GATE 2015 (Set01)
Question 1 
any paper  
much paper  
no paper  
a few paper 
Question 2 
Acquiescence – Submission  
Wheedle – Roundabout  
Flippancy – Lightness  
Profligate – Extravagant 
Wheedle = Use endearments (or) flattery to persuade some to do something
Flippancy = Lack of respect (or) seriousness
Profligate = Recklessly extravagant
Question 3 
0.20  
0.25  
0.30  
0.33 
Total outcomes = ^{4}C_{1} × ^{5}C_{1} = 20
Probability = Favourable outcomes/ Total outcomes = 4/20 = 0.20
Question 4 
She had a terrible time at the party.
 
She had a horrible time at the party.  
She had a terrific time at the party
 
She had a terrifying time at the party 
→ Horrible and terrible means fearful.
→ Terrific = Wonderful
→ So, C is the Answer.
Question 5 
Statement I alone is sufficient, but statement II alone is not sufficient.  
Statement II alone is sufficient, but statement I alone is not sufficient.
 
Both statements together are sufficient, but neither statement alone is sufficient.  
Statement I and II together are not sufficient.

Statement I:
Each step is 3/4 foot high.
No. of steps = 9/(3/4) = 12 steps
Statement I alone is sufficient.
Statement II:
Each step is 1 foot wide.
Statement II alone is not sufficient.
Question 6 
32  
33  
34  
35 
There are 40 males in Electrical Engineering.
⇒ Number of females in Electrical Engineering = 4/5 × 40 = 32
Total number of students in Electrical Engineering = 40+32 = 72
This constitutes 20% of the strength of college.
Number of students in Civil department = 30/100 × 72 = 108
Number of female students in Civil department = 4/(4+5) × 108 = 48
Number of students in Mechanical department = 10/20 × 72 = 36
Number of female students in Mechanical department = 4/(4+5) × 36 = 16
Required Difference = 48  16 = 32
Question 7 
took shelter in a thick jungle  
open indiscriminate fire
 
took to flight
 
unconditionally surrendered 
Question 8 
Only relation I is true  
Only relation II is true  
Relations II and III are true.  
Relations I and III are true. 
50% of students have a chance pass in atleast two
(1  m)pc + (1  p)mc + (1  c) mp + mpc = 0.5 (ii)
40% of students have a chance of passing exactly two
(1  m)pc + (1  p)mc + (1  c)mp = 0.4 (iii)
From equation (ii) and (iii) we can get
mpc = 0.1
⇒ m*p*c = 1/10
Statement (III) is correct.
→ Simplify eq (i), we get ⇒ p+c+m  (mp+mc+pc) + mpc = 0.75
⇒ p+c+m  (mp+mc+pc) = 0.65 (iv) → Simplify equation (iii), we get
⇒ pc + mc + mp  3mpc = 0.4
From (iv) and (v)
p + c +m  0.7 = 0.65
⇒ p + c + m = 1.35 = 27/20
Statement (I) is correct.
Question 9 
Statements I and II follow  
Statements I and III follow
 
Statements II and III follow  
All statements follow 
Question 10 
2.290  
2.970  
6.795  
8.795 
Total marks obtained = 21×2+15×3+11×1+23×2+31×5=299
Average marks = 299/44 = 6.795
Question 11 
Viable prefixes appear only at the bottom of the stack and not inside
 
Viable prefixes appear only at the top of the stack and not inside  
The stack contains only a set of viable prefixes  
The stack never contains viable prefixes

Question 12 
2 3 1 4  
3 4 2 1
 
3 1 4 2  
3 1 2 4 
Equivalence class partitioning ⇒ is a software testing technique that divides the input data of a software unit into partitions of equivalent data from which test cases can be derived, which is nothing but black box testing Hence B1.
Volume testing ⇒ volume testing refers to testing a software application with certain amount of data which is nothing but system testing C2.
Alpha testing ⇒ Alpha testing is simulated or actual operation testing by potential user/customers, which is nothing but performance testing. D4.
Question 13 
Either S1 or S2
 
