## Nielit Scientific Assistance IT 15-10-2017

Question 1 |

What will be the output of following?

main()

{

static int a=3;

printf("%d",a--);

if(a)

main();

}

main()

{

static int a=3;

printf("%d",a--);

if(a)

main();

}

3 | |

3 2 1 | |

3 3 3 | |

Program will fall in continuous loop and print 3 |

Question 1 Explanation:

The variable is static variable, the value is retained during multiple function calls. Initial value is 3

“A--” is post decrement so it will print “3”

if(2) condition is true and main() function will call again , Here the “a” value is 2.

“A--” is post decrement so it will print “2”

if(1) condition is true and main() function will call again Here the “a” value is 1.

“A--” is post decrement so it will print “1”

if(0) condition is false it won’t call main() function

“A--” is post decrement so it will print “3”

if(2) condition is true and main() function will call again , Here the “a” value is 2.

“A--” is post decrement so it will print “2”

if(1) condition is true and main() function will call again Here the “a” value is 1.

“A--” is post decrement so it will print “1”

if(0) condition is false it won’t call main() function

Question 2 |

The height of a binary tree is the maximum number of edges in any root to leaf path. The maximum number of nodes in a binary tree of height h is

2 ^{h} | |

2 ^{h-1} -1 | |

2 ^{ h+1} -1 | |

2 ^{h+1} |

Question 2 Explanation:

● The number of nodes n in a full binary tree, is at least n = 2 h + 1 and at most n = 2

● The number of leaf nodes l in a perfect binary tree, is l = ( n + 1 )/2 because the number of non-leaf (a.k.a. internal) nodes

● This means that a perfect binary tree with l leaves has n = 2 l − 1 nodes. ● In a balanced full binary tree, h = ⎡ log

● In a perfect full binary tree, l = 2

● The maximum possible number of null links (i.e., absent children of the nodes) in a complete binary tree of n nodes is ( n + 1 ) , where only 1 node exists in bottom-most level to the far left.

●The number of internal nodes in a complete binary tree of n nodes is ⎣ n/2⎦ .

^{h+1}− 1 , where h is the height of the tree. A tree consisting of only a root node has a height of 0.● The number of leaf nodes l in a perfect binary tree, is l = ( n + 1 )/2 because the number of non-leaf (a.k.a. internal) nodes

● This means that a perfect binary tree with l leaves has n = 2 l − 1 nodes. ● In a balanced full binary tree, h = ⎡ log

_{2 }(l)⎤ + 1 = ⎡ log_{2}((n + 1 )/2)⎤ + 1 = ⎡ log_{ 2}(n + 1 )⎤● In a perfect full binary tree, l = 2

^{h}thus n = 2^{ h+1}− 1● The maximum possible number of null links (i.e., absent children of the nodes) in a complete binary tree of n nodes is ( n + 1 ) , where only 1 node exists in bottom-most level to the far left.

●The number of internal nodes in a complete binary tree of n nodes is ⎣ n/2⎦ .

Question 3 |

Web links are stored within the page itself and when you wish to 'jump' to the page that is linked, we select the hotspot or anchor. This technique is called

Hypertext | |

hypermedia | |

Both (A) and (B) | |

Anchoring |

Question 3 Explanation:

Web links are stored within the page itself and when you wish to 'jump' to the page that is linked, we select the hotspot or anchor. This technique is called hypertext or hypermedia.

Question 4 |

If the objects focus on the problem domain, then we are concerned with

Object Oriented Analysis | |

Object Oriented Design | |

Object Oriented Analysis and design | |

None of the above |

Question 4 Explanation:

→ The purpose of any analysis activity in the software life cycle is to create a model of the system's functional requirements that is independent of implementation constraints.

→ The main difference between object-oriented analysis and other forms of analysis is that by the object-oriented approach we organize requirements around objects, which integrate both behaviors (processes) and states (data) modeled after real world objects that the system interacts with. In other or traditional analysis methodologies, the two aspects: processes and data are considered separately.

→ For example, data may be modeled by ER diagrams, and behaviors by flow charts or structure charts.

→ The main difference between object-oriented analysis and other forms of analysis is that by the object-oriented approach we organize requirements around objects, which integrate both behaviors (processes) and states (data) modeled after real world objects that the system interacts with. In other or traditional analysis methodologies, the two aspects: processes and data are considered separately.

→ For example, data may be modeled by ER diagrams, and behaviors by flow charts or structure charts.

Question 5 |

Which of the following is not a form of main memory?

Instruction cache | |

Instruction register | |

Instruction Opcode | |

Translation lookaside buffer |

Question 5 Explanation:

In computing, an opcode (abbreviated from operation code, also known as instruction syllable, instruction parcel or opstring) is the portion of a machine language instruction that specifies the
operation to be performed. Beside the opcode itself, most instructions also specify the data they will process, in the form of operands. In addition to opcodes used in the instruction set
architectures of various CPUs, which are hardware devices, they can also be used in abstract computing machines as part of their byte code specifications.

Note: Instruction opcode not in a form of main memory.

Note: Instruction opcode not in a form of main memory.

Question 6 |

which of the following is useful in traversing a given graph by breadth first search?

Stack | |

Set | |

List | |

Queue |

Question 6 Explanation:

Queue can be generally thought as horizontal in structure i.e, breadth/width can be attributed to it - BFS, whereas Stack is visualized as a vertical structure and hence has depth - DFS.

Question 7 |

M is a square matrix of order 'n' and its determinant value is 5, If all the elements of M are multiplied by 2, its determinant value becomes 40. he value of 'n' is

2 | |

3 | |

5 | |

4 |

Question 7 Explanation:

M has n rows. If all the elements of a row are multiplied by , the determinant value becomes 2*5. Multiplying all the n rows by 2, will make the determinant value 2

^{n} *5=40. Solving n= 3.Question 8 |

Is null an object?

yes | |

No | |

Sometimes yes | |

None of these |

Question 8 Explanation:

If null were an Object, it would support the methods of java.lang.Object such as equals().

However, this is not the case - any method invocation on a null results in a NullPointerException.

There is also a special null type, the type of the expression null, which has no name. Because the null type has no name, it is impossible to declare a variable of the null type or to cast to the null type. The null reference is the only possible value of an expression of null type. The null reference can always be cast to any reference type. In practice, the programmer can ignore the null type and just pretend that null is merely a special literal that can be of any reference type

However, this is not the case - any method invocation on a null results in a NullPointerException.

