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Interview MCQ Set 1

1. It is possible to make array of booleans.
a) True
b) False

View Answer

Answer: a [Reason:] Arrays of strings, numbers, booleans can be made.

2. What is the output of the given code?

string_array = ["a","e","i","o","u"]
print string_array

a) [“a”,”e”,”i”,”o”,”u”].
b) Error
c) Vowels
d) None of the mentioned

View Answer

Answer: a [Reason:] The array is a string array.

Output:
["a","e","i","o","u"]

3. What is the output of the given code?

string_array = ["a","e","i","o","u"]
print string_array[3]

a) [“a”,”e”,”i”,”o”,”u”].
b) Error
c) o
d) None of the mentioned

View Answer

Answer: c [Reason:] The array is a string array and the index is 3 so ‘o’ will be the output.

Output:
o

4. What is the output of the given code?

string_array = ["a","e","i","o","u"]
boolean_array = ["True","False"]
puts string_array[3]
puts boolean_array

a) [“a”,”e”,”i”,”o”,”u”].
b) Error
c) o
True
False
d) None of the mentioned

View Answer

Answer: c [Reason:] The array is a string array and the index is 3 so ‘o’ will be the output and then the boolean_array will get printed.

Output:
o
True
False

5. What is the output of the given code?

string_array = ["a","e","i","o","u"]
boolean_array = ["True","False"]
puts string_array[3]
puts boolean_array[1]

a) [“a”,”e”,”i”,”o”,”u”].
b) Error
c) o
False
d) None of the mentioned

View Answer

Answer: c [Reason:] The array is a string array and the index is 3 so ‘o’ will be the output and then the boolean_array[1] = false will get printed.

Output:
o
False

6. What is the output of the given code?

a=[1,2,3,4,5]
b=[1,2,4,6,8]
if a[3]==b[2]
    print "Equal"
end

a) Equal
b) Error
c) 4
d) None of the mentioned

View Answer

Answer: a [Reason:] a[3]=4 and b[2]=4 hence it will print equal according to the given if condition.

Output:
Equal

7. What is the output of the given code?

a=[1,2,3,4,5]
b=[1,2,3,4,5]
if a==b
    print "Equal"
else
    print "Not equal"
end

a) Equal
b) Error
c) Not equal
d) None of the mentioned

View Answer

Answer: a [Reason:] Elements of both the array are same hence they are equal.

Output:
Equal

8. What is the output of the given code?

a=["hey", "ruby", "language"]
b=["hey", "ruby", "language"]
if a==b
    print "Equal"
else
    print "Not equal"
end

a) Equal
b) Error
c) Not equal
d) None of the mentioned

View Answer

Answer: a [Reason:] Elements of both the array are same and in same sequence hence they are equal.

Output:
Equal

9. What is the output of the given code?

a=["hey", "ruby", "language"]
b=["hey", "language", "ruby"]
if a==b
    print "Equal"
else
    print "Not equal"
end

a) Equal
b) Error
c) Not equal
d) None of the mentioned

View Answer

Answer: c [Reason:] Elements of both the array are same but not in same sequence hence they are not equal.

Output:
Not equal

10. What is the output of the given code?

a=["hey", "ruby", "language"]
b=[1, 2, 3]
puts b[1]
puts a[2]

a) 3 ruby
b) Error
c) 2
language
d) None of the mentioned

View Answer

Answer: c [Reason:] b[1]=2 and a[2]=language hence these both will get printed.

Output: 2 language

Interview MCQ Set 2

1. Find the order of the group G = <Z12*, ×>?
a) 4
b) 5
c) 6
d) 2

View Answer

Answer: a [Reason:] It can be obtained using Euler Phi function, i.e. f(n).

2. Find the order of the group G = <Z21*, ×>?
a) 12
b) 8
c) 13
d) 11

View Answer

Answer: a [Reason:] |G| = f(21) = f(3) × f(7) = 2 × 6 =12 There are 12 elements in this group: G = <Z21*, ×> = {1, 2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20}. All are relatively prime with 21.

3. Find the order of group G= <Z20*, x>
a) 6
b) 9
c) 10
d) 8

View Answer

Answer: d [Reason:] |G| = f(20) = f(4) × f(5) = f(22) × f(5) = (22-21)(51-50) = 8. G = <Z20 *, x> = { 1, 3, 7, 9, 11, 13, 17, 19 }.

4. Find the order of group G= <Z7*, x>
a) 6
b) 4
c) 3
d) 5

View Answer

Answer: a [Reason:] |G| = f(7) = (71-70) = 6 G = <Z20, x> = { 1, 2, 3, 4, 5, 6 }.

5. “In the group G = <Zn*, ×>, when the order of an element is the same as order of the group (i.e. f(n)), that element is called the Non – primitive root of the group.”
a) True
b) False

View Answer

Answer: b [Reason:] Such a group is called the primitive root of the group.

