## GATE 2017(set-02)

Question 1 |

The representation of the value of a 16-bit unsigned integer X in hexadecimal number system is BCA9. The representation of the value of X in octal number system is

136251 | |

736251 | |

571247 | |

136252 | |

136251 |

Question 1 Explanation:

X=(BCA9)

Each hexadecimal digit is equal to a 4-bit binary number. So convert

X=(BCA9)

Divide the binary data into groups 3 bits each because each octal digit is represented by 3-bit binary number.

X=(001 011 110 010 101 001)

Note: Two zeroes added at host significant position to make number bits of a multiple of 3. (16+2=18)

X=(1 3 6 2 5 1)

_{16}Each hexadecimal digit is equal to a 4-bit binary number. So convert

X=(BCA9)

^{16}to binaryDivide the binary data into groups 3 bits each because each octal digit is represented by 3-bit binary number.

X=(001 011 110 010 101 001)

_{2}Note: Two zeroes added at host significant position to make number bits of a multiple of 3. (16+2=18)

X=(1 3 6 2 5 1)

_{8}Question 2 |

P→(ii),Q→(iv),R→(i),S→(iii) | |

P→(ii),Q→(i),R→(iv) S→(iii) | |

P→(ii),Q→(iv),R→(iii),S→(i) | |

P→(iii),Q→(iv),R→(i),S→(ii) |

Question 2 Explanation:

Static char var; ⇾ A variable located in Data Section of memory

P→(ii),Q→(iv),R→(i),S→(iii)

P→(ii),Q→(iv),R→(i),S→(iii)

Question 3 |

P→(iii),Q→(iv),R→(i),S→(ii) | |

P→(iv),Q→(iii),R→(i),S→(ii) | |

P→(iii),Q→(iv),R→(ii),S→(i) | |

P→(iv),Q→(iii),R→(ii),S→(i) |

Question 3 Explanation:

In this problem, we have to find Average case of different algorithms

→ Tower of Hanoi with n disks takes θ(2

It is a mathematical game or puzzle. It consists of three rods and a number of disks of different sizes, which can slide onto any rod. The puzzle starts with the disks in a neat stack in ascending order of size on one rod, the smallest at the top, thus making a conical shape.

The objective of the puzzle is to move the entire stack to another rod, obeying the following simple rules:

1. Only one disk can be moved at a time.

2. Each move consists of taking the upper disk from one of the stacks and placing it on top of another stack.

3. No disk may be placed on top of a smaller disk.

With 3 disks, the puzzle can be solved in 7 moves. The minimal number of moves required to solve a Tower of Hanoi puzzle is 2

→ Binary Search given n sorted numbers takes Ɵ(log

→ Heap sort given n numbers of the worst case takes Ɵ(n log n)

→ Addition of two n×n matrices takes Ɵ(n

→ Tower of Hanoi with n disks takes θ(2

^{n}) timeIt is a mathematical game or puzzle. It consists of three rods and a number of disks of different sizes, which can slide onto any rod. The puzzle starts with the disks in a neat stack in ascending order of size on one rod, the smallest at the top, thus making a conical shape.

The objective of the puzzle is to move the entire stack to another rod, obeying the following simple rules:

1. Only one disk can be moved at a time.

2. Each move consists of taking the upper disk from one of the stacks and placing it on top of another stack.

3. No disk may be placed on top of a smaller disk.

With 3 disks, the puzzle can be solved in 7 moves. The minimal number of moves required to solve a Tower of Hanoi puzzle is 2

^{n}-1, where n is the number of disks.→ Binary Search given n sorted numbers takes Ɵ(log

_{2}n)→ Heap sort given n numbers of the worst case takes Ɵ(n log n)

→ Addition of two n×n matrices takes Ɵ(n

^{2})Question 4 |

I, II and IV only | |

I and III only | |

II and IV only | |

I only |

Question 4 Explanation:

Since CFL is closed under UNION so L

CFL is not closed under complementation; So L

L

CFL is not closed under INTERSECTION, so L

_{1}∪L_{2}is CFL, is a true statement.CFL is not closed under complementation; So L

_{1}compliment may or may not be CFL. Hence is Context free, is a false statement.L

_{1}– R means and Regular language is closed under compliment, so is also a regular language, so we have now L_{1}∩R . Regular language is closed with intersection with any language, i.e. L∩R is same type as L. So L_{1}∩R is context free.CFL is not closed under INTERSECTION, so L

_{1}∩ L_{2}may or may not be CFL and hence IV^{th}is false.Question 5 |

P→(ii),Q→(iii),R→(iv),S→(i) | |

P→(ii),Q→(i),R→(iii),S→(iv) | |

P→(iii),Q→(iv),R→(i),S→(ii) | |

P→(i),Q→(iv),R→(ii),S→(iii) |

Question 5 Explanation:

Character stream is input to lexical analyzer which produces tokens as output. So Q → (iv).

Token stream is forwarded as input to Syntax analyzer which produces syntax tree as output. So, S → (ii).

Syntax tree is the input for the semantic analyzer, So P → (iii).

Intermediate representation is input for Code generator. So R → (i).

Token stream is forwarded as input to Syntax analyzer which produces syntax tree as output. So, S → (ii).

Syntax tree is the input for the semantic analyzer, So P → (iii).

Intermediate representation is input for Code generator. So R → (i).

Question 6 |

I only | |

II only | |

III only | |

II and III only |

Question 6 Explanation:

Canonical LR is more powerful than SLR as every grammar which can be parsed by SLR parser, can also be parsed by CLR parser. The power in increasing order is:

LR(0) < SLR < LALR < CLR

Hence only I is true.

LR(0) < SLR < LALR < CLR

Hence only I is true.

Question 7 |

I and II only | |

III only | |

IV only | |

III and IV only |

Question 7 Explanation:

First of all, you need to know about process and threads. A process, in the simplest terms, is an executing program. One or more threads run in the context of the process. A thread is the basic unit to which the operating system allocates processor time. A thread can execute any part of the process code, including parts currently being executed by another thread.

Each thread has its own stack, register and PC. So here address space that is shared by all thread for a single process.

Each thread has its own stack, register and PC. So here address space that is shared by all thread for a single process.

Question 8 |

In a file allocation system, which of the following allocation scheme(s) can be used if no external fragmentation is allowed?
I. Contiguous
II. Linked
III. Indexed

I and III only | |

II only | |

III only | |

II and III only |

Question 8 Explanation:

→ Contiguous Allocation:

Advantage:

i) Both sequential and direct access is possible.

ii) Extremely fast.

Disadvantage:

i) Suffers from both internal and external fragmentation.

ii) Increasing file size is difficult, because it depends on the availability of contiguous memory at a particular instance.

Linked list allocation:

Advantage:

i) Flexible in terms of size.

ii) Does not suffers from external fragmentation.

Disadvantage:

i) Allocation is slower.

ii) Does not support random or direct access.

iii) Pointers require some extra overhead.

→ Indexed allocation:

Advantage:

i) Support direct access, so provides fast access to the file blocks.

ii) No external fragmentation.

Disadvantage:

i) The pointer overhead for indexed allocation is greater than linked allocation.

ii) Inefficient in terms of memory utilization.

Advantage:

i) Both sequential and direct access is possible.

ii) Extremely fast.