There is also a special null type, the type of the expression null, which has no name. Because the null type has no name, it is impossible to declare a variable of the null type or to cast to the null type. The null reference is the only possible value of an expression of null type. The null reference can always be cast to any reference type. In practice, the programmer can ignore the null type and just pretend that null is merely a special literal that can be of any reference type

Question 9 |

Which of the following scheduling algorithm could result in starvation?

Priority | |

Round Robin | |

FCFS | |

none of the above |

Question 9 Explanation:

Priority scheduling algorithm could result starvation. Starvation is nothing but indefinite blocking.

Question 10 |

Which of the following is a desirable property of module?

Independency | |

low cohesiveness | |

High coupling | |

Multifunctional |

Question 10 Explanation:

The goal of software engineering is high cohesion and low coupling. The desirable property of module is independency.

Question 11 |

Which of these events will be generated if we close an applet's window?

ActionEvent | |

ComponentEvent | |

AdjustmentEvent | |

WindowEvent |

Question 11 Explanation:

A low level event that indicates that a window has changed its status. This low-level event is generated by a Window object when it is opened, closed, activated, deactivated, iconified, or deiconified, or when focus is transferred into or out of the Window.

Question 12 |

If queue is implemented using arrays, what would be the worst run time complexity of queue and dequeue operations?

O(n), O(n) | |

O(n), O(1) | |

O(1), O(n) | |

O(1), O(1) |

Question 12 Explanation:

Everyone thought that answer in worst case time is O(n) for both the cases but using circular queue, we can implement in constant amount of time.

Question 13 |

The following program fragment prints

int i=5;

do

{

putchar(i+100);

printf("%d",i--);

}

while(i);

int i=5;

do

{

putchar(i+100);

printf("%d",i--);

}

while(i);

i5h4g3f2el | |

14h3g2f1e0 | |

An error message | |

None of the above |

Question 13 Explanation:

Here, i=5 and putchar(i+100) it means actually putchar(5+100) ⇒ putchar(105);

It will print the ASCII equivalent of 105 which is lower case ' i '. The printf statement prints the current value of i. i.e. 5 and then decrements it. So, h4 will be printed in the next pass. This continues until ' i ' becomes 0, at which point the loop gets terminated.

It will print the ASCII equivalent of 105 which is lower case ' i '. The printf statement prints the current value of i. i.e. 5 and then decrements it. So, h4 will be printed in the next pass. This continues until ' i ' becomes 0, at which point the loop gets terminated.

Question 14 |

The running time of an algorithm T(n),where 'n' is the input size, is given by

T(n)=8T(n/2)+qn, if n>1

=p, if n=1

Where p,q are constants. The order of this algorithm is

T(n)=8T(n/2)+qn, if n>1

=p, if n=1

Where p,q are constants. The order of this algorithm is

n ^{2} | |

n ^{n} | |

n ^{3} | |

N |

Question 14 Explanation:

Apply master theorem.

a=8,b=2,k=0,p=0;

So, according to masters theorem, a>b

=O(n^log

=O(n

a=8,b=2,k=0,p=0;

So, according to masters theorem, a>b

^{k}=O(n^log

_{2}^{ 8} )=O(n

^{ 3} )Question 15 |

Consider a system having 'm' resources of the same type. these resources are shared by 3 processes A,B,C; which have peak time demands of 3,4,6 respectively. The
minimum value of 'm' that ensures that deadlock will never occur is

11 | |

12 | |

13 | |

14 |

Question 15 Explanation:

A requires 3, B-4, C-6;

→ If A have 2, B have 3, C have 5 then deadlock will occur i.e., 2+3+5=10.

→ If we have one extra resource then deadlock will not occur i.e., 10+1=11.

→ If we have equal (or) more than 11 resources then deadlock will never occur.

→ If A have 2, B have 3, C have 5 then deadlock will occur i.e., 2+3+5=10.

→ If we have one extra resource then deadlock will not occur i.e., 10+1=11.

→ If we have equal (or) more than 11 resources then deadlock will never occur.

Question 16 |

For a database relation R(a,b,c,d) where the domains of a,b,c and d include only

atomic values, only the following functional dependencies and those that can be inferred

from them hold

A → c

B → d

The relation is in

atomic values, only the following functional dependencies and those that can be inferred

from them hold

A → c

B → d

The relation is in

1 normal form but not in 2 normal form | |

2 normal form but not in 3 normal form | |

3 normal form | |

None of these |

Question 16 Explanation:

Candidate key is ab.

Since all a, b, c, d are atomic. So the relation is in 1NF.

Checking the FD's

a → c

b → d

We can see that there is partial dependencies. So it is not 2NF.

Since all a, b, c, d are atomic. So the relation is in 1NF.

Checking the FD's

a → c

b → d

We can see that there is partial dependencies. So it is not 2NF.

Question 17 |

The convergence of the bisection method is

Cubic | |

Quadratic | |

Linear | |

none |

Question 17 Explanation:

The convergence of the bisection method is very slow. Although the error, in general, does not decrease monotonically, the average rate of convergence is 1/2 and so, slightly changing the definition of order of convergence, it is possible to say that the method converges linearly with rate 1/2.

Note:For the bisection you simply have that ε

Note:For the bisection you simply have that ε

_{ i+1} /ε_{i} =1/2, so, by definition the order of convergence is 1 (linearly).Question 18 |

A pipeline is having speed up factor as 10 and operating with efficiency of 80%. what will be the number of stages in the pipeline?

10 | |

8 | |

13 | |

None |

Question 18 Explanation:

Efficiency of a pipeline(E k ) = Speed Up / No of stages

No of Stages =Speed Up / E

No of Stages =Speed Up / E

_{ k} = 10/0.8 =12.5(approximately 13)Question 19 |

In a ripple counter using edge-triggered JK flip-flops, the pulse input is applied to

Clock input of all flip flops | |

J and K input of one flip flop | |

J and K input of all flip flops | |

Clock input of one flip flops |

Question 19 Explanation:

In a ripple counter using edge triggered JK flip-flops, the pulse input is applied to the clock input of one flip flops.

Question 20 |

The number of possible binary trees with 4 nodes is

12 | |

13 | |

14 | |

15 |

Question 20 Explanation:

There is one binary tree with one node. There are two differently shaped trees with two nodes. There are 14 different (shaped) binary trees with four nodes. These different trees are shown below.

Question 21 |

A program p calls two subprograms P1 and P2. P1 can fail 50% time and P2 can fail 40% times. The program P can fail

50% | |

10% | |

60% | |

70% |

Question 21 Explanation:

Program P fails when either P1 fails or P2 fails, i.e., failure of P1 + failure of P2 .

