Subject Books for UGC NET Computer Science

Friends,

I found a comprehensive list of suggested books for preparing different subjects under computer Science. You may check it and even suggest me if you came across better books than I have mentioned here.

S.No.	Book Title	Author
1.	Matrices	AR Vasista
2.	Higher Engineering Maths	BS Grewal
3.	Probability and Statistics	Miller and Freund
4.	Discrete Mathematical Structures	JP Trembly and Manohar
5.	Discrete Maths	Schaum’s Series
6.	Discrete Mathematical Structures	Kolman, Busby and Ross
7.	Graph Theory	Narsingh Deo
8.	Numerical Methods	Golden Series
9.	Numerical Methods	BS Grewal
10.	Numerical Methods	SS Sastry
11.	Numerical Methods	Jain and Iyengar
12.	Discrete Mathematics	Mott, Kandel and Baker
13.	Automata	Ullman and Hopcroft
14.	Introduction to Computer Theory	Daniel Cohen 2nd Edition
15.	Theory of ComputerScience 2nd Edition	Chandrasekaran & Mishra
16.	Design Analysis of Comp. Algorithms	Aho, Hopcroft and Ullman
17.	Computer Algorithms	Horowitz and Sahni
18.	Introduction to Algorithms	Thomas Cormon, Charles Ronald
19.	Switching and Finite Automata Theory	Kohavi
20.	Digital Logic and Computer Design	Morris Mano
21.	Digital Principles and Applications	Leach and Malvino
22.	Computer System Architecture	Mano
23.	Computer Org. and Architecture	William Stallings
24.	Computer Organization	Hamacher, Vranesic &Zaky
25.	Computer Architecture and Organization	Hayes
26.	Programming Languages	Pratt and Zelkewitz
27.	Compilers	Aho, Ullman and Sethi
28.	Principles of Compiler Design	Aho and Ullman
29.	Fundamentals of Data Structures	Horowitz. Sahni
30.	Introduction to DS with Applications	Trembly and Sorenson
31.	Data Structures	Schaum Series
32.	DS and Programming in C	Kruse, Tondo and Leung
33.	C Programming	Venugopal
34.	C Programming Language	Kernighan and Ritchie
35.	System Programming	Donovan
36.	System Programming and OS	Dhamdhere
37.	Operating Systems	Dietel
38.	Operating Systems	William Stallings
39.	Operating Systems	Milan Milenkovic
40.	OS Concepts	Galvin and Silberschatz
41.	Computer Networks	Andrew Tanenbaum
42.	Data Communications	William Stallings
43.	Engg. Approach to Comp. Networks	S Keshav
44.	Introduction to Switching Theory & Logic Design	FJ Hill and Peterson
45.	Database Systems	Korth and Sudhershan
46.	Fundamentals of DBMS	Elmasri, Navathe
47.	An Introduction to DBMS	CJ Date
48.	DBMS 3rd Edition	Ramakrishnan, Gehrke
49.	An Introduction to DBMS	Bipin C Desai
50.	Multiple Choice Questions in Computer Science	Timothy J Williams

Software Testing Techniques

Here, we will discuss some important software testing techniques. In exams like UGC NET, questions are often asked from this topic.

There are four levels of software testing: Unit >> Integration >> System >> Acceptance.

Usually, Black Box Testing method is used in Acceptance Testing. Acceptance Testing is performed after System Testing and before making the system available for actual use.

Smoke Testing, also known as “Build Verification Testing", comprises of a non-exhaustive set of tests that aim at ensuring that the most important functions work, but none of them in depth. It helps in exposing integration and major problems early in the cycle.

Functional testing ensures that the requirements are properly satisfied by the application. During functional testing, Black Box Testing technique is used. It is performed during the levels of System Testing and Acceptance Testing. It has a high possibility of redundant testing.

Security Testing intends to uncover vulnerabilities of the system and determine that its data and resources are protected from possible intruders.
Example of a basic security test: Click the BACK button of the browser (Check if you are asked to log in again or if you are provided the logged-in application.)

Regression testing intends to ensure that changes (enhancements or defect fixes) to the software have not adversely affected it.Regression testing can be performed during any level of testing but it is mostly relevant during System Testing.

Unit Testing is performed by using the White Box Testing method. This type of testing is performed prior to Integration Testing, by software developers themselves or their peers.

