13 |
The octal equivalent of hexadecimal (A.B)16 is:
47.21 | |
12.74 | |
12.71 | |
17.21 | |
None of the above |
Question 13 Explanation:

Question 14 |
A reduced state table has 18 rows. The minimum number of Flips flops needed to implement the sequential machine is :
18 | |
9 | |
5 | |
4 |
Question 14 Explanation:
Mod value of a counter is the number of unique states.
Let K = Number of states =18.
Number of flip-flops is the smallest number which is greater than or equal to L log2K
log218 = 4.1.
L=5>=4.1.
So number of flip-flops needed is 5.
Let K = Number of states =18.
Number of flip-flops is the smallest number which is greater than or equal to L log2K
log218 = 4.1.
L=5>=4.1.
So number of flip-flops needed is 5.
Question 15 |
What is the value of ‘b’ after the execution of the following code statements :
c=10;
b=+ +c+ + +c;
20 | |
22 | |
23 | |
None |
Question 15 Explanation:
It will give error message because they are not given in proper syntax.
Suppose when we are taken (++c)+(++c);
→ It will give result 23 because pre/post increment will evaluate from right to left.
(++c)+(++c); /* It will increment by one. It means 11 */
(++c)+(++c); /* Before c=11 and now we are incrementing by 1. It will become 23. */
→ It will give result 23 because pre/post increment will evaluate from right to left.
(++c)+(++c); /* It will increment by one. It means 11 */
(++c)+(++c); /* Before c=11 and now we are incrementing by 1. It will become 23. */
Question 16 |
Which of the following does not represent a valid storage class in ‘c’ ?
automatic | |
static | |
union | |
extern |
Question 16 Explanation:
→ Storage classes are auto,static,register an extern.
→ Union is not a storage class.
→ Union is not a storage class.
Question 17 |
The friend functions are used in situations where :
We want to have access to unrelated classes | |
Dynamic binding is required | |
Exchange of data between classes to take place | |
None of the above |
Question 17 Explanation:
A friend function of a class is defined outside that class scope but it has the right to access all private and protected members of the class. Even though the prototypes for friend functions appear in the class definition, friends are not member functions.
1. A function can only be declared a friend by a class itself.
2. It can have access to all members of the class, even private ones.
1. A function can only be declared a friend by a class itself.
2. It can have access to all members of the class, even private ones.
Question 18 |
(i) DML includes a query language based on both relation algebra and tuple calculus
(ii) DML includes a query language based on tuple calculus
(iii) DML includes a query language based on relational algebra
(iv) DML includes a query language based on none of the relational algebra and tuple calculus
Which one is correct ?
(ii) DML includes a query language based on tuple calculus
(iii) DML includes a query language based on relational algebra
(iv) DML includes a query language based on none of the relational algebra and tuple calculus
Which one is correct ?
(i) only | |
(ii) only | |
(iii) only | |
(iv) only |
Question 18 Explanation:
→ DML(Data Manipulation Language) performs to SELECT,UPDATE,INSERT and DELETE operations.
→ DML includes a query language based on both relation algebra and tuple calculus
→ DML includes a query language based on both relation algebra and tuple calculus
Question 19 |
Suppose it takes 100 ns to access a page table and 20 ns to access associative memory with a 90% hit rate, the average access time equals :
20 ns | |
28 ns
| |
90 ns | |
100 ns |
Question 19 Explanation:
Given data,
-- Access page table time=100 ns
-- Associate memory=20 ns
-- hit ratio=90% = 0.9
-- Miss ratio=1-hit ration
= 10% =0.1
-- Average Access Time=?
Step-1: AAT= Hit Ratio*Access page table + Miss Ratio*Associate memory
= 0.9*20 + 0.1*100
= 28 ns
-- Access page table time=100 ns
-- Associate memory=20 ns
-- hit ratio=90% = 0.9
-- Miss ratio=1-hit ration
= 10% =0.1
-- Average Access Time=?
Step-1: AAT= Hit Ratio*Access page table + Miss Ratio*Associate memory
= 0.9*20 + 0.1*100
= 28 ns
Question 20 |
There exists a construct which returns a value ‘true’ if the argument subquery is :
empty | |
non-empty | |
in error | |
none of the above |
Question 20 Explanation:
The exists construct returns the value true if the argument subquery is nonempty.