But this will also contain the case when both P1 and P2 fails at the same time, i.e., failure of P1 ∩ failure of P2 , since this case will be already be counted on ( P1+P2 ).

Therefore, our final answer will be failure of P1 + failure of P2 - (failure of P1 ∩ failure of P2)

But this will also contain the case when both P1 and P2 fails at the same time, i.e., failure of P1 ∩ failure of P2 , since this case will be already be counted on ( P1+P2 ).

Therefore, our final answer will be failure of P1 + failure of P2 - (failure of P1 ∩ failure of P2)

Question 22 |

In time division switches if each memory access takes 100 ns and one frame period is 125 us, then the maximum number of lines that can be supported is

625 lines | |

1250 lines | |

2300 lines | |

318 lines |

Question 22 Explanation:

In time division switches, 2NT = 1 frame period, where T = memory access time.

Number of lines=125*1000/100*2=625 (1us =1000ns)

Number of lines=125*1000/100*2=625 (1us =1000ns)

Question 23 |

Which two are valid constructors for Thread?

1. Thread(Runnable r, String name)

2. Thread()

3. Thread(int priority)

4.Thread(Runnable r, ThreadGroup g)

5.Thread(Runnable r, int priority)

1. Thread(Runnable r, String name)

2. Thread()

3. Thread(int priority)

4.Thread(Runnable r, ThreadGroup g)

5.Thread(Runnable r, int priority)

1 and 3 | |

2 and 4 | |

1 and 2 | |

2 and 5 |

Question 23 Explanation:

(1) and (2) are both valid constructors for Thread.

(3), (4), and (5) are not legal Thread constructors

(3), (4), and (5) are not legal Thread constructors

Question 24 |

Which of the following object types are generally autonomous, meaning that they can exhibit some behavior without being operated upon by another object.

Passive | |

Active | |

Both A) and B) | |

None of the mentioned |

Question 24 Explanation:

● An active object is one that encompasses its own thread of control.

● Active objects are generally autonomous, meaning that they exhibit some behaviour without being operated upon by another object

● Passive objects can only undergo a state change when explicitly acted upon

● Active objects are generally autonomous, meaning that they exhibit some behaviour without being operated upon by another object

● Passive objects can only undergo a state change when explicitly acted upon

Question 25 |

The address of a class B host o be split into subnets with a 3 bit subnet number.What is the maximum number of subnets and maximum number of hosts in each subnet?

8 subnets and 262141 hosts | |

6 subnets and 262141 hosts | |

6 subnets and 1022 hosts | |

8 subnets and 1024 hosts | |

None of the Above |

Question 25 Explanation:

For subnetting, bits are taken from Host-Id.

→ Class-B addressing is given

→ Class-B addressing is given

Question 26 |

Which level of RAID refers to disk mirroring with block striping?

RAID level 1 | |

RAID level 2 | |

RAID level 0 | |

RAID level 3 |

Question 26 Explanation:

● The standard RAID levels comprise a basic set of RAID (redundant array of independent disks) configurations that employ the techniques of striping, mirroring, or parity to create
large reliable data stores from multiple general-purpose computer hard disk drives (HDDs).

●RAID 0: Stripping

● RAID 1: Mirroring

● RAID 2: Hamming code for error detection

● RAID 3: Byte-level striping with a dedicated parity disk

● RAID 4: Block-level striping with block-level striping with two parity blocks parity

● RAID 5: Block-level striping with distributed parity

● RAID 6: Block-level striping with double distributed parity.

●RAID 0: Stripping

● RAID 1: Mirroring

● RAID 2: Hamming code for error detection

● RAID 3: Byte-level striping with a dedicated parity disk

● RAID 4: Block-level striping with block-level striping with two parity blocks parity

● RAID 5: Block-level striping with distributed parity

● RAID 6: Block-level striping with double distributed parity.

Question 27 |

which of the following desired features are beyond the capability of relational algebra?

Aggregate Computation | |

Multiplication | |

Finding transitive closure | |

All of the above |

Question 27 Explanation:

● Aggregate Computation is a a function to a collection of values to generate a single result.

● Multiplication means cartesian product

●Transitive closure:

Given a domain D, let binary relation R be a subset of D×D. The transitive closure R

∀ x∀y∀z((x, y ) ∈ R

● Multiplication means cartesian product

●Transitive closure:

Given a domain D, let binary relation R be a subset of D×D. The transitive closure R

^{ +}of R is the smallest subset of D×D that contains R and satisfies the following condition:∀ x∀y∀z((x, y ) ∈ R

^{+}⋀ ( y, z ) ∈ R^{ +}⇒ ( x, z ) ∈ R^{+})Question 28 |

What is 'basis of Encapsulation'?

Object | |

Class | |

Method | |

All of the above |

Question 28 Explanation:

Encapsulation is the mechanism that binds together code and data it manipulates, and keeps both safe from outside interface and misuse. Class, which contains data members and methods is used to implement Encapsulation.

Question 29 |

Consider the following declaration

int a, *b=&a, **c=&b;

a=4;

**c=5;

If the statement

b=(int *)**c

is appended to the above program fragment then

int a, *b=&a, **c=&b;

a=4;

**c=5;

If the statement

b=(int *)**c

is appended to the above program fragment then

Value of b becomes 5 | |

value of b will be the address of c | |

value of b is unaffected | |

none of these |

Question 29 Explanation:

In the given program fragment, three variables are there

Ordinary variable -a // value will be stored

Pointer variable -b // address of variable will be stored

Pointer to pointer variable(double pointer) -c // address of pointer variable will stored.

a=4 means storing/assigning value to “a”.

**c means value at(value at(c)) =value at(value at(&b)) (c holds address of pointer “b”)

=value at(&a) (b holds the address of “a”)

Memory location of variable “a”, value 5 is

stored/assigned into the variable.

Given statement is b=(int *)**c. So the value of b becomes 5.

Ordinary variable -a // value will be stored

Pointer variable -b // address of variable will be stored

Pointer to pointer variable(double pointer) -c // address of pointer variable will stored.

a=4 means storing/assigning value to “a”.

**c means value at(value at(c)) =value at(value at(&b)) (c holds address of pointer “b”)

=value at(&a) (b holds the address of “a”)

Memory location of variable “a”, value 5 is

stored/assigned into the variable.

Given statement is b=(int *)**c. So the value of b becomes 5.