Either S2 or S3  
Only S2 and S3  
All of S1, S2 and S3

Question 14 
T(n) = 2T (n/2) + cn
 
T(n) = T(n – 1) + T(1) + cn  
T(n) = 2T (n – 2) + cn
 
T(n) = T(n/2) + cn

Question 15 
I only  
III only  
III and IV only  
I and IV only 
This one is true, because L_{1} is context free which is nothing but recursive, recursive language is closed under complement hence true.
2 ⇒ (complement of L_{2}) is recursive
If L_{2} and both are recursive enumerable then is recursive
Hence option 2 is false
3 ⇒ is context free
Which is false because context free language does not closed under complement
4 ⇒ ∪L_{2} is recursive enumerable
⇒ recursive
Every recursive language is also recursive enumerable
L_{2} ⇒ recursive enumerable
∪ L_{2} ⇒ recursive enumerable
Because recursive enumerable language closed under union.
Question 16 
III only  
I and III only  
I and IV only  
II and IV only 
The sequence number of the subsequent segment depends on the number of 8byte characters in the current segment.
S2: TRUE
Depending on the value of α or Estimated RTT it may or may not be greater than 1.
S3: FALSE
It is the size of the receiver's buffer that's never changed. Receive Window is the part of the receiver's buffer that's changing all the time depending on the processing capability at the receiver's side and the network traffic.
S4: TRUE
The number of unacknowledged bytes that A sends cannot exceed the size of the receiver's window. But if it can't exceed the receiver's window, then it surely has no way to exceed the receiver's buffer as the window size is always less than or equal to the buffer size. On the other hand, for urgent messages, the sender CAN send it in even though the receiver's buffer is full.
Question 17 
3  
4  
5  
6 
If we execute P1 process after P2 process, then B = 4
If we did preemption between P1 & P2 processes, then B = 2 (Preemption have done from P1 to P2) or B = 3 (Preemption have done from P2 to P1). So, among 2 & 3 values, only one value will be saved in B. So, total no. of distinct values that B can possibly take after the execution is 3.
Question 18 
0, 1, 3, 7, 15, 14, 12, 8, 0  
0, 1, 3, 5, 7, 9, 11, 13, 15, 0  
0, 2, 4, 6, 8, 10, 12, 14, 0  
0, 8, 12, 14, 15, 7, 3, 1, 0 
The state sequence is 0,8,12,14,15,7,3,1,0.
Question 19 
Checksum  
Source address  
Time to Live (TTL)  
Length 
Option A (Checksum) needs to be updated by each visited Router since TTL Value is modified.
Option D (Length) also modified whenever there is a need of performing the fragmentation process.
Option B (Source Address) can’t be modified by an IP router. Only NAT can modify it.
Question 20 
the selection operation in relational algebra  
the selection operation in relational algebra, except that SELECT in SQL retains duplicates
 
the projection operation in relational algebra  
the projection operation in relational algebra, except that SELECT in SQL retains duplicates

Question 21 
5  
6  
7  
8 
l_{11} = 2 (1)
l_{11}u_{12} = 2
u_{12} = 2/2
u_{12} = 1  (2)
l_{21} = 4  (3)
l_{21}u_{12}+l_{22} = 9
l_{22} = 9  l_{21}u_{12} = 9  4 × 1 = 5
Question 22 
Piii, Qii, Riv, Si
 