Disadvantage:

i) Suffers from both internal and external fragmentation.

ii) Increasing file size is difficult, because it depends on the availability of contiguous memory at a particular instance.

Linked list allocation:

Advantage:

i) Flexible in terms of size.

ii) Does not suffers from external fragmentation.

Disadvantage:

i) Allocation is slower.

ii) Does not support random or direct access.

iii) Pointers require some extra overhead.

→ Indexed allocation:

Advantage:

i) Support direct access, so provides fast access to the file blocks.

ii) No external fragmentation.

Disadvantage:

i) The pointer overhead for indexed allocation is greater than linked allocation.

ii) Inefficient in terms of memory utilization.

Question 9 |

I and IV only | |

I, II and III only | |

I, II and IV only | |

II, III and IV only |

Question 9 Explanation:

I: RIP uses distance vector routing. “TRUE”

RIP is one of the oldest DVR protocol which employ the hop count as a routing metric.

II: RIP packets are sent using UDP. “TRUE”

RIP uses the UDP as its transport protocol, and is assigned the reserved port no 520.

III: OSPF packets are sent using TCP. “FASLE”

OSPF encapsulates its data directly into IP Packets and does not use either TCP or UDP.

IV: OSPF operation is based on link state routing. “TRUE”

OSPF is a routing protocol which uses link state routing (LSR) and works within a single autonomous system.

Hence correct is answer “C”.

RIP is one of the oldest DVR protocol which employ the hop count as a routing metric.

II: RIP packets are sent using UDP. “TRUE”

RIP uses the UDP as its transport protocol, and is assigned the reserved port no 520.

III: OSPF packets are sent using TCP. “FASLE”

OSPF encapsulates its data directly into IP Packets and does not use either TCP or UDP.

IV: OSPF operation is based on link state routing. “TRUE”

OSPF is a routing protocol which uses link state routing (LSR) and works within a single autonomous system.

Hence correct is answer “C”.

Question 11 |

Let p, q, r denote the statements “

*It is raining*”, “*It is cold*”, and “*It is pleasant*”, respectively. Then the statement “*It is not raining and it is pleasant, and it is not pleasant only if it is raining and it is cold*” is represented by(¬p∧r)∧(¬r→(p∧q)) | |

(¬p∧r)∧((p∧q)→¬r) | |

(¬p∧r)∨((p∧q)→¬r) | |

(¬p∧r)∨(r→(p∧q)) |

Question 11 Explanation:

p: It is raining

q: It is cold

r: It is pleasant

“If it is not raining and it is pleasant, and it is not pleasant only if it is raining and it is cold.”

We can divide the statement into two parts with “Conjunction”.

i.e., ¬r→(p∧q) ⇾②

From ① & ②, the given statement can be represented as

q: It is cold

r: It is pleasant

“If it is not raining and it is pleasant, and it is not pleasant only if it is raining and it is cold.”

We can divide the statement into two parts with “Conjunction”.

i.e., ¬r→(p∧q) ⇾②

From ① & ②, the given statement can be represented as

Question 12 |

1.45×10 ^{1} | |

1.45×10 ^{-1} | |

2.27×10 ^{-1} | |

2.27×10 ^{1} |

Question 12 Explanation:

For single-precision floating-point representation decimal value is equal to (-1)

^{5}×1.M×2

^{(E-127)}

S=0

E=(01111100)

_{2}=(124). So E – 127 = - 3

1.M=1.11011010…0

=2

^{0}+2

^{(-1)}+2

^{(-1)}+2

^{(-4)}+2

^{(-5)}+2

^{(-7)}

=1+0.5+0.25+0.06+0.03+0.007

≈1.847

(-1)

^{5}×1.M×2

^{(E-127)}=(-1)

^{0}×1.847×2

^{(-3)}

≈0.231

≈2.3×10

^{(-1)}

Question 13 |

I only | |

II only | |

Both I and II | |

Neither I nor II |

Question 13 Explanation:

1. Next pointer of front node points to the rear node.
2. Next pointer of rear points to the front node.

So if we consider circular linked list the next of 44 is 11.

If you place pointer on next of front node (11) then to reach 44 (last node), we have to traverse the entire list.

For delete O(1), for insert O(n).

It is clearly known that next pointer of rear node points to the front node.

Hence, only II is true.

So if we consider circular linked list the next of 44 is 11.

If you place pointer on next of front node (11) then to reach 44 (last node), we have to traverse the entire list.

For delete O(1), for insert O(n).

It is clearly known that next pointer of rear node points to the front node.

Hence, only II is true.

Question 14 |

0, 0 | |

0, 1 | |

1, 0 | |

1, 1 |

Question 14 Explanation:

printxy (int x, int y)

{

int *ptr;

x = 0;

ptr = &x;

y = *ptr;

*ptr = 1;

}

printxy (1, 1)

{

int *ptr;

x = 0;

ptr = &x;

y = *ptr;

*ptr = 1;

}

printxy (1, 1)

Question 15 |

MNOPQR | |

NQMPOR | |

QMNROP | |

POQNMR |

Question 15 Explanation:

The possible order of visiting the nodes in Breadth First Search Algorithm, implementing using Queue Data Structure is

(Do it by option Elimination)

(a) MNOPQR – MNO is not the proper order R must come in between.

(b) NQMPOR – QMP is not the order O is the child of N.

(C) QMNROP – M is not the child of Q, so QMN is false.

(D) POQNMR – P → OQ → NMR is the correct sequence. Hence Option (D).

(Do it by option Elimination)

(a) MNOPQR – MNO is not the proper order R must come in between.

(b) NQMPOR – QMP is not the order O is the child of N.

(C) QMNROP – M is not the child of Q, so QMN is false.

(D) POQNMR – P → OQ → NMR is the correct sequence. Hence Option (D).

Question 16 |

{a ^{m} b^{n}│m≥n,n>0} | |

{a ^{m} b^{n}│m≥n,n≥0} | |

{a ^{m} b^{n}│m>n,n≥0} | |

{a ^{m} b^{n}│m>n,n>0} |

Question 16 Explanation:

The production X→ aX | a can generate any number of a’s ≥ 1 and the production Y→ aYb | ϵ will generate equal number of a’s and b’s.

So the production S→XY can generate any number of a’s (≥1) in the beginning (due to X) and then Y will generate equal number of a’s and b’s.

So, the number of a’s will always be greater than number of b’s and number of b’s must be greater than or equal to 0 (as Y → ϵ, so number of b’s can be zero also).

Hence the language is {a

So the production S→XY can generate any number of a’s (≥1) in the beginning (due to X) and then Y will generate equal number of a’s and b’s.

So, the number of a’s will always be greater than number of b’s and number of b’s must be greater than or equal to 0 (as Y → ϵ, so number of b’s can be zero also).

Hence the language is {a

^{m}b^{n}│m>n,n≥0}.Question 17 |

An ER model of a database consists of entity types A and B. These are connected by a relationship R which does not have its own attribute. Under which one of the following conditions, can the relational table for R be merged with that of A?

Relationship R is one-to-many and the participation of A in R is total. | |

Relationship R is one-to-many and the participation of A in R is partial. | |

Relationship R is many-to-one and the participation of A in R is total. | |

Relationship R is one-to-many and the participation of A in R is partial. |

Question 17 Explanation:

The relational table for R be merged that of A, if the relationship R is Many-to-one and the participation of A in R is total.

