Gate-2003

Question 1
 
A
Xy
B
ex
C
ln(1+x)
D
Xx
Question 1 Explanation: 
P = P * (x/i)
S = S+P
Iteration 1:
P=1; i=1; S=1
P=x
S=1+x
Iteration 2:
P=x; S=1+x; i=2
P = x * x/2 = x2/2
Iteration 3:
P=x2/2; S=1+x+x2/2; i=3
P = (x2/2)(x/3) = x3/6
S = 1 + x + x2/2 + x3/6
Continue upto n Then f( ) returns:
S = 1 + x/1 + x2/2⋅1 + x3/3⋅2 + ...
= 1 + x/1! + x2/2! + x3/3! + ... + xn/n!
= ex
Question 2
 
A
I, II, and IV only
B
II, III, and IV only
C
II and IV only
D
IV only
Question 2 Explanation: 
i) A[2] can be consider as a pointer and this will not give any compile-time error.
ii) A[2][3] This results an integer, no error will come.
iii) B[1] is a base address of an array. This will not be changed it will result a compile time error.
iv) B[2][3] This also results an integer. No error will come.
Question 3
 
A
B
C
D
Question 3 Explanation: 
P(A)=1, P(B)=1/2
P(A/B) = P(A∩B)/P(B)
= P(A)⋅P(B)/P(B) (consider P(A), P(B) are two independent events)
= 1
P(B/A) = P(B∩A)/P(A)
= P(B)⋅P(A)/P(A)
= 1/2
Question 4
Let A be a sequence of 8 distinct integers sorted in ascending order. How many distinct pairs of sequences, B and C are there such that (i) each is sorted in ascending order, (ii) B has 5 and C has 3 elements, and (iii) the result of merging B and C gives A?  
A
2
B
30
C
56
D
256
Question 4 Explanation: 
A can have sequence of 8 distinct integers which are sorted in ascending order.
→ If we are pick 3 elements from 8 sequence integers then remaining 5 elements are already in ascending order. After merging these elements then it gives A.
→ No. of possibilities of choosing 8 elements from total of 8 = 8C3
= 8!/3!5!
= 8 * 7
= 56
Question 5
n couples are invited to a party with the condition that every husband should be accompanied by his wife. However, a wife need not be accompanied by her husband. The number of different gatherings possible at the party is
A
B
C
D
Question 5 Explanation: 
The possibilities to attend party is
i) Both husband and wife comes
ii) Only wife comes
iii) Both are not come
The no. of different gatherings possible at party is
= 3 * 3 * 3 * 3 * ... n times
= 3n
Question 6
 
A
n – k + 1
B
n – k
C
n – k – 1
D
n – k – 2
Question 6 Explanation: 
A binary search tree consists of n distinct elements. Let consider on left subtree, it consists of (k-1) elements. Then right subtree consists of (n-k) elements. From this we c an write recursive function as T(k-1)*(n-k) i.e.,
Question 7
Consider the set Σ* of all strings over the alphabet Σ = {0, 1}. Σ* with the concatenation operator for strings  
A
does not form a group
B
forms a non-commutative group
C
does not have a right identity element
D
forms a group if the empty string is removed from Σ*
Question 7 Explanation: 
In the concatenation '∊' is the identity element. And given one is not a group because no element has inverse element.
→ To perform concatenation with the given set can result a Monoid and it follows the property of closure, associativity and consists of identity element.
Question 8
Let G be an arbitrary graph with n nodes and k components. If a vertex is removed from G, the number of components in the resultant graph must necessarily lie between  
A
k and n
B
k – 1 and k + 1
C
k – 1 and n – 1
D
k + 1 and n – k
Question 8 Explanation: 
While a vertex is removed from a graph then that can be itself be forms a new component. The minimum number of components is k-1.
If a vertex is removed then it results that all the components are also be disconnected. So removal can create (n-1) components.
Question 9
Assuming all numbers are in 2’s complement representation, which of the following numbers is divisible by 11111011?  
A
11100111
B
11100100
C
11010111
D
11011011
Question 9 Explanation: 
Given: Binary numbers = 11111011
MSB bit is '1' then all numbers are negative
1's complement = 00000100
2's complement = 00000100 + 00000001 = 00000101 = -5
(A) 11100111 - (-25)10
(B) 11100100 - (-28)10
(C) 11010111 - (-41)10
(D) 11011011 - (-37)10
Answer: Option A (-25 is divisible by -5)
Question 10
 
A
I and II only
B
II and III only
C
III only
D
All the three
Question 10 Explanation: 
I is belongs to the Data hazard.
II is belongs to the Control hazard.
III is belongs to the Structural hazard.
→ Hazards are the problems with the instruction pipeline in CPU micro architectures.
Question 11
Consider an array multiplier for multiplying two n bit numbers. If each gate in the circuit has a unit delay, the total delay of the multiplier is  
A
Θ (1)
B
Θ (log n)
C
Θ (n)
D
Θ (n2)
Question 11 Explanation: 
Each bit in Multiplier is ANDed with a bit in Multiplicand which produce n n-bit numbers. The multiplication takes n units of time. The n n-bit numbers are added by using (n-1) n-bit adders. The time taken by (n-1) n-bit adders is k*(n-1) units.
The total time is n+kn-k = Θ(n)
Question 12
Ram and Shyam have been asked to show that a certain problem Π is NP-complete. Ram shows a polynomial time reduction from the 3-SAT problem to Π, and Shyam shows a polynomial time reduction from Π to 3-SAT. Which of the following can be inferred from these reductions?
A
Π is NP-hard but not NP-complete
B
Π is in NP, but is not NP-complete
C
Π is NP-complete
D
Π is neither NP-hard, nor in NP
Question 12 Explanation: 
Note: As per Present syllabus, it is not required.
Question 13
   
