Gate-1993

Question 1
 
A
(0, 0,α )
B
(α,0,0)
C
(0,0,1)
D
(0, ,0 α )
E
Both B and D
       Engineering Mathematics       Linear Algebra
Question 1 Explanation: 
Since, the given matrix is an upper triangular one, all eigen values are A. And hence A - λI = A.
So the question as has
(A - λI)X = 0
AX = 0

What x1, x2, x3 are suitable?
Which means:
x1 times column 1 + x2 times column 2 + x3 times column 3 = zero vector
Since α is not equal to zero, so x3 must be necessarily zero to get zero vector.
Hence, only (B) and (D) satisfies.
Question 2
       
A
linear
B
non-linear
C
homogeneous
D
of degree two
       Engineering Mathematics       Calculus
Question 2 Explanation: 
Note: Out of syllabus.

In this DE, degree is 1 then this represent linear equation.
Question 3
Simpson’s rule for integration gives exact result when f (x) is a polynomial of degree
A
1
B
2
C
3
D
4
       Engineering Mathematics       Simphson\'s Rule
Question 3 Explanation: 
Note: Out of syllabus.
Question 4
Which of the following is (are) valid FORTRAN 77 statement(s)?
A
DO 13 I = 1
B
A = DIM ***7
C
READ = 15.0
D
GO TO 3 = 10
       Computer Organization       FORTRAN
Question 4 Explanation: 
Note: Out of syllabus.
Question 5
   
A
B
C
D
       Engineering Mathematics       Calculus
Question 5 Explanation: 
Note: Out of syllabus.
Question 6
Which of the following improper integrals is (are) convergent?
A
B
C
D
        Engineering Mathematics       Calculus
Question 7
The function f(x,y) = x2y - 3xy + 2y + x has
A
no local extremum
B
one local minimum but no local maximum
C
one local maximum but no local minimum
D
one local minimum and one local maximum
        Engineering Mathematics       Functions
Question 7 Explanation: 
Note: Out of syllabus.
Question 8
   
A
1
        Engineering Mathematics       Calculus
Question 8 Explanation: 
Since the given expression is in 0/0 form, so we can apply L-Hospital rule.
Question 9
       
A
Out of syllabus.
        Engineering Mathematics       Calculus
Question 10
 
A
Out of syllabus.
        Engineering Mathematics       Vectors
Question 11
Given the differential equation, y ′ = x − y with the initial condition y (0) = 0. The value of y (0.1) calculated numerically upto the third place of decimal by the second order Runga-Kutta method with step size h = 0.1 is ________  
A
Out of syllabus.
        Engineering Mathematics       R-K Method
Question 12
 
A
Out of syllabus.
        Engineering Mathematics       FORTRAN
Question 13
 
A
1/3
        Engineering Mathematics       Calculus
Question 13 Explanation: 
Question 14
 
A
A4 = I
        Engineering Mathematics       Linear Algebra
Question 14 Explanation: 
Let λ be eigen value, then characteristic equation will be
(1-λ) (-1-λ) (i-λ) (-i-λ)
= (λ2-1) (λ2+1)
= λ4-1
Characteristic equation is λ4-1 = 0.
According to Cayley-Hamilton theorem, every matrix satisfies its own characteristic equation, so
A4 = I
Question 15
 
A
Out of syllabus.
        Engineering Mathematics       Vector Algebra
Question 16
 
A
Out of syllabus.
        Engineering Mathematics       Calculus
Question 17
   
A
Out of syllabus.
        Engineering Mathematics       Functions
Question 18
 
A
exclusive OR
B
exclusive NOR
C
NAND
D
NOR
E
None of the above
       Digital Logic Design       Logic Gates
Question 18 Explanation: 

So finally, we can write
Question 19
 
A
B
C
D
E
       Theory of Computation       Finite Automata
Question 20
 
A
Shift Register
B
Mod-3 Counter
C
Mod-6 Counter
D
Mod-2 Counter
E
Both A and C
       Digital Logic Design       J-K flip flop
Question 20 Explanation: 

Circuit behaves as shift register and mod-6 counter. Note that this is the Johnson counter which is the application of shift register. And Johnson counter is mod-2N counter.
Question 21
Assume that each character code consists of 8 bits. The number of characters that can be transmitted per second through an asynchronous serial line at 2400 baud rate, and with two stop bits, is
A
109
B
216
C
218
D
219
E
240
       Computer Organization       Serial Communication
Question 21 Explanation: 
Total bit per character
= 8 bit data + 2 stop bit + 1 start bit
= 11 bits
No. of characters = 2400/11 = 218.18
Since, it is asked for transmitted characters we take floor and answer is 218.
Question 22
 
A
(a) 6.625, (b) (45E)H
       Digital Logic Design       Number System
Question 22 Explanation: 
(a) 1*22 + 1*21 + 0*20 + 1*2-1 + 0*2-2 + 1*2-3
= 4 + 2 + 0 + 0.5 + 0 + 0.125
= 6.625
(b) 1118 mod 16 = E, quotient = 69
69 mod 16 = 5, quotient = 4
4 mod 16 = 4
Writing the mods result in reverse order gives (45E)H.
Question 23

A ROM is used to store the Truth table for a binary multiple unit that will multiply two 4-bit numbers. The size of the ROM (number of words × number of  bits) that is required to accommodate the Truth table is M words × N bits. Write the values of M and N.

