Network-Security
Question 1 |
The value of 351 mod 5 is ______.
3 | |
5 | |
2 | |
1 |
Question 1 Explanation:
351 mod 5
⇒ 31 = 3 ⇒ 3 mod 5 = 3
32 ⇒ 9 mod 5 = 4
33 ⇒ 27 mod 5 = 2
34 ⇒ 81 mod 5 = 1
35 ⇒ 243 mod 5 = 3
For every four numbers sequence is repeating.
So, (51 % 4) = 3
⇒ 33 = 27
⇒ 27 mod 5 = 2
⇒ 31 = 3 ⇒ 3 mod 5 = 3
32 ⇒ 9 mod 5 = 4
33 ⇒ 27 mod 5 = 2
34 ⇒ 81 mod 5 = 1
35 ⇒ 243 mod 5 = 3
For every four numbers sequence is repeating.
So, (51 % 4) = 3
⇒ 33 = 27
⇒ 27 mod 5 = 2
Question 2 |
Suppose that everyone in a group of N people wants to communicate secretly with the N–1 others using symmetric key cryptographic system. The communication between any two persons should not be decodable by the others in the group. The number of keys required in the system as a whole to satisfy the confidentiality requirement is
2N | |
N(N – 1) | |
N(N – 1)/2 | |
(N – 1)2 |
Question 2 Explanation:
For N people, the number of unique pairs can be calculated using the combination formula (N 2), which is given by:
(N 2) = N(N – 1)/2
The number of keys required in the system as a whole to satisfy the confidentiality requirement is:
N(N – 1)/2
(N 2) = N(N – 1)/2
The number of keys required in the system as a whole to satisfy the confidentiality requirement is:
N(N – 1)/2
Question 3 |
In an RSA cryptosystem, the value of the public modulus parameter n is 3007. If it is also known that Φ(n) = 2880, where Φ() denotes Euler's Quotient Function, then the prime factor of n which is greater than 50 is ______.
107 | |
97 | |
45 | |
92 |
Question 3 Explanation:
It can be solved by Hit and trial method in less time.
n = 3007, fi(n) = 2880 → fi(n) = (p – 1) (q – 1),
where p, q are prime factor of n.
The unit place of n is 7, it is a prime number and factor will be
1.7=7
11*17
21*37
31*47
….
31*97 =>3007
n = 3007 => 31*97
Therefore, 31 & 97 are the two prime numbers, which is satisfying the condition and 97 is greater than 50.
So, 97 is the correct answer.
Other methods:
When ϕ(n) is given when n=pq where p and q are prime numbers, then we have
ϕ(n) = (p−1)(q−1) = pq−(p+q)+1
But pq=n,
therefore, ϕ(n) = n−(p+q)+1 and p+q = n+1−ϕ(n).
Now, p and q are the roots of the equation,
x2 − (p+q)x + pq = (x-p)(x-q)
Substituting for p+q and pq in the above equation
x2 - (n+1-ϕ(n))x + n
n = 3007, fi(n) = 2880 → fi(n) = (p – 1) (q – 1),
where p, q are prime factor of n.
The unit place of n is 7, it is a prime number and factor will be
1.7=7
11*17
21*37
31*47
….
31*97 =>3007
n = 3007 => 31*97
Therefore, 31 & 97 are the two prime numbers, which is satisfying the condition and 97 is greater than 50.
So, 97 is the correct answer.
Other methods:
When ϕ(n) is given when n=pq where p and q are prime numbers, then we have
ϕ(n) = (p−1)(q−1) = pq−(p+q)+1
But pq=n,
therefore, ϕ(n) = n−(p+q)+1 and p+q = n+1−ϕ(n).
Now, p and q are the roots of the equation,
x2 − (p+q)x + pq = (x-p)(x-q)
Substituting for p+q and pq in the above equation
x2 - (n+1-ϕ(n))x + n
Question 4 |
Using public key cryptography, X adds a digital signature to message M, encrypts <M, σ>, and sends it to Y, where it is decrypted. Which one of the following sequences of keys is used for the operations?
Encryption: X’s private key followed by Y’s private key; Decryption: X’s public key followed by Y’s public key | |
Encryption: X’s private key followed by Y’s public key; Decryption: X’s public key followed by Y’s private key | |
Encryption: X’s public key followed by Y’s private key; Decryption: Y’s public key followed by X’s private key | |
Encryption: X’s private key followed by Y’s public key; Decryption: Y’s private key followed by X’s public key |
Question 4 Explanation:
Encryption: Source has to encrypt with its private key for forming Digital signature for Authentication. Source has to encrypt the (M, σ) with Y’s public key to send it confidentially.