A
L is recursive
B
L is recursively enumerable but not recursive
C
L is not recursively enumerable
D
Whether L is recursive or not will be known after we find out if P = NP
Question 13 Explanation: 
Here, we have two possibilities, whether
P = NP (or) P != NP
→ If P=NP then L=(0+1)* which is recular, then it is recursive.
→ If P!=NP then L becomes ɸ which is also regular, then it is recursive.
So, finally L is recursive.
Question 14
The regular expression 0*(10*)* denotes the same set as  
A
(1*0)*1*
B
0+(0+10)*
C
(0+1)*10(0+1)*
D
None of the above
Question 14 Explanation: 
Both (A) and the given expression generates all strings over Σ.
Option (B) and (C) doesn't generate 11.
Question 15
If the strings of a language L can be effectively enumerated in lexicographic (i.e., alphabetic) order, which of the following statements is true?  
A
L is necessarily finite
B
L is regular but not necessarily finite
C
L is context free but not necessarily regular
D
L is recursive but not necessarily context free
Question 15 Explanation: 
The given language L is to be recursively enumerable. The TM which accepts the language which is in lexicographic order. If the language is not in lexicograhic order which is not accepted by TM.
The give 'L' is recursive but not necessarily context free.
Question 16
Which of the following suffices to convert an arbitrary CFG to an LL(1) grammar?  
A
Removing left recursion alone
B
Factoring the grammar alone
C
Removing left recursion and factoring the grammar
D
None of the above
Question 16 Explanation: 
Left recursion removing (or) factoring the given grammar are not sufficient to convert an arbitrary CFG to an LL(1) grammar.
To convert an arbitrary CFG to an LL(1) grammar we need to remove the left recursion and as well as left factoring without that we cannot convert.
Question 17
Assume that the SLR parser for a grammar G has n1 states and the LALR parser for G has n2 states. The relationship between n1 and n2 is  
A
n1 is necessarily less than n2
B
n1 is necessarily equal to n2
C
n1 is necessarily greater than n2
D
None of the above
Question 17 Explanation: 
No. of states in SLR and LALR are equal and no. of states in SLR and LALR are less than or equal to LR(1).
Question 18
In a bottom-up evaluation of a syntax directed definition, inherited attributes can  
A
always be evaluated
B
be evaluated only if the definition is L-attributed
C
be evaluated only if the definition has synthesized attributes
D
never be evaluated
Question 18 Explanation: 
L-Attributed grammar can able to inherits either inherited attributes (or) synthesized attributes.
L-Attributed definitions are a class of syntax directed definitions whose attributes can be evaluated by a single traversal of the parse-tree.
Question 19
Suppose the numbers 7, 5, 1, 8, 3, 6, 0, 9, 4, 2 are inserted in that order into an initially empty binary search tree. The binary search tree uses the usual ordering on natural numbers. What is the in-order traversal sequence of the resultant tree?
A
7 5 1 0 3 2 4 6 8 9
B
0 2 4 3 1 6 5 9 8 7
C
0 1 2 3 4 5 6 7 8 9
D
9 8 6 4 2 3 0 1 5 7
Question 19 Explanation: 

Inorder: 0 1 2 3 4 5 6 7 8 9
Question 20
 
A
I and II
B
I and III
C
II and III
D
I, II, and III
Question 20 Explanation: 

Which is true by considering leading ordered term present in polynomial expression.