Integration Testing: The purpose of this level of testing is to expose possible faults in the interaction between integrated units. Different approaches are:
Big Bang:  all or most of the units of the code are combined together and tested at one go. This approach is used when the testing team receives the entire software in a bundle. Big Bang Integration tests only the interactions between the units while System Testing tests the entire system.

Top Down: top level units are tested first and (step by step ) lower level units are tested later. This approach is followed with top down development. Test Stubs are needed to simulate lower level units which may not be available during the initial phases.

Bottom Up: bottom level units are tested first and upper level units step by step after that. This approach is used when bottom up development approach is followed. Test Drivers are needed to simulate higher level units which may not be available during the initial phases.

Sandwich/Hybrid: A combination of Top Down and Bottom Up approaches.

System Testing: A complete, integrated system/software is tested. The purpose is to evaluate the system’s compliance with the specified requirements. Usually, Black Box Testing method is used. independent Testers perform System Testing.

Acceptance Testing: This testing method does not usually follow a strict procedure and is not scripted but is rather ad-hoc. It is performed after System Testing and before making the system available for actual use.

Black Box Testing is also known as Behavioral Testing. There are some techniques that can be used for designing black box tests:

Equivalence partitioning: Involves dividing input values into valid and invalid partitions and selecting representative values from each partition as test data.

Boundary Value Analysis: Involves determination of boundaries for input values and selecting values that are at the boundaries and just inside/outside of the boundaries as test data.

Cause Effect Graphing: Identifies the cases (input conditions) and effects (output conditions), producing a Cause-Effect Graph, and generating test cases accordingly.

White Box Testing (Code-Based Testing or Structural Testing): The internal structure/design/implementation of the item being tested is known to the tester. It is mainly applied to Unit Testing.

Gray Box Testing: The internal structure is partially known. This involves having access to internal data structures and algorithms for purposes of designing the test cases, but testing at the user, or black-box level. it is primarily useful in Integration Testing.

7 Steps to success in UGC NET Exam

Hello Friends!

Cracking UGC NET examination, especially in a subject like Computer Science, is not a piece of cake. But at the same time, it is not an uphill task, which only selected few in the world can do.

Many of the candidates, after failing in 1-2 attempts, start thinking that it is not their cup of tea. I too failed to crack the exam, not just once or twice, but even after 7 attempts. However, my misery lasted only a few days. Every time the notification for UGC NET was out, I always decided to give it a shot. I must admit that every time I missed the sweet success, I felt low on confidence. Still I had a gut feeling that I was doing better than my last attempts and inch by inch, I was getting closer towards my goal.

Now, after qualifying UGC NET exam twice, I have gained more confidence in my teaching abilities. I have also realised that the way I planned and studied for getting success in NET exam was a kind of mixed bag. Means, if I had planned in a better manner and had devoted more time to practice, then it cannot take one more than 2 years to crack the exam!!

Here, I want to share with you my experience and my topsy-turvy journey towards this goal. If you do not want to spend 3-4 years to qualify this exam (like I did) and qualify the exam in less than two years, then, get ready to follow the steps (with determination) I am mentioning here:

Step 1:
Buy/procure the standard text books for different subjects plus one or two good MCQ books on computer Sci. (At the end, I will list names of some of good books). You need not to buy books on all subjects, focus on maximum 9-10 subjects, since you are not supposed to answer every question right and you are never expected to score 90% and above ;-)

Step 2:
Download all previous question papers (of all 3 papers) from the following website links of UGC:

http://ugcnetonline.in/news.php
http://www.cbsenet.nic.in
http://www.ugc.ac.in/net/oldqpyr.aspx?sub=87
http://www.ugc.ac.in/net/oldqpyr.aspx?sub=00

Also download all available answer keys of different exams (from June' 12 to June' 14, you will find ans keys of all 3 papers at UGCNetonline site, for NET exam December 2014 onward, please check CBSENET site).

DO TAKE PRINTOUT of all papers. It is better to use printed paper than reading it on your screen. However, to save your time and energy, better focus on question papers of last 4-5 years only and leave very old papers. For example, patter of question papers of 2004 and 2005 is very old and not many of the questions of the old pattern are asked nowadays.

