Gate1993
Question 1 
(0, 0,α )  
(α,0,0)  
(0,0,1)  
(0, ,0 α )  
Both B and D 
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 x_{1}, x_{2}, x_{3} are suitable?
Which means:
x_{1} times column 1 + x_{2} times column 2 + x_{3} times column 3 = zero vector
Since α is not equal to zero, so x_{3} must be necessarily zero to get zero vector.
Hence, only (B) and (D) satisfies.
So the question as has
(A  λI)X = 0
AX = 0
What x_{1}, x_{2}, x_{3} are suitable?
Which means:
x_{1} times column 1 + x_{2} times column 2 + x_{3} times column 3 = zero vector
Since α is not equal to zero, so x_{3} must be necessarily zero to get zero vector.
Hence, only (B) and (D) satisfies.
Question 2 
linear  
nonlinear  
homogeneous  
of degree two

Question 2 Explanation:
Note: Out of syllabus.
In this DE, degree is 1 then this represent linear equation.
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
1  
2  
3  
4 
Question 3 Explanation:
Note: Out of syllabus.
Question 4 
Which of the following is (are) valid FORTRAN 77 statement(s)?
DO 13 I = 1  
A = DIM ***7  
READ = 15.0  
GO TO 3 = 10 
Question 4 Explanation:
Note: Out of syllabus.
Question 6 
Which of the following improper integrals is (are) convergent?
Question 7 
The function f(x,y) = x^{2}y  3xy + 2y + x has
no local extremum  
one local minimum but no local maximum  
one local maximum but no local minimum  
one local minimum and one local maximum 
Question 7 Explanation:
Note: Out of syllabus.
Question 8 
1 
Question 8 Explanation:
Since the given expression is in 0/0 form, so we can apply LHospital rule.
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 RungaKutta method with step size h = 0.1 is ________
Out of syllabus. 
Question 14 
A^{4} = I 
Question 14 Explanation:
Let λ be eigen value, then characteristic equation will be
(1λ) (1λ) (iλ) (iλ)
= (λ^{2}1) (λ^{2}+1)
= λ^{4}1
Characteristic equation is λ^{4}1 = 0.
According to CayleyHamilton theorem, every matrix satisfies its own characteristic equation, so
A^{4} = I
(1λ) (1λ) (iλ) (iλ)
= (λ^{2}1) (λ^{2}+1)
= λ^{4}1
Characteristic equation is λ^{4}1 = 0.
According to CayleyHamilton theorem, every matrix satisfies its own characteristic equation, so
A^{4} = I
Question 18 
exclusive OR  
exclusive NOR  
NAND  
NOR  
None of the above 
Question 18 Explanation:
So finally, we can write
Question 20 
Shift Register  
Mod3 Counter  
Mod6 Counter  
Mod2 Counter  
Both A and C 
Question 20 Explanation:
Circuit behaves as shift register and mod6 counter. Note that this is the Johnson counter which is the application of shift register. And Johnson counter is mod2N 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
109  
216  
218  
219  
240 
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.
= 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) 6.625, (b) (45E)_{H} 
Question 22 Explanation:
(a) 1*2^{2} + 1*2^{1} + 0*2^{0} + 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}.
= 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 4bit 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.
M=256, N = 8 
Question 23 Explanation:
Input will consist of 8 bit (two 4bit numbers) = 2^{8} address.
Output will be of 8 bits.
So memory will be of 2^{8} × 8.
So, M = 256, N = 8.
Output will be of 8 bits.
So memory will be of 2^{8} × 8.
So, M = 256, N = 8.
Question 24 
P = 12.5, Q = 2.5×10^{6} 
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 × 10^{6} bps
So,
P = 12.5, Q = 2.5×10^{6}
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 × 10^{6} bps
So,
P = 12.5, Q = 2.5×10^{6}
Question 25 
90% 
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%
= 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 
1, because m is a local variable in P  
0, because m is the actual parameter that corresponds to the formal parameter in p
 