Question 30 |

How many 2-input multiplexers are required to construct a 2

^{ 10} input multiplexer?1023 | |

31 | |

10 | |

127 |

Question 30 Explanation:

2

Level-1 has 2

Similarly,

Level-2 has 256 MUX.

Level-3 has 128 MUX.

Level-4 has 64 MUX.

Level-5 has 32 MUX.

Level-6 has 16 MUX.

Level-7 has 8 MUX.

Level-8 has 4 MUX.

Level-9 has 2 MUX.

Level-10 has 1 MUX.

Total number of Multiplexers=

512+256+128+64+32+16+8+4+2+1=1023

^{10}x1 MUX has 2^{10}inputs.Level-1 has 2

^{9}(=512) 2x1 multiplexers which take 2*2^{9}= 2^{10}inputs and produces 512 outputs.Similarly,

Level-2 has 256 MUX.

Level-3 has 128 MUX.

Level-4 has 64 MUX.

Level-5 has 32 MUX.

Level-6 has 16 MUX.

Level-7 has 8 MUX.

Level-8 has 4 MUX.

Level-9 has 2 MUX.

Level-10 has 1 MUX.

Total number of Multiplexers=

512+256+128+64+32+16+8+4+2+1=1023

Question 31 |

When transaction T

_{i}requests a data item currently held by T_{ j} , T_{ j} is allowed to wait only if it has a timestamp smaller than that of T_{ j} (that is, T_{ i} older than T_{ j} ). Otherwise, T_{ i} is rolled back(dies). this isWait-die | |

Wait-wound | |

Wound-Wait | |

Wait |

Question 31 Explanation:

Wait-Die method

In this method, if a transaction requests to lock a resource (data item), which is already held with a conflicting lock by another transaction, then one of the two possibilities may occur −

● If TS(T

● If TS(T

This method allows the older transaction to wait but kills the younger one.

In this method, if a transaction requests to lock a resource (data item), which is already held with a conflicting lock by another transaction, then one of the two possibilities may occur −

● If TS(T

_{i} ) < TS(T_{ j} ) − that is T_{ i} , which is requesting a conflicting lock, is older than T_{ j} − then T_{ i} is allowed to wait until the data-item is available.● If TS(T

_{i} ) > TS(t_{ j} ) − that is T_{ i} is younger than T_{ j} − then T_{i} dies. T_{i} is restarted later with a random delay but with the same timestamp.This method allows the older transaction to wait but kills the younger one.

Question 32 |

The maximum data rate of a channel of 3000 Hz bandwidth and SNR of 30 db is

60000 | |

15000 | |

30000 | |

3000 |

Question 32 Explanation:

Maximum number of bps = B log

=3000*log

=3000*4.95

=14850. So approximately, 15000

_{2}(1 + SNR).=3000*log

_{2}(1+30)=3000*4.95

=14850. So approximately, 15000

Question 33 |

The number of columns in a state table for a sequential circuit with 'm' flip flops and 'n' input is

m+n | |

m+2n | |

2m+n | |

2m+2n |

Question 33 Explanation:

Its 2m+2n because. If there are m flip-flops, there should be 2m nodes. If there are n inputs, then each node will have 2n.

Question 34 |

The solution of the recurrence relation a

_{r} =a_{ r-1} +2a_{ r-2} with a_{ 0} =2,a_{ 1} =7a _{r} =(3) ^{r} +(1)^{ r} | |

2a _{r} =(2)^{ r} /3 -(1)^{ r} | |

a _{ r} =3^{ r+1} -(-1)^{ r} | |

a _{r} =3(2)^{ r} -(-1)^{ r} |

Question 34 Explanation:

Given the recurrence relation a

For r =2, a

For r=3,a

From the above options , Substitute the r values 0,1,2,3 then option -D gives the solution to recurrence relation.

_{ r} =a_{ r-1} +2a_{ r-2}and a_{ 0} =2,a_{ 1} =7.For r =2, a

_{2}= a_{ 2-1} +2a_{ 2-2} =a_{1} +2a_{ 0} =7+2*2=7+4=11For r=3,a

_{ 3}= a_{ 3-1} +2a_{ 3-2} =a_{ 2} +2a_{ 1} =11+2*7=11+14=25From the above options , Substitute the r values 0,1,2,3 then option -D gives the solution to recurrence relation.

Question 35 |

In a 10 bit computer instruction format, the size of address field is 3 bits. Thecomputer uses expanding OPcode technique and has 4 two address instructions and 16
one address instructions. The number of zero address instructions it can support is

256 | |

356 | |

640 | |

56 |

Question 35 Explanation:

No. of possible instruction encoding =2

No. of encoding taken by two-address instructions =4×2

No. of encoding taken by one-address instructions =16×2

So, no. of possible zero-address instructions =1024−(256+128)=640

^{10} =1024No. of encoding taken by two-address instructions =4×2

^{3} ×2^{3} =256No. of encoding taken by one-address instructions =16×2

^{3} =128So, no. of possible zero-address instructions =1024−(256+128)=640

Question 36 |

The average search time for hashing with linear probing will be less if the load factor

Is far less than one | |

Equals one | |

Is far greater than one | |

None of the above |

Question 36 Explanation:

Load factor is the ratio of number of records that are currently present and the total number of records that can be present. If the load factor is less, free space will be more. This means probability of collision is less. So, the search time will be less.

Question 37 |

The question is based on the following program fragment.

f(int Y[10], int x)

{

int u,j,k;

i=0;j=9;

do

{

k=(i+j)/2;

if(Y[k]
i=k;

else

j=k;

}

while((Y[k]!=x) && (i
if(Y[k] ==x)

printf("x is in the array");

else

printf("x is not in the array");

}

On which of the following contents of 'Y' and 'x' does the program fail?

f(int Y[10], int x)

{

int u,j,k;

i=0;j=9;

do

{

k=(i+j)/2;

if(Y[k]

else

j=k;

}

while((Y[k]!=x) && (i

printf("x is in the array");

else

printf("x is not in the array");

}

On which of the following contents of 'Y' and 'x' does the program fail?

Y is [1 2 3 4 5 6 7 8 9 10] and x<10 | |

Y is [1 3 5 7 9 11 13 15 17 19] and x<1 | |

Y is [ 2 2 2 2 2 2 2 2 2 2] and x>2 | |

Y is [ 2 4 6 8 10 12 14 16 18 20] and 2 |

Question 37 Explanation:

This above code is similar to binary search

First iteration:

I=0,j=9

K=(0+9)/2=4

Elements with duplicate values of “2” the condition if(Y[k]
Second iteration :

I=0, j=k=4

K= (0+4)/2=2.