Question 18 |

I only | |

II only | |

Both I and II | |

Neither I nor II |

Question 18 Explanation:

A connected UDP socket, the result of calling connection a UDP socket, Connect () specifying the remote address.

I. A connected UDP socket can be used to communicate with only one peer.

A DNS client can be configured to use one or more servers, normally by listing the IP addresses of the servers in the file /etc/resolv.conf. If a single server is listed, the client can call connect, but if multiple servers are listed the client cannot call connect.

II. A process with a connected UDP socket can call connect function again for that socket for one of two reasons:

(a) To specify a new IP address and port.

(b) To unconnect the socket.

Hence, the correct answer is (B).

I. A connected UDP socket can be used to communicate with only one peer.

A DNS client can be configured to use one or more servers, normally by listing the IP addresses of the servers in the file /etc/resolv.conf. If a single server is listed, the client can call connect, but if multiple servers are listed the client cannot call connect.

II. A process with a connected UDP socket can call connect function again for that socket for one of two reasons:

(a) To specify a new IP address and port.

(b) To unconnect the socket.

Hence, the correct answer is (B).

Question 19 |

0 | |

1 | |

2 | |

3 |

Question 19 Explanation:

Therefore, no. of additional records deleted from the table T1 is 0 (zero).

Question 20 |

The maximum number of IPv4 router addresses that can be listed in the record route (RR) option field of an IPv4 header is _________.

9 | |

10 | |

11 | |

12 |

Question 20 Explanation:

A record route option is used to record the internet router that handles the datagram. It can be used for debugging and management purpose.

In IPv4 header, 40 bytes are reserved for OPTIONS.

For Record Route to stores, 1 byte is used to store type of option, 1 byte for length and 1 byte for pointer. Out of 40 bytes, 37 bytes are left.

Each IP4 address takes 32 bits or 4 bytes.

Therefore, it can store at most floor (37/4) = 9 router addresses.

Hence correct answer is 9 router address.

In IPv4 header, 40 bytes are reserved for OPTIONS.

For Record Route to stores, 1 byte is used to store type of option, 1 byte for length and 1 byte for pointer. Out of 40 bytes, 37 bytes are left.

Each IP4 address takes 32 bits or 4 bytes.

Therefore, it can store at most floor (37/4) = 9 router addresses.

Hence correct answer is 9 router address.

Question 21 |

0 | |

1 | |

2 | |

3 |

Question 21 Explanation:

R={(a,a),(a,b),(a,c),(a,d),(a,e),(b,b),(b,c),(b,e),(c,c),(c,e),(d,d),(d,e),(e,e)}

As per the definition of lattice, each pair should have GLB, LUB.

The given ‘R’ has GLB, LUB for each and every pair.

So, no need to add extra pair.

Thus no. of required pair such that Hasse diagram to become lattice is “0”.

As per the definition of lattice, each pair should have GLB, LUB.

The given ‘R’ has GLB, LUB for each and every pair.

So, no need to add extra pair.

Thus no. of required pair such that Hasse diagram to become lattice is “0”.

Question 22 |

2 | |

3 | |

4 | |

5 |

Question 22 Explanation:

R

_{2}→R

_{2}+R

_{1}

The number of non-zero rows of P + Q = 2,

So Rank = 2

Note: “Rank” is the number of independent vectors.

Method-1:

Each vector is a row in matrix.

Echelon form of a matrix has no. of zeroes increasing each rows.

The total non-zero rows left give the rank.

Method-2:

Find determinant of matrix, for 3×5, if determinant is ‘0’, the max rank can be 2. If determinant of any n×n is non-zero, then rows proceed with (n-1)×(n-1).

Question 23 |

G is an undirected graph with n vertices and 25 edges such that each vertex of G has degree at least 3. Then the maximum possible value of n is ___________.

16 | |

17 | |

18 | |

19 |

Question 23 Explanation:

An undirected graph ‘G’ has ‘n’ vertices & 25 edges.

Degree of each vertex ≥3

|v|=2|E|

The relation between max and min degree of graph are

m≤2|E|/|v| ≤M

Given minimum degree = 3

So 3≤2 |E|/|v|

3|v|≤2|E|

3(n)≤2(25)

n≤50/3

n≤16.6

(n=16)

Degree of each vertex ≥3

|v|=2|E|

The relation between max and min degree of graph are

m≤2|E|/|v| ≤M

Given minimum degree = 3

So 3≤2 |E|/|v|

3|v|≤2|E|

3(n)≤2(25)

n≤50/3

n≤16.6

(n=16)

Question 24 |

Consider a quadratic equation x

^{2 }-13x+36=0 with coefficients in a base b. The solutions of this equation in the same base b are x=5 and x=6. Then b = _____.8 | |

9 | |

10 | |

11 |

Question 24 Explanation:

x

Generally if a, b are roots.

(x-a)(x-b)=0

x

Given that x=5,x=6 are roots of ①

So a+b=13

ab=36 (with same base ‘b’)

i.e., (5)

Convert them into decimal value

5

6

13

11=b+3

b=8

Now check with ab=36

5

Convert them into decimals

5

30=b×3+6

24=b×3

b=8

∴ The required base = 8

^{2}-13x+36=0 ⇾①Generally if a, b are roots.

(x-a)(x-b)=0

x

^{2}-(a+b)x+ab=0Given that x=5,x=6 are roots of ①

So a+b=13

ab=36 (with same base ‘b’)

i.e., (5)

_{b}+(6)_{b}=(13)_{b}Convert them into decimal value

5

_{b}=5_{10}6

_{10}=6_{10}13

_{b}=b+311=b+3

b=8

Now check with ab=36

5

_{b}×6_{b}=36_{b}Convert them into decimals

5

_{b}×6_{b}=(b×3)+6_{10}30=b×3+6

24=b×3

b=8

∴ The required base = 8

Question 25 |

The minimum possible number of a deterministic finite automation that accepts the regular language L={w

_{1}aw_{2}|w_{1}, w_{2}∈{a,b}*,|w_{1}|=2,|w_{2}|≥3} is _____.8 | |

9 | |

10 | |

11 |

Question 25 Explanation:

|w

|w

w

So the required DFA is

_{1}|=2 means the length of w_{1}is two. So we have four possibilities of w_{1}={aa,ab,ba,bb}.|w

_{2}|≥3 means the w_{2}will have at least three length string from {a,b}.w

_{2}will have {aaa,aab,aba,abb,baa,bab,bba,bbb,……….}So the required DFA is

Question 26 |

4/5 | |

5/6 | |

7/8 | |

11/12 |

Question 26 Explanation:

Probability that ‘P’ applies for a job =1/4=P(p) ⇾①

Probability that ‘P’ applies for the job given that Q applies for the job =P(p/q)=1/2 ⇾ ②

Probability that ‘Q’ applies for the job, given that ‘P’ applies for the job =P(p/q)=1/3 ⇾ ③

Bayes Theorem:

(P(A/B)=(P(B/A)∙P(A))/P(B) ;P(A/B)=P(A∩B)/P(B))

⇒ P(p/q)=(P(q/p)∙P(p))/p(q)