A
M=256, N = 8
       A ROM is used to store the Truth table for a binar       Truth Table
Question 23 Explanation: 
Input will consist of 8 bit (two 4-bit numbers) = 28 address.
Output will be of 8 bits.
So memory will be of 28 × 8.
So, M = 256, N = 8.
Question 24
 
A
P = 12.5, Q = 2.5×106
       Operating Systems       Memory Management
Question 24 Explanation: 
RPM = 2400
So, in DOS, the disk rotates 2400 times.
Average latency is the time for half a rotation
= 0.5×60/2400 s
= 12.5 ms
In one full rotation, entire data in a track can be transferred. Track storage capacity = 62500 bits
So, disk transfer rate
= 62500 × 2400/60
= 2.5 × 106 bps
So,
P = 12.5, Q = 2.5×106
Question 25
 
A
90%
       Computer Organization       iNTERRUPTS
Question 25 Explanation: 
Time to service an interrupt
= saving state of CPU + ISR execution + restoring of CPU state
= (80 + 10 + 10) × 10-6
= 100 μs
For every 1ms an interrupt occurs which is served for 100 μs.
1ms = 1000μs
Thus, for every 1000μs, (1000 - 100) = 900 μs of main program and 100μs of interrupt overhead exists.
Thus, 900/1000 is usage of CPU to execute main program .
∴ % of CPU to execute main program is (900/1000) × 100 = 90%
Question 26
 
A
1, because m is a local variable in P
B
0, because m is the actual parameter that corresponds to the formal parameter in p
C
0, because both x and y are just reference to m, and y has the value 0
D
1, because both x and y are just references to m which gets modified in procedure P
E
none of the above
       Programming       Programming
Question 26 Explanation: 
0, because global m is not modified, m is just passed to formal argument of P.
Question 27
   
A
0, because n is the actual parameter corresponding to x in procedure Q.
B
0, because n is the actual parameter to y in procedure Q.
C
1, because n is the actual parameter corresponding to x in procedure Q.
D
1, because n is the actual parameter corresponding to y in procedure Q.
E
none of the above
       Programming       Programming
Question 27 Explanation: 
0, because n is just passed to formal parameters of Q and no modification in global n.
Question 28
   
A
PARAM, P, Q
B
PARAM, P
C
PARAM, Q
D
P, Q
E
none of the above
       Programming       Programming
Question 28 Explanation: 
Since m is defined global it is visible inside all the procedures.
Question 29
 
A
exchanges a and b
B
doubles a and stores in b
C
doubles b and stores in a
D
leaves a and b unchanged
E
none of the above
       Programming       Programming
Question 29 Explanation: 
Exchanges a and b.
Let us consider a=5; b=2
a := a+b = 5+2 = 7
b := a-b = 7-2 = 5
a := a-b = 7-5 = 2
O/P: a=2; b=5
Question 30
 