Third iteration:

I=0, j=k=2

K= (0+2)/2=1

Fourth iteration:

I=0, j=k=1

K= (0+1)/2=0

The while condition is Y[k] != x && i < j is true

● The above program doesn’t work for the cases where element to be searched is the last element of Y[ ] or greater than the last element (or maximum element) in Y[ ].

● For such cases, program goes in an infinite loop because i is assigned value as k in all iterations, and i never becomes equal to or greater than j.

● So while condition never becomes false.

First iteration:

I=0,j=9

K=(0+9)/2=4

Elements with duplicate values of “2” the condition if(Y[k]

I=0, j=k=4

K= (0+4)/2=2.

Third iteration:

I=0, j=k=2

K= (0+2)/2=1

Fourth iteration:

I=0, j=k=1

K= (0+1)/2=0

The while condition is Y[k] != x && i < j is true

● The above program doesn’t work for the cases where element to be searched is the last element of Y[ ] or greater than the last element (or maximum element) in Y[ ].

● For such cases, program goes in an infinite loop because i is assigned value as k in all iterations, and i never becomes equal to or greater than j.

● So while condition never becomes false.

Question 38 |

An algorithm is made up of two modules M1 and M2. If order of M1 is f(n) and M2 is g(n) then the order of algorithm is

max(f(n),g(n)) | |

min(f(n),g(n)) | |

f(n)+g(n) | |

f(n) * g(n) |

Question 38 Explanation:

From the given statement, algorithm consists of two modules M1 and M2.

f(n) is order of M1

g(n) is order of M2.

In order to find the order of the algorithm, there are three possible cases with f(n) and g(n)

Case-1 : if f(n) > g(n)

In this case we take O(f(n)) the complexity of the algorithm as g(n) is a lower order term, we can ignore this one .

Case-2 : f(n) < g(n)

In this case we take O(g(n)) the complexity of the algorithm as f(n) is a lower order term, we can ignore this one.

Case-3: f(n) = g(n)

Time Complexity can be either O(g(n)) or O(f(n)) (which is equal asymptotically). So the order of the algorithm is max(f(n), g(n)).

f(n) is order of M1

g(n) is order of M2.

In order to find the order of the algorithm, there are three possible cases with f(n) and g(n)

Case-1 : if f(n) > g(n)

In this case we take O(f(n)) the complexity of the algorithm as g(n) is a lower order term, we can ignore this one .

Case-2 : f(n) < g(n)

In this case we take O(g(n)) the complexity of the algorithm as f(n) is a lower order term, we can ignore this one.

Case-3: f(n) = g(n)

Time Complexity can be either O(g(n)) or O(f(n)) (which is equal asymptotically). So the order of the algorithm is max(f(n), g(n)).

Question 39 |

By open domain CASE tools we mean

tools available in open domain | |

Software packages which can be downloaded from the internet | |

Software packages to aid each phase of the systems analysis and design which an be downloaded free of cost from the internet | |

Source codes of CASE tools |

Question 39 Explanation:

● Tools are also in the open domain which can be downloaded and used.

● They do not usually have very good user interfaces.

● They do not usually have very good user interfaces.

Question 40 |

Bit stuffing refers to

Inserting a '0' in user data stream to differentiate it with a flag | |

Inserting a '0' in flag stream to avoid ambiguity | |

Appending a nibble to the flag sequence | |

Appending a nibble to the user data stream |

Question 40 Explanation:

●The term "bit stuffing" broadly refers to a technique whereby extra bits are added to a data stream, which do not themselves carry any information, but either assist in management of the communications or deal with other issues.

●The receiver knows how to detect and remove or disregard the stuffed bits.

●The receiver knows how to detect and remove or disregard the stuffed bits.

Question 41 |

Assume transaction A holds a shared lock R. If transaction B also requests for a shared lock on R. It will

result in deadlock situation | |

immediately be granted | |

immediately be rejected | |

be granted as soon as it is released by A |

Question 41 Explanation:

● Shared locks exist when two transactions are granted read access.

● One transaction gets the shared lock on data and when the second transaction requests the same data it is also given a shared lock.

● Both transactions are in a read-only mode, updating the data is not allowed until the shared lock is released.

● There is no conflict with the shared lock because nothing is being updated.

● Shared locks last as long as they need to last; it depends on the level of the transaction that holds the lock.

● One transaction gets the shared lock on data and when the second transaction requests the same data it is also given a shared lock.

● Both transactions are in a read-only mode, updating the data is not allowed until the shared lock is released.

● There is no conflict with the shared lock because nothing is being updated.

● Shared locks last as long as they need to last; it depends on the level of the transaction that holds the lock.

Question 42 |

Given relations R(w,x) and S(y,z) the result of SELECT DISTINCT w,x from R,S

R has no duplicates and S is non empty | |

R and S have no duplicates | |

S has no duplicates and R is non empty | |

R and S has the same number of tuples |

Question 42 Explanation:

r has no duplicate, if r can have duplicates it can be remove in the final state. s in non-empty if s is empty then r*s becomes empty.

Question 43 |

A decimal has 25 digits. the number of bits needed for its equivalent binary representation is approximately

50 | |

74 | |

40 | |

60 | |

None of these |

Question 43 Explanation:

Consider three digits(1,2,3) of decimal numbers.Maximum number, we can generate by that three digits are 10

Then, Decimal number has 25 digits, so maximum number is 10

Similarly, in the binary representation with “n” bits the maximum number is 2

So we can write 10

log

n log

n=25*(3.322) [ log

n=83

^{ 3} -1 which is 999.Then, Decimal number has 25 digits, so maximum number is 10

^{ 25} -1Similarly, in the binary representation with “n” bits the maximum number is 2

^{n} -1So we can write 10

^{ 25} –1 = 2^{ n} – 1 --->10^{25} = 2^{ n}After taking log^{2} on both sideslog

_{2} 2^{ n} =log_{ 2} 10^{ 25}n log

_{2} 2=25 log_{ 2} 10n=25*(3.322) [ log

_{2}2=1 & log_{ 2} 10 =3.322]n=83

Question 44 |

Match list I with list II and select the correct answer using the codes given below the lists.