2n×2n can't be written as Θ(2n)
So, this is False.
Question 21
 
A
I, II and IV only
B
I and IV only
C
II, III and IV only
D
I, III and IV only
Question 21 Explanation: 
I) a → b → e → g → h → f (✔️)
II) a → b → f → e (✖️)
III) a → b → f → h → g → e (✔️)
IV) a → f → g → h → b → e (✔️)
Question 22
The usual Θ(n2) implementation of Insertion Sort to sort an array uses linear search to identify the position where an element is to be inserted into the already sorted part of the array. If, instead, we use binary search to identify the position, the worst case running time will  
A
remain Θ(n2)
B
become Θ(n (log n)2)
C
become Θ(n log n)
D
become Θ(n)
Question 22 Explanation: 
While using Insertion sort to sort array by using linear search then time complexity = Θ(n2)
Instaed, linear search use binary search then (log n) will be the worst case time complexity of binary search and performing n swaps to place an element in right position for the corresponding n elements
i.e., n×(logn+n)
Θ((n×logn)+n2)
Θ(n2)
Remains same.
Question 23
In a heap with n elements with the smallest element at the root, the 7th smallest element can be found in time  
A
Θ(n log n)
B
Θ(n)
C
Θ(log n)
D
Θ(1)
Question 23 Explanation: 
The 7th smallest elements can be present in any of 7 levels. Then total possible elements can be present is seven levels is
1 + 2 + 4 + 6 + 8 + 16 + 32
Which is constant then we can find the 7th smallest element in Θ(1) time.
Question 24
Which of the following statements is FALSE?  
A
In statically typed languages, each variable in a program has a fixed type
B
In un-typed languages, values do not have any types
C
In dynamically typed languages, variables have no types
D
In all statically typed languages, each variable in a program is associated with values of only a single type during the execution of the program
Question 24 Explanation: 
Dynamic typed languages are those languages in which variable must necessarily be defined before they are used. Then dynamic typed languages have types.
Question 25
Using a larger block size in a fixed block size file system leads to
A
better disk throughput but poorer disk space utilization
B
better disk throughput and better disk space utilization
C
poorer disk throughput but better disk space utilization
D
poorer disk throughput and poorer disk space utilization
Question 25 Explanation: 
While using a larger block size means that contains less number of blocks then that results better throughput. This can be implemented in a fixed block size then the space utilization is not upto the mark. So the statement results better disk throughput but poorer disk space utilization.
Question 26
In a system with 32 bit virtual addresses and 1KB page size, use of one-level page tables for virtual to physical address translation is not practical because of  
A
the large amount of internal fragmentation
B
the large amount of external fragmentation
C
the large memory overhead in maintaining page tables
D
the large computation overhead in the translation process
Question 26 Explanation: 
Page size = 1KB
Virtual address = 32 bit = 232
No. of page level entries = 232 / 210
= 222
= 4M (Too large size)
Question 27
Which of the following assertions is FALSE about the Internet Protocol (IP)?  
A
It is possible for a computer to have multiple IP addresses
B
IP packets from the same source to the same destination can take different routes in the network
C
IP ensures that a packet is discarded if it is unable to reach its destination within a given number of hops
D
The packet source cannot set the route of an outgoing packets; the route is determined only by the routing tables in the routers on the way
Question 27 Explanation: 
Because in strict source routing or loose source routing path is set by the source not by router and main task of router is to check outgoing path with the help of forwarding table inside it.
Question 28
Which of the following functionalities must be implemented by a transport protocol over and above the network protocol?  
A
Recovery from packet losses
B
Detection of duplicate packets
C
Packet delivery in the correct order
D
End to end connectivity
Question 28 Explanation: 
End to end connectivity is the required functionality provided by Transport protocol.
Question 29
Which of the following scenarios may lead to an irrecoverable error in a database system?
A
A transaction writes a data item after it is read by an uncommitted transaction
B
A transaction reads a data item after it is read by an uncommitted transaction
C
A transaction reads a data item after it is written by a committed transaction
D
A transaction reads a data item after it is written by an uncommitted transaction
Question 29 Explanation: 
Irrecoverable error occurs when a transaction reads a data item after it is written by uncommitted transaction.
Question 30
 
A
B
C
D
Question 30 Explanation: 
If we want to get distinct elements then we need to perform cross product in between the relations r1, r2, .... rm.
Question 31
Let (S, ≤) be a partial order with two minimal elements a and b, and a maximum element c. Let P: S → {True, False} be a predicate defined on S. Suppose that P(a) = True, P(b) = False and P(x) ⇒ P(y) for all x, y ∈ S satisfying x≤y, where stands for logical implication. Which of the following statements CANNOT be true?    
A
P(x) = True for all x ∈ S such that x ≠ b
B
P(x) = False for all x ∈ S such that x ≠ a and x ≠ c
C
P(x) = False for all x ∈ S such that b ≤ x and x ≠ c
D
P(x) = False for all x ∈ S such that a ≤ x and b ≤ x
Question 31 Explanation: 
c is the maximum element.
a or b the minimal element in set.
P(a) = True for all x ∈ S such that a ≤ x and b ≤ x.
Option D is False.
Question 32
Which of the following is a valid first order formula? (Here α and β are first order formulae with x as their only free variable)  
A
((∀x)[α] ⇒ (∀x)[β]) ⇒ (∀x)[α⇒β]
B
(∀x)[α] ⇒ (∃x)[α ∧ β]
C
((∀x)[α ∨ β] ⇒ (∃x)[α] ⇒ (∀x)[α]
D
(∀x)[α ⇒ β] ⇒ ((∀x)[α] ⇒ (∀x)[β])
Question 32 Explanation: 
Option D is valid.
Here, α, β are holding values of x. Then and RHS saying that α holding the value of x and β is holding value of x.
Then LHS ⇒ RHS.
Question 33
 
A
I1 satisfies α, I2 does not
B
I2 satisfies α, I1 does not
C
Neither I2 nor I1 satisfies α
D
Both I1 and I2 satisfy α
Question 33 Explanation: 
Given that:
(∀x)[Px ⇔ (∀y)[Qxy ⇔ ¬Qyy]] ⇒(∀x)[¬Px]
Qyy is always true, because y divide y, then ¬Qyy is false.
∀x[(P(x) ⇔ ∀y [Qxy ⇔ False]]
∀y [Qxy ⇔ False] can be written as ∀y[¬axy]
⇒(∀x)[P(x) ⇔ ∀y[¬Qxy]]
Here, ¬Qxy says that y doesnot divides x, which is not always be true.
For example, if x=y then it is false then ∀y[¬Qxy] is not true for all values of y.
⇒(∀x)[P(x) ⇔ False]
⇒(∀x)[¬P(x) = RHS]
LHS = RHS
⇒ Irrespective of x, whether x is prime of composite number I1 and I2 satisfies α.
Question 34
m identical balls are to be placed in n distinct bags. You are given that m ≥ kn, where k is a natural number ≥1. In how many ways can the balls be placed in the bags if each bag must contain at least k balls?
A
B
C
D
Question 34 Explanation: 
Since we want at least k balls in each bag, so first we put kn balls into bags, k balls in each bag. Now we are left with m - kn balls, and we have to put them into n bags such that each bag may receive 0 or more balls. So applying theorem 2 of stars and bars with m - nk stars and n bars, we get number of ways to be