1. exists r ⇔ r ≠ ∅
2. not exists r ⇔ r = ∅
1. exists r ⇔ r ≠ ∅
2. not exists r ⇔ r = ∅
Question 21 |
There exists a construct which returns a value ‘true’ if the argument subquery is :
empty | |
non-empty | |
in error | |
none of the above |
Question 21 Explanation:
The exists construct returns the value true if the argument subquery is nonempty.
1. exists r ⇔ r ≠ ∅
2. not exists r ⇔ r = ∅
1. exists r ⇔ r ≠ ∅
2. not exists r ⇔ r = ∅
Question 22 |
Which construct in SQL is used to test whether a subquery has any tuples in its result ?
UNIQUE | |
EXISTS | |
GROUP BY
| |
EXCEPT |
Question 22 Explanation:
EXISTS:The EXISTS function in SQL is used to check whether the result of a correlated
nested query is empty (contains no tuples) or not.
UNIQUE: UNIQUE(Q), which returns TRUE if there are no duplicate tuples in the result of query ; otherwise, it returns FALSE. This can be
used to test whether the result of a nested query is a set or a multiset.
GROUP BY: The GROUP BY clause specifies the grouping attributes, which should also appear in the SELECT clause, so that the value resulting from applying each aggregate function to a group of tuples appears along with the value of the grouping attribute(s).
EXCEPT: The EXCEPT query is like the set difference operation in SQL. The difference between EXCEPT and EXCEPT ALL is EXCEPT do not allow duplicates while EXCEPT ALL allows duplicates in the result of a query.
Hence correct answer is OPTION(B)
UNIQUE: UNIQUE(Q), which returns TRUE if there are no duplicate tuples in the result of query ; otherwise, it returns FALSE. This can be
used to test whether the result of a nested query is a set or a multiset.
GROUP BY: The GROUP BY clause specifies the grouping attributes, which should also appear in the SELECT clause, so that the value resulting from applying each aggregate function to a group of tuples appears along with the value of the grouping attribute(s).
EXCEPT: The EXCEPT query is like the set difference operation in SQL. The difference between EXCEPT and EXCEPT ALL is EXCEPT do not allow duplicates while EXCEPT ALL allows duplicates in the result of a query.
Hence correct answer is OPTION(B)
Question 23 |
Which two of the following are equivalent for an undirected graph G ?
(i) G is a tree
(ii) There is at least one path between any two distinct vertices of G
(iii) G contains no cycles and has (n-1) edges
(iv) G has n edges
(i) G is a tree
(ii) There is at least one path between any two distinct vertices of G
(iii) G contains no cycles and has (n-1) edges
(iv) G has n edges
(i) and (ii) | |
(i) and (iii) | |
(i) and (iv) | |
(ii) and (iii) |
Question 23 Explanation:
Option-A: FALSE because undirected graph will be cycle also.
Option-B: TRUE because there is at least one path between any two distinct vertices of G. otherwise we are no calling as undirected graph.
Option-C: TRUE. If G contains no cycles and has (n-1) edges.
Option-D: FALSE
Option-B: TRUE because there is at least one path between any two distinct vertices of G. otherwise we are no calling as undirected graph.
Option-C: TRUE. If G contains no cycles and has (n-1) edges.
Option-D: FALSE
Question 24 |
In a B tree of order m with p nodes the average number of splits is at most
1/⌈(m/2)⌉-1) | |
(⌈(m/2)⌉-1) | |
1/⌈m/2⌉ | |
None |
Question 24 Explanation:
Leaf Nodes=p
keys=m
To split the node, assume that following the insertion of the new key, p has the format m,A0,(K1,A1),...,(Km,Am), and Kii+1, 1<=i
The node is split into two nodes, p and q, with the following formats:
node p: ⌈(m/2)⌉-1, A0(K1A1<`/sub>), ..., (K⌈(m/2)⌉-1, A⌈(m/2)⌉-1)
node q: m-⌈(m/2)⌉, A⌈(m/2)⌉, (K⌈(m/2)⌉+1, A⌈(m/2)⌉+1), ..., (Km,Am)
The remaining key, K⌈(m/2)⌉-1, and a pointer to the new node,q, from a tuple(K⌈(m/2)⌉,q).