Decryption: Destination Y has to decrypt first with its private key, then decrypt using source public key.
Question 5 |
A layer-4 firewall (a device that can look at all protocol headers up to the transport layer) CANNOT
block entire HTTP traffic during 9:00PM and 5:00AM | |
block all ICMP traffic | |
stop incoming traffic from a specific IP address but allow outgoing traffic to the same IP address | |
block TCP traffic from a specific user on a multi-user system during 9:00PM and 5:00AM |
Question 5 Explanation:
(A) It is possible to block entire HTTP traffic by blocking port no.80.
(B) Possible because it is network layer protocol.
(C) Possible because SP address is present in Network layer.
(D) Not possible, because to block specific user, we need user id which is present in Application layer.
(B) Possible because it is network layer protocol.
(C) Possible because SP address is present in Network layer.
(D) Not possible, because to block specific user, we need user id which is present in Application layer.
Question 6 |
The total number of keys required for a set of n individuals to be able to communicate with each other using secret key and public key crypto-systems, respectively are:
n(n-1) and 2n | |
2n and n(n-1)/2 | |
n(n-1)/2 and 2n | |
n(n-1)/2 and n |
Question 6 Explanation:
For private key crypto, a key used for encryption as well as decryption. So, no. of keys required for n individuals is same as no. of communication link between any two individuals which is
nC2 = n(n-1)/2
In case of public key, each sender has its own public key as well as private key. So, no. of keys are 2n.
nC2 = n(n-1)/2
In case of public key, each sender has its own public key as well as private key. So, no. of keys are 2n.
Question 7 |
Consider the following two statements:
(i) A hash function (these are often used for computing digital signatures) is an injective function.
(ii) An encryption technique such as DES performs a permutation on the elements of its input alphabet.
Which one of the following options is valid for the above two statements?
Both are false | |
Statement (i) is true and the other is false | |
Statement (ii) is true and the other is false | |
Both are true |
Question 7 Explanation:
i) Hash function is many to one function. It is not one-one (or) injective.
ii) It uses the P-Box permutation.
Statement-I is false, II is true.
ii) It uses the P-Box permutation.
Statement-I is false, II is true.
Question 8 |
Your are given the following four bytes :
10100011 00110111 11101001 10101011Which of the following are substrings of the base 64 encoding of the above four bytes?
zdp | |
fpq | |
qwA | |
oze |
Question 8 Explanation:
You are given the following four bytes:
10100011 00110111 11101001 10101011
So, in total we have 32 bits. And for base 64 we need 6 digits of binary no. to represent one digit of base 64 no.
So lets padd 4 bits on RHS, so that total digits will become 36 and we can separate then as group of 6 digits each.
Now, the longest substring will be from checking option is 'fpq'.
10100011 00110111 11101001 10101011
So, in total we have 32 bits. And for base 64 we need 6 digits of binary no. to represent one digit of base 64 no.
So lets padd 4 bits on RHS, so that total digits will become 36 and we can separate then as group of 6 digits each.
Now, the longest substring will be from checking option is 'fpq'.
Question 9 |
In the RSA public key cryptosystem, the private and public keys are (e, n) and (d, n) respectively, where n = p*q and p and q are large primes. Besides, n is public and p and q are private. Let M be an integer such that 0 < M < n and f(n) = (p- 1)(q-1). Now consider the following equations.
I. M’= Me mod n
M = (M’)d mod n
II. ed ≡ 1 mod n
III. ed ≡ 1 mod f(n)
IV. M’= Me mod f(n)
M = (M’)d mod f(n)
Which of the above equations correctly represent RSA cryptosystem?
I and II | |
I and III | |
II and IV | |
III and IV |
Question 9 Explanation:
To generate the encryption and decryption keys, we can proceed as follows.
1. Generate randomly two “large” primes p and q.
2. Compute n=pq and ∅=(p-1)(q-1).
3. Choose a number e so that
gcd(e,∅)=1
4. Find the multiplicative inverse of e modulo ∅, i.e., find d so that
ed≡1 (mod ∅)
This can be done efficiently using Euclid’s Extended Algorithm.
The encryption public key is KE=(n,e) and the decryption private key is KD=(n,d).
