Ans .
(B) Piggy backing
Ans .
(C) Application
Ans .
(D) 1000 bauds \ sec & 4000 bps
Bit rate is the number of bits per second. Baud rate is the number of signal units per second. Baud rate is less than or equal to the bit rate. Baud rate = 1000 bauds per second (baud/s) Bit rate = 1000 x 4 = 4000 bps
Ans .
(A) Ground
VLF (Very Low Frequency) LF (Low Frequency) In general there are three type of propagation.1.Ground Propagation 2.Sky propagation 3.Line-of-sight propagation. In ground propagation, radio waves travel through the lowest portion of the atmosphere. These low-frequency signals travel in all directions from the transmitting antenna.
Ans .
(C) 103
Basic RSA Algorithm: 1. Choose two primes, p & q. 2. Compute n=p*q and z=(p-1)*(q-1). 3. Choose a number relatively prime to z and call it d. 4. Find e such that e*d=1modz. Given p=13, q=31 & d=7, e=? n=p*q=403 z=(p-1)(q-1)=360 e*d=1modz 7e=1mod360, then 7e must be 361, 721, 1081, 1441, etc. Dividing each of these in turn by 7 to see which is divisible by 7, we find that 721/7 = 103, hence e = 103.
Ans .
(A) Number of components that can call or pass control to component A.
Ans .
(C) Cohesion
Cohesion is the indication of the relationship within module. Coupling is the indication of the relationships between modules. While designing you should strive for high cohesion. While designing you should strive for low coupling.
Ans .
(D) all of the above
Ans .
(C) Requirement design
Ans .
(B) Beta testing
Ans .
(A) Integrity constraint
Ans .
(B) 27th record
Ans .
(A) I, II, III & IV
Ans .
(B) Constraints
Ans .
(B) Passwords to prohibit unauthorised access to files.
Ans .
(B) ptr is a pointer to an array of 10 integers.
Ans .
(D) none of the above
Ans .
(D) ::
Ans .
(B) Inheritance
Ans .
(A) y = (int)(x+0.5)
Ans .
(D) -24
Ans .
(C) Only rear pointer changes
Ans .
(C) Mathematical Induction
Ans .
(C) t-1, 2t-1
Ans .
(D) m + n – 1
Ans .
(D) S1 is correct and S2 is correct.
Ans .
(D) S→aaaA|aaA|aA|λ, A→aAb|B, B→Bb|λ
Ans .
(C) S1 is correct and S2 is not correct.
Ans .
(D) Relocation
Ans .
(A) Leftmost derivation
Ans .
(C) Boolean sums and Boolean products and interchanging 0's & 1's
Ans .
(D) None of the above
6|292
6|48| 4
6|8| 0
6|1| 2
6|1| 1
Solution: (1204)6
So here, x=6
Ans .
(D) xyz’ + xy’z’ + x’yz’
Explanation:
Use Boolean identities to expand the product and simplify.
F(x, y, z)=(x + y)z’
=xz’+yz’ Distributive law
=x1z’+1yz’ Identity law
=x(y+y’)z’+(x+x’)yz’ Unit property
=xyz’+xy’z’+xyz’+x’yz’ Distributive law
=xyz’+xy’z’+x’yz’ Idempotent law
Ans .
(C) I - false & II - true
Ans .
(B) O(x2)
Big-O notation of f(x)=(x+1) log(x2+1) + 3x2
Note (x+1) is O(x) and x2+1≤2x2 when x>1
So, log x2+1≤log(2x2)=log 2+ log x2=log 2+ 2 log x
≤ 3 log x if x >2
Thus, log x2+1 is O(log x)
The first part of f(x) is O(x log x)
Also, 3x2 is O(x2)
So, f(x) is O(max(x log x, x2))=O(x2) as x log x ≤ x2 for x > 1
Ans .
(B) n-t
Ans .
(A) 6x + 7, 6x + 11
fog(x)=f(g(x))=f(3x+2)=2(3x+2)+3=6x+7
gof(x)=g(f(x))=g(2x+3)=3(2x+3)+2=6x+11
Ans .
(A) P(n,r)(n+1) / (n+1-r)
p(n, r) = n!/(n-r)!
p( n+1, r) = (n+1)!/(n+1-r)!
= (n+1) n! /(n+1-r) (n-r)!
= P(n, r)(n+1)/(n+1-r)
Ans .
(C) K3,3 or K5
Kuratowski’s Theorem: A graph is non-planar if and only if it contains a subgraph that is homeomorphic to either K5 or K3,3.
Ans .
(D) I, III & IV
Ans .
(D) Web pages
Ans .
(D) iii i iv ii
Ans .
(B) ATM
Ans .
(B) ANSI
Ans .
(B) Business application, Internal format conversion, EDI translator, EDI envelope
Ans .
(B) 8, 2, 7 and 5
T EP Priority After T
00 P1 P1[0] P2[-] P3[-] P4[-]
01 P2 P1[1] P2[0] P3[-] P4[-]
02 P1 P1[1] P2[X] P3[2] P4[-] ........P2 completed at 2
03 P3 P1[2] P2[X] P3[2] P4[2]
04 P4 P1[3] P2[X] P3[3] P4[2]
05 P1 P1[3] P2[X] P3[4] P4[X] .......P4 completed at 5
06 P3 P1[4] P2[X] P3[4] P4[X]
07 P1 P1[4] P2[X] P3[X] P4[X] .......P3 completed at 7
08 -- P1[X] P2[X] P3[X] P4[X] .......P1 completed at 8
Ans .
(C) 16 bits
VirtualAddress = 32 bits
PageSize = 4KB = 12 bits
therefore : VPNTag = 20 bits, OffsetTag = 12 bits
TLBEntryLength = VPNTag = 20 bits
TotalTLBEntries = 64,
4-way implies 64/ 4 = 16 sets = 4 bits
therefore : TLBIndex = 4 bits
TLBTag = TLBEntryLength - TLBIndex = 20 - 4 = 16 bits
Ans .
(A) 236 and 208 cylinders
SSTF
Initial head position =53
The closest queue to initial head position=65
head moves from 53 to 65=12
head moves from 65 to 67=2
head moves from 67 to 37=30
head moves from 37 to 14=23
head moves from 14 to 98=84
head moves from 98 to 122=24
head moves from 122 to 124=2
head moves from 124 to 183=59
Total head movement=236
SCAN
Initial head position=53 and moving towards 0
head moves from 53 to 37=16
head moves from 37 to 14=23
head moves from 14 to 65=51
head moves from 65 to 67=2
head moves from 67 to 98=31
head moves from 98 to 122=24
head moves from 122 to 124=2
head moves from 124 to 183=59
Total head movement=208
Ans .
(A) 332.8 KB
block size=29x23 bits (512 bytes)
disk size=1.3x230x23 bits (1.3 GB)
n=1.3x233/212=1.3x221 bits=1.3x210x28x23 bits=1.3x28 Kilo Bytes=332.8 KB
Ans .
(B) Fair Preemptive Scheduling
Linux uses two process-scheduling algorithms:
1. A time-sharing algorithm for fair preemptive scheduling between multiple processes
2. A real-time algorithm for tasks where absolute priorities are more important than fairness