0, because both x and y are just reference to m, and y has the value 0  
1, because both x and y are just references to m which gets modified in procedure P  
none of the above 
Question 26 Explanation:
0, because global m is not modified, m is just passed to formal argument of P.
Question 27 
0, because n is the actual parameter corresponding to x in procedure Q.  
0, because n is the actual parameter to y in procedure Q.  
1, because n is the actual parameter corresponding to x in procedure Q.  
1, because n is the actual parameter corresponding to y in procedure Q.  
none of the above 
Question 27 Explanation:
0, because n is just passed to formal parameters of Q and no modification in global n.
Question 28 
PARAM, P, Q  
PARAM, P  
PARAM, Q  
P, Q  
none of the above 
Question 28 Explanation:
Since m is defined global it is visible inside all the procedures.
Question 29 
exchanges a and b  
doubles a and stores in b  
doubles b and stores in a  
leaves a and b unchanged  
none of the above 
Question 29 Explanation:
Exchanges a and b.
Let us consider a=5; b=2
a := a+b = 5+2 = 7
b := ab = 72 = 5
a := ab = 75 = 2
O/P: a=2; b=5
Let us consider a=5; b=2
a := a+b = 5+2 = 7
b := ab = 72 = 5
a := ab = 75 = 2
O/P: a=2; b=5
Question 30 
The program leads to compile time error  
The program leads to run time error  
The program outputs 5.2  
The program produces error relating to nil pointer dereferencing  
None of the above 
Question 30 Explanation:
Note: Out of syllabus.
Question 31 
A simple twopass assembler does the following in the first pass:
It allocates space for the literals.  
It computes the total length of the program
 
It builds the symbol table for the symbols and their values.  
It generates code for all the load and store register instructions.  
A, B and C 
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.
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:
text editor  
assembler  
linker  
loader
 
none of the above 
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
at a fixed address in main memory  
at a fixed location on the disk  
anywhere on the disk  
at a fixed location on the system disk  
anywhere on the system disk 
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?
7  
9  
13, 15  
13  
15 
Question 34 Explanation:
A requires 3, B4, C6;
→ 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.
→ 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?
7  
9  
13, 15  
13  
15 
Question 35 Explanation:
A requires 3, B4, C6;
→ 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.
→ 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 
4  
10  
11  
12  
None of the above 
Question 36 Explanation:
Algorithm: Round robin; TQ: 1
p is departure at 11.
p is departure at 11.
Question 37 
Consider a simple connected graph G with n vertices and nedges (n>2). Then, which of the following statements are true?
G has no cycles.  
The graph obtained by removing any edge from G is not connected.
 
G has at least one cycle.  
The graph obtained by removing any two edges from G is not connected.  
Both C and D. 
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
For example let us consider, n=3
Question 38 
The proposition p ∧(~p ∨ q) is:
a tautology  
logically equivalent to p ∧ q  
logically equivalent to p ∨ q  
a contradiction  
none of the above 
Question 38 Explanation:
p ∧(~p ∨ q)
(p ∧ ~p) ∨ (p ∧ q)
F ∨ (p ∧ q)
(p ∧ q)
(p ∧ ~p) ∨ (p ∧ q)
F ∨ (p ∧ q)
(p ∧ q)
Question 39 
Let S be an infinite set and S_{1} ..., S_{n} be sets such that S_{1} ∪ S_{2} ∪ ... ∪ S_{n} = S. Then,
at least one of the set S_{i} is a finite set  
not more than one of the set S_{i} can be finite  
at least one of the sets S_{i} is an infinite set  
None of the above 
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:
2^{2n}  
2^{n2}  
(2^{n})^{2}  
(2^{2})^{n} 
Question 40 Explanation:
Question 41 
The lessthan relation, <, on reals is
a partial ordering since it is asymmetric and reflexive  
a partial ordering since it is antisymmetric and reflexive  
not a partial ordering because it is not asymmetric and not reflexive  
not a partial ordering because it is not antisymmetric and reflexive
 
none of the above 
Question 41 Explanation:
Relation < is:
1) not reflexive
2) irreflexive
3) not symmetric
4) Asymmetric
5) Antisymmetric
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 oneone mappings (injections) from A to B, when m < n, is:
m^{n}  
^{n}P_{m}  
^{m}C_{n}  
^{n}C_{m} 
Question 42 Explanation:
Let,
A oneone function 'f' assigns each element a_{i} of A a distinct element, b_{j}=f(a_{i}) of B_{i} for a, there are n choices, for a_{2} there are n1 choices, for a_{m} there are (n(m1)) choices.
i.e.,
A oneone function 'f' assigns each element a_{i} of A a distinct element, b_{j}=f(a_{i}) of B_{i} for a, there are n choices, for a_{2} there are n1 choices, for a_{m} there are (n(m1)) choices.
i.e.,
Question 43 
O(n)  
O(n^{2})  
O(n^{3})  
O(3n^{2})  
O(1.5n^{2})  
B, C, D and E 
Question 43 Explanation:
⇒ In this 'n' is constant. So, n is added to n times itself which is O(n^{2}).
Hence, (a) is wrong. And rest (B), (C), (D), (E) are correct.
There are 43 questions to complete.