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1. cryptography-questions-answers-message-authentication-codes-q1
Which figure implements confidentiality and authentication?

View Answer

Answer: a [Reason:] Figure a implements both confidentiality and authentication.

2. cryptography-questions-answers-message-authentication-codes-q1
Which figure implements confidentiality, authentication and signature?

View Answer

Answer: d [Reason:] Figure a implements confidentiality, authentication and signature.

3. cryptography-questions-answers-message-authentication-codes-q3
The above figure implements authentication and signature only?
a) True
b) False

View Answer

Answer: a [Reason:] The system does not implement confidentiality.

4. cryptography-questions-answers-message-authentication-codes-q4
The above figure represents internal error control.
a) True
b) False

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Answer: b [Reason:] The system is that of external error control.

5. Another name for Message authentication codes is
a) cryptographic codebreak
b) cryptographic codesum
c) cryptographic checksum
d) cryptographic checkbreak

View Answer

Answer: c [Reason:] Cryptographic checksum is another name for MACs.

6. The following figure has authentication tied to the plaintext-
cryptography-questions-answers-message-authentication-codes-q6
a) True
b) False

View Answer

Answer: b [Reason:] The figure has authentication tied to the ciphertext.

7. For a 150-bit message and a 10-bit MAC, how many values are the MAC value dependent on?
a) 2140
b) 2150
c) 215
d) 210

View Answer

Answer: a [Reason:] 2150/210 = 2140.

8. Confidentiality can only be provided if we perform message encryption before the MAC generation.
a) True
b) False

View Answer

Answer: b [Reason:] Confidentiality can be provided even if we perform message encryption after the MAC generation.

9. MACs are also called
a) testword
b) checkword
c) testbits
d) none of the mentioned

View Answer

Answer: d [Reason:] Another term for MACs are tags(or check sum).

10. For a 100 bit key and a 32 bit tag, how many possible keys can be produced in the 3rd round?
a) 24
b) 232
c) 216
d) 264

View Answer

Answer: a [Reason:] First round: 100 – 32 = 68 Second round: 68 – 32 = 36. Third round: 36 – 32 = 4. Therefore 24 keys can be produced by the third round.

11. MAC is a
a) one-to-one mapping
b) many-to-one mapping
c) onto mapping
d) none of the mentioned

View Answer

Answer: b [Reason:] MACs are many to one mapping, which makes it tougher for the intruder for cryptanalysis.

12. For an n-bit tag and a k-bit key, the level of effort required for brute force attack on a MAC algorithm is
a) 2k
b) 2n
c) min(2k,2n)
d) 2k/2n

View Answer

Answer: c [Reason:] The level of effort required for brute force attack on a MAC algorithm is min(2k,2n).