Gate 2011
Question 2 
9  
8  
512  
258 
The max Mod values is 2n.
So 2^{n} ≥ 258 ⇒ n = 9
Question 3 
Question 4 
On perthread basis, the OS maintains only CPU register state  
The OS does not maintain a separate stack for each thread  
On perthread basis, the OS does not maintain virtual memory state  
On per thread basis, the OS maintains only scheduling and accounting information

B) False, OS do maintain a separate stack for each thread.
C) True
D) False
Question 5 
K4 is planar while Q3 is not  
Both K4 and Q3 are planar  
Q3 is planar while K4 is not  
Neither K4 not Q3 is planar 
Both the above graphs are planar.
Question 6 
R = 0  
R < 0  
R ≥ 0  
R > 0 
So the answer will be R≥0.
Question 7 
Finite state automata  
Deterministic pushdown automata  
NonDeterministic pushdown automata  
Turing machine 
Question 8 
21 ns  
30 ns  
23 ns  
35 ns 
EA = p × page fault service time + (1 – p) × Memory access time
=1/10^{6}×10×106+(11/106)×20 ≅29.9 ns
Question 9 
Immediate Addressing  
Register Addressing  
Register Indirect Scaled Addressing  
Base Indexed Addressing 
Question 10 
char c[] = "GATE2011" ; char *p =c; printf ( "%s" , p + p[3]  p[1]) ; 
GATE2011  
E2011  
2011  
011 
p[3]  p[1] = 4, and p+4 will be pointing to the fifth position in the array 'c'. So printf starts printing from 2 and prints 2011.
Question 11 
Option A: It violates the property of complete binary tree.
Option C: 8 is greater than 5. The root value is always greater than his children. So, the above tree is violating the property of max heap
Option D: 10 is greater than 8. The root value is always greater than his children. So, the above tree is violating the property of max heap
Question 12 
The algorithm uses dynamic programming paradigm  
The algorithm has a linear complexity and uses branch and bound paradigm  
The algorithm has a nonlinear polynomial complexity and uses branch and bound paradigm  
The algorithm uses divide and conquer paradigm. 
2. Dynamic programming is a technique used to avoid computing multiple time the same subproblem in a recursive algorithm.
Note: It is neither backtracking nor branch and bound because we are not branching anywhere in the solution space. The algorithm is not divide and conquer as we are not dividing the problem and then merging the solution
Question 13 
P ∩ Q  
P – Q  
Σ* – P  
Σ* – Q 
since regular languages are closed under complementation
Question 14 
parsing of the program  
the code generation
 
the lexical analysis of the program  
dataflow analysis 
Question 15 
block entire HTTP traffic during 9:00PM and 5:00AM  
block all ICMP traffic  
stop incoming traffic from a specific IP address but allow outgoing traffic to the same IP address  
block TCP traffic from a specific user on a multiuser system during 9:00PM and 5:00AM 
(B) Possible because it is network layer protocol.
(C) Possible because SP address is present in Network layer.
(D) Not possible, because to block specific user, we need user id which is present in Application layer.
Question 16 
1/3  
1/4  
1/2  
2/3 
Question 17 
m1: Send an email from a mail client to a mail server m2: Download an email from mailbox server to a mail client m3: Checking email in a web browserWhich is the application level protocol used in each activity?
m1:HTTP m2: SMTP m3: POP  
m1:SMTP m2:FTP m3:HTTP  
m1: SMTP m2: POP m3: HTTP  
m1: POP m2: SMTP m3: IMAP 
Question 18 
4000  
5000  
4333  
4667 
Question 19 
(t_{1}) > (t_{2})  
(t_{1}) = (t_{2})  
(t_{1}) < (t_{2})  
Nothing can be said about the relation between t_{1} and t_{2} 
Question 20 
234.25  
932.50  
287.80  
122.40 
Question 21 
Deterministic finite automata (DFA) and Nondeterministic finite automata (NFA)  
Deterministic push down automata (DPDA) and Nondeterministic push down automata (NFDA)  
Deterministic singletape Turning machine and Nondeterministic single tape Turning machine  
Singletape Turning machine and multitape Turning machine 
Hence answer is (B)
Question 22 
Embed web objects from different sites into the same page  
Refresh the page automatically after a specified interval  
Automatically redirect to another page upon download  
Display the client time as part of the page 
Question 23 
Functional Requirements
 