6. In the order of group G= <Z20*, x>, what is the order of element 17?
a) 16
b) 4
c) 11
d) 6

View Answer

Answer: b [Reason:] 17 17 9 13 1 ord(17) = 4 n? 1 2 3 4 5 6 7 order

7. The order of group G= <Z9, x> , primitive roots of the group are –
a) 8 , Primitive roots- 2,3
b) 6 , Primitive roots- 5
c) 6 , Primitive roots- 2,5
d) 6 , Primitive roots- 5,7

View Answer

Answer: c [Reason:] |G| = f(9) = (32-31) = 6 G = <Z20, x> = { 1, 2, 4, 5, 7, 8 }.

8. Which among the following values: 17, 20, 38, and 50, does not have primitive roots in the group G = <Zn*, ×>?
a) 17
b) 20
c) 38
d) 50

View Answer

Answer: b [Reason:] The group G = <Zn*, ×> has primitive roots only if n is 2, 4, pt, or 2pt ‘p’ is an odd prime and‘t’ is an integer. G = <Z17*, ×> has primitive roots, 17 is a prime. G = <Z20*, ×> has no primitive roots. G = <Z38*, ×> has primitive roots, 38 = 2 × 19 prime. G = <Z50*, ×> has primitive roots, 50 = 2 × 52 and 5 is a prime.

9. Find the number of primitive roots of G=<Z11*, x>?
a) 5
b) 6
c) 4
d) 10

View Answer

Answer: c [Reason:] Number of primitive roots = f(f(11))=f((111-110)) = f(10) = f(2). f(5) = (21-20)(51-50) = 1 x 4 = 4 The primitive roots of this set {2, 6, 7, 8}.

10. Find the primitive roots of G=<Z11*, x>?.
a) {2, 6, 8}
b) {2, 5, 8}
c) {3, 4, 7, 8}
d) {2, 6, 7, 8}

View Answer

Answer: d [Reason:] Number of primitive roots = f(f(11))=f((111-110)) = f(10) = f(2). f(5) = (21-20)(51-50) = 1 x 4 = 4 The primitive roots of this set {2, 6, 7, and 8}.

11. “If a group has primitive roots, it is a cyclic group”
a) True
b) False

View Answer

Answer: a [Reason:] Yes, a group which has primitive roots is a cyclic group.

12. Find the primitive roots of G = <Z10*, ×>.
a) {2, 6, 8}
b) {3,6 ,9}
c) {3, 7, 8}
d) {3, 7}

View Answer

Answer: c [Reason:] Number of primitive roots = f(f(11))=f((111-110)) = f(10) = f(2). f(5) = (21-20)(51-50) = 1 x 4 = 4 The primitive roots of this set are {3, 7}.

13. “The group G = <Zp*, ×> is always cyclic.”
a) True
b) False

View Answer

Answer: a [Reason:] G = <Zp*, ×> is always cyclic.

Interview MCQ Set 3

1. What is the symbol used for mass?
a) M
b) L
c) A
d) V

View Answer

Answer: a [Reason:] Unit of mass is kg.

2. What is the symbol used for length?
a) W
b) M
c) L
d) K

View Answer

Answer: c [Reason:] Unit of length is meter.

3. What is the symbol used for time?
a) M
b) H
c) K
d) T

View Answer

Answer: d [Reason:] Unit of time is second.

4. What is the symbol used for gravity?
a) g
b) h
c) k
d) v

View Answer

Answer: a [Reason:] Unit of gravity is m/s2.

5. What is the symbol used for dynamic viscosity?
a) σ
b) µ
c) λ
d) δ

View Answer

Answer: b [Reason:] Unit of dynamic viscosity is kg/m s.

6. What is the symbol used for thermal conductivity?
a) v
b) g
c) h
d) k

View Answer

Answer: d [Reason:] Unit of thermal conductivity is W/m Degree.

7. What is the symbol used for specific heat?
a) C S
b) C R
c) C P
d) C M

View Answer

Answer: c [Reason:] Unit of specific heat is k J/kg degree.

8. What is the symbol used for convective film coefficient?
a) v
b) h
c) σ
d) µ

View Answer

Answer: b [Reason:] Unit of convective film coefficient is W/m2 degree.

9. What is the symbol used for kinematic viscosity?
a) v
b) λ
c) µ
d) σ

View Answer

Answer: a [Reason:] Unit of kinematic viscosity is m2/s.

10. What is the symbol used for coefficient of volumetric expansion?
a) λ
b) µ
c) δ
d) β

View Answer

Answer: d [Reason:] Unit of coefficient of volumetric expansion is per degree.

Interview MCQ Set 4

1. In terms of the size of the network the correct order (ascending) is –
a) PAN, MAN, LAN, WAN
b) LAN, MAN, WAN, PAN
c) PAN, LAN, MAN, WAN
d) LAN, PAN, MAN, WAN

View Answer

Answer: c [Reason:] Personal Area Network, Local Area Network, Metropolitan Area Network, Wide Area Network is the correct order in terms of size of the network from smallest to largest.

2. How many layers are there in the OSI reference model?
a) 4
b) 5
c) 6
d) 7

View Answer

Answer: d [Reason:] The 7 layers are : Physical Layer, Data Link Layer, Network Layer, Transport Layer, Session Layer, Presentation Layer, Application Layer.