⇒ 1/2=(1/3×1/4)/p(q)

p(q)=1/12×2=1/(6 ) (P(q)=1/6) ⇾ ④

From Bayes,

P(p/q)=(P(p∩q))/(P(q))

1/2=P(p∩q)/(1⁄6)

(p(p∩q)=1/12)

We need to find out the “probability that ‘P’ does not apply for the job given that q does not apply for the job =P(p'/q')

From Bayes theorem,

P(p'/q')=(P(p'∩q'))/P(q') ⇾⑤

We know, p(A∩B)=P(A)+P(B)-P(A∪B)

also (P(A'∩B')=1-P(A∪B))

P(p'∩q')=1-P(p∪q)

=1-(P(p)+P(q)-P(p∩q))

=1-(P(p)+P(q)-P(p)∙P(q))

=1-(1/4+1/6-1/12)

=1-(10/24-2/24)

=1-(8/24)

=2/3

(P(p'∩q')=2/3) ⇾⑥

Substitute in ⑤,

P(p'⁄q')=(2⁄3)/(1-P(q))=(2⁄3)/(1-1/6)=(2⁄3)/(5⁄6)=4/5

(P(p'/q')=4/5)

Probability that ‘P’ applies for the job given that Q applies for the job =P(p/q)=1/2 ⇾ ②

Probability that ‘Q’ applies for the job, given that ‘P’ applies for the job =P(p/q)=1/3 ⇾ ③

Bayes Theorem:

(P(A/B)=(P(B/A)∙P(A))/P(B) ;P(A/B)=P(A∩B)/P(B))

⇒ P(p/q)=(P(q/p)∙P(p))/p(q)

⇒ 1/2=(1/3×1/4)/p(q)

p(q)=1/12×2=1/(6 ) (P(q)=1/6) ⇾ ④

From Bayes,

P(p/q)=(P(p∩q))/(P(q))

1/2=P(p∩q)/(1⁄6)

(p(p∩q)=1/12)

We need to find out the “probability that ‘P’ does not apply for the job given that q does not apply for the job =P(p'/q')

From Bayes theorem,

P(p'/q')=(P(p'∩q'))/P(q') ⇾⑤

We know, p(A∩B)=P(A)+P(B)-P(A∪B)

also (P(A'∩B')=1-P(A∪B))

P(p'∩q')=1-P(p∪q)

=1-(P(p)+P(q)-P(p∩q))

=1-(P(p)+P(q)-P(p)∙P(q))

=1-(1/4+1/6-1/12)

=1-(10/24-2/24)

=1-(8/24)

=2/3

(P(p'∩q')=2/3) ⇾⑥

Substitute in ⑤,

P(p'⁄q')=(2⁄3)/(1-P(q))=(2⁄3)/(1-1/6)=(2⁄3)/(5⁄6)=4/5

(P(p'/q')=4/5)

Question 27 |

If w, x, y, z are Boolean variables, then which one of the following is INCORRECT?

wx+w(x+y)+x(x+y)=x+wy | |

(w+y)(wxy+wyz)=wxy+wyz |

Question 27 Explanation:

Option-A:

wx+w(x+y)+x(x+y)

=(wx+wx)+wy+(x+xy)

=wx+wy+x(1+y)

=wx+wy+x

=(w+1)x+wy

=x+wy

Option-B:

Option-C:

Option-D:

(w+y)(wxy+wyz)=wxy+wyz+wxy+wyz

=wxy+wyz

wx+w(x+y)+x(x+y)

=(wx+wx)+wy+(x+xy)

=wx+wy+x(1+y)

=wx+wy+x

=(w+1)x+wy

=x+wy

Option-B:

Option-C:

Option-D:

(w+y)(wxy+wyz)=wxy+wyz+wxy+wyz

=wxy+wyz

Question 28 |

Given f(w, x, y, z) = Σ

_{m}(0, 1, 2, 3, 7, 8, 10) + Σ_{d}(5, 6, 11, 15), where d represents the don’t-care condition in Karnaugh maps. Which of the following is a minimum product-of-sums (POS) form of f(w, x, y, z)?Question 28 Explanation:

f(w,x,y,z)=Σ

K-Map for the function f is

Consider maxterms in K-map to represent function in product-of-sums (POS) form

f(w,x,y,z)=(w'+z')(x'+z)

_{m}(0,1,2,3,7,8,10)+Σ_{d}(5,6,11,15)K-Map for the function f is

Consider maxterms in K-map to represent function in product-of-sums (POS) form

f(w,x,y,z)=(w'+z')(x'+z)

Question 29 |

In a two-level cache system, the access times of L

_{1}and L_{2}caches are 1 and 8 clock cycles, respectively. The miss penalty from the L_{2}cache to main memory is 18 clock cycles. The miss rate of L_{1}cache is twice that of L_{2}. The average memory access time (AMAT) of this cache system is 2 cycles. The miss rates of L_{1}and L_{2}respectively are:0.111 and 0.056 | |

0.056 and 0.111 | |

0.0892 and 0.1784 | |

0.1784 and 0.0892 |

Question 29 Explanation:

The Average memory access time is,

AMAT = (L

= L

We can write, L

AMAT = L

By taking L

= (L

AMAT = L

We access L

L

= Hit_rate_of_L

By taking Access time of L

= Access time of L

We know, MissRate of L

So, the above formula becomes,

L

It is given,

access time of L

access time of L

miss penalty of L

AMAT = 2.

Let, miss rate of L

Substituting the above values,

L

L

AMAT = L

2 = 1 + (2*x) (8+18*x)

36*x

By solving the above quadratic equation we get,

x = Miss rate of L

Miss rate of L

AMAT = (L

_{1}hit rate)*(L_{1}access time) + (L_{1}miss rate)*(L_{1}access time + L_{1}miss penalty)= L

_{1}hit rate * L_{1}access time + L_{1}miss rate * L_{1}access time + L_{1}miss rate * L_{1}miss penaltyWe can write, L

_{1}miss rate = 1 - L_{1}hit rateAMAT = L

_{1}hit rate * L_{1}access time + (1 - L_{1}hit rate) * L_{1}access time + L_{1}miss rate * L_{1}miss penaltyBy taking L

_{1}access time common,= (L

_{1}hit rate + 1 - L_{1}hit rate)* L_{1}access time + L_{1}miss rate * L_{1}miss penaltyAMAT = L

_{1}access time + L_{1}miss rate * L_{1}miss penaltyWe access L

_{2}only when there is a miss in L_{1}. So, L_{1}miss penalty is nothing but the effective time taken to access L_{2}.L

_{1}_miss_penalty = Hit_rate_of_L_{2}* Access time of L_{2}+ MissRate of L_{2}*(Access time of L_{2}+ miss penalty L_{2})= Hit_rate_of_L

_{2}* Access time of L_{2}+ MissRate of L_{2}*Access time of L_{2}+ MissRate of L_{2}* miss penalty L_{2}By taking Access time of L

_{2}common we get,= Access time of L

_{2}* (Hit_rate_of_L_{2}+ MissRate of L_{2}) + MissRate of L_{2}* miss penalty L_{2}We know, MissRate of L

_{2}= 1 - Hit_rate_of_L_{2}→ Hit_rate_of_L_{2}+ MissRate of L_{2}= 1So, the above formula becomes,

L

_{1}_miss_penalty = Access time of L_{2}+ (MissRate of L_{2}* miss penalty L_{2})It is given,

access time of L

_{1}= 1,access time of L

_{2}= 8,miss penalty of L

_{2}= 18,AMAT = 2.