Step 3:
Study the questions of previous papers thoroughly. Remember, this step is very important to understand where you stand now and how much time and preparation you will need. So, do this religiously and devote next 2-3 days in reading the questions and try to understand what is asked by the examiner. You will surely gain a good understanding about 3 things:

1. What are the subjects I must focus on
2. What are the subjects I can ignore or I can't study
3. What are the important topics of each subject from the point of view of NET exam

There are definitely some subjects of you will have a weak understanding. Rather than wasting your time and energy on these topics, better you focus on strengthening the subjects you have a flair as well as you can prepare. For example, I was not comfortable with Image Processing, Computer Graphics and ANN (Artificial Neural Network), so I left them out. For such topics, one would be tempted to not to miss them. I have a tip for these. For questions beyond your comprehension, just memorise their right answer. Who knows, the same question is asked in next exam :-)

Step 4:
Select one or two subjects (at max) to begin with and start reading their text book. For next 7-10 days, read only the books on the selected 1-2 topics. ALWAYS KEEP A HIGHLIGHTER OR A GOOD PENCIL with you. Any statement or concept which you may find important, then immediately highlight the points in the book itself. (You will be able to do this easily if you did step 3 right)

After finishing the book, go to
step 5. After step 5 is finished, come back to this step and follow the similar exercise for next 1-2 topics in the sequence.

Step 5:
Now, start solving the question papers you downloaded earlier. (Here, you need to solve only questions of the subject you studied in step 4). Eureka!! You will find that you are able to understand the question better now than previously-before reading the text book. (Still, there will be many questions you will not be able to solve). Mark your answers with a pencil then compare your answers with official ans. keys of UGC. For all the questions, you were not able to solve, go to step 6:

Step 6:
Research the tricky questions on Net (Internet dear, not UGC NET ;-) ). Take the help of Google Baba as well as your Facebook group of UGC to find the solution of the question. Stay determined to find the solution till you have not found the satisfactory explanation of the question). Who knows if the same question is asked in the exam itself, then you will be happy for your homework.

Step 7:
The final step is highly important-perseverance. Come what may, stay determined to get through your goal at any cost. Follow all 6 steps, especially step 5 and 6 in a loop, with all your determination. If you fail in one attempt, hold your emotions and bounce back to fight for the next exam.

After taking exam of one session, say June, do not sit idle and wait for the result. Continue with other subjects, solve the questions of your last exam. If you were not able to succeed, you will be left with lesser time to prepare for the exam of the next session. If you are already following your routine, nothing will be unexpected for you.

Stay focused, determined and relaxed also. Enjoy the feeling of putting your best foot forward. Pray to God to keep you on track. Nobody can stop you from achieving your share of success and joys in this world.

ALL THE BEST!!

Useful Q & A on Data Communication & Networks - Part 2

Hello Friends,

Based on the types of questions being asked in UGC NET, I have compiled some important questions on the topic of Data Communication & Networks. Please not just read this post, but also understand the concept behind the given answers. I have given explanation of these questions to help this process simple for you. Since Data Communication & Networks is one of the subjects on which lots of questions are asked, do make a note of preparing this subject well.

Please share your ideas and doubts also after reading the post.......

Q 1. What is the bucket size, when the overlapping and collision occur at same time?

Ans.: If there is only one entry possible in the bucket, when the collision occurs, there is no way to accommodate the colliding value. This results in the overlapping of values.

Q 2. Using RSA algorithm, what is the value of cipher text C, if the plain text M = 5 and p=3, q=11 & d=7 ?

A. 33            B. 5              C. 25            D. 26

Here p and q will be used to calculate public keys. Steps for transmission:

1. calculate n = p * q = 3 * 11 = 33

2. calculate z = ( p - 1 ) * ( q - 1 ) = ( 3 - 1 ) * ( 11 - 1 ) = 20

3. choose a prime number k, such that k is co-prime to z, i.e, z is not divisible by k. We have several choices for k: 7, 11, 13, 17, 19 (we cannot use 5, because 20 is divisible by 5).

Let's pick k=7 (smaller k, "less math").

4. the numbers n = 33 and k = 7 become the Server's public key.

5. the Server has to calculate it's secret key. Here is how.

6. k * j = 1 ( mod z )

7.)  7 * j = 1 ( mod 20 )

8.) ( 7 * j ) / 20 = ? with the remainder of 1 (the "?" here means: "something, but don't wory about it"; we are only interested in the remainder).