NonFunctional Requirements
 
Goals of Implementation
 
Algorithms for Software Implementation 
Question 24 
. Interrupt from CPU temperature sensor (raises interrupt if CPU temperature is too high) . Interrupt from Mouse(raises interrupt if the mouse is moved or a button is pressed) . Interrupt from Keyboard(raises interrupt when a key is pressed or released) . Interrupt from Hard Disk(raises interrupt when a disk read is completed) Which one of these will be handled at the HIGHEST priority?
Interrupt from Hard Dist  
Interrupt from Mouse  
Interrupt from Keyboard  
Interrupt from CPU temperature sensor 
Therefore, we can say that priorities are,
CPU temperature sensor > Hard disk > Mouse > Keyboard.
Question 25 
1. Registration_Num: Unique registration number of each registered student 2. UID: Unique identity number, unique at the national level for each citizen 3. BankAccount_Num: Unique account number at the bank. A student can have multiple accounts or join accounts. This attribute stores the primary account number. 4. Name: Name of the student 5. Hostel_Room: Room number of the hostel Which one of the following option is INCORRECT?
BankAccount_Num is a candidate key  
Registration_Num can be a primary key
 
UID is a candidate key if all students are from the same country  
If S is a superkey such that S∩UID is NULL then S∪UID is also a superkey 
Question 26 
f_{3}, f_{2}, f_{4}, f_{1}  
f_{3}, f_{2}, f_{1}, f_{4}  
f_{2}, f_{3}, f_{1}, f_{4}  
f_{2}, f_{3}, f_{4}, f_{1} 
→ Divide functions into 2 categories
1. Polynomial functions
2. Exponential functions
Above 4 functions we have only one exponential function is f_{1} (n) = 2n . So, It’s value is higher than to rest of the functions.
Substitute log on both sides then we get an ascending order is f_{3}, f_{2}, f_{4}.
Question 27 
248000  
44000  
19000  
25000 
p = 10
q = 100
r = 20
s = 5
t = 80
Total no. of matrix multiplication for (n+1) matrices,
^{2n}C_{n}/(n+1)
Here,
n+1 = 4
n = 3
So, total no. of matrix multiplication possible is,
^{6}C_{3}/4 = ∠6/∠3×∠3×4 = 5
So, total 5 cases are possible,
Case 1:
(((M_{1} × M_{2}) × M_{3}) × M_{4})
∴ Total no. of scalar multiplications needed is,
20000 + 4000 + 1000 = 25000
Case 2:
((M_{1} × M_{2}) × (M_{3} × M_{4}))
∴ Total no. of scalar multiplications needed is,
20000 + 8000 + 16000 = 44000
Case 3:
(M_{1} ×(M_{2} ×(M_{3} × M_{4})))
∴ Total scalar multiplications,
160000 + 80000 + 8000 = 248000
Case 4:
((M_{1} ×(M_{2} ×M _{3})) × M_{4})
∴ Total scalar multiplications,
10000 + 5000 + 4000 = 19000
Case 5:
(M_{1} × ((M_{2} × M_{3}) × M_{4}))
∴ Total scalar multiplications,
10000 + 40000 + 80000 = 130000
Hence, the minimum number of scalar multiplications is, 19000.
Question 28 
Ordered indexing will always outperform hashing for both queries  
Hashing will always outperform ordered indexing for both queries  
Hashing will outperform ordered indexing on Q1, but not on Q2  
Hashing will outperform ordered indexing on Q2, but not on Q1. 
For example, consider B^{+} tree, once you have searched a key in B^{+}; you can find range of values via the block pointers pointing to another block of values on the leaf node level.
Question 29 
1, 4, 3  
3, 7, 3  
7, 3, 2
 