Let, miss rate of L

_{2}= x. Since it is given that L_{1}miss rate is twice that of 12 miss rate, L_{1}miss rate = 2 * x.Substituting the above values,

L

_{1}_miss_penalty = Access time of L_{2}+ (MissRate of L_{2}* miss penalty L_{2})L

_{1}_miss_penalty = 8 + (x*18)AMAT = L

_{1}access time + L_{1}miss rate * L_{1}miss penalty2 = 1 + (2*x) (8+18*x)

36*x

^{2}+ 16*x -1 = 0By solving the above quadratic equation we get,

x = Miss rate of L

_{2}= 0.056Miss rate of L

_{1}= 2*x = 0.111Question 30 |

Θ(loglogn) | |

Θ(logn) | |

Θ(√n) | |

Θ(n) |

Question 30 Explanation:

T(n)=2T(√n)+1

(or)

T(n)=2T(n

Here Assume n=2

T(2

T(2

Assume T(2

S(k)=2S(k/2)+1

Apply Master’s theorem Case-1

a=2,b=2

S(k)=k

S(k)=θ(k')

but S(k)=T(2

T(2

but n=2

k=logn

T(n)=θ(logn)

(or)

T(n)=2T(n

^{(1⁄2)})+1Here Assume n=2

^{k}T(2

^{k})=2T(2^{k})^{(1⁄2)}+1T(2

^{k})=2T(2^{(k/2)})+1Assume T(2

^{k})=S(k)S(k)=2S(k/2)+1

Apply Master’s theorem Case-1

a=2,b=2

S(k)=k

^{(log22 )}S(k)=θ(k')

but S(k)=T(2

^{k})T(2

^{k})=θ(k')but n=2

^{k}k=logn

T(n)=θ(logn)

Question 31 |

Nβ(1-β) | |

Nβ | |

N(1-β) | |

Not expressible in terms of N and β alone |

Question 31 Explanation:

For a discrete random variable X,

Given g

Expectation (i.e., mean) of a binomial distribution will be np.

The polynomial function ,

given

Mean of Binomial distribution of b(x

The probability Mass function,

Given:

Given g

_{y}(z)=(1-β+βz)^{N}⇾ it is a binomial distribution like (x+y)^{n}Expectation (i.e., mean) of a binomial distribution will be np.

The polynomial function ,

given

Mean of Binomial distribution of b(x

_{j},n,p)=The probability Mass function,

Given:

Question 32 |

Question 32 Explanation:

Consider the production (given below) which has left recursion.

S→Sα | β

The equivalent production (after removing left recursion) is

S→βS

S

Hence after removing left recursion from: E→ E-T | T, here α = -T and β = T

E→TE

E

After removing left recursion from: T→T+F | F, here α=+F and β=F

T→FT

T

Replace E

We have,

E→TX

X→-TX | ϵ

T→FY

Y→+FY | ϵ

F→(E)| id

S→Sα | β

The equivalent production (after removing left recursion) is

S→βS

_{1}S

_{1}→αS_{1}| ϵHence after removing left recursion from: E→ E-T | T, here α = -T and β = T

E→TE

_{1}E

_{1}→ -TE_{1}| ϵAfter removing left recursion from: T→T+F | F, here α=+F and β=F

T→FT

_{1}T

_{1}→ +FT_{1}| ϵReplace E

_{1}= X and T_{1}= YWe have,

E→TX

X→-TX | ϵ

T→FY

Y→+FY | ϵ

F→(E)| id

Question 33 |

Safe, Deadlocked | |

Safe, Not Deadlocked | |

Not Safe, Deadlocked | |

Not Safe, Not Deadlocked |

Question 33 Explanation:

Available: (9 - (3 + 1 + 3)) = 2, P3 can be satisfied, New available = 3 + 2 = 5

Now, P2 can be satisfied. New available: 5 + 1 = 6.

Now, P1 can be satisfied. Thus safe sequence: P3→ P2 → P1

That is not deadlocked.

Question 34 |

p=3 and q=1 | |

p=3 and q=2 | |

p=4 and q=1 | |

p=4 and q=2 |

Question 34 Explanation:

Hamming distance of a code is minimum distance between any two code words.

Minimum Distance =p=3

Error bits that can be corrected =(p-1)/2=(3-1)/2=1

∴ p=3 and q=1

Minimum Distance =p=3

Error bits that can be corrected =(p-1)/2=(3-1)/2=1

∴ p=3 and q=1

Question 35 |

Consider two hosts X and Y, connected by a single direct link of rate 10

^{6 }bits/sec. The distance between the two hosts is 10,000 km and the propagation speed along the link is 2 × 10^{8}m/sec. Host X sends a file of 50,000 bytes as one large message to host Y continuously. Let the transmission and propagation delays be p milliseconds and q milliseconds, respectively. Then the values of p and q arep=50 and q=100 | |

p=50 and q=400 | |

p=100 and q=50 | |

p=400 and q=50 |

Question 35 Explanation:

Given Data:

B=10

L=50000 Bytes

d=10000 Km = 10

v=8×10

Transmission time,

P = L/B = 50000×8bits/ 10

Propagation time,

q = d/v = 10

B=10

^{6}bits/secL=50000 Bytes

d=10000 Km = 10

^{7}mv=8×10

^{8}m/secTransmission time,

P = L/B = 50000×8bits/ 10

^{6}bits/sec = 0.4sec = 400msecPropagation time,

q = d/v = 10

^{7}m / 2×10^{8}m/s = 0.05sec = 50 msecQuestion 36 |

The pre-order traversal of a binary search tree is given by 12, 8, 6, 2, 7, 9, 10, 16, 15, 19, 17, 20. Then the post-order traversal of this tree is:

2, 6, 7, 8, 9, 10, 12, 15, 16, 17, 19, 20 | |

2, 7, 6, 10, 9, 8, 15, 17, 20, 19, 16, 12 | |

7, 2, 6, 8, 9, 10, 20, 17, 19, 15, 16, 12 | |

7, 6, 2, 10, 9, 8, 15, 16, 17, 20, 19, 12 |

Question 36 Explanation:

from these 2 orders, we can say 12 is the root & 8 is the root of left subtree & 16 is the root of right subtree.

From 2, 6, 7 we can identify 6 is the root from preorder traversal and from 9, 10 → 9 is root.

From <17, 19, 20>, 19 as root.

Hence, 2,7,6,10,9,8 |,15,17,20,19,16 |12 is the post-order traversal.

Question 37 |

(q==r) && (r==0) | |

(x>0) && (r== x) && (y>0) | |

(q==0) && (r==x) && (y>0) | |

(q==0) && (y>0) |

Question 37 Explanation:

Divide x by y.

x, y, q, r are unsigned integers.

while (r >= y)

{

r = r – y;

q = q + 1;

}

Loop terminates in a state satisfying the condition

x = = (y * q + r)

y ⇒ Dividend = Divisor * Quotient + Remainder

So to divide a number with repeated subtractions, the Quotient should be initialized to 0 and it must be incremented for every subtraction.