Since we selected (on purpose) to work with small numbers, we can easily conclude that 21 / 20 gives "something" with the remainder of 1. So, 7 * j = 21, and j = 3. This is our secret key. We MUST NOT give this key to anyone. Now, server sends n=33 and k=7 back to the Browser

Q 3. If a packet arrive with an M-bit value is ‘1’ and a fragmentation offset value ‘0’, then it is ______ fragment.
(A) First                                 (B) Middle                         (C) Last         (D) All of the above

Answer is A. If the M bit is 0, it means that there are no more fragments. If the M bit is 1, it means that there is at least one more fragment. This fragment can be the first one or a middle one, but not the last one. Offset value '0' always means this is the first fragment. To determine the position of the first byte in a particular fragment, we would multiply given offset value with 8. For example, offset is 200, then the position of first byte in that fragment is 1600.

Fragment Offset field solves the problem of sequencing fragments by indicating to the recipient device where in the overall message each particular fragment should be placed. The field is 13 bits wide, so the offset can be from 0 to 8191. Fragments are specified in units of 8 bytes, which is why fragment length must be a multiple of 8. Uncoincidentally, 8191 * 8 is 65,528, just about the maximum size allowed for an IP datagram.

IMP: When an MTU requirement forces a datagram to be fragmented, it is split into several smaller IP datagrams, each containing part of the original. The header of the original datagram is changed into the header of the first fragment, and new headers are created for the other fragments. Each is set to the same Identification value to mark them as part of the same original datagram. The Fragment Offset of each is set to the location where the fragment belongs in the original. The More Fragments field is set to 1 for all fragments but the last, to let the recipient know when it has received all the fragments.

Q 4. In a network of LANs connected by bridges, packets are sent from one LAN to another through intermediate bridges. Since more than one path may exist between two LANs, packets may have to be routed through multiple bridges. Why is the spanning tree algorithm used for bridge-routing? (GATE 2005)

(a) For shortest path routing between LANs         (b) For avoiding loops in the routing paths

(c) For fault tolerance                                      (d) For minimizing collisions

SOLUTION: Spanning tree is a protocol that allows the bridges to exchange information so that only one of them will handle a given message that is being sent between two computers within the network. This  protocol prevents the condition known as a BRIDGE LOOP. It is typical to add a second bridge between two network segments as a backup in case the primary bridge fails (both bridges need to have some way to understand which bridge is the primary one). To do this, they have a separate path connection just between the bridges in which they exchange information, using bridge protocol data units (BPDUs).

The program in each bridge that allows it to determine how to use the protocol is known as the spanning tree algorithm. The algorithm is specifically constructed to avoid bridge loops (for a bridge using only the most efficient path when faced with multiple paths). If the best path fails, the algorithm recalculates the network and finds the next best route.

Q 5. How many 8-bit characters can be transmitted per second over a 9600 baud serial communication link using asynchronous mode of transmission with one start bit, eight data bits, two stop bits and one parity bit?
(1) 600        (2) 800        (3) 876        (4) 1200

For 9600 baud, 1 bit=1/9600=0.104mS. Each char would require 11 bits. That means, to transmit one char, it would take 1.144 mS. So, applying the formula 1000/1.144 (mS in one Sec/transmission time for one char)=approx 875. Nearest answer is (3)

Q 6. The single stage network is also called

A) one sided network                             B) two sided network

C) recirculating network                      D) pipeline network

Single-Stage Network is a single stage of switching elements (SEs) existing between the inputs and the outputs of the network. Data is circulated a number of times around the network.

Q 7. If a class B network on the Internet has a subnet mask of 255.255.248.0, what is the maximum number of hosts per subnet?
(a) 1022 (b) 1023 (c) 2046 (d) 2047

Explanation: Convert the subnet mask into binary format.

255.255.248.0 = 11111111.11111111.11111000.00000000

Number of 1's in the subnet mask indicates the Network-ID and the Subnet-ID part. Number of 0's in the subnet mask indicates the Host-ID part. Maximum number of Hosts per subnet = 2¹¹ = 2048, where 11 = Number of 0's in the Subnet Mask. Out of 2048 values, 2 addresses are reserved, hence we remove them (2048-2 = 2046). Note: In the host part of the address:- all bits as 1 is reserved as broadcast address and all bits as 0 is used as network address of subnet.