. So option (B) is correct
Question 35
 
A
B
C
D
Question 35 Explanation: 

Considering floor value for square root of numbers.
Successive root number of numbers are in the series 3, 5, 7, ... like 3 numbers from 1... 4, 5 numbers 5-9 and so on.
Question 36
How many perfect matching are there in a complete graph of 6 vertices?  
A
15
B
24
C
30
D
60
Question 36 Explanation: 
We have formula to find no. of perfect matchings in complete graphs of 2n vertices,
(2n)!/n!×2n
Given, 2n = 6 ⇒ n = 3
So, finally, 6!/3!×23 = 15
Question 37
 
A
g(h(D)) ⊆ D
B
g(h(D)) ⊇ D
C
g(h(D)) ∩ D = ɸ
D
g(h(D)) ∩ (B—D) ≠ ɸ
Question 37 Explanation: 
f: A→B ba an injective (one-to-one)
→ g: 2A→2B be also one to one function and g(C) = f(x)|x∈C}, for all subsets C of A.
The range of this function is n(2A).
→ h: 2B→2A it is not a one to one function and given h(D) = {x|x∈A, f(x)∈D}, for all subsets D of B.
The range of this function is also n(2A).
→ The function g(h(D)) also have the range n(2A) that implies n(A)≤n(B), i.e., n(2A) is less than n(2B).
Then this result is g(h(D))⊆D.
Question 38
 
A
0
B
1
C
2
D
3
Question 38 Explanation: 
Total possible values = 32 = 9 i.e., (a,a), (b,b), (c,c), (a,b), (a,c), (b,a), (b,c), (c,a), (c,b)
In those (x, y) = (b,c) & (c,b) are the possible solution for the corresponding equations.
(x, y) = (b,c) ⇒ (a*b)+(a*c) ⇒ (b*b)+(c*c)
⇒ (b) + (c) ⇒ c + b
⇒ c (✔️) ⇒ c (✔️)
(x,y) = (c,b) ⇒ (a*c)+(a*b) ⇒ (b*c)+(c*b)
⇒ c+b ⇒ a+c
⇒ c (✔️) ⇒ c (✔️)
Question 39
 
A
273757
B
283858
C
293959
D
210510710
Question 39 Explanation: 
The answers can have only three possibilities of sequences i.e., "a", "a", "a", and there is no multiplies of 7 and 9. So eliminates option A & C.
And f(S) = 23 = 8
So answer is 283858.
Question 40
 
A
3
B
4
C
5
D
6
Question 40 Explanation: 
The minimum degree of G = minv∈V {degree(v)}
|E| ≤ 3|v| - 6
Based on handshaking lemma, the minimum degree is (min×|v|)/2
⇒ (min×|v|)/2 ≤ 3|v| - 6
Checking the options lets take min=6
(6×|v|)/2 ≤ 3|v| - 6
0 ≤ -6 (Not satisfied)
And which is inconsistent.
Question 41

A
0
B
1
C
2
D
infinitely many
Question 41 Explanation: 

This is in the form AX = B

⇒ R(AB) < n [If we want infinitely many solution]
then -1+5α = 0
5α = 1
α = 1/5 There is only one value of α. System can have infinitely many solutions.
Question 42
 
A
1.3, 0.6, and 0.6 respectively
B
0.6, 0.6, and 1.3 respectively
C
1.3, 1.3, and 0.6 respectively
D
1.3, 0.6, and 1.3 respectively
Question 42 Explanation: 
Note: Out of syllabus.
Question 43
 
A
2-40
B
2-9
C
222
D
231
Question 43 Explanation: 
Largest gap will be in between two most largest numbers.
The largest number is 1.111111111× 262-31 = (2−2−9)×231
Second largest number is 1.111111110×262-31 = (2−2-8)×231
Difference = (2−2−9)×231 - (2−2-8)×231
= (2-8−2−9) ×231
= 2−9×231
= 222
Question 44

A
5
B
6
C
7
D
8
Question 44 Explanation: 
Let q is the initial state.

q0 ← Number of zeros is one more than number of ones.
q1 ← Number of ones is one more than number of zeros.
q00 ← Number of zeros is two more than number of ones.
q11 ← Number of ones is two more than number of zeros.
Question 45
   
A
(11, 9)
B
(9, 13)
C
(9, 10)
D
(11, 11)
Question 45 Explanation: 
For SOP,

⇒ w'y' + z'wx' + xyz'
Total 8 literals are there.
For POS,

⇒ (z' + w')(z' + y')(w' + x')(x + z + w)
Total 9 literals are there.
Question 46

A
A + B, and A – B, but not A + 1
B
A + B, and A + 1, but not A – B
C
A + B, but not A – B or A + 1
D
A + B, and A – B, and A + 1
Question 46 Explanation: 
The circuits performs
1) A+B when K=0 and C0 = 0. It is binary adder which performs addition of two binary numbers.
2) A - B = A+ B' + 1 when K=1 and C0 = 1 ;
Here XOR gates produce B' if K=1. Since 1⊕b= b'.
"1" in (A+B+1) is coming from C0.
Note: 2's complement of B is (B'+1). 3) A+1 when B=0, K=0, C0= 1.
Increments A.
Question 47
 