→ This is to be inserted into the parent of p. Before attempting this the nodes p and q are written to disk.
keys=m
To split the node, assume that following the insertion of the new key, p has the format m,A0,(K1,A1),...,(Km,Am), and Ki
node p: ⌈(m/2)⌉-1, A0(K1A1<`/sub>), ..., (K⌈(m/2)⌉-1, A⌈(m/2)⌉-1)
node q: m-⌈(m/2)⌉, A⌈(m/2)⌉, (K⌈(m/2)⌉+1, A⌈(m/2)⌉+1), ..., (Km,Am)
The remaining key, K⌈(m/2)⌉-1, and a pointer to the new node,q, from a tuple(K⌈(m/2)⌉,q).
→ This is to be inserted into the parent of p. Before attempting this the nodes p and q are written to disk.
Question 25 |
The propositional formula given by the tree :


∧ ∨ x2 ∨ x1¬x1¬x1 | |
(x2∨ ¬x2) ∧ (x1 ∨ x2) | |
(¬x1∨ x2) ∧ (¬x1∨ x2) | |
None |
Question 25 Explanation:
By default we are using inorder traversal to evaluate it.
Inorder traversal starts from left,root and right.
Inorder traversal = (¬x1∨ x2) ∧ (¬x1∨ x2)
Inorder traversal starts from left,root and right.
Inorder traversal = (¬x1∨ x2) ∧ (¬x1∨ x2)
Question 26 |
Queue is a __________ list.
LIFO | |
LILO | |
FILO | |
FIFO |
Question 26 Explanation:
Queue is FIFO(First In First Out) and Stack is LIFO(Last In First Out).
Question 27 |
In a full binary tree of height k, there are __________ internal nodes.
2k-1
| |
2k-1 | |
2k | |
2k+1 |
Question 27 Explanation:
→ In a full binary tree of height k, there are 2k-1 internal nodes.
→ A Binary Tree is full if every node has 0 or 2 children.
→ In a Full Binary, number of leaf nodes is number of internal nodes plus 1
L = I + 1
Where,
L = Number of leaf nodes,
I = Number of internal nodes.
→ A Binary Tree is full if every node has 0 or 2 children.
→ In a Full Binary, number of leaf nodes is number of internal nodes plus 1
L = I + 1
Where,
L = Number of leaf nodes,
I = Number of internal nodes.
Question 28 |
A binary tree is said to have heap property if the elements along any path :
from leaf to root are non–increasing | |
from leaf to root are non–decreasing | |
from root to leaf are non–decreasing | |
from root to leaf are non–increasing |
Question 28 Explanation:
A heap is a complete binary tree. A binary tree is said to have heap property if the elements along any path from root to leaf are non–increasing
Question 29 |
X.25 protocol consists of :
Physical and Frame levels | |
Frame and Packet levels | |
Physical, Frame and Packet levels | |
None of the above |
Question 29 Explanation:
→ X.25 originally defined three basic protocol levels or architectural layers.
1. Physical layer(bits)
2. Data link layer(frames)
3. Network layer(packet)
→ X.25 protocol consists of Physical, Frame and Packet levels.
1. Physical layer(bits)
2. Data link layer(frames)
3. Network layer(packet)
→ X.25 protocol consists of Physical, Frame and Packet levels.
Question 30 |
GSM/CDMA systems :
are limited to very low speed data | |
require no local loop wires | |
are predominantly used for voice | |
all of the above | |
Option A & C |
Question 30 Explanation:
→ GSM offers very low speed data service because mobility management in wide area network is the main focus in GSM.
→ GSM/CDMA systems are predominantly used for voice.
→ GSM/CDMA systems are predominantly used for voice.
Question 31 |
Usually information security in a network is achieved by :
Layering | |
Cryptography | |
Grade of service | |
None of the above |
Question 31 Explanation:
Usually information security in a network is achieved by cryptography because very difficult to decrypt messages.