The encryption function is
E(M)=Me mod n
The decryption function is
D(M)=Md mod n
1. Generate randomly two “large” primes p and q.
2. Compute n=pq and ∅=(p-1)(q-1).
3. Choose a number e so that
gcd(e,∅)=1
4. Find the multiplicative inverse of e modulo ∅, i.e., find d so that
ed≡1 (mod ∅)
This can be done efficiently using Euclid’s Extended Algorithm.
The encryption public key is KE=(n,e) and the decryption private key is KD=(n,d).
The encryption function is
E(M)=Me mod n
The decryption function is
D(M)=Md mod n
Question 10 |
Which of the following statement(s) is TRUE?
1. A hash function takes a message of arbitrary length and generates a fixed length code.
2. A hash function takes a message of fixed length and generates a code of variable length.
3. A hash function may give the same hash value for distinct messages.
1 only | |
2 and 3 only | |
1 and 3 only | |
2 only |
Question 10 Explanation:
(1) A hash function takes a message of arbitrary length and generates a fixed length code. So this is correct.
(2) Statement-2 is wrong, refer statement-1.
(3) Statement-3 is correct, for example hash function N%10, this will generate same values for 1 as well as 2!
(2) Statement-2 is wrong, refer statement-1.
(3) Statement-3 is correct, for example hash function N%10, this will generate same values for 1 as well as 2!
Question 11 |
A sender is employing public key cryptography to send a secret message to a receiver. Which one of the following statements is TRUE?
Sender encrypts using receiver’s public key | |
Sender encrypts using his own public key | |
Receiver decrypts using sender’s public key | |
Receiver decrypts using his own public key |
Question 11 Explanation:
Sender can encrypts using the receiver public key and receiver decrypts it using his own private key.
Question 12 |
In cryptography, the following uses transposition cyphers and the keyword is LAYER. Encrypt the following message. (Spaces are omitted during encryption)
WELCOME TO NETWORK SECURITY!
WMEKREETSILTWETCOOCYONRU! | |
EETSICOOCYWMEKRONRU!LTWET | |
LTWETONRU!WMEKRCOOCYEETSI | |
ONRU!COOCYLTWETEETSIWMEKR |
Question 12 Explanation:
Question 13 |
Avalanche effect in cryptography
Is desirable property of the cryptographic algorithm | |
Is undesirable property of the cryptographic algorithm | |
Has no effect on the encryption algorithm | |
None of the above |
Question 13 Explanation:
→ In cryptography, the avalanche effect is the desirable property of cryptographic algorithms, typically block cyphers and cryptographic hash functions, wherein if input is changed slightly (for example, flipping a single bit), the output changes significantly (e.g., half the output bits flip).
→ In the case of high-quality block cyphers, such a small change in either the key or the plaintext should cause a drastic change in the ciphertext.
→ In the case of high-quality block cyphers, such a small change in either the key or the plaintext should cause a drastic change in the ciphertext.
Question 14 |
What is one advantage of setting up a DMZ(Demilitarized Zone) with two firewalls?
You can control where traffic goes in the three networks | |
You can do stateful packet filtering | |
You can do load balancing | |
Improve network performance |
Question 14 Explanation:
→ The most secure approach is to use two firewalls to create a DMZ.
1. The first firewall (also called the "front-end" or "perimeter" firewall) must be configured to allow traffic destined to the DMZ only.
2. The second firewall (also called "back-end" or "internal" firewall) only allows traffic from the DMZ to the internal network.
This setup is considered more secure since two devices would need to be compromised. There is even more protection if the two firewalls are provided by two different vendors because it makes it less likely that both devices suffer from the same security vulnerabilities.
1. The first firewall (also called the "front-end" or "perimeter" firewall) must be configured to allow traffic destined to the DMZ only.
2. The second firewall (also called "back-end" or "internal" firewall) only allows traffic from the DMZ to the internal network.
This setup is considered more secure since two devices would need to be compromised. There is even more protection if the two firewalls are provided by two different vendors because it makes it less likely that both devices suffer from the same security vulnerabilities.
Question 15 |
Which one of the following algorithm is not used in asymmetric key cryptography?
RSA Algorithm | |
Diffie-Hellman Algorithm | |
Electronic Code Book Algorithm | |
None of the above |
Question 15 Explanation:
Question 16 |
SSL is not responsible for
Mutual authentication of client & server | |
Secret communication | |
Data Integrity protection | |
Error detection and correction |
Question 16 Explanation:
SSL (Secure Sockets Layer) is the standard security technology for establishing an encrypted link between a web server and a browser. This link ensures that all data passed between the web server and browsers remain private and integral. SSL is an industry standard and is used by millions of websites in the protection of their online transactions with their customers.