A
1
B
2
C
3
D
4
Question 47 Explanation: 

⇒ a will always be equal to A.
Question 48

A
the number of 0 bits in A0
B
the number of 1 bits in A0
C
A0
D
8
Question 48 Explanation: 
B is to be increments when a is moved to carry.
The code is counting the number of 1 bits in A0
Question 49
   
A
RRC A, #1
B
NOP ; no operation
C
LRC A, #1 ; left rotate A through carry flag by one bit
D
ADD A, #1
Question 49 Explanation: 
Initially, the 8 bits will be,
a7, a6, a5, a4, a3 , a2, a1, a0
Now right rotate it once,
C0, a7, a6, a5, a4, a3 , a2, a1, now a0 is the new carry.
Now again right rotate it,
a0C0, a7, a6, a5, a4, a3 , a2

So after 8 rotations,
a6, a5, a4, a3 , a2, a1, a0C0 and carry is a7
Now, one more rotation will restore the original value of A0.
Hence, answer is Option (A).
Question 50
   
A
1
B
5
C
7
D
8
Question 50 Explanation: 

There are possible: 7 strings
Question 51
   
A
G is not ambiguous
B
There exist x, y ∈ L(G) such that xy ∉ L(G)
C
There is a deterministic pushdown automaton that accepts L(G)
D
We can find a deterministic finite state automaton that accepts L(G)
Question 51 Explanation: 
a) False
We can derive ϵ with more than one parse tree,

So ambiguous.
b) False
Let take x=aabb and y=ab then xy=aabbab we can produce it,

c) True
Because the language generated is no. of a's = no' of b's. So DPDA exist for this language.
d) Not possible.
Infinite memory needed to count 'a' for no. of 'b'.
Question 52
 
A
L1 ∈ P and L2 is finite
B
L1 ∈ NP and L2 ∈ P
C
L1 is undecidable and L2 is decidable
D
L1 is recursively enumerable and L2 is recursive
Question 52 Explanation: 
L1 is polynomial time reducible to L2.
Now if L2 is decidable then L1 should also be decidable. Hence, option (c) is wrong.
Question 53
 
A
M does not halt on any string in (0+1)+
B
M does not halt on any string in (00+1)*
C
M halts on all strings ending in a 0
D
M halts on all strings ending in a 1
Question 53 Explanation: 

Try for any string, it will not Halt for any string other than ϵ. Hence, option (A) is correct.
Question 54
   
A
B
C
D
Question 55
 
A
L1 = {0,1}* - L
B
L1 = {0,1}*
C
L1 ⊆ L
D
L1 = L
Question 55 Explanation: 
As in the question said,

As in above NFA language,
L1 is {0,1}*.
Question 56
 
A
{S'→e S} and {S'→ε}
B
{S'→e S} and { }
C
{S'→ε} and {S'→ε}
D
{S'→e S, S'→ε} and {S'→ε}
Question 56 Explanation: 
First(S) = {1,a}
First(S') = {e,ε}
First(E) = {b}
Follow(S') = {e,$}
Only when 'First' contains ε, we need to consider FOLLOW for getting the parse table entry.

Hence, option (D) is correct.
Question 57
 
A
LL(1)
B
SLR(1) but not LL(1)
C
LALR(1) but not SLR(1)
D
LR(1) but not LALR(1)
Question 57 Explanation: 

Hence, it is LL(1).
Question 58
 
A
9 + 5 + 2
B
9 5 + 2 +
C
9 5 2 + +
D
+ + 9 5 2
Question 58 Explanation: 

Now traverse the tree and whatever comes first to print, just print it.
Answer will be 9 5 + 2 +.
Question 59

A
X = Y + Z
B
t1 = Y + Z; X = t1
C
t1= Y; t2 = t1 + Z; X = t2
D
t1 = Y; t2 = Z; t3 = t1 + t2; X = t3
Question 59 Explanation: 
Question 60
A program consists of two modules executed sequentially. Let f1(t) and f2(t) respectively denote the probability density functions of time taken to execute the two modules. The probability density function of the overall time taken to execute the program is given by
A
B
C
D
Question 60 Explanation: 
f1(t) and f2(t) are executed sequentially.
→ They representing the probability density functions of time taken to execute.
→ f1 can be executed in 'x' time.
f2 can be executed in 't-x' time.
→ The probability density function =
Question 61

A
B
C
D
Question 61 Explanation: 
Probability of inverse (ai, aj(i Probability of expected no. of inversions = (1/2) × (n(n-1)/2) = n(n-1)/4
Question 62
 
A
Θ(n2)
B
Θ(n log n)
C
Θ(n1.5)
D
Θ(n)
Question 62 Explanation: 
Here the inputs are to be restricted to 1...n with atmost 'n' inversions. Then the worst case time complexity of inversion sort reduces to Θ(n).
Question 63
   
