GATE 2015 (Set-01)
Question 1 |
Didn’t you buy _________ when you went shopping?
any paper | |
much paper | |
no paper | |
a few paper |
Question 2 |
Which of the following combinations is incorrect?
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 |
Given set A = {2, 3, 4, 5} and Set B = {11, 12, 13, 14, 15}, two numbers are randomly selected, one from each set. What is probability that the sum of the two numbers equals 16?
0.20 | |
0.25 | |
0.30 | |
0.33 |
Total outcomes = 4C1 × 5C1 = 20
Probability = Favourable outcomes/ Total outcomes = 4/20 = 0.20
Question 4 |
Which of the following options is the closest in meaning to the sentence below?
She enjoyed herself immensely at the party.
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 |
Based on the given statements, select the most appropriate option to solve the given question.
If two floors in a certain building are 9 feet apart, how many steps are there in a set of stairs that extends from the first floor to the second floor of the building?
Statements:-
(I) Each step is 3/4 foot high.
(II) Each step is 1 foot wide.
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 |
The pie chart below has the breakup of the number of students from different departments in an engineering college for the year 2012. The proportion of male to female students in each department is 5:4. There are 40 males in Electrical Engineering. What is the difference between numbers of female students in the Civil department and the female students in the Mechanical department?

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 |
Select the alternative meaning of the underlined part of the sentence.
The chain snatchers took to their heels when the police party arrived.
took shelter in a thick jungle | |
open indiscriminate fire
| |
took to flight
| |
unconditionally surrendered |
Question 8 |
The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c respectively. Of these subjects, the student has 75% chance of passing in at least one, a 50% chance of passing in at least two and a 40% chance of passing in exactly two. Following relations are drawn in m, p, c:
-
(I) p + m + c = 27/20
(II) p + m + c = 13/20
(III) (p) × (m) × (c) = 1/10
Only relation I is true | |
Only relation II is true | |
Relations II and III are true. | |
Relations I and III are true. |
= 1 - (1 - m) (1 - p) (1 - c) = 0.75 ----(i)
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 |
The given statement is followed by some courses of action. Assuming the statement to be true, decide the correct option.
Statement:
There has been a significant drop in the water level in the lakes supplying water to the city.
Course of action:
-
(I) The water supply authority should impose a partial cut in supply to tackle the situation.
(II) The government should appeal to all the residents through mass media for minimal use of water.
(III)The government should ban the water supply in lower areas.
Statements I and II follow | |
Statements I and III follow
| |
Statements II and III follow | |
All statements follow |
Banning of water in lower areas is not the solution of the problem.
Question 10 |
The number of students in a class who have answered correctly, wrongly, or not attempted each question in an exam, are listed in the table below. The marks for each question are also listed. There is no negative or partial marking.

What is the average of the marks obtained by the class in the examination?
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 |
Which one of the following is True at any valid state in shift-reduce parsing?
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
|
A viable prefixes is prefix of the handle and so can never extend to the right of handle, i.e., top of stack.
So set of viable prefixes is in stack.
Question 12 |
Match the following:
(P) Condition coverage (i) Black-box testing (Q) Equivalence class partitioning (ii) System testing (R) Volume testing (iii) White-box testing (S) Alpha testing (iv) Performance testing
P-ii, Q-iii, R-i, S-iv | |
P-iii, Q-iv, R-ii, S-i | |
P-iii, Q-ii, R-iv, S-ii | |
P-iii, Q-i, R-ii, S-iv |
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 B-1.
Volume testing ⇒ volume testing refers to testing a software application with certain amount of data which is nothing but system testing C-2.
Alpha testing ⇒ Alpha testing is simulated or actual operation testing by potential user/customers, which is nothing but performance testing. D-4.
Question 13 |
For computers based on three-address instruction formats, each address field can be used to specify which of the following:
-
(S1) A memory operand
(S2) A processor register
(S3) An implied accumulator register
Either S1 or S2
| |
Either S2 or S3 | |
Only S2 and S3 | |
All of S1, S2 and S3
|
So as the question asks what can be specified using the address fields, implied accumulator register can't be represented in address field.
So, S3 is wrong.
Hence Option A is the correct answer.
Question 14 |
Which one of the following is the recurrence equation for the worst case time complexity of the Quicksort algorithm for sorting n(≥ 2) numbers? In the recurrence equations given in the options below, c is a constant.
T(n) = 2T(n/2) + cn
| |
T(n) = T(n – 1) + T(1) + cn | |
T(n) = 2T(n – 1) + cn
| |
T(n) = T(n/2) + cn
|
Hence recurrence relation T(n) = T(n - 1) + T(1) + cn.
Question 15 |
For any two languages L1 and L2 such that L1 is context-free and L2 is recursively enumerable but not recursive, which of the following is/are necessarily true?