So initially q=0 which represents

x = 0 + r ⇒ x = r

and y must be a positive value (>0).

x, y, q, r are unsigned integers.

while (r >= y)

{

r = r – y;

q = q + 1;

}

Loop terminates in a state satisfying the condition

x = = (y * q + r)

y ⇒ Dividend = Divisor * Quotient + Remainder

So to divide a number with repeated subtractions, the Quotient should be initialized to 0 and it must be incremented for every subtraction.

So initially q=0 which represents

x = 0 + r ⇒ x = r

and y must be a positive value (>0).

Question 38 |

Θ(n√n) | |

Θ(n ^{2} ) | |

Θ(n logn) | |

Θ(n ^{2} logn) |

Question 38 Explanation:

We can solve iterative programs time complexity with the help of rollback method.

int fun(int n)

{

int i, j;

for (i = 1; i <= n ; i++) /* It is independent loop. So, it takes O(n) time */

{

for (j = 1; j < n; j += i) /* It is dependent loop, It always incrementing with the help of i. It will take approximately O(log n) times*/

{

printf("%d %d", i, j); /* It takes O(1) time

}

}

}

So, the overall complexity of the program is θ(n logn) times.

int fun(int n)

{

int i, j;

for (i = 1; i <= n ; i++) /* It is independent loop. So, it takes O(n) time */

{

for (j = 1; j < n; j += i) /* It is dependent loop, It always incrementing with the help of i. It will take approximately O(log n) times*/

{

printf("%d %d", i, j); /* It takes O(1) time

}

}

}

So, the overall complexity of the program is θ(n logn) times.

Question 39 |

∅ | |

{q _{0},q_{1},q_{3}} | |

{q _{0},q_{1},q_{2} } | |

{q _{0},q_{2},q_{3} } |

Question 39 Explanation:

Extended transition function describes, what happens when we start in any state and follow any sequence of inputs. If δ is our transition function, then the extended transition function is denoted by δ. The extended transition function is a function that takes a state q and a string w and returns a state p (the state that the automaton reaches when starting in state q and processing the sequence of inputs w).

The starting state is q

With epsilon transition we reach to q

From q

From q

From q

So state q

The starting state is q

_{2}, from q_{2}, transition with input “a” is dead so we have to use epsilon transition to go to other state.With epsilon transition we reach to q

_{0}, at q_{0}we have a transition with input symbol “a” so we reach to state q_{1}.From q

_{1}, we can take transition with symbol “b” and reach state q_{2}but from q_{2}, again we have no further transition with symbol “a” as input, so we have to take another transition from state q_{1}, that is, the epsilon transition which goes to state q_{2}.From q

_{2}we reach to state q_{0}and read input “b” and then read input “a” and reach state q_{1}. So q_{1}is one of the state of extended transition function.From q

_{1}we can reach q_{2}by using epsilon transition and from q_{2}we can reach q_{0}with epsilon move so state q_{2}and q_{0}are also part of extended transition function.So state q

_{0},q_{1},q_{2}.Question 40 |

I, II and IV only | |

II and III only | |

I and IV only | |

III and IV only |

Question 40 Explanation:

L

L

L

L

_{1}={a^{p}│p is a prime number} is a context sensitive language. So I is false.L

_{2}={a^{n}b^{m}c^{2m}│n≥0,m≥0} is a context free as we have to do only one comparison (between b’s and c’s) which can be done by using PDA, so L^{2}is Context free and II is true.L

_{3}={a^{n}b^{n}c^{2n}│n≥0} is context sensitive. The reason it has more than one comparison (at a time) as we have to compare number of a’s, b’s and c’s. So this cannot be done using PDA. Hence III is false.L

_{4}={a^{n}b^{n}│n≥1} is Context free (as well as Deterministic context free). We can define the transition of PDA in a deterministic manner. In beginning push all a’s in stack and when b’s comes pop one “a” for one “b”. If input and stack both are empty then accept.Question 41 |

I and IV only | |

II and III only | |

II, III and IV only | |

III and IV only |

Question 41 Explanation:

Since membership problem for regular language is decidable, so I is decidable.

Emptiness problem for Context free language is decidable, so II is decidable.

Completeness problem (i.e. L(G)=Σ* for a CFG G) is undecidable.

Membership problem for recursive enumerable language (as language of Turing Machine is recursive enumerable) is undecidable. So IV is undecidable.

Emptiness problem for Context free language is decidable, so II is decidable.

Completeness problem (i.e. L(G)=Σ* for a CFG G) is undecidable.

Membership problem for recursive enumerable language (as language of Turing Machine is recursive enumerable) is undecidable. So IV is undecidable.

Question 43 |

3 | |

4 | |

5 | |

6 |

Question 43 Explanation:

Swap (&x, &y) exchanges the contents of x & y.

①⇒ 1st for i = 0 <= 4 a[0] < a[1] ≃ 3<5 so perform swapping

done=

①⇒ no swap (3, 1)

②⇾ perform swap (1, 4)

①⇒ perform swap (1, 6)

①⇒ perform swap (1, 2)

①⇒ (done is still 0)

for i = 5 >= 1 a[5] > a[4] ≃ 1>2 – false. So, no swapping to be done

②⇾ no swap (6, 2)

③⇾ swap (4, 6)

①⇒ Swap (3, 6)

①⇒ Swap (5, 6)

⇒ Done is still 0. So while loop executes again.

①⇾ no swap (6, 5)

done = 1

②⇾ No swap (5, 3)

③⇾Swap (3, 4)

So, array [3] = 3

①⇒ 1st for i = 0 <= 4 a[0] < a[1] ≃ 3<5 so perform swapping

done=

①⇒ no swap (3, 1)

②⇾ perform swap (1, 4)

①⇒ perform swap (1, 6)

①⇒ perform swap (1, 2)

①⇒ (done is still 0)

for i = 5 >= 1 a[5] > a[4] ≃ 1>2 – false. So, no swapping to be done

②⇾ no swap (6, 2)

③⇾ swap (4, 6)

①⇒ Swap (3, 6)

①⇒ Swap (5, 6)

⇒ Done is still 0. So while loop executes again.

①⇾ no swap (6, 5)

done = 1

②⇾ No swap (5, 3)

③⇾Swap (3, 4)

So, array [3] = 3

Question 44 |

54 | |

55 | |

56 | |

57 |

Question 44 Explanation:

From the given transactions T

=8!/(4×3×2×4×3×2)

=(8×7×6×5×4×3×2×1)/(4×3×2×4×3×2)

=70

Following two conflict actions are possible:

∴# Permutations=4×3=12

#permutations =4×1=4

∴ Total no. of conflict serial schedules possible =70-12-4=54

_{1}and T_{2}, total number of schedules possible =(4+4)!/4!4!=8!/(4×3×2×4×3×2)

=(8×7×6×5×4×3×2×1)/(4×3×2×4×3×2)

=70

Following two conflict actions are possible:

∴# Permutations=4×3=12

#permutations =4×1=4

∴ Total no. of conflict serial schedules possible =70-12-4=54

Question 45 |

4.72 | |

4.73 | |

4.74 | |

4.75 |

Question 45 Explanation:

Since nothing is given whether it is hierarchical memory or simultaneous memory, by default we consider it as hierarchical memory.

