11. The
golden ratio Ï• and its conjugate Ï•’ both satisfy the equation
(A) x3 – x – 1 = 0
(B) x3 + x – 1 = 0
(C) x2 – x – 1 = 0
(D) x2 + x – 1 = 0
Answer: C
Explanation:
The golden ratio (Greek letter Ï• (phi)) is a
special number approximately equal to 1.618. The Golden Ratio is found when we
divide a line into two parts so that the longer part divided by the smaller
part is also equal to the whole length divided by the longer part.
a/b = (a+b)/a = 1.618...= Ï•
It is exactly equal to (1+√5)/2
12. The
solution of recurrence relation, T(n) = 2T(floor (√n)) + logn is
(A) O(n log log logn)
(B) O(n log logn)
(C) O(log logn)
(D) O(logn log logn)
Answer: D
13. In
any n-element heap, the number of nodes of height h is
(A) less than equal to [n/2h]
(B) greater than [n/2h]
(C) greater than [n/2h+1]
(D) less than equal to [n/2h+1]
Answer: D
14. A
data file of 1,00,000 characters contains only the characters g-l, with the
frequencies as indicated in table:
using the variable-length code by Huffman
codes, the file can be encoded with
(A) 2,52,000 bits
(B) 2,64,000 bits
(C) 2,46,000 bits
(D) 2,24,000 bits
Answer: D
Explanation:
A binary code encodes each character as a
binary string or codeword. We would like to find a binary code that encodes the
file using as few bits as possible, ie., compresses it as much as possible.
In a fixed-length code each codeword has the
same length. In a variable-length code codewords may have different lengths.
Here are examples of fixed and variable legth codes for our problem (note that
a fixedlength code must have at least 3 bits per codeword).
The fixed length-code requires 3,00,000 bits
to store the file.
The variable-length code uses only
(45x1+13x3+12x3+16x3+9x4+5x4)x1000=2,24,000
bits.
15. A
vertex cover of an undirected graph G(V, E) is a subset V1 ⊆ V vertices such that
(A) Each pair of vertices in V1 is
connected by an edge
(B) If (u, v) ∈ E then u ∈ V1 and v ∈ V1
(C) If (u, v) ∈ E then u ∈ V1 or v ∈ V1
(D) All pairs of vertices in V1
are not connected by an edge
Answer: C
16. In
a fully connected mesh network with n devices, there are ................ physical
channels to link all devices.
(A) n(n–1)/2
(B) n(n+1)/2
(C) 2n
(D) 2n+1
Answer: A
17. The
baud rate of a signal is 600 baud/second. If each signal unit carries 6 bits,
then the bit rate of a signal is .................
(A) 3600
(B) 100
(C) 6/600
(D) None of the above
Answer: A
Explanation:
Bit rate=baud rate X bit
18. Match
the following:
List - I
a. Data link layer
b. Network layer
c. Transport layer
d. Application layer
List - II
i. Flow control
ii. Node to node delivery
iii. Mail services
iv. Routing
Codes:
a
b c d
(A) ii i iv iii
(B) ii
iv i iii
(C) ii i iii iv
(D) ii iv iii i
Answer: B
19. An
image is 1024∗800
pixels with 3 bytes/pixel. Assume the image is uncompressed. How long does it
take to transmit it over a 10-Mbps Ethernet?
(A) 196.6 seconds
(B) 19.66 seconds
(C) 1.966 seconds
(D) 0.1966 seconds
Answer: C
Explanation:
Time taken to
transmit=(1024x800x3x8)/(10x1000x1000)=1.966 seconds
20. The
.............. measures the relative strengths of two signals or a signal at two
different points.
(A) frequency
(B) attenuation
(C) throughput
(D) decibel
Answer: D
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