Q 8. Which of the following system calls results in the sending of SYN packets?(a) socket (b) bind (c) listen (d) connect

The connect system call is normally called by the client process to connect to the server process. The socket system call creates a new socket and assigns the protocol and resources to the created socket descriptor. The bind system call associates a local network transport address with a socket. For a client process, it is not mandatory to issue a bind call. The kernel takes care of doing an implicit binding when the client process issues the connect system call. 

It is often necessary for a server process to issue an explicit bind request before it can accept connections or start communication with clients. The listen call indicates to the protocol that the server process is ready to accept any new incoming connections on the socket. There is a limit on the number of connections that can be queued up, after which any further connection requests are ignored.

What is the maximum size of data that the application layer can pass on to the TCP layer below?
Q 9. (a) Any size (b) 2¹⁶ bytes-size of TCP header     (c) 2¹⁶ bytes   (d) 1500 bytes

Which of the following transport layer protocols is used to support electronic mail?

(a) SMTP (b) IP  (c) TCP  (d) UDP

Q 10. When a host on network A sends a message to a host on network B, which address does the router look at ?
(a) Port                  (b) IP           (c) Physical              (d) Subnet mask

Q 11. The ....................... field in the SNMP PDU, reports an error in a response message.

(A) Error index         (B) Error status       (C) Set request                  (D) Agent index

Error status: An integer value that is used in a Response-PDU to tell the requesting SNMP entity the result of its request. A value of zero indicates that no error occurred; the other values indicate what sort of error happened.

Error Index: When Error Status is non-zero, this field contains a pointer that specifies which object generated the error. Always zero in a request.

Paper 1 Notes Series: Number systems (ICT)

Friends!

As I mentioned in one of my previous posts, there will be 6 questions on ICT (Information & Communication technology). Most of times, there are 1-2 questions on the topic of number systems and conversions between different numbers systems. Here, I am going to guide you about how to solve these questions without difficulty. 

Read on....
What is number system?

Number system is a way to represent numbers. There are 2 kinds of systems-1. Non-positional 2. Positional. The best example of non-positional system is Roman number system. In a positional system, a value of each digit in a number can be determined using
1. The digit
2. The position of the digit in the number
3. The base of the number system ("base" is the total number of digits available in the number system).

The examples of positional system are:

1. Binary (Base 2. Digits used : 0, 1)
2. Octal (Base 8. Digits used : 0 to 7)
3. Decimal (Base 10. Digits used : 0 to 9)
4. Hexadecimal (Base 16. Digits used : 0 to 9, Letters used : A- F)

Decimal number system has base 10 as it uses 10 digits from 0 to 9. The successive positions to the left of the decimal point represent units, tens, hundreds, thousands and so on.

Each position in a positional number system represents a "x" power of the base (e.g. 2). Example 2^x where x represents the last position - 1. Similar rule can be applied to Octal and Hexadecimal also.

NOTE: The understanding of the above rules will help you in learning how to convert a number from one base to the another.

There are many methods or techniques which can be used to convert numbers from one base to another. For example:

1. Decimal to Other Base System
2. Other Base System to Decimal
3. Other Base System to Non-Decimal
4. Shortcut method - Binary to Octal
5. Shortcut method - Octal to Binary
6. Shortcut method - Binary to Hexadecimal
7. Shortcut method - Hexadecimal to Binary

The easiest way to understand how the conversion techniques work is to learn a few basic tricks:-

1. When converting a decimal number to others, apply division method

2. When converting a non-decimal number to decimal, apply addition method.

3. Under addition method, we first obtain the value of RHS by first multiplying it with base raised to power X, then adding it with the value at next position.

4. Under division method, we always divide the number in question with the base of the target number system. Its steps are:

Step 1 - Divide the decimal number to be converted by the value of the target base.

Step 2 - Get the remainder from Step 1 as the rightmost digit (least significant digit) of new base number.

Step 3 - Divide the quotient of the previous divide by the new base.

Step 4 - Record the remainder from Step 3 as the next digit (to the left) of the new base number.

Repeat the above steps till the number becomes less than the base of the target. In this case, we will stop when number becomes 1, which is less than the divisor.