1 2 3 4 | |

3 2 4 1 | |

2 3 1 4 | |

1 4 2 3 |

Question 44 Explanation:

● A stack-organized computer does not use an address field for the instructions ADD and MUL. The PUSH and POP instructions, however, need an address field to specify the operand that communicates with the stack.

● One-address instructions use an implied accumulator (AC) register for all data manipulation.

● Two address instructions are the most common in commercial computers. Here each address field can specify either a processor register or a memory word.

● Computers with three-address instruction formats can use each address field to specify either a processor register or a memory operand.

● One-address instructions use an implied accumulator (AC) register for all data manipulation.

● Two address instructions are the most common in commercial computers. Here each address field can specify either a processor register or a memory word.

● Computers with three-address instruction formats can use each address field to specify either a processor register or a memory operand.

Question 45 |

Which of the definitions generates the same languages as L, where

L={ x

i. E→ xEy |xy

ii.xy|x

iii. x

L={ x

^{n} y^{n} , n>=1 }i. E→ xEy |xy

ii.xy|x

^{+} x yy^{ +}iii. x

^{+} y^{+}i only | |

i and ii only | |

ii and iii only | |

ii only |

Question 45 Explanation:

The language L={ x

For n=1, xy

n=2,x

n=3,x

In the above language , there are equal number of x’s followed by y’s

Definition-1:E→ xEy |xy → xEy → xxEyy → xxxEyyy → xxxxyyyy

Which is nothing but equal number of x’s followed by equal number of y’s

X

Y

Definition -2 :

The expression xy|x

The definition-2 won’t give equal number of x’s followed by y’s

It generates any number of x’s followed by any number of y’s.

Definition -3 :

The expression x

The definition-3 won’t give equal number of x’s followed by y’s

It generates any number of x’s followed by any number of y’s.

^{n} y^{ n} , n>=1 }For n=1, xy

n=2,x

^{2}y^{ 2}n=3,x

^{3} y^{ 3}In the above language , there are equal number of x’s followed by y’s

Definition-1:E→ xEy |xy → xEy → xxEyy → xxxEyyy → xxxxyyyy

Which is nothing but equal number of x’s followed by equal number of y’s

X

^{+} means one or more x’sY

^{+} means one or more y’sDefinition -2 :

The expression xy|x

^{+} x yy^{ +} gives xy, xxyy , xxxxyy , xxyyyyy and so onThe definition-2 won’t give equal number of x’s followed by y’s

It generates any number of x’s followed by any number of y’s.

Definition -3 :

The expression x

^{ +} y^{ +} gives xy, xxy,xxxy,xyy,xyyy and so on strings.The definition-3 won’t give equal number of x’s followed by y’s

It generates any number of x’s followed by any number of y’s.

Question 46 |

The address sequence generated by tracing a particular program executing in a pure demand paging system with 100 records per page, with 1 free main memory fram is recorded as follows. What is the number of page faults?
0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0370

15,4 | |

6,4 | |

7,2 | |

4,6 |

Question 46 Explanation:

→ Page are fitted in frames, so first we need to determine the pages but given request are just the record request in decimal. We can assume that first page to address from 0000 to 0099 and page 2 contains records from 0100 to 0199 and so on (it is given in question that each page contains 100 records) and so on. So page request string is

01, 02, 04, 04, 05, 05, 05, 01, 02, 02, 02, 03, 03.

Clearly 7 page faults.

01, 02, 04, 04, 05, 05, 05, 01, 02, 02, 02, 03, 03.

Clearly 7 page faults.

Question 47 |

If the channel is band limited to 6 kHz and signal to noise ratio is 16, what would be the capacity of channel?

16.15kbps | |

23.24 kbps | |

40.12 kbps | |

24.74 kbps |

Question 47 Explanation:

Shannon Capacity (Noisy Channel):

In presence of Gaussian band-limited white noise, Shannon Hartley theorem gives the

maximum data rate capacity C = B log

=6log

=6*log

=6*4.08=24.48

In presence of Gaussian band-limited white noise, Shannon Hartley theorem gives the

maximum data rate capacity C = B log

_{2}(1+ S/N)=6log

_{2} (1+16)=6*log

_{2} (17)=6*4.08=24.48

Question 48 |

An ideal op-amp is an ideal

Voltage controlled current source | |

voltage controlled voltage source | |

current controlled current source | |

current controlled voltage source |

Question 48 Explanation:

An ideal op-amp lead us to an op-amp with input resistance is infinite, so no current flows into either input terminal (the “current rule”) and that the differential input offset voltage is zero (the “voltage rule”).

Ideal operational amplification will have

➝ High input impedance ( R

➝ R

➝ Infinite voltage gain

➝ So, it is a voltage controlled voltage source device.

Ideal operational amplification will have

➝ High input impedance ( R

_{in}= ∞ ) and low output impedance ( R_{out}= 0 )➝ R

_{in}= ∞ , so zero input current➝ Infinite voltage gain

➝ So, it is a voltage controlled voltage source device.

Question 49 |

A carrier A

_{c} Cos(w_{ C} )t is frequency modulated by a signal E_{m} Cos(w_{m} )t. The modulation index is m_{f} . the expression for the resulting FM signal isA _{c}, Cos[w_{ c} t+ m _{f} Sin(w _{m} )t] | |

A _{c} Cos[w_{ c} t+ m_{ f} Cos(w _{m} )t] | |

A _{c} Cos[w _{c} t+ π m_{ f} Sin w _{m} t] | |

A _{c} Cos[w_{ c} t+ 2 π m_{ f} E_{ m} Cos(w_{ m} )t/w_{ m} ] |

Question 49 Explanation:

Question 50 |

A sequential circuit using D flip flop and logic gates is shown in figure, Where X and Y are the inputs and Z is the output. The circuit is

S-R flip flop with inputs X=R and Y=S | |

S-R flip flop with inputs X=S and Y=R | |

J-K flip flop with inputs X=J and Y=K | |

J-K flip flop with X=k and Y=J |

Question 50 Explanation:

Here, PI=Present Input

PS=Present state

NS=Next State

Q

D=X’Z+YZ

PS=Present state

NS=Next State

Q

_{n+1} =D=f(PI,PS)=f(x,y,Q_{ n} )D=X’Z+YZ

Question 51 |

A 4 bit ripple counter and a 4 bit synchronous counter are made using flip flops having a propagation delay of 10ns each. If the worst case delay in the ripple counter and the synchronous counter be R and S respectively, then

R=10 ns, S=40ns | |

R=40ns, S=10ns | |

R=10ns, S=30ns | |

R=30 ns, S=10ns |

Question 51 Explanation:

● A ripple counter is an asynchronous counter where only the first flip-flop is clocked by an external clock. All subsequent flip-flops are clocked by the output of the preceding flip-flop.