Question 32 |
The linker :
is similar to interpreter | |
uses source code as its input | |
is required to create a load module | |
none of the above |
Question 32 Explanation:
The linker is required to create a load module. Object modules generated by assembler that contains unresolved external references are resolved for two or more object module by a Linker.
Question 33 |
In which addressing mode the operand is given explicitly in the instruction itself ?
Absolute mode | |
Immediate mode | |
Indirect mode | |
Index mode |
Question 33 Explanation:
Immediate addressing mode the operand is given explicitly in the instruction itself.
Question 34 |
A compiler that runs on one machine and produces code for a different machine is called:
Cross compilation | |
One pass compilation | |
Two pass compilation | |
None of the above |
Question 34 Explanation:
Cross compiler: The compiler used to compile a source code for different kinds platform.
Note: We have an one and two pass assemblers but not compilers.
Note: We have an one and two pass assemblers but not compilers.
Question 35 |
Any syntactic construct that can be described by a regular expression can also be described by a :
Context sensitive grammar | |
Non context free grammar | |
Context free grammar | |
None of the above |
Question 35 Explanation:
Context-free Grammar (CFG) can be recognized by finite state automata and linear bounded automata(LBA).
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Question 36 |
Find the odd man out :
tail | |
cut | |
wart | |
sed |
Question 36 Explanation:
tail,cut and sed belongs to one category.
wart→ caused by a virus. So, it is odd one.
wart→ caused by a virus. So, it is odd one.
Question 37 |
Which of the following changes permission to deny write permission to group and others?
Chmod go-w filex | |
Chmod go w filex | |
Chmod go=w filex | |
None of the above |
Question 37 Explanation:
Key terms in chmod command:
‘+' to add permissions.
‘-' to remove permissions.
‘=' to assign permissions absolutely.
‘r’ to read permission
‘w’ to write permission
‘x’ to execute permissions
‘g’ to group permissions
‘o’ to others permissions
chmod go -w filex → Removing write permission to group and others
chmod go=w filex → Giving all rwx permissions to groups and others
‘+' to add permissions.
‘-' to remove permissions.
‘=' to assign permissions absolutely.
‘r’ to read permission
‘w’ to write permission
‘x’ to execute permissions
‘g’ to group permissions
‘o’ to others permissions
chmod go -w filex → Removing write permission to group and others
chmod go=w filex → Giving all rwx permissions to groups and others
Question 38 |
Variable partition memory management technique with compaction results in :
Reduction of fragmentation | |
Minimal wastage | |
Segment sharing | |
None of the above |
Question 38 Explanation:
Variable partition memory management technique with compaction results in reduction of fragmentation
Question 39 |
Capability Maturity Model is meant for :
Product | |
Process | |
Product and Process | |
None of the above |
Question 39 Explanation:
CMM stands for Capability Maturity Model is a process model which specifies the process improvement approach in software development.
CMM levels:
1. Initial
2. Repeatable
3. Defined
4. Managed
5. Optimizing
CMM levels:
1. Initial
2. Repeatable
3. Defined
4. Managed
5. Optimizing
Question 40 |
In the light of software engineering software consists of :
Programs | |
Data | |
Documentation | |
All of the above |
Question 40 Explanation:
In the light of software engineering software consists of programs,data and documentation.
Question 41 |
Which one of the following ISO standard is used for software process ?
ISO 9000 | |
ISO 9001 | |
ISO 9003 | |
ISO 9000–3 |
Question 41 Explanation:
ISO 9003 is "Quality systems-Model for quality assurance in final inspection and test".
ISO 9001 is a non-industry specific certification.
ISO 9000 refers to the family of standards including ISO 9000, ISO 9001, ISO 9004 and ISO 19011.
ISO 9001 is a non-industry specific certification.
ISO 9000 refers to the family of standards including ISO 9000, ISO 9001, ISO 9004 and ISO 19011.
Question 42 |
Which of the following is used for test data generation ?
White Box | |
Black Box | |
Boundary value analysis | |
All of the above |
Question 42 Explanation:
Boundary testing is the process of testing between extreme ends or boundaries between partitions of the input values. It is used to test data generation.