It is utilized to encrypt Web traffic using Hypertext Transfer Protocol (HTTP) and to authenticate Web servers, and to encrypt communications between Web browsers and Web servers etc.
So, other than error detection and correction, all options are correct.
It is utilized to encrypt Web traffic using Hypertext Transfer Protocol (HTTP) and to authenticate Web servers, and to encrypt communications between Web browsers and Web servers etc.
So, other than error detection and correction, all options are correct.
Question 17 |
The standard for certificates used on internet is
X.25 | |
X.301 | |
X.409 | |
X.509 |
Question 17 Explanation:
In cryptography, X.509 is a standard that defines the format of public key certificates. X.509 certificates are used in many Internet protocols, including TLS/SSL, which is the basis for HTTPS, the secure protocol for browsing the web.
Question 18 |
Hashed message is signed by a sender using
his public key | |
his private key | |
receiver’s public key | |
receiver’s private key |
Question 18 Explanation:
After the generation of hashed message that needs to be transmitted over a network, it is signed or encrypted using the private key of the sender which also generates the digital signatures. The message is transmitted along with the digital signatures which ensures Authentication and Non-repudiation.
Question 19 |
MD5 is a widely used hash function for producing hash value of
64 bits | |
128 bits | |
512 bits | |
1024 bits |
Question 19 Explanation:
→ The Message Digest(MD5) algorithm is a widely used hash function producing a 128-bit hash value. Although MD5 was initially designed to be used as a cryptographic hash function, it has been found to suffer from extensive vulnerabilities.
→ It can still be used as a checksum to verify data integrity, but only against unintentional corruption. It remains suitable for other non-cryptographic purposes, for example for determining the partition for a particular key in a partitioned database.
→ It can still be used as a checksum to verify data integrity, but only against unintentional corruption. It remains suitable for other non-cryptographic purposes, for example for determining the partition for a particular key in a partitioned database.
Question 20 |
Pretty Good Privacy (PGP) is used in
Browser security | |
FTP security | |
Email security | |
None of the above |
Question 20 Explanation:
→ Pretty Good Privacy (PGP) is an encryption program that provides cryptographic privacy and authentication for data communication. PGP is used for signing, encrypting, and decrypting texts, e-mails, files, directories, and whole disk partitions and to increase the security of e-mail communications. Phil Zimmermann developed PGP in 1991.
→ It uses Digital Signatures to provide authentication and integrity checking basically to check if the message is actually sent by the person or entity claimed to be the sender.
→ It uses Digital Signatures to provide authentication and integrity checking basically to check if the message is actually sent by the person or entity claimed to be the sender.
Question 21 |
Decrypt the message “WTAAD” using the Caesar Cipher with key = 15.
LIPPS
| |
HELLO
| |
OLLEH | |
DAATW |
Question 21 Explanation:
We decrypt one character at a time. Each character is shifted 15 characters up Letter W is decrypted to H shifted 15 characters up. Letter W is decrypted to H. Letter T is decrypted to E. The first A is decrypted to L. The second A is decrypted to L And finally D is The second A is decrypted to L. And, finally, D is decrypted to O. The plain text is HELLO.
Question 22 |
Encrypt the Message “HELLO MY DEARZ” using Transposition Cipher with
HLLEO YM AEDRZ | |
EHOLL ZYM RAED | |
ELHL MDOY AZER | |
ELHL DOMY ZAER |
Question 22 Explanation:
Step-1: According to key we have to divide into number of character blocks.
Here, key size = 4. So, character block size is 4.
Step-2: Remove the spaces in the message and write into sequential order.
Step-3: Get cipher text according to ascending order is ELHL MDOY AZER.
Here, key size = 4. So, character block size is 4.
Step-2: Remove the spaces in the message and write into sequential order.
Step-3: Get cipher text according to ascending order is ELHL MDOY AZER.
Question 23 |
Transparent mode firewall is a:
Layer 2 firewall | |
Layer 3 firewall | |
Layer 4 firewall | |
Layer 7 firewall |
Question 23 Explanation:
→ Transparent mode firewall (or routed firewall) is a routed hop and acts as a default gateway for hosts that connect to one of its screened subnets.
→ A transparent firewall, on the other hand, is a Layer 2 firewall that acts like a "bump in the wire," or a "stealth firewall," and is not seen as a router hop to connected devices.
→ A transparent firewall, on the other hand, is a Layer 2 firewall that acts like a "bump in the wire," or a "stealth firewall," and is not seen as a router hop to connected devices.