Pi, Qii, Riv, Siii
 
Pii, Qiii, Riv, Si  
Pii, Qi, Riii, Siv 
Floydwarshall always changes it distance at each iteration which is nothing but dynamic programming.
Merge sort in merge sort first we always divide and merge to perform sorting hence divide and conquer.
Hamiltonian circuit used to reach the entire vertex once, if some vertex is repeating in its path it will backtrack.
Question 23 
5  
6  
7  
8 
But f_{2} will swap the value of 'b' and 'c' because f_{2} is call by reference. So finally the value of
a=4
b=6
c=5
So, answer will be
c  a  b
5  4  6 = 5
Question 24 
h(x)/g(x)  
1/x  
g(x)/h(x)
 
x/(1x)^{2}

Replace x by h(x) in (1), replacing x by g(x) in (2),
g(h(x))=1h(x)=1x/x1=1/x1
h(g(x))=g(x)/g(x)1=1x/x
⇒ g(h(x))/h(g(x))=x/(x1)(1x)=(x/x1)/1x=h(x)/g(x)
Question 25 
63 and 6, respectively  
64 and 5, respectively  
32 and 6, respectively
 
31 and 5, respectively 
2^{h+1}  1 = 2^{5+1}  1 = 63
Minimum number of nodes in a binary tree of height h is
h + 1 = 5 + 1 = 6
Question 26 
2036, 2036, 2036  
2012, 4, 2204  
2036, 10, 10  
2012, 4, 6 
⇒ x [4] [3] can represents that x is a 2dimensional array.
⇒ x+3 = (Address of x) + 3 * 4 * 3 [3×4×3 is inner dimention]
= 2000 + 36
= 2036
⇒ *(x+3) also returns the address i.e., 2036.
The '*' represents 1  D but x is starting at 2036.
⇒ *(x+3)+3 = *(Address of x + 2 * 3 * 4) + 3
= *(2000 + 24) +3
= *(2024) + 3 ['*' will change from 2D to 1D]
= 2024 + 3 * 4
= 2024 + 12
= 2036
Question 27 
1  
2  
3  
4 
Question 28 
3.2  
3.3  
3.4  
3.5 
Time taken to complete one cycle = (1 / 2.5 G) seconds
Since it is given that average number of cycles per instruction = 4, the time taken for completing one instruction = (4 / 2.5 G) = 1.6 ns
In the pipelined case we know in the ideal case CPI = 1, and the clock speed = 2 GHz.
Time taken for one instruction in the pipelined case = (1 / 2 G) = 0.5 ns
Speedup = 1.6/0.5 = 3.2
Question 29 
8  
9  
10  
11 
The given expression:
q+r/3+s−t∗5+u∗v/w
t1=r/3;
t2=t∗5;
t3=u∗v;
t4=t3/w;
t5=q+t1;
t6=t5+s;
t7=t6−t2;
t8=t7+t4;
So in total we need 8 temporary variables. If it was not mentioned as static single assignment then answer would have been 3 because we can reuse the same temporary variable several times.
Question 30 
pr = 0  
pr = 1  
0 < pr ≤ 1/5  
1/5 < pr < 1 
Let A be the event that an element (x,y,z)∈ L^{3} satisfies x ∨(y∧z) = (x∨y) ∧ (x∨z) Since q∨(r∧s) = q∨p = q
and (q∨r)∧(q∨s)=t∧t=t q∨(r∧s)≠(q∨r)∧(q∨s)
Therefore, (x,y,z) = (q,r,s),(q,s,r),(r,q,s),(r,s,q),(s,r,q),(s,q,r)
i.e., 3! = 6 elements will not satisfy distributive law and all other (x,y,z) choices satisfy distributive law
n(A) = 1256 = 119
∴ required probability is 119/125
⇒ 1/5
Question 31 
10110  
10010  
01010  
01001 
Question 32 
{r = qx+y ∧ r  
{x = qy+r ∧ r  
{y = qx+r ∧ 0  
{q+1 
⇒ x = qy + r
Question 33 
An algorithm performs (log N)^{1/2} find operations, N insert operations, (log N)^{1/2} delete operations, and (log N)^{1/2} decreasekey operations on a set of data items with keys drawn from a linearly ordered set. For a delete operation, a pointer is provided to the record that must be deleted. For the decreasekey operation, a pointer is provided to the record that has its key decreased. Which one of the following data structures is the most suited for the algorithm to use, if the goal is to achieve the best total asymptotic complexity considering all the operations?
Unsorted array  
Minheap  
Sorted array  
Sorted doubly linked list 
Question 34 
40, 30, 20, 10, 15, 16, 17, 8, 4, 35  
40, 35, 20, 10, 30, 16, 17, 8, 4, 15  
40, 30, 20, 10, 35, 16, 17, 8, 4, 15  
40, 35, 20, 10, 15, 16, 17, 8, 4, 30 
Heapification:
Array representation of above maxheap is (BFS)
40,35,20,10,30,16,17,8,4,15
Question 35 
a=6,b=4  
a=4,b=6
 