● Synchronous Counters are so called because the clock input of all the individual flip-flops within the counter are all clocked together at the same time by the same clock signal.

● In the ripple counter, each flip flop will depending upon the precede flip flop and propagation delay is 10ns.So the propagation delay of 4-bit ripple counter is 4*10=40ns.

● In the synchronous counter,one clock input is enough so the propagation delay is 10ns.

● Synchronous Counters are so called because the clock input of all the individual flip-flops within the counter are all clocked together at the same time by the same clock signal.

● In the ripple counter, each flip flop will depending upon the precede flip flop and propagation delay is 10ns.So the propagation delay of 4-bit ripple counter is 4*10=40ns.

● In the synchronous counter,one clock input is enough so the propagation delay is 10ns.

Question 52 |

In MOSFET fabrication, the channel length is delayed during the process of

Isolation oxide growth | |

Channel stop implantation | |

Ply-Silicon gate patterning | |

Lithography step leading to the contact pad |

Question 52 Explanation:

In MOSFET fabrication channel length is defined during Poly-Silicon gate patterning process

Question 53 |

The open loop transfer function of a feedback control system is G(s).H(s)=1/(S+1)

^{3} . The gain margin of the system is2 | |

4 | |

8 | |

16 |

Question 53 Explanation:

The open loop transfer function of a feedback control system is G(s).H(s)=1/(S+1)

Gain margin=?

→ Here, the formula is -3tan

= -60

W

G(s).H(s)=1/(S+1)

|G(s).H(s)|=X= 1/(1+W

=1⁄8

G(s).H(s)=1/X

= 8

^{3}Gain margin=?

→ Here, the formula is -3tan

^{ -1} (W^{ pc} ) = -180^{0}tan^{-1} (W^{pc} ) = -180^{ 0} /3= -60

^{0}W

^{pc} =√3 rad/secG(s).H(s)=1/(S+1)

^{ 3}|G(s).H(s)|=X= 1/(1+W

_{pc}^{2} )=1⁄8

G(s).H(s)=1/X

= 8

Question 54 |

The final value theorem is used to find the

Steady state value of the system output | |

Initial value of the system output | |

Transient behaviour of the system output | |

None of these |

Question 54 Explanation:

The final value theorem is used to find the steady state value of the system output

Question 55 |

For the discrete signal x[n]=a

^{n} u[n], a>0 the z-transform is(z+a)/z | |

(z-z)/z | |

z/(z-a) | |

z/(z+a) |

Question 55 Explanation:

Let x[n] be causal signal given by x[n] = a

→ The Z-Transform of x[n] is given by

Region of Convergence for a discrete time signal x[n] is dened as a continuous region in z plane where the Z-Transform converges. In order to determine RoC, it is convenient to represent the Z-Transform as a is rational X(z)=P(z)/Q(z)

1. The roots of the equation P(z) = 0 correspond to the ’zeros’ of X(z)

2. The roots of the equation Q(z) = 0 correspond to the ’poles’ of X(z)

3. The RoC of the Z-transform depends on the convergence of the polynomials P(z) and Q(z).

^{n} u[n]→ The Z-Transform of x[n] is given by

__Region of Convergence(RoC)__Region of Convergence for a discrete time signal x[n] is dened as a continuous region in z plane where the Z-Transform converges. In order to determine RoC, it is convenient to represent the Z-Transform as a is rational X(z)=P(z)/Q(z)

1. The roots of the equation P(z) = 0 correspond to the ’zeros’ of X(z)

2. The roots of the equation Q(z) = 0 correspond to the ’poles’ of X(z)

3. The RoC of the Z-transform depends on the convergence of the polynomials P(z) and Q(z).

Question 56 |

For a periodic signal v(t)=30 sin 100t+10cos 300t +6 sin (500t + π /4), the fundamental frequency in rad/s

100 | |

300 | |

500 | |

None of the above |

Question 56 Explanation:

Given, the signal

V (t) = 3 0 sin 100t + 10 cos 300t + 6 sin (500t + π/4)

So, we have

ω

ω

ω

∴ The respective time periods are

T

T

T

So, the fundamental time period of the signal is

LCM (T

as T

∴ The fundamental frequency, ω

V (t) = 3 0 sin 100t + 10 cos 300t + 6 sin (500t + π/4)

So, we have

ω

_{1}= 100 radsω

_{2}= 300 radsω

_{3}= 500 rads∴ The respective time periods are

T

_{1}= 2π/ω_{1}= 2π/100 secT

_{2}= 2π/ω_{2}= 2π/300 secT

_{3}= 2π/ω_{3}= 2π/500 secSo, the fundamental time period of the signal is

LCM (T

_{1}, T_{ 2}, T_{ 3}) =LCM (2π,2π,2π)/HCF (100,300,500)as T

_{0}=2π/100∴ The fundamental frequency, ω

_{0}=2π/T_{ 0}= 100 rad/sQuestion 57 |

If the number of bits per sample in a PCM system is increased from a n to n+1, the improvement in signal to quantization noise ratio will be

3 dB | |

6 db | |

2n dB | |

n dB |

Question 57 Explanation:

(S/N

= (1.76 + 6 n)

(SQNR)

(SQNR)

(SQNR)

So for every one bit increase in bits per sample will result is 6 dB improvement in signal to quantization ratio.

_{q})_{dB}= (1.76 + 6 n)

_{dB}(SQNR)

_{ 1}= 1 .76 + 6 n(SQNR)

_{ 2}= 1 .76 + 6 n(n + 1 ) = 1 .76 + 6 n + 6(SQNR)

_{2}− (SQNR)_{1}= 1 .76 + 6 n + 6 − 1 .76 − 6 n = 6 dBSo for every one bit increase in bits per sample will result is 6 dB improvement in signal to quantization ratio.

Question 58 |

At 100% modulation, the power in each sideband is__ of that of carrier.

50% | |

40% | |

60% | |

25% |

Question 58 Explanation:

Power in sidebands may be calculated as modulation index.

→ Each sideband is 25%. Total modulation index=1.

→ Each sideband is 25%. Total modulation index=1.

Question 59 |

Twelve 1 Ω resistances are used as edges to form a cube. The resistance between two diagonally opposite corners of the cube is

5/6 Ω | |

1/6 Ω | |

6/5 Ω | |

3/2 Ω |

Question 59 Explanation:

Apply Ohms law, the resistance is equal to the voltage divided by the current. The resulting equivalent resistance is 5/6 Ω.