A
A heap can be used but not a balanced binary search tree
B
A balanced binary search tree can be used but not a heap
C
Both balanced binary search tree and heap can be used
D
Neither balanced binary search tree nor heap can be used
Question 63 Explanation: 
→ In heap deletion takes O(log n).
Insertion of an element takes O(n).
→ In balanced primary tree deletion takes O(log n).
Insertion also takes O(log n).
Question 64
Let S be a stack of size n ≥1. Starting with the empty stack, suppose we push the first n natural numbers in sequence, and then perform n pop operations. Assume that Push and Pop operations take X seconds each, and Y seconds elapse between the end of one such stack operation and the start of the next operation. For m ≥1, define the stack-life of m as the time elapsed from the end of Push(m) to the start of the pop operation that removes m from S. The average stack-life of an element of this stack is  
A
n(X + Y)
B
3Y + 2X
C
n(X + Y) - X
D
Y + 2X
Question 64 Explanation: 
Life time of last element present in the stack = Y
(After push into stack then immediately popped)
Life time of (n-1) element = Y + X + Y + X + Y = 2X + 3Y
Life time of (n-2) element = (n-1) + 2X + 2Y = 2X + 3Y + 2X + 2Y = 4X + 5Y
Life time of 1's element = 2(n-1)X + (2n-1)Y
Life time of all elements is ⇒
2X(1+2+3+...+n-1)+Y(1+3+5+...+(2n-1))
⇒ 2X(n(n-1) /2) +Y((n/2)(1+2n-1))
⇒ n(n(X+Y)-X))
Avg. of n numbers = n(n(X+Y)-X)/n = n(X+Y)-X
Question 65
   
A
B
C
D
None of the above
Question 65 Explanation: 
Given Tree is

Insert G at root level:
Question 66
The cube root of a natural number n is defined as the largest natural number m such that m3≤n. The complexity of computing the cube root of n (n is represented in binary notation) is  
A
O(n) but not O(n0.5)
B
O(n0.5 but not O((log n)k) for any constant k>0
C
O((log n)k) for some constant k>0, but not O((log log n)m) for any constant m>0
D
O((log log n)k) for some constant k>0.5, but not O((log log n)0.5)
Question 66 Explanation: 
Time complexity to search using binary search is O(log n). The cube root involves to serach again O(log n) times in worst case. So time taken to find cube root is O(log2n) ... (I)
The time complexity is never less than O(log n) because to represent a binary search will takes O(log n) ... (II)
From (I), option A and B False
From (II), option D False.
Question 67
 
A
The number of edges in the shortest paths from v1 to all vertices of G
B
G1 is connected
C
V1 forms a clique in G
D
G1 is a tree
Question 67 Explanation: 
Calculate the minimum weight of the graph by using single source shortest path, if the weight of the graph is 0 then G is Connected, then that is the shortest path. In case the weight of the graph is not 0 then the graph G is disconnected.
Question 68
 
A
29
B
31
C
38
D
41
Question 68 Explanation: 

1+2+3+2+8+4+4+2+5 =31
Question 69
The following are the starting and ending times of activities A, B, C, D, E, F, G and H respectively in chronological order: as bs cs ae ds ce es fs be de gs ee fe hs ge he. Here, xs denotes the starting time and xe denotes the ending time of activity X. We need to schedule the activities in a set of rooms available to us. An activity can be scheduled in a room only if the room is reserved for the activity for its entire duration. What is the minimum number of rooms required?
A
3
B
4
C
5
D
6
Question 69 Explanation: 
Given order is “as bs cs ae ds ce es fs be de gs ee fe hs ge he
Where xs→Starting time, xe→ Ending time.
R1→for a
R2→for b
R3→for c
ae, R1 is free
R1→for d
ce, R3 is free
R3→for e
R4→for f
be, R2 is free
de, R1 is free
R1→for g
ee, R3 is free
fe, R4 is free
R2→for h
ge, R1 is free
he, R2 is free
Here, total we used is: R1, R2, R3, R4.
Therefore, total no. of rooms we required = 4(minimum).
Question 70

A
A[j,k] ≤ n
B
If A[j,j] ≥ n - 1, then G has a Hamiltonian cycle
C
If there exists a path from j to k, A[j,k] contains the longest path length from j to k
D
If there exists a path from j to k, every simple path from j to k contains at most A[j,k] edges
Question 70 Explanation: 

Here, we have two functionalities i.e., from j to k, there exists a path then it results otherwise 0.
If there exist a path then it results
A[j,k] = max(A[j,k], A[j,i]+A[i,k])
i.e., if there exists a path from j to k, every simple path from j to k contains the atmost A[j,k] edges.
Question 71
 
A
(∀x) [(∃y) [B(x,y) ∧ C(y)] ⇒ A(x)] ∧ ¬(∃x)[B(x,x)]
B
(∀x) [(∀y) [B(x,y) ∧ C(y)] ⇒ A(x)] ∧ ¬(∃x)[B(x,x)]
C
(∀x) [(∃y) [B(x,y) ∧ C(y)] ⇒ A(x)] ∨ ¬(∃x)[B(x,x)]
D
(∀x) [(∀y) [B(x,y) ∧ C(y)] ⇒ A(x)] ∧ (∃x)[B(x,x)]
Question 71 Explanation: 
Note: This is not in gate syllabus. Please ignore this question.
Question 72
A
((P ∨ R) ∧ (Q ∨ ¬R)) ⇒ (P ∨ Q) is logically valid
B
(P ∨ Q) ⇒ ((P ∨ R) ∧ (Q ∨ ¬R)) is logically valid
C
(P ∨ Q) is satisfiable if and only if (P∨R) ∧ (Q∨¬R) is satisfiable
D
(P ∨ Q) ⇒ FALSE if and only if both P and Q are unsatisfiable
Question 72 Explanation: 
(P ∨ Q) ⇒ ((P ∨ R) ∧ (Q ∨ ¬R))