1. Here, we give an example of conversion from decimal to binary:

Step	Operation	Result	Remainder
Step 1	35 / 2	17	1
Step 2	17 / 2	8	1
Step 3	8 / 2	4	0
Step 4	4 / 2	2	0
Step 5	2 / 2	1	0
Step 6	1/2	0	1

Now, arrange the remainder in the reverse order so that the first remainder becomes the least significant digit (LSD) and the last remainder becomes the most significant digit (MSD).

Thus the binary of 35 is 100011_2.

The similar steps will be applied when you convert a decimal no. to octal or hexadecimal. The only thing that will change will be the divisor, i.e. it will be the base of target number system.

2. Now, steps for converting From other base system to Decimal System

Step 1 - Determine the column (positional) value of each digit, starting from the rightmost digit first.

Step 2 - Multiply the digit with the base raised to power X (starting from 0, then increment it by 1 at each digit position of source number system.)

Step 3 - Sum the products calculated in Step 2. The total is the equivalent value in decimal.

_{Here is an example to make it simple to understand.}

_{We take the same number we obtained in previous example, i.e. 100011 to convert it back into decimal:}

Step	Binary Number	Decimal Number
Step 1	1000112	( (1 x 25) + (0 x 24) + (0 x 23) + (0 x 22) + (1 x 21) + (1 x 20))10
Step 2	1000112	(32 + 0 + 0 + 0 + 2 + 1)10
Step 3	1000112	3510

3. When converting from other Base Systems to Non-Decimal System, steps will be a little different. 

Step 1 - Convert the original number to a decimal number (base 10).

Step 2 - Convert the decimal number so obtained to the new base number. Example:

Octal Number : 25₈

Calculating Binary Equivalent:

Step 1 : Convert to Decimal

Step	Octal Number	Decimal Number
Step 1	258	((2 x 81) + (5 x 80))10
Step 2	258	(16 + 5 )10
Step 3	258	2110

Octal Number : 25₈ = Decimal Number : 21₁₀

Step 2 : Convert Decimal to Binary

Step	Operation	Result	Remainder
Step 1	21 / 2	10	1
Step 2	10 / 2	5	0
Step 3	5 / 2	2	1
Step 4	2 / 2	1	0
Step 5	1 / 2	0	1

Decimal Number : 21₁₀ = Binary Number : 10101₂

Octal Number : 25₈ = Binary Number : 10101₂

4. Shortcut method - Binary to Octal

Step 1 - Divide the binary digits into groups of three (starting from the right).

Step 2 - Convert each group of three binary digits to one octal digit.

Example: Binary Number: 10101₂

Calculating Octal Equivalent:

Step	Binary Number	Octal Number
Step 1	101012	010   101
Step 2	101012	28   58
Step 3	101012	258

Binary Number: 10101₂ = Octal Number : 25₈

_{5. Shortcut method - Octal to Binary}

Step 1 - Convert each octal digit to a 3 digit binary number (the octal digits may be treated as decimal for this conversion).

Step 2 - Combine all the resulting binary groups (of 3 digits each) into a single binary number.

Example: Octal Number : 25₈

Calculating Binary Equivalent:

Step	Octal Number	Binary Number
Step 1	258	210 510
Step 2	258	0102 1012
Step 3	258	0101012

Octal Number: 25₈ = Binary Number : 10101₂

_{6. Shortcut method - Binary to Hexadecimal}

Step 1 - Divide the binary digits into groups of four (starting from the right).

Step 2 - Convert each group of four binary digits to one hexadecimal symbol.

Example: Binary Number: 10101₂

Calculating hexadecimal Equivalent:

Step	Binary Number	Hexadecimal Number
Step 1	101012	0001 0101
Step 2	101012	110 510
Step 3	101012	1516

Binary Number: 10101₂ = Hexadecimal Number : 15₁₆

_{7. Shortcut method - Hexadecimal to Binary}

Step 1 - Convert each hexadecimal digit to a 4 digit binary number (the hexadecimal digits may be treated as decimal for this conversion).

Step 2 - Combine all the resulting binary groups (of 4 digits each) into a single binary number.

Example: Hexadecimal Number: 15₁₆

Calculating Binary Equivalent:

Step	Hexadecimal Number	Binary Number
Step 1	1516	110 510
Step 2	1516	00012 01012
Step 3	1516	000101012

Hexadecimal Number: 15₁₆ = Binary Number : 10101₂

_{I hope this post will help you understand the concept and you will be able to answer any question on Number System without mistake.}

Best of Luck for December' 15 Exam!!