Hierarchical memory (Default case):

For 2-level memory: The formula for average memory access time = h

For 3-level memory: h

t

Instruction fetch happens from I-cache whereas operand fetch happens from D-cache. Using that we need to calculate the instruction fetch time (Using I-cache and L

The equation for instruction fetch time = t

=2+0.2*8+0.2*0.1*90 = 5.4ns

Operand fetch time = t

The average read access time = 0.6*5.4+0.4*3.7 = 4.72ns

Hierarchical memory (Default case):

For 2-level memory: The formula for average memory access time = h

_{1}t_{1}+ (1-h_{1})(t_{1}+t_{2}), this can be simplified as t_{1}+(1-h_{1})t_{2}For 3-level memory: h

_{1}t_{1}+(1-h_{1})(t_{1}+h_{2}t_{2}+(1-h_{2})(t_{2}+t_{3})) this can be simplified ast

_{1}+ (1-h_{1})t_{2}+ (1-h_{1})(1-h_{2})t_{3}Instruction fetch happens from I-cache whereas operand fetch happens from D-cache. Using that we need to calculate the instruction fetch time (Using I-cache and L

_{2}--cache) and operand fetch time (Using D-cache and L_{2}-cache) separately. Then calculate 0.6 (instruction fetch time) + 0.4(operand fetch time) to find the average read access time.The equation for instruction fetch time = t

_{1}+ (1-h_{1}) t_{2}+ (1-h_{1})(1-h_{2}) t_{3}=2+0.2*8+0.2*0.1*90 = 5.4ns

Operand fetch time = t

_{1}+(1-h_{1})t_{2}+ (1-h_{1})(1-h_{2})t_{3}= 2+0.1*8+0.1*0.1*90 = 3.7nsThe average read access time = 0.6*5.4+0.4*3.7 = 4.72ns

Question 46 |

7 | |

8 | |

9 | |

10 |

Question 46 Explanation:

In the given database table top_scorer no players are there from ‘Spain’. So, the query ① results 0 and ALL (empty) is always TRUE.

The query ② selects the goals of the players those who are belongs to ‘Germany’. So, it results in ANY (16, 14, 11, 10).

So, the outer most query results the player names from top_scorer, who have more goals.

Since, the minimum goal by the ‘Germany’ player is 10, it returns the following 7 rows.

Question 48 |

If a random variable X has a Poisson distribution with mean 5, then the expectation E[(X+2)

^{2}] equals _________.54 | |

55 | |

56 | |

57 |

Question 48 Explanation:

In Poisson distribution:

Mean = Variance

E(X)=E(X

E(X

So, E[(X+2)

= E(X

= 54

Mean = Variance

E(X)=E(X

^{2}) - (E(X))^{2}= 5E(X

^{2}) = 5+(E(X))^{2}= 5+25 = 30So, E[(X+2)

^{2}] = E[X^{2}+ 4 +4X]= E(X

^{2})+4+4E(X) = 30+4+4×5= 54

Question 49 |

In a B

^{+}tree, if the search-key value is 8 bytes long, the block size is 512 bytes and the block pointer size is 2 bytes, then the maximum order of the B^{+}tree is _________.52 | |

53 | |

54 | |

55 |

Question 49 Explanation:

Given,

Block size = 512 bytes

Block pointer = 2 bytes

Search key = 8 bytes

Let Maximum order of B

n(P

n(2)+(n-1)(8)≤512

2n+8n-8≤512

10n≤520

n≤520/10

(n=52)

Block size = 512 bytes

Block pointer = 2 bytes

Search key = 8 bytes

Let Maximum order of B

^{+}tree = nn(P

_{b})+(n-1)(k)≤512n(2)+(n-1)(8)≤512

2n+8n-8≤512

10n≤520

n≤520/10

(n=52)

Question 50 |

225 | |

226 | |

227 | |

228 |

Question 50 Explanation:

General procedure to solve Huffman coding problem

Step-1: Arrange into either descending/ ascending order according to that profits.

Step-2: Apply optimal merge pattern procedure.

Step-3: Make left sub-tree value either 0 or 1, for right sub-tree, vice-versa.

=2×0.34+2×0.22+2×0.19+3×0.17+3×0.08

=2.25

∴ So, for 100 characters, 2.25* 100 = 225

Question 51 |

29 | |

30 | |

31 | |

32 |

Question 51 Explanation:

Waiting Time = 0 + (33 - 5) + (40 - 2) + (49 - 12) + (51 - 9) = 145

Average waiting time: 145/5 = 29

Question 52 |

If the characteristic polynomial of a 3×3 matrix M over R (the set of real numbers) is λ

^{3}-4 λ^{2}+aλ+30, a∈R, and one eigenvalue of M is 2, then the largest among the absolute values of the eigenvalues of M is ________.5 | |

6 | |

7 | |

8 |

Question 52 Explanation:

For a 3×3 matrix ‘M’, the characteristic equation |A – λI| is

λ

One eigen value is ‘2’, so substitute it

2

8-16+2a+30=0

2a=-22

a=-11

Substitute in ①,

λ

So ① can be written as

(λ-2)(λ

(λ-2)(λ

(λ-2)(λ-3)(λ-5)=0

λ=2,3,5

Max λ=5

λ

^{3}-4λ^{2}+aλ+30=0 ⇾①One eigen value is ‘2’, so substitute it

2

^{3}-4(2)^{2}+a(2)+30=08-16+2a+30=0

2a=-22

a=-11

Substitute in ①,

λ

^{3}-4λ^{2}-11+30=0So ① can be written as

(λ-2)(λ

^{2}-2λ-15)=0(λ-2)(λ

^{2}-5λ+3λ-15)=0(λ-2)(λ-3)(λ-5)=0

λ=2,3,5

Max λ=5

Question 53 |

Consider a machine with a byte addressable main memory of 2

^{32}bytes divided into blocks of size 32 bytes. Assume that a direct mapped cache having 512 cache lines is used with this machine. The size of the tag field in bits is _____________.18 | |

19 | |

20 | |

21 |

Question 53 Explanation:

Given that it is a byte addressable memory and the main memory is of size 2

So the physical address is 32 bits long.

Each block is of size 32(=2

Also given that there are 512(=2

When it is directed mapped cache, the physical address can be divided as

(Tag bits + bits for block/LINE number + bits for block offset)

So, tag bits + 9 + 5 = 32

Tag bits = 32 - 14 = 18

^{32}Bytes.So the physical address is 32 bits long.

Each block is of size 32(=2

^{5}) Bytes. So block offset 5.Also given that there are 512(=2

^{9}) cache lines, since it is a direct mapped cache we need 9 bits for the LINE number.When it is directed mapped cache, the physical address can be divided as

(Tag bits + bits for block/LINE number + bits for block offset)

So, tag bits + 9 + 5 = 32

Tag bits = 32 - 14 = 18

Question 54 |

0 | |

1 | |

2 | |

3 |

Question 54 Explanation:

n=++m; (pre)

n=11

n1=m++ (post)

Here, n1=11, but m will be updated to 12.

n-- & --n1 decrements 11 to 10 & subtracting gives us zero.