a=3,b=5
 
a=5,b=3 
By properties,
⇒ 6=1+a and 7=a4b
⇒ a=5 ⇒ 7=54b
⇒ b=3
Question 36 
1  
0  
1  
2 
Then the difference between the d(u) and d(v) is not more than '1'.
In the option 'D' the difference is given as '2' it is not possible in the undirected graph.
Question 37 
Both commutative and associative  
Commutative but not associative  
Not commutative but associative  
Neither commutative nor associative 
Question 40 
Consider a disk pack with a seek time of 4 milliseconds and rotational speed of 10000 rotations per minute (RPM). It has 600 sectors per track and each sector can store 512 bytes of data. Consider a file stored in the disk. The file contains 2000 sectors. Assume that every sector access necessitates a seek, and the average rotational latency for accessing each sector is half of the time for one complete rotation. The total time (in milliseconds) needed to read the entire file is _________.
14020  
14021  
14022  
14023 
Seek time = 4ms
60s→ 10000 rotations
∴ Rotational latency =1/2×6ms=3ms
1track→600sectors
⇒6ms←600sectors (1 rotation means 600 sectors (or)1 track)
1 track→6ms/600=0.01ms
2000sector→2000(0.01)=20ms
∴total time needed to read the entire file is
=2000(4+3)+20
=8000+6000+20
=14020ms
Question 43 
n^{3}  
n(log n) ^{2}  
nlog n  
nlog (log n) 
int i, j, k, p, q = 0;
for (i=1; i
return q;
}
∴ nlog(logn)
Question 44 
a_{n2}+a_{n1}+2^{n2}
 
a_{n2}+2a_{n1}+2^{n2}  
2a_{n2}+a_{n1}+2^{n2}  
2a_{n2}+2a_{n1}+2^{n2} 
So, a_{1=0 For string of length 2, Similarly, a3 = 3 a4= 8 Only (A) will satisfy the above values.}
Question 45 
p, s, u  
r, s, u  
r, u  
q, v 
A variable is live at some point if it holds a value that may be needed in the future, of equivalently if its value may be read before the next time the variable is written to.
→ '1' can be assigned by the p and s and there is no intermediate use of them before that.
→ And p and s are not to be live in the both 2 & 3.
→ And q also assigned to u not live in 2 & 3.
→ And v is live at 3 not at 2.
→ u is live at 3 and also at 2, if we consider a path of length 0 from 28.
Finally r, u is the correct one.
Question 46 
13  
12  
15  
16 
∴ At the end of 12 milliseconds, 1st instance of T_{3} will complete its execution.
Question 48 
A positive edgetriggered D flipflop is connected to a positive edgetriggered JK flipflop as follows. The Q output of the D flipflop is connected to both the J and K inputs of the JK flipflop, while the Q output of the JK flipflop is connected to the input of the D flipflop. Initially, the output of the D flipflop is set to logic one and the output of the JK flipflop is cleared. Which one of the following is the bit sequence (including the initial state) generated at the Q output of the JK flipflop when the flipflops are connected to a freerunning common clock? Assume that J = K = 1 is the toggle mode and J = K = 0 is the stateholding mode of the JK flipflop. Both the flipflops have nonzero propagation delays.
0110110...  
0100100...  
011101110...  
011001100...