I only | |
III only | |
III and IV only | |
I and IV only |

This one is true, because L1 is context free which is nothing but recursive, recursive language is closed under complement hence true.
2 ⇒

If L2 and


Hence option 2 is false
3 ⇒

Which is false because context free language does not closed under complement
4 ⇒


Every recursive language is also recursive enumerable
L2 ⇒ 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 8-byte 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 |

l11 = 2 -----(1)
l11u12 = 2
u12 = 2/2
u12 = 1 ----- (2)
l21 = 4 ----- (3)
l21u12+l22 = 9
l22 = 9 - l21u12 = 9 - 4 × 1 = 5
Question 22 |
P-iii, Q-ii, R-iv, S-i
| |
P-i, Q-ii, R-iv, S-iii
| |
P-ii, Q-iii, R-iv, S-i | |
P-ii, Q-i, R-iii, S-iv |
Floyd-warshall 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 f2 will swap the value of 'b' and 'c' because f2 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/(1-x)2
|
Replace x by h(x) in (1), replacing x by g(x) in (2),
g(h(x))=1-h(x)=1-x/x-1=-1/x-1
h(g(x))=g(x)/g(x)-1=1-x/-x
⇒ g(h(x))/h(g(x))=x/(x-1)(1-x)=(x/x-1)/1-x=h(x)/g(x)
Question 25 |
63 and 6, respectively | |
64 and 5, respectively | |
32 and 6, respectively
| |
31 and 5, respectively |
2h+1 - 1 = 25+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 2-dimensional 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 re-use 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)∈ L3 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) = 125-6 = 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 decrease-key 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 decrease-key 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 | |
Min-heap | |
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 max-heap 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=a-4b
⇒ a=5 ⇒ -7=5-4b
⇒ 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 39 |
69 | |
70 | |
71 | |
72 |

⇒ Total sum = 10 + 9 + 2 + 15 + 7 + 16 + 4 + 6 = 69
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 |
n3 | |
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 |
an-2+an-1+2n-2
| |
an-2+2an-1+2n-2 | |
2an-2+an-1+2n-2 | |
2an-2+2an-1+2n-2 |
So, a1 = 0
For string of length 2,
a2 = 1
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 2-8.
Finally r, u is the correct one.
Question 46 |
13 | |
12 | |
15 | |
16 |

∴ At the end of 12 milliseconds, 1st instance of T3 will complete its execution.
Question 48 |
A positive edge-triggered D flip-flop is connected to a positive edge-triggered JK flipflop as follows. The Q output of the D flip-flop is connected to both the J and K inputs of the JK flip-flop, while the Q output of the JK flip-flop is connected to the input of the D flip-flop. Initially, the output of the D flip-flop is set to logic one and the output of the JK flip-flop is cleared. Which one of the following is the bit sequence (including the initial state) generated at the Q output of the JK flip-flop when the flip-flops are connected to a free-running common clock? Assume that J = K = 1 is the toggle mode and J = K = 0 is the state-holding mode of the JK flip-flop. Both the flip-flops have non-zero propagation delays.
0110110... | |
0100100... | |
011101110... | |
011001100...
|

The characteristic equations are
QDN=D=QJK

The state table and state transition diagram are as follows:

Consider QDQJK=10 as initial state because in the options QJK=0 is the initial state of JK flip-flop.
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
∴3|R|=2|E|⇒|R|=2/3|E| ------(3)
Substituting (2), (3) in (1), we get
10+2/3|E|=|E|+2⇒|E|/3=8⇒|E|=24
Question 50 |
Consider S2, S3 and S4. Time is divided into fixed-size 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 S1, S2, S3 and S4 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 |
S2→0.2
S3→0.3
S4→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 = 140-130 = 10
Question 52 |
Suppose that the stop-and-wait 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 |
Tp = 20 ms
η ≥ 50%
For η≥50%⇒L≥BR
⇒ L=64×103×2×20×10-3
=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 First-In-First 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.
Question 55 |
0.99 | |
1.00 | |
2.00 | |
3.00 |

=2-1/1(2)+3-2/2(3)+4-3/3(4)+…+100-99/99(100)
=1/1-1/2+1/2-1/3+1/3…+1/98-1/99+1/99-1/100
=1-1/100
=99/100
=0.99