1, 2, 3 
Question 30 
4.0
 
2.5  
1.1  
3.0 
Clock cycle time of pipelining
= max (5ns, 6ns, 11ns, 8ns) + 1ns
= 11 + 1
= 12ns
Time without pipelining = 5 + 6 + 11 + 8 = 30 ns
∴ Speedup = 30/12 = 2.5
Question 31 
k+1  
n+1  
2^{n+1}  
2^{k+1} 
So lets check of n = 2,
L = a_{2k}, k>0
Since k>0 than zero.
So L is the language accepting even no. of a's except 'ε'.
So DFA will be,
So, no. of states required is 2+1 = 3.
So for a^{nk}, (n+1) states will be required.
Question 32 
4864 bits  
6144 bits  
6656 bits  
5376 bits 
So we need 8 bits for indexing the 256 blocks in the cache. And since a block is 32 bytes we need 5 word bits to address each byte.
So out of remaining (32  8  5), 19 bits should be tag bits.
So tag entry size = 19 + 1 (valid bit) + 1 (modified bit) = 21 bits
∴ Total size of metadata = 21 × Number blocks = 21 × 256 = 5376 bits
Question 33 
0.50 s  
1.50 s  
1.25 s  
1.00 s 
= 10ms + (60/12000)+0
For 100 libraries = 100(10ms + (60/12000)+0) = 1.5seconds
Question 34 
(B) False, as it accepts string 'b', which is not accepted by original DFA.
(C) Same reason as (B).
(D) False, as it accepts string 'bba' which is not accepted by the given DFA.
Question 35 
/* This function computes the roots of a quadratic equation a.x^2 + b.x + c = . The function stores two real roots in *root1 and *root2 and returns the status of validity of roots. It handles four different kinds of cases. (i) When coefficient a is zero irrespective of discriminant (ii) When discreminant is positive (iii) When discriminant is zero (iv) When discriminant is negative. Only in case (ii) and (iii) the stored roots are valid. Otherwise 0 is stored in roots. The function returns 0 when the roots are valid and 1 otherwise. The function also ensures root1 >= root2 int get_QuadRoots( float a, float b, float c, float *root1, float *root2); */
A software test engineer is assigned the job of doing black box testing. He comes up with the following test cases, many of which are redundant.
Which one of the following option provide the set of nonredundant tests using equivalence class partitioning approach from input perspective for black box testing?T1, T2, T3, T6  
T1, T3, T4, T5  
T2, T4, T5, T6  
T2, T3, T4, T5 
T1 and T2 checking same condition a = 0 hence, any one of T1 and T2 is redundant.
T3, T4: in both case discriminant (D) = b2 – 4ac = 0. Hence any one of it is
T5 : D > 0
T6 : D < 0
Question 36 
Borrower Bank_Manager Loan_Amount Ramesh Sunderajan 10000.00 Suresh Ramgopal 5000.00 Mahesh Sunderajan 7000.00What is the output of the following SQL query?
SELECT Count(*) FROM ( ( SELECT Borrower, Bank_Manager FROM Loan_Records) AS S NATURAL JOIN ( SELECT Bank_Manager, Loan_Amount FROM Loan_Records) AS T );
3  
9  
5  
6 
Question 37 
⇒ 7 ↓ (3 ↑ (4 ↑ 3)) ↓ 2
⇒ 7 ↓ (3 ↑ (4 ↑ 3))) ↓ 2 as ↓ is left associative
Question 38 
Push Down Automate (PDA) can be used to recognize L1 and L2  
L1 is a regular language  
All the three languages are context free  
Turing machines can be used to recognize all the languages 
L2: context free language
L3: context sensitive language
Question 39 
Initialize the address register Initialize the count to 500 LOOP: Load a byte from device Store in memory at address given by address register Increment the address register Decrement the count If count != 0 go to LOOP
Assume that each statement in this program is equivalent to machine instruction which takes one clock cycle to execute if it is a nonload/store instruction. The loadstore instructions take two clock cycles to execute. The designer of the system also has an alternate approach of using DMA controller to implement the same transfer. The DMA controller requires 20 clock cycles for initialization and other overheads. Each DMA transfer cycle takes two clock cycles to transfer one byte of data from the device to the memory. What is the approximate speedup when the DMA controller based design is used in place of the interrupt driven program based inputoutput?
3.4  
4.4  
5.1  
6.7

using DMA = 20 + 500 × 2 = 1020
Required speed up = 3502/1020=3.4
Question 40 
0  
1  
n!  
Question 41 
P(x) being true means that x is a prime number  
P(x) being true means that x is a number other than 1
 
P(x) is always true irrespective of the value of x
 
P(x) being true means that x has exactly two factors other than 1 and x 
This is the definition of prime nos.
Question 43 
SELECT Y FROM T WHERE X=7;`
127
 
255  
129  
257 
Question 44 
Index position of mode of X in X is the same as the index position of mode of Y in Y.  
Index position of median of X in X is the same as the index position of median of Y in Y.
 