Question 24 |
Using public key cryptography, X adds a digital signature σ to a message M, encrypts (M,σ) and sends it to Y, where it is decrypted. Which one of the following sequence of keys is used for operations?
Encryption : X’s private key followed by Y’s private key. Decryption : X’s public key followed by Y’s public key. | |
Encryption : X’s private key followed by Y’s public key; Decryption : X’s public key followed by Y’s private key | |
Encryption : X’s private key followed by Y’s public key; Decryption : Y’s private key followed by X’s public key. | |
Encryption : X’s public key followed by Y’s private key; Decryption : Y’s public key followed by X’s private key. |
Question 24 Explanation:
Encryption: Source has to encrypt with its private key for forming Digital signature for Authentication. Source has to encrypt the (M, σ) with Y’s public key to send it confidentially. Decryption: Destination Y has to decrypt first with its private key, then decrypt using source public key.
Question 25 |
Use of IPSEC in tunnel mode results in
IP packet with same header
| |
IP packet with new header
| |
IP packet without header
| |
No changes in IP packet |
Question 25 Explanation:
IPSec can be used to create VPN Tunnels to end-to-end IP Traffic (also called as IPSec Transport mode) or site-to-site IPSec Tunnels (between two VPN Gateways, also known as IPSec Tunnel mode).
IPSec Tunnel mode: In IPSec Tunnel mode, the original IP packet (IP header and the Data payload) is encapsulated within another packet.
In IPSec tunnel mode the original IP Datagram from is encapsulated with an AH (provides no confidentiality by encryption) or ESP (provides encryption) header and an additional IP header. The IP addresses of the newly added outer IP header are that of the VPN Gateways.
The traffic between the two VPN Gateways appears to be from the two gateways (in a new IP datagram), with the original IP datagram is encrypted (in case of ESP) inside IPSec packet.
IPSec Tunnel mode: In IPSec Tunnel mode, the original IP packet (IP header and the Data payload) is encapsulated within another packet.
In IPSec tunnel mode the original IP Datagram from is encapsulated with an AH (provides no confidentiality by encryption) or ESP (provides encryption) header and an additional IP header. The IP addresses of the newly added outer IP header are that of the VPN Gateways.
The traffic between the two VPN Gateways appears to be from the two gateways (in a new IP datagram), with the original IP datagram is encrypted (in case of ESP) inside IPSec packet.
Question 26 |
In a system an RSA algorithm with p=5 and q=11, is implemented for data security. What is the value of the decryption key if the value of the encryption key is 27?
3 | |
7 | |
27 | |
40 |
Question 26 Explanation:
The keys for the RSA algorithm are generated the following way:
1. Choose two distinct prime numbers p and q.
2. Compute n = pq.
3. Compute λ(n) = lcm(λ(p), λ(q)) = lcm(p − 1, q − 1), where λ is Carmichael's totient function. Choose an integer e such that 1 < e < λ(n) and gcd(e, λ(n)) = 1; i.e., e and λ(n) are coprime.
4. Determine d as d ≡ e−1 (mod λ(n)); i.e., d is the modular multiplicative inverse of e modulo λ(n). This means: solve for d the equation d⋅e ≡ 1 (mod λ(n)).
Given two prime numbers are p = 5 and q = 11, encryption key, e = 27
n = p * q = 5 * 11 = 55
λ(n)= (p-1) * (q-1) = 4 * 10 = 40
Let the value of decryption key be ‘d’ such that:
e * d mod λ(n) = 1
27 * d mod 40 = 1
d = 3
1. Choose two distinct prime numbers p and q.
2. Compute n = pq.
3. Compute λ(n) = lcm(λ(p), λ(q)) = lcm(p − 1, q − 1), where λ is Carmichael's totient function. Choose an integer e such that 1 < e < λ(n) and gcd(e, λ(n)) = 1; i.e., e and λ(n) are coprime.
4. Determine d as d ≡ e−1 (mod λ(n)); i.e., d is the modular multiplicative inverse of e modulo λ(n). This means: solve for d the equation d⋅e ≡ 1 (mod λ(n)).
Given two prime numbers are p = 5 and q = 11, encryption key, e = 27
n = p * q = 5 * 11 = 55
λ(n)= (p-1) * (q-1) = 4 * 10 = 40
Let the value of decryption key be ‘d’ such that:
e * d mod λ(n) = 1
27 * d mod 40 = 1
d = 3
There are 26 questions to complete.