It is may be True (or) False depending on values. So this is not valid.
Question 73
 
A
115, 220
B
25, 220
C
25, 15
D
115, 105
Question 73 Explanation: 
P(i+j)
P(100+5) = P(105)
→void P(105)
{   
int i=10;   
print (x+10);  ⇒ 105+10=115 prints
  i=200;  
  j = 20;  
  print (x);  ⇒ x=105 prints
}  
115, 105 prints
Question 74
 
A
115, 220
B
25, 220
C
25, 15
D
115, 105
Question 74 Explanation: 
In dynamic,
In void P(x)
{  int i = 10;
   print(x + 10);   ⇒ 105+10=115 prints
 
print (x);   ⇒ print x=220;
Question 75
 
A
1 2 1
B
2 1 1
C
2 1 2
D
2 2 2
Question 75 Explanation: 
Because of using dynamic binding it results a values such as 2 2 2.
Note: The given question is not in the present syllabus
Question 76
Which of the following is NOT an advantage of using shared, dynamically linked libraries as opposed to using statically linked libraries?  
A
Smaller sizes of executable files
B
Lesser overall page fault rate in the system
C
Faster program startup
D
Existing programs need not be re-linked to take advantage of newer versions of libraries
Question 76 Explanation: 
Dynamic link libraries takes more time in program setup (in loading and linking phase to set up the global offset table and load and link the required libraries).
Question 77
A uni-processor computer system only has two processes, both of which alternate 10 ms CPU bursts with 90 ms I/O bursts. Both the processes were created at nearly the same time. The I/O of both processes can proceed in parallel. Which of the following scheduling strategies will result in the least CPU utilization (over a long period of time) for this system?
A
First come first served scheduling
B
Shortest remaining time first scheduling
C
Static priority scheduling with different priorities for the two processes
D
Round robin scheduling with a time quantum of 5 ms
Question 77 Explanation: 
We have two processes P and Q. These process have 10ms CPU burst time and 90ms I/O bursts.
(i) FCFS:
0-10: Process 1 can execute
10-20: Process 2 can execute
100-110: Process 1 Terminate
110-120: Process 2 Terminate
CPU utilization = 20/100 [In every 100ms it utilizes]
=20%
(ii) SRTF: can process P and Q same as FCFS
then CPU utilization = 20%
(iii) Round robin: with TQ-5
0-5: Process P1
5-10: Process P2
10-15: Process P1
15-20: Process P2
105-110: Process P1
110-115: Process P2
CPU utilization = 20/105 = 19.5
Question 78

A
1.5 ns
B
2 ns
C
3 ns
D
4 ns
Question 78 Explanation: 
The possibilities are = (TLB Hit * Cache Hit) + (TLB Hit * Cache Miss)
(TLB Miss * Cache Hit) + (TLB Miss * Cache Miss)
= (0.96*0.9*2)+(0.96*0.1+12)
(0.04*0.9*2)+(0.04*0.1*32)
= 3.8
= 4 (approximately)
Question 79

A
8 KB
B
12 KB
C
16 KB
D
20 KB
Question 79 Explanation: 
Breakup of given addresses into bit form:-
32bits are broken up as 10bits (L2) | 10bits (L1) | 12bits (offset)
First code page:
0x00000000 = 0000 0000 00 | 00 0000 0000 | 0000 0000 0000
So next code page will start from
0x00001000 = 0000 0000 00 | 00 0000 0001 | 0000 0000 0000
First data page:
0x00400000 = 0000 0000 01 | 00 0000 0000 | 0000 0000 0000
So next data page will start from
0x00401000 = 0000 0000 01 | 00 0000 0001 | 0000 0000 0000
Only one stack page:
0xFFFFF000 = 1111 1111 11 | 11 1111 1111 | 0000 0000 0000
Now, for second level page table, we will just require 1 Page
which will contain following 3 distinct entries i.e. 0000 0000 00, 0000 0000 01, 1111 1111 11.
Now, for each of these distinct entries, we will have 1-1 page in Level-1.
Hence, we will have in total 4 pages and page size = 212 = 4KB.
Therefore, Memory required to store page table = 4*4KB = 16KB.
Question 80
A
P(S) at W, V(S) at X, P(T) at Y, V(T) at Z, S and T initially 1
B
P(S) at W, V(T) at X, P(T) at Y, V(S) at Z, S initially 1, and T initially 0
C
P(S) at W, V(T) at X, P(T) at Y, V(S) at Z, S and T initially 1
D
P(S) at W, V(S) at X, P(T) at Y, V(T) at Z, S initially 1, and T initially 0
Question 80 Explanation: 
Option B is correct.
Process P1 will be executed first and then Process P2 can be executed next.
At the process P: W=P(S)
X=V(T)
At the process Q: Y=P(T)
Z=V(S)
Here, S=1, T=0 then the process P executes first and then Q, and both can run on process alternate way start with P.
Question 81