A
The program leads to compile time error
B
The program leads to run time error
C
The program outputs 5.2
D
The program produces error relating to nil pointer dereferencing
E
None of the above
       Compiler Design       cOMPILERS
Question 30 Explanation: 
Note: Out of syllabus.
Question 31
A simple two-pass assembler does the following in the first pass:
A
It allocates space for the literals.
B
It computes the total length of the program
C
It builds the symbol table for the symbols and their values.
D
It generates code for all the load and store register instructions.
E
A, B and C
       Computer Organization       aSSEMBLERS
Question 31 Explanation: 
Pass 1:
1) Assign address to all statements in the program.
2) Save the values assigned to all tables for use in pass 2.
3) Perform some processing of assembler directives.
Question 32
A part of the system software, which under all circumstances must reside in the main memory, is:
A
text editor
B
assembler
C
linker
D
loader
E
none of the above
       Compiler Design       gENERAL
Question 32 Explanation: 
In a program the loader that can loads the object of the program from secondary memory into the main memory to execute the corresponding program. Then the loader is to be resides in the main memory.
Question 33
The root directory of a disk should be placed
A
at a fixed address in main memory
B
at a fixed location on the disk
C
anywhere on the disk
D
at a fixed location on the system disk
E
anywhere on the system disk
       Algorithms
Question 33 Explanation: 
Root directory can points to the various user directories. Then they will be stored in a way that user can't be easily modify them. Then they should be at fixed location on the disk.
Question 34
Consider a system having m resources of the same type. These resources are shared by 3 processes A, B and C, which have peak demands of 3, 4 and 6 respectively. For what value of m deadlock will not occur?
A
7
B
9
C
13, 15
D
13
E
15
       Operating Systems       Deadlocks
Question 34 Explanation: 
A requires 3, B-4, C-6;
→ If A have 2, B have 3, C have 5 then deadlock will occur i.e., 2+3+5=10.
→ If we have one extra resource then deadlock will not occur i.e., 10+1=11.
→ If we have equal (or) more than 11 resources then deadlock will never occur.
Question 35
Consider a system having m resources of the same type. These resources are shared by 3 processes A, B and C, which have peak demands of 3, 4 and 6 respectively. For what value of m deadlock will not occur?
A
7
B
9
C
13, 15
D
13
E
15
       Operating Systems       Deadlock
Question 35 Explanation: 
A requires 3, B-4, C-6;
→ If A have 2, B have 3, C have 5 then deadlock will occur i.e., 2+3+5=10.
→ If we have one extra resource then deadlock will not occur i.e., 10+1=11.
→ If we have equal (or) more than 11 resources then deadlock will never occur.
Question 36
 
A
4
B
10
C
11
D
12
E
None of the above
       Operating Systems       Cpu Scheduling
Question 36 Explanation: 
Algorithm: Round robin; TQ: 1

p is departure at 11.
Question 37
Consider a simple connected graph G with n vertices and n-edges (n>2). Then, which of the following statements are true?
A
G has no cycles.
B
The graph obtained by removing any edge from G is not connected.
C
G has at least one cycle.
D
The graph obtained by removing any two edges from G is not connected.
E
Both C and D.
       Algorithms       Graphs
Question 37 Explanation: 
If a graph have n vertices and n edges (n>2) then it is to be cyclic graph. Then it have atleast one cycle and if we remove two edges then it is not connected.
For example let us consider, n=3
Question 38
The proposition p ∧(~p ∨ q) is:
A
a tautology
B
logically equivalent to p ∧ q
C
logically equivalent to p ∨ q
D
a contradiction
E
none of the above
       Engineering Mathematics        Prepositional Logic
Question 38 Explanation: 
p ∧(~p ∨ q)
(p ∧ ~p) ∨ (p ∧ q)
F ∨ (p ∧ q)
(p ∧ q)
Question 39
Let S be an infinite set and S1 ..., Sn be sets such that S1 ∪ S2 ∪ ... ∪ Sn = S. Then,
A
at least one of the set Si is a finite set
B
not more than one of the set Si can be finite
C
at least one of the sets Si is an infinite set
D
None of the above
       Engineering Mathematics       Set Theory
Question 39 Explanation: 
Given sets are finite union of sets. One set must be infinite to make whole thing to be infinite.
Question 40
Let A be a finite set of size n. The number of elements in the power set of A × A is:
A
22n
B
2n2
C
(2n)2
D
(22)n
       Engineering Mathematics       Set Theory
Question 40 Explanation: 

Question 41
The less-than relation, <, on reals is
A
a partial ordering since it is asymmetric and reflexive
B
a partial ordering since it is antisymmetric and reflexive
C
not a partial ordering because it is not asymmetric and not reflexive
D
not a partial ordering because it is not antisymmetric and reflexive
E
none of the above
       Engineering Mathematics       Relations
Question 41 Explanation: 
Relation < is:
1) not reflexive
2) irreflexive
3) not symmetric
4) Asymmetric
5) Antisymmetric
Question 42
Let A and B be sets with cardinalities m and n respectively. The number of one-one mappings (injections) from A to B, when m < n, is:
A
mn
B
nPm
C
mCn
D
nCm
       Engineering Mathematics       Functions
Question 42 Explanation: 
Let,

A one-one function 'f' assigns each element ai of A a distinct element, bj=f(ai) of Bi for a, there are n choices, for a2 there are n-1 choices, for am there are (n-(m-1)) choices.
i.e.,
Question 43
 
A
O(n)
B
O(n2)
C
O(n3)
D
O(3n2)
E
O(1.5n2)
F
B, C, D and E
       Algorithms       Time Complexity
Question 43 Explanation: 

⇒ In this 'n' is constant. So, n is added to n times itself which is O(n2).
Hence, (a) is wrong. And rest (B), (C), (D), (E) are correct.
There are 43 questions to complete.