Question 43 |
Reverse engineering is the process which deals with :
Size measurement | |
Cost measurement | |
Design recovery | |
All of the above |
Question 43 Explanation:
Design recovery from source code is done during reverse engineering, re-engineering and reusability.
Question 44 |
The spacing between character pairs is called :
Kerning | |
x-height | |
intercap | |
serif |
Question 44 Explanation:
→ kerning is the process of adjusting the spacing between characters in a proportional font, usually to achieve a visually pleasing result.
→ Kerning adjusts the space between individual letter forms, while tracking (letter-spacing) adjusts spacing uniformly over a range of characters.
→ Kerning adjusts the space between individual letter forms, while tracking (letter-spacing) adjusts spacing uniformly over a range of characters.
Question 45 |
when compared with analog cellular systems, an advantage of digital TDMA cellular system is that :
it is less complicated | |
it requires less of computer memory | |
it conserves spectrum bandwidth | |
it costs less |
Question 45 Explanation:
when compared with analog cellular systems, an advantage of digital TDMA cellular system is that it conserves spectrum bandwidth.
Question 46 |
E–commerce includes :
B2C | |
B2B | |
C2C | |
All of the above |
Question 46 Explanation:
E–commerce includes:
Consumer-to-business (C2B) is a business model in which consumers (individuals) create value and businesses consume that value.
Business-to-government (B2G) (or) "public sector marketing": Which encompasses marketing products and services to various government levels through integrated marketing communications techniques such as strategic public relations, branding, marketing communications (marcom), advertising, and web-based communications.
Customer to customer (C2C) markets provide an innovative way to allow customers to interact with each other.
Business-to-business (B2B) is a situation where one business makes a commercial transaction with another.
Consumer-to-business (C2B) is a business model in which consumers (individuals) create value and businesses consume that value.
Business-to-government (B2G) (or) "public sector marketing": Which encompasses marketing products and services to various government levels through integrated marketing communications techniques such as strategic public relations, branding, marketing communications (marcom), advertising, and web-based communications.
Customer to customer (C2C) markets provide an innovative way to allow customers to interact with each other.
Business-to-business (B2B) is a situation where one business makes a commercial transaction with another.
Question 47 |
A clustering technique that permits a convenient graphical display is :
partition based clustering | |
probabilistic model based clustering | |
hierarchical clustering | |
agglomerative clustering |
Question 47 Explanation:
→ Hierarchical methods of cluster analysis permit a convenient graphical display in which the entire sequence of merging(or splitting) because of its tree like nature, the display has the name of dendrogram.
Types:
1. Agglomerative: This is a "bottom-up" approach: each observation starts in its own cluster, and pairs of clusters are merged as one moves up the hierarchy.
2. Divisive: This is a "top-down" approach: all observations start in one cluster, and splits are performed recursively as one moves down the hierarchy.
Types:
1. Agglomerative: This is a "bottom-up" approach: each observation starts in its own cluster, and pairs of clusters are merged as one moves up the hierarchy.
2. Divisive: This is a "top-down" approach: all observations start in one cluster, and splits are performed recursively as one moves down the hierarchy.
Question 48 |
After sending a message ,the sender should not be able to, at a later date, deny having sent the message, is referred to as :
Authenticity | |
Non–Repudiability | |
Auditability | |
Repudiability |
Question 48 Explanation:
After sending a message ,the sender should not be able to, at a later date, deny having sent the message, is referred to as Non–Repudiation.
Question 49 |
The device which connects dissimilar LANs of different topologies using different sets of communication protocols so that information can flow from one to another is called :
Router | |
Bridge | |
Gateway | |
Switch |
Question 49 Explanation:
→ Gateway allows data to flow from one discrete network to another.
→ Gateways are distinct from routers or switches in that they communicate using more than one protocol and can operate at any of the seven layers of the open systems interconnection model (OSI).
→ Gateways are distinct from routers or switches in that they communicate using more than one protocol and can operate at any of the seven layers of the open systems interconnection model (OSI).
Question 50 |
We can not delete the __________ icon but we can made it invisible.
Recycle | |
My computer | |
Internet explorer | |
None of the above
|
Question 50 Explanation:
We can not delete the recycle icon but we can made it invisible but we can delete my computer and internet explorer icon completely.
There are 50 questions to complete.