A
P(S) at W, V(S) at X, P(T) at Y, V(T) at Z, S and T initially 1
B
P(S) at W, V(T) at X, P(T) at Y, V(S) at Z, S and T initially 1
C
P(S) at W, V(S) at X, P(S) at Y, V(S) at Z, S initially 1
D
V(S) at W, V(T) at X, P(S) at Y, P(T) at Z, S and T initially 1
Question 81 Explanation: 
Here to print the required output only one semaphore is enough, if we initialize two at a time then they will run concurrently and leads the processing error.
Question 82
The subnet mask for a particular network is 255.255.31.0. Which of the following pairs of IP addresses could belong to this network?  
A
172.57.88.62 and 172.56.87.233
B
10.35.28.2 and 10.35.29.4
C
191.203.31.87 and 191.234.31.88
D
128.8.129.43 and 128.8.161.55
Question 82 Explanation: 
To find whether hosts belong to same network or not , we have to find their net id, if net id is same then hosts belong to same network and net id can be find by ANDing subnet mask and IP address.
128.8.129.43 (Bitwise AND) 255.255.31.0 = 128.8.1.0
128.8.161.55 (Bitwise AND) 255.255.31.0 = 128.8.1.0
Question 83
A 2km long broadcast LAN has 107 bps bandwidth and uses CSMA/CD. The signal travels along the wire at 108 m/s. What is the minimum packet size that can be used on this network?  
A
50 bytes
B
100 bytes
C
200 bytes
D
None of the above
Question 83 Explanation: 
Minimum packet size for a CSMA/CD LAN is the frame which cover whole RTT(round trip time). i.e. Tt=2Tp
d= 2 km = 2 x 103 m, v = 2 x 108 m/s, B= 107
Tp= d / v = 2 x 103 /(2 x 108 ) seconds= 10-5 seconds
Let L bits be minimum size of frame, then Tt=t L / B = L / 107 seconds
Now, Tt=2Tp
L/107 = 2 x 10-5 = 200 bits = (200 / 8) bytes = 25 bytes
Question 84
Host A is sending data to host B over a full duplex link. A and B are using the sliding window protocol for flow control. The send and receive window sizes are 5 packets each. Data packets (sent only from A to B) are all 1000 bytes long and the transmission time for such a packet is 50 µs Acknowledgement packets (sent only from B to A) are very small and require negligible transmission time. The propagation delay over the link is 200 µs. What is the maximum achievable throughput in this communication?    
A
7.69 × 106 bps
B
11.11 × 106 bps
C
12.33 × 106 bps
D
15.00 × 106 bps
Question 84 Explanation: 
Given, Tt= 50 μs, Tp = 200 μs, L= 1000 bytes, N= 5,
Transmission rate , Tt = L / B.W
Therefore, B.W. = L / Tt = 1000 bytes/ 50 μs = 8000 bits / 50 μs=160 Mbps
Efficiency = N / 1 + 2a, where a = Tp / Tt
Efficiency = 5 * 50 / (50+400)=250/450=5/9
Maximum achievable throughput= Efficiency * B.W = (5/9)*160 Mbps = 88.88 Mbps = = 11.11 x 106 bytes per second
*Actual option should be in bytes per second.
Question 85
 
A
in second normal form but not in third normal form
B
in third normal form but not in BCNF
C
in BCNF
D
in none of the above
Question 85 Explanation: 
Three FD's are valid from the above set of FD's for the given relation.
Date_of_Birth → Age
Name → Roll_number
Roll_number → Name
Candidate keys for the above are:
(Date_of_Birth, Name) and (Date_of_Birth, Roll_number)
Clearly, there is a partial dependency,
Date_of_Birth → Age
So, it is only in 1NF.
Question 86
 
A
Names of students who have got an A grade in all courses taught by Korth
B
Names of students who have got an A grade in all courses
C
Names of students who have got an A grade in at least one of the courses taught by Korth
D
in none of the above
Question 86 Explanation: 
The query results a names of students who got an A grade in at least one of the courses taught by korth.
Question 87
   
A
The schedule is serializable as T2; T3; T1
B
The schedule is serializable as T2; T1; T3
C
The schedule is serializable as T3; T2; T1
D
The schedule is not serializable
Question 87 Explanation: 
Precedence graph:

Cycle formed so not serializable.
Question 88
 
A
{m|m ≤ n, (∃i)[m=i!]}
B
{m|m ≤ n, (∃i)[m=i2]}
C
{m|m ≤ n, m is prime}
D
{ }
Question 88 Explanation: 
Take n=4, so Two log_n=4
Now Trace the code,
for (k=3; k<=n; k++)
A[k]=0; // A[3]=0
A[4]=0
for (k=2; k<=Two log_n; k++)
for(j=k+1; j<=n; j++)
A[j] = A[j] // (j%k); // A[3] = 0 // I=1
A[4] = 0 // I=1
for (j=3; j<=n; j++)
if (!A[j]) printf("%d", j);
// if (!1) means if (0), so printf will never execute
Hence, Option (D) is the answer.
Question 89
 
A
12 7 6
B
22 12 11
C
14 6 6
D
7 6 6
Question 89 Explanation: 
In main function x=5; it is global array
p(&x) it goes to P( ) function
y=5
x=5+2=7;
Q(x)
z=7
z=7+5=12(Print+z→I)
comes to P( )
*y=7-1=6
x=7(Print x→II)
comes to main ( ),
print x=*y=6 (print x→III)
Output: 12 7 6
Question 90
 
A
the list is empty or has exactly one element
B
the elements in the list are sorted in non-decreasing order of data value
C
the elements in the list are sorted in non-increasing order of data value
D
not all elements in the list have the same data value
Question 90 Explanation: 
It return a value '1' when the elements in the list are presented in sorted order and non-decreasing order of data value.
There are 90 questions to complete.