3. Logical Addressing and Routing are functions of which layer?
a) Physical Layer
b) Transport Layer
c) Data Link Layer
d) Network Layer

View Answer

Answer: d [Reason:] . Logical Addressing and Routing are functions of the Network Layer.

4. Flow Control and Error Control are functions of which layer?
a) Physical Layer
b) Application Layer
c) Data Link Layer
d) Network Layer

View Answer

Answer: c [Reason:] Flow Control and Error Control are functions of the Data Link Layer.

5. Dialog Control and Synchronization are function of which layer?
a) Presentation Layer
b) Application Layer
c) Session Layer
d) Data Link Layer

View Answer

Answer: c [Reason:] Dialog Control and Synchronization are function of the Session Layer.

6.Encryption and Compression are functions of which OSI layer?
a) Presentation Layer
b) Application Layer
c) Session Layer
d) Data Link Layer

View Answer

Answer: a [Reason:] Encryption and Compression are functions of the Presentation Layer.

7. File, Transfer, Access and Management (FTAM) is a function of which layer ?
a) Presentation Layer
b) Application Layer
c) Session Layer
d) Data Link Layer

View Answer

Answer: b [Reason:] FTAM is an Application Layer function.

8. How many layers are present in the TCP/IP Reference model?
a) 6
b) 7
c) 5
d) 4

View Answer

Answer: d [Reason:] There are 4 layers in the TCP/IP reference model : Link, Internet, Transport and Application.

Interview MCQ Set 5

1. In AES, to make the s-box, we apply the transformation –
b’i = bi XOR b(i+4) XOR b(i+5) XOR b(i+6) XOR b(i+7) XOR ci
What is ci in this transformation?
a) ci is the ith bit of byte c with value 0x63
b) ci is the ith bit of byte c with value 0x25
c) ci is the ith bit of byte c with value 0x8F
d) ci is the ith bit of byte c with value 0x8A

View Answer

Answer: a [Reason:] ci is the ith bit of byte c with value 0x63 i.e, c = 01100011

2. The inverse s-box permutation follows,
b’i = b(i+2) XOR b(i+5) XOR b(i+7) XOR di
Here di is –
a) di is the ith bit of a byte ‘d’ whose hex value is 0x15
b) di is the ith bit of a byte ‘d’ whose hex value is 0x05
c) di is the ith bit of a byte ‘d’ whose hex value is 0x25
d) di is the ith bit of a byte ‘d’ whose hex value is 0x51

View Answer

Answer: b [Reason:] The value of ‘d’ is 0x05.

3. What is the block size in the Simplified AES algorithm?
a) 8 bits
b) 40 bits
c) 16 bits
d) 36 bits

View Answer

Answer: b [Reason:] The block size for the AES algorithm is 16 bits.

4. Which function can be used in AES multiplication –
a) m(x)=x7+x4+x3
b) m(x)=x8+x4+x3+x+1
c) m(x)=x8+x3+x2+x+1
d) m(x)=x8+x5+x3+x

View Answer

Answer: b [Reason:] m(x)=x8+x4+x3+x+1 stands for 100011011 which is an irreducible polynomial. Others are not irreducible polynomials.

5. In the DES algorithm the Round Input is 32 bits, which is expanded to 48 bits via ______________
a) Scaling of the existing bits
b) Duplication of the existing bits
c) Addition of zeros
d) Addition of ones

View Answer

Answer: a [Reason:] The round key is 48 bits. The input is 32 bits. This input is first expanded to 48 bits (permutation plus an expansion), that involves duplication of 16 of the bits.

6. Using Differential Crypt-analysis, the minimum computations required to decipher the DES algorithm is-
a) 256
b) 243
c) 255
d) 247

View Answer

Answer: d [Reason:] Differential Crypt-analysis requires only 247 computations to decipher the DES algorithm.

7. Using SDES, the Plaintext for the Ciphertext 00001111, given that the key is 1111111111 is,
a) 01100111
b) 00001010
c) 11111111
d) 01101101

View Answer

Answer: c [Reason:] Perform the SDES Decryption algorithm and compute the cipher text.

8. 12. For p = 11 and q = 17 and choose e=7. Apply RSA algorithm where Cipher message=11 and thus find the plain text.
a) 88
b) 122
c) 143
d) 111

View Answer

Answer: a [Reason:] n = pq = 11 × 19 = 187. C=M^e mod n ; C=11^23 mod 187 ; C = 88 mod 187.

9. Reduce the following Big-O notation. O[ ax7 + 3 x3 + sin(x)] =
a) O[ax7].
b) O[sin(x)].
c) O[x7].
d) O[x7 + x3].

View Answer

Answer: c [Reason:] O[ ax7 + 3 x3 + sin(x)] = O(ax7) = O(x7).

10. Reduce the following Big-O notation. O[ en + an10] =
a) O[ an10].
b) O[ n10 ].
c) O[ en ].
d) O[ en+ n10 ].

View Answer

Answer: c [Reason:] O[ en + an10 ] = O[ en ].