Question 55 |

1 | |

2 | |

4 | |

6 |

Question 55 Explanation:

char * C = “GATECSIT2017”;

char * P = C;

(int) strlen (C + 2[P] – 6[P] – 1)

C + 2[P] - 6[P] – 1 ∵2[P] ≃ P[2]

100 + P[2] – P[6] – 1

100 + T – I – 1

100 + 84 – 73 – 1 ASCII values T – 84, I – 73

100 + 11 – 1

= 110

(int) strlen (110)

strlen (17) ≃ 2

char * P = C;

(int) strlen (C + 2[P] – 6[P] – 1)

C + 2[P] - 6[P] – 1 ∵2[P] ≃ P[2]

100 + P[2] – P[6] – 1

100 + T – I – 1

100 + 84 – 73 – 1 ASCII values T – 84, I – 73

100 + 11 – 1

= 110

(int) strlen (110)

strlen (17) ≃ 2

Question 56 |

Choose the option with words that are not synonyms.

aversion, dislike | |

luminous, radiant | |

plunder, loot | |

yielding, resistant |

Question 56 Explanation:

Yeilding means produce (or) provide (a natural, agricultural or industrial product)

Resistant means offering a resistance to something or someone.

Resistant means offering a resistance to something or someone.

Question 57 |

Saturn is __________ to be seen on a clear night with the naked eye.

enough bright | |

bright enough | |

as enough bright | |

bright as enough |

Question 57 Explanation:

In this question enough is used with bright, where bright is an adjective.

With adjectives and adverbs enough is comes after an adjectives and adverbs.

So, here we use bright enough to complete the sentence.

With adjectives and adverbs enough is comes after an adjectives and adverbs.

So, here we use bright enough to complete the sentence.

Question 58 |

There are five buildings called V, W, X, Y and Z in a row (not necessarily in that order). V is to the West of W, Z is to the East of X and the West of V, W is to the West of Y. Which is the building in the middle?

V | |

W | |

X | |

Y |

Question 58 Explanation:

V is to west of W = VW .........(i)

Z is to East of X and west of V = XZV ........(ii)

W is to west of Y = WY .........(iii)

From (i) and (ii) ⇒ VWY .........(iv)

From (ii) and (iv) ⇒ XZVWY

→While building V is in the middle.

Z is to East of X and west of V = XZV ........(ii)

W is to west of Y = WY .........(iii)

From (i) and (ii) ⇒ VWY .........(iv)

From (ii) and (iv) ⇒ XZVWY

→While building V is in the middle.

Question 59 |

A test has twenty questions worth 100 marks in total. There are two types of questions. Multiple choice questions are worth 3 marks each and essay questions are worth 11 marks each. How many multiple choice questions does the exam have?

12 | |

15 | |

18 | |

19 |

Question 59 Explanation:

No. of 3 marks questions = X say

No. of 4 marks questions = Y say

No. of questions X+Y = 20

Total no. of marks = 100

⇒ 3X+11Y = 100

Option I: If X=12; Y=8 ⇒ 3(12)+11(8) ⇒ 36+88 ≠ 100

Option II: If X=15; Y=5 ⇒ 3(15)+11(5) ⇒ 100 = 100

Option III: If X=18; Y=2 ⇒ 3(18)+11(2) ≠ 100

Option IV: If X=19; Y=1 ⇒ 3(19)+11(1) ≠ 100

No. of 4 marks questions = Y say

No. of questions X+Y = 20

Total no. of marks = 100

⇒ 3X+11Y = 100

Option I: If X=12; Y=8 ⇒ 3(12)+11(8) ⇒ 36+88 ≠ 100

Option II: If X=15; Y=5 ⇒ 3(15)+11(5) ⇒ 100 = 100

Option III: If X=18; Y=2 ⇒ 3(18)+11(2) ≠ 100

Option IV: If X=19; Y=1 ⇒ 3(19)+11(1) ≠ 100

Question 60 |

There are 3 red socks, 4 green socks and 3 blue socks. You choose 2 socks. The probability that they are of the same colour is

1⁄5 | |

7⁄30 | |

1⁄4 | |

4⁄15 |

Question 60 Explanation:

Totally there are 3 red socks, 4 green socks, 3 blue socks. In those we need to select 2 socks

The probability of selecting same colour is

⇒

⇒ 3/45 + 6/45 + 3/45

⇒ 12/45

= 4/15

The probability of selecting same colour is

⇒

^{3}C_{2}/^{10}C_{2}*^{4}C_{2}/^{10}C_{2}*^{3}C_{2}/^{10}C_{2}⇒ 3/45 + 6/45 + 3/45

⇒ 12/45

= 4/15

Question 61 |

“We lived in a culture that denied any merit to literary works, considering them important only when they are handmaidens to something seemingly more urgent – namely ideology. This was a country where all gestures, even the most private, were interpreted in political terms.”
The author’s belief that ideology is not as important as literature is revealed by the word:

‘culture’ | |

‘seemingly’ | |

‘urgent’ | |

‘political’ |

Question 61 Explanation:

Seemingly means that "according to the facts as one knows them" as far as one knows.

(or)

As to give the impression of having a certain quality.

(or)

As to give the impression of having a certain quality.

Question 62 |

There are three boxes. One contains apples, another contains oranges and the last one contains both apple and oranges. All three are known to be incorrectly labeled. If you are permitted to open just one box and then pull out and inspect only one fruit, which box would you open to determine the contents of all three boxes?

The box labeled ‘Apples’ | |

The box labeled ‘Apples and Oranges’ | |

The box labeled ‘Oranges’ | |

Cannot be determined |

Question 62 Explanation:

Correct Answer is option B i.e., the box labeled Apples an Oranges.

If we open a box labeled as apples and oranges that contains either apples (or) oranges:

Case I: If it contains apples, then the box labeled oranges can contain apples and oranges and box labeled apples can contain oranges.

Case II: If it contains oranges, then box labeled apples can contains apples and oranges and box labeled oranges can contains apples.

If we open a box labeled as apples and oranges that contains either apples (or) oranges:

Case I: If it contains apples, then the box labeled oranges can contain apples and oranges and box labeled apples can contain oranges.

Case II: If it contains oranges, then box labeled apples can contains apples and oranges and box labeled oranges can contains apples.

Question 63 |

X is a 30 digit number starting with the digit 4 followed by the digit 7. Then the number X

^{3}will have90 digits | |

91 digits | |

92 digits | |

93 digits |

Question 63 Explanation:

X is a 30 digit number starting with digit 4 and followed by the digit 7

X = 4777......... (29 times (7))

X = 4.777........ ×10

X = 5×10

X

Total no. of digits = 3+87 [125=3 digit, 10

= 90

X = 4777......... (29 times (7))

X = 4.777........ ×10

^{29}X = 5×10

^{29}(4.777 rounded as 5)X

^{3}= 125×10^{87}Total no. of digits = 3+87 [125=3 digit, 10

^{87}= 87 digit]= 90

Question 64 |

The number of roots of e

^{x}+ 0.5x^{2}– 2 in the range [-5, 5] is0 | |

1 | |

2 | |

3 |

Question 64 Explanation:

e

e

If we draw a graph for e

^{x}+0.5x^{2}-2=0e

^{x}=-0.5x^{2}+2If we draw a graph for e

^{x}and -0.5x^{2}+2 in between [-5, 5]
There are 65 questions to complete.