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Online MCQ Assignment
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1. The number of rounds in RC5 can range from 0 to _____________
a) 127
b) 63
c) 255
d) 31

View Answer

Answer: c [Reason:] Number of Rounds range from 0 to 255.

2. Multiply the polynomials P1 = (x5 + x2 + x) by P2 = (x7 + x4 + x3 + x2 + x) in GF(28) with irreducible polynomial (x8 + x4 + x3 + x + 1). The result is
a) x4+ x3+ x+1
b) x5+ x3+ x2+x+1
c) x5+ x4+ x3+x+1
d) x5+ x3+ x2+x

View Answer

Answer: b [Reason:] On performing polynomial multiplication we get with respect to modulus (x8 + x4 + x3 + x + 1) we get x5+ x3+ x2+x+1.

3. RC5 uses 2 magic constants to define their subkeys. These are
a) Base of natural Logarithm and Golden ratio
b) Base of natural Logarithm and Pi
c) Golden Ratio and Pi
d) Pi and Golden Ration

View Answer

Answer: a [Reason:] The initialize operations makes use of magic constants defined as follows: P_w=Odd[(e-2) 2w] Q_w=Odd[(φ-1) 2w].

4. How many keys are used in the ANSI X9.17 PRNG?
a) 3
b) 2
c) 4
d) 6

View Answer

Answer: a [Reason:] 2 keys are used for the 3DES Encrypt Decrypt Encrypt method.

5. The GCD of x5+ x4+ x3– x2 – x + 1 and x3 + x2+ x + 1 over GF(3) is
a) 1
b) x
c) x + 1
d) x2 + 1

View Answer

Answer: c [Reason:]The GCD ofx5+ x4+ x3 – x2 – x + 1 and x3 + x2+ x + 1 over GF(3) is x + 1.

6. The property that any extracted subsequence should pass the test for randomness is
a) Scalability
b) Uniformity
c) Stability
d) Consistency

View Answer

Answer: a [Reason:] Scalability is the property where any extracted subsequence should pass the test for randomness.

7. Maurer’s Universal Statistical Test is
a) a test to determine whether the number of runs of ones and zeros of various lengths is as expecte
b) a test to detect if a sequence can be significantly compressed without loss of information
c) a test to determine whether the number of ones and zeros in a sequence is approx the same as in a truely random o/p
d) a test for the check of evenness in the 1s and 0s

View Answer

Answer: b [Reason:] Maurer’s Universal Statistical Test is a test to detect if a sequence can be significantly compressed without loss of information.

8. Find the first 8 bits for Blum Blum Shub Bit Generator when seed = 101355 and n = 192649.
a) 10101010
b) 11100010
c) 11001011
d) 11001110

View Answer

Answer: d [Reason:] The blum blum shub algorithm is as follows Xo = s^2 mod n for i=1 to 8 Xi = X(i-1)^2 mod n Bi = Xi mod 2 Using this we compute the bits as – 11001110.

9. Which one of the following is not a RC5 mode of operation?
a) RC5 block cipher
b) RC5-Cipher Block Chaining
c) RC5-Cipher Padding
d) RC5-CipherText Stealing

View Answer

Answer: c [Reason:] RFC 2040 [BALD96] defines four different modes of operation: 1. RC5 block cipher 2. RC5-CBC 3. RC5-CBC-Pad 4. RC5-CTS.

10. On Encrypting “thepepsiisintherefrigerator” using Vignere Cipher System using the keyword “HUMOR” we get cipher text
a) abqdnwewuwjphfvrrtrfznsdokvl
b) abqdvmwuwjphfvvyyrfznydokvl
c) tbqyrvmwuwjphfvvyyrfznydokvl
d) baiuvmwuwjphfoeiyrfznydokvl

View Answer

Answer: b [Reason:] Cipher text:= Ci = Pi + ki mod m (mod 26).

11. Consider the cipher text message:
YJIHX RVHKK KSKHK IQQEV IFLRK QUZVA EVFYZ RVFBX UKGBP KYVVB QTAJK TGBQO ISGHU CWIKX QUXIH DUGIU LMWKG CHXJV WEKIH HEHGR EXXSF DMIIL UPSLW UPSLW AJKTR WTOWP IVXBW NPTGW EKBYU SBQWS

Relative Frequencies
3 7 2 2 5 5 7 9 11 4 14 4 2 1 3 4 6 5 6 5 7 10 9 8 4 2

The Index of Coincidence is
a) 0.065
b) 0.048
c) 0.067
d) 0.044

View Answer

Answer: d [Reason:] Number of letters = 145.From this, IC=0.0438697 .This is very strong evidence that the message came from a polyalphabetic ciphering scheme.

12. The plain text for the cipher text RRLMWBKASPDH. Given the encryption key
[(17 17 5 | 21 18 21 | 2 2 19)]
is –
a) PAYMOREMONEY
b) PAYINGRANSOM
c) GETINTHECAB
d) THEEXPRESSID

View Answer

Answer: a [Reason:] Calculated the Deciphering matrix K and then perform the hill cipher algorithm.

13. The Inverse S-box value for byte stored in cell (D,2)
a) 0x5F
b) 0x2D
c) 0x7F
d) 0x5D

View Answer

Answer: c [Reason:] We first find the multiplicative inverse of 0xD2. And then perform the matrix transformation to get 0x7F.

14. Division of (131B6C3) base 16 by (lA2F) base 16 yeilds
a) 1AD
b) DAD
c) BAD
d) 9AD

View Answer

Answer: d [Reason:] Base 16 division to be followed where A-F stand for 10-15.

15. How many rounds does the AES-192 perform?
a) 10
b) 12
c) 14
d) 16

View Answer

Answer: b [Reason:] AES 192 performs 12 rounds.