The characteristic equations are
Q_{DN}=D=Q_{JK}
The state table and state transition diagram are as follows:
Consider Q_{D}Q_{JK}=10 as initial state because in the options Q_{JK}=0 is the initial state of JK flipflop.
The state sequence is
0 → 1 → 1 → 0 → 1 → 1
∴ Option (a) is the answer.
Question 49 
24  
25  
26  
27 
V+R=E+2 (1) where V, E, R are respectively number of vertices, edges and faces (regions)
Given V=10 (2) and number of edges on each face is three
∴3R=2E⇒R=2/3E (3)
Substituting (2), (3) in (1), we get
10+2/3E=E+2⇒E/3=8⇒E=24
Question 50 
Consider S_{2}, S_{3} and S_{4}. Time is divided into fixedsize slots, and a node can begin its transmission only at the beginning of a slot. A collision is said to have occurred if more than one node transmit in the same slot. The probabilities of generation of a frame in a time slot by S_{1}, S_{2}, S_{3} and S_{4} are 0.1, 0.2, 0.3 and 0.4, respectively. The probability of sending a frame in the first slot without any collision by any of these four stations is _________.
0.4404  
0.463  
0.464  
0.465 
S_{2}→0.2
S_{3}→0.3
S_{4}→0.4
The probability of sending a frame without any collision by any of these stations is
Question 51 
Suppose the following disk request sequence (track numbers) for a disk with 100 tracks is given: 45, 20, 90, 10, 50, 60, 80, 25, 70. Assume that the initial position of the R/W head is on track 50. The additional distance that will be traversed by the R/W head when the Shortest Seek Time First (SSTF) algorithm is used compared to the SCAN (Elevator) algorithm (assuming that SCAN algorithm moves towards 100 when it starts execution) is _________ tracks.
10  
11  
12  
13 
∴ Total Head moments = 10+10+10+10+10+55+20+5+10 = 140
SSTF:
∴ Total Head moments = 5+15+10+10+10+65+5+10 = 130
∴ Additional distance that will be traversed by R/w head is = 140130 = 10
Question 52 
Suppose that the stopandwait protocol is used on a link with a bit rate of 64 kilobits per second and 20 milliseconds propagation delay. Assume that the transmission time for the acknowledgment and the processing time at nodes are negligible. Then the minimum frame size in bytes to achieve a link utilization of at least 50% is _________.
320  
321  
322  
323 
T_{p} = 20 ms
η ≥ 50%
For η≥50%⇒L≥BR
⇒ L=64×10^{3}×2×20×10^{3}
=2560bits
=320bytes
Question 53 
Consider a main memory with five page frames and the following sequence of page references: 3, 8, 2, 3, 9, 1, 6, 3, 8, 9, 3, 6, 2, 1, 3. Which one of the following is true with respect to page replacement policies FirstInFirst Out (FIFO) and Least Recently Used (LRU)?
Both incur the same number of page faults
 
FIFO incurs 2 more page faults than LRU
 
LRU incurs 2 more page faults than FIFO  
FIFO incurs 1 more page faults than LRU 
∴ Number of page faults = 9
∴ Number of page faults = 9
Question 54 
Both {f} and {g} are functionally complete  
Only {f} is functionally complete
 
Only {g} is functionally complete
 
Neither {f} nor {g} is functionally complete 
f(X,X,X)=X'XX'+XX'+X'X'
=0+0+X'
=X'
Similarly, f(Y,Y,Y)=Y' and f(X,Z,Z)=Z'
f(Y',Y',Z')=(Y')'Y'Z'+Y'(Y')'+(Y')'(Z')'
=YY'Z'+Y'Y+YZ
=0+0+YZ
=YZ
We have derived NOT and AND. So f(X,Y,Z) is functionally complete.
g(X,Y,Z)=X'YZ+X'YZ'+XY
g(X,X,X)=X'XX+X'XZ'+XX
=0+0+X
=X
Similarly, g(Y,Y,Y)=Y and g(Z,Z,Z)=Z
NOT is not derived. Hence, g is not functionally complete.