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CSE - Questions and solutions set-4

Computer science and engineering(CSE) - Question & solution set-4
Question:-
Option (A)

25, 9 and 57 respectively

Option (B)

-6, -6 and -6 respectively

Option (C)

-7, -7 and -7 respectively

Option (D)

-25, -9 and -57 respectively

Correct Option:
C
Question:-
Option (A)
1 unit
Option (B)
4 unit
Option (C)
3 unit
Option (D)
2 unit
Correct Option:
D
Question Solution:
This can be reduced to A' + B which can be generated by one nor and one or gates. Let see how
A' + B = A'(B + B') + B
= A'B + A'B' + B
=(A' + 1)B + A'B'
=A'B' + B
=> A and B as input to nor gate and the resultant output being ored with B
Question:-
Option (A)
5
Option (B)
7
Option (C)
6
Option (D)
None of these
Correct Option:
B
Question Solution:
Best way to judge is that even in worst conditions atleast one process have enough resources to finish his task.
Since process require 3 USB ports, hence worst case will be when all process acquires 2 USB ports and are wainting for 3rd USB port.
If no one receives third port then this is deadlock.
Therefore
2*n < 15
or n < 15/2
or n < 7.5
therefore n = 7
At least one process always has three U.S.B port.
Question:-
Option (A)
2n/3 ternary digits.
Option (B)
n/log23
Option (C)
2n/3, ternary digits
Option (D)
none of these
Correct Option:
B
Question Solution:
Þ n × (log22/log23)

Þ n/log23

because log22=1
Question:-
Option (A)
1/2
Option (B)
1/6
Option (C)
1/4
Option (D)
1/3
Correct Option:
C
Question Solution:
Probability of even number =3/6 =1/2

Probability of odd number =3/6 =1/2

Both are independent so probability=1/2.1/2 =1/4
Question:-
Option (A)
7, 8
Option (B)
19, 20
Option (C)
18, 17
Option (D)
17, 18
Correct Option:
D
Question Solution:
logical 128 pages = 27

+ 1024 words = 210

= 217 words

Hence logical add. = 17 bits long.

Physical 256 fromes = 28

+ 1024 words = 210

= 218 words

Hence

Physical add. = 18 bits long.