μ_{y} = aμ_{x} + b  
σ_{y} = aσ_{x} + b 
(σ_{y})^{2} is variance so,
y_{i} = a * x_{i} + b
(σ_{y})^{2} = a^{2 }(σ_{x})^{2}
⇒ σ_{y} = a σ_{x}
Hence option (D) is incorrect.
Question 45 
1/5  
4/25  
1/4  
2/5 
(2,1) (3,2) (4,3) (5,4).
So only 4 possibilities are there and sample space will be,
^{5}C_{1} × ^{4}C_{1} = 20
So probability = 4/20 = 1/5
Question 46 
The preemptive shortest job first scheduling algorithm is used. Scheduling is carried out only at arrival or completion of processes. What is the average waiting time for the three processes?
5.0 ms  
4.33 ms  
6.33 ms  
7.33 ms 
CT = Completion time
TAT = Turn Around Time
WT = Waiting Time
TAT = CT  AT
WT = TAT  BT
Gantt Chart using preemptive shortest job first scheduling algorithm,
Avg. WT = 4+0+11/3 = 5ns
Question 47 
2  
9  
5  
3 
Load R2, b ; R2 ← M[b]
Sub R1, R2 ; R1 ← R1 – R2
Load R2, c ; R2 ← M[c]
Load R3, d ; R3 ← M[d]
Add R2, R3 ; R2 ← R2 + R3
Load R3, e ; R3 ← M[e]
Sub R3, 3 ; R3 ← R3 – R2
Add R1, R3 ; R1 ← R1 + R3
Total 3 Registers are required minimum
Question 49 
unsigned int foo(unsigned int n, unsigned int r) { if (n > 0) return (n%r + foo (n/r, r )); else return 0; } 
345  
12  
5  
3 
∴ 5+4+3+0 = 12
Question 50 
3  
4  
5  
6 
So total no. of distinct output (states) are 4.
Question 51 
000  
001  
010  
011 
So, after 010 it moves to 011.
Question 52 
1/12(11n^{2}5n)  
n^{2} – n + 1  
6n – 11  
2n + 1 
Cost of MST,
= 3+4+6+8 = 21
Only option (B) satisfies it.
Question 53 
11
 
25  
31  
41 
Now MST of above graph is,
∴ The length of path from v_{5} to v_{6} in the MST is,
8+4+3+6+10 = 31
Question 54 
(3, 2, 0, 2, 5)  
(3, 2, 0, 2, 6)  
(7, 2, 0, 2, 5)  
(7, 2, 0, 2, 6) 
Question 55 
3  
9  
10  
∞ 
N2 has made entry ∞
N4 has the distance of 8 to N1
N3 has the distance of 2 to N4
So 2 + 8 = 10
Question 56 
P^{2} = Q^{3}R^{2}  
Q^{2} = PR  
Q^{2} = R^{3}P  
R = P^{2}Q^{2} 
∴ P = b^{k}, Q = b^{2k}, R = b^{3k}
Now, Q^{2} = b^{4k} = b^{3k} b^{k} = PR
Question 57 
to visit  
having to visit  
visiting  
for a visit 
Question 58 
hyperbolic  
restrained  
argumentative  
indifferent 
Question 59 
merge  
split  
collect  
separate 
Question 60 
Incomprehensible  
Indelible
 
Inextricable  
Infallible 
Question 61 
P  
Q  
R  
S 
Question 62 
5  
4  
7  
6 
Question 63 
4  
5  
6  
7 
2800 = 4n + e; n = normal
300 = 10n + 2; e = excess/pending
n = 100/3, e = 8000/3
5 days: 500 ×(5⋅100/3)+ 8000/3
Minimum possible = 6
Question 64 
7.58 litres  
7.84 litres  
7 litres  
7.29 litres 
Question 65 
how to write a letter of condolence  
what emotional stages are passed through in the healing process  
what the leading causes of death are  
how to give support to a grieving friend 