R=E/I

R=(2*(1/2)V)/3A

= ⅚

Note: This is the cube structure consisting of 12 resistors electrically connected between the 8 vertices. Each resistor is 1 Ω, but any value can be used so long as they are all the same.

R=E/I

R=(2*(1/2)V)/3A

= ⅚

Note: This is the cube structure consisting of 12 resistors electrically connected between the 8 vertices. Each resistor is 1 Ω, but any value can be used so long as they are all the same.

Question 60 |

A solution for the differential equation x'(t) + 2x(t) = δ (t) with initial condition x(~0)=0

e _{-2t} u(t) | |

e _{2t} u(t) | |

e _{-t} u(t) | |

e _{t} u(t) |

There are 60 questions to complete.

## Scientific Assistance CS 15-10-2017

Question 1 |

The difference between the compound interest and the simple earned at the end of 3 rd year on a sum of money at a rate of 10% per annul is Rs. 77.5. What is the sum?

Rs. 3,500 | |

Rs. 2,500 | |

Rs. 3,000 | |

Rs. 2,000 |

Question 1 Explanation:

Let principal be ‘x’

S.I = P T R / 100

S.I = X * 3 * 10 / 100

C.I = [ P (1 + (R / 100)

C.I = [ X (1+ (10 / 100)

C.I = 331X / 1000

Difference between Compound interest and Simple interest is 77.5

( 331X / 1000 ) - ( 3X / 10) = 77.5

31x / 1000 = 77.5

x = 2500.

S.I = P T R / 100

S.I = X * 3 * 10 / 100

C.I = [ P (1 + (R / 100)

^{T}) - P]C.I = [ X (1+ (10 / 100)

^{3}) - X]C.I = 331X / 1000

Difference between Compound interest and Simple interest is 77.5

( 331X / 1000 ) - ( 3X / 10) = 77.5

31x / 1000 = 77.5

x = 2500.

Question 2 |

Aamir and Birju can cut 5000g of wood in 20 min. Birju and Charles can cut 5000g of wood in 40 min. Charles and Aamir cut 5 kg of wood in 30 min. How much time Charles will take to cut 5 kg wood alone?

120 min | |

48 min | |

240 min | |

120 min |

Question 2 Explanation:

Here in this question 5000g means 5kg only

So from that question Aamir (A) Birju (B) Charles (C)

Let Aamir, Birju and Charles alone can complete the work in A, B and C minutes respectively, then

1 min. work of (A+B)=1/20, (B+C)=1/40 and (C+A)=1/30

A + B + C = 13/240

1 min. work of C = (A+B+C) - (A+B) = 13/240 - 1/20 = 1/240

So C alone can complete the work in 240 min.

So from that question Aamir (A) Birju (B) Charles (C)

Let Aamir, Birju and Charles alone can complete the work in A, B and C minutes respectively, then

1 min. work of (A+B)=1/20, (B+C)=1/40 and (C+A)=1/30

A + B + C = 13/240

1 min. work of C = (A+B+C) - (A+B) = 13/240 - 1/20 = 1/240

So C alone can complete the work in 240 min.

Question 3 |

An alloy contains copper and zinc in the ratio 5 : 3 and another contains copper and tin in the ratio 8 : 5, If equal weights of the two are melted together to form a 3 rd alloy, find the weight of tin per kg in the new alloy.

40/129 | |

5/13 | |

5/26 | |

28/5 |

Question 3 Explanation:

Amount of copper in first alloy is 5x/8 and zinc is 3x/8.

Amount of copper in second alloy is 8x/13 and tin is 5x/13.

So after mixing tin is available in 2 alloy only that means in 3rd alloy 5x/13/2x

Tin present in 3rd alloy is 5/26.

Amount of copper in second alloy is 8x/13 and tin is 5x/13.

So after mixing tin is available in 2 alloy only that means in 3rd alloy 5x/13/2x

Tin present in 3rd alloy is 5/26.

Question 4 |

X is a whole number. If the only common factors of X and X2 are 1 and X, then X is _______

1 | |

a perfect square | |

an odd number | |

a prime number |

Question 4 Explanation:

A number that is divisible only by itself and 1 is called prime number.

Question 5 |

Line AB is 24 metres in length and tangent to the inner one of the two concentric circles at point C, Points A and B lie on the circumference of the outer circle. It is known that the radii of the two circles are integers. The radius of the outer circle is

13m | |

5 m | |

7 m | |

4m |

Question 5 Explanation:

Given data,

AB=24cm

CB=12cm

θC=5cm(Assume)

θB=5

^{2} +12

^{ 2} =169

θB=13cm

Note: They forgot to mention the value 5. If we are considering value=5 then answer should be A.

Question 6 |

Monisha is working with a real estate agent to find a location for the kids’ toy store she plans to open in her town. She is looking for a place that is either in the centre or not too far from the centre of town. It should also be attractive for the right kind of footfall too. Which of the following locations should Monisha’s agent call to her attention?

a storefront in a new high-rise building near the train station in the center of town whose occupants are mainly young, childless professionals who use the train to commute to their offices each day | |

a little shop three blocks away from the town’s main street, located across the street from an elementary school and next door to an ice cream store | |

a stand-alone storefront on a quiet residential street ten blocks away from the town’s center | |

a storefront in a small a strip mall located on the outskirts of town that is also occupied by a pharmacy and a dry cleaner |

Question 6 Explanation:

This option is both near the center of town and in a location where children and their parents are sure to be around. This is the only option that meets both of Monisha’s requirements.

Question 7 |

Reading is a psycholinguistic guessing game. To read critically is a skill as it is a demanding process. One must slow down one’s reading and, with a pencil in hand, perform specific operations on the text. Mark up the text with reactions, conclusions and questions. When one reads, one becomes an active participant.

critical reading is a slow, dull, but essential process. | |

the best critical reading happens at critical times in a person’s life. | |

readers should get in the habit of questioning the truth of what they read. | |

critical reading requires thoughtful and careful attention |

Question 7 Explanation:

Option-A FALSE: The author never says that reading is dull.

Option-B,C FALSE: Not related to given paragraph.

Option-D TRUE: The author stresses the need to read critically by performing thoughtful and careful operations on the text.

Option-B,C FALSE: Not related to given paragraph.

Option-D TRUE: The author stresses the need to read critically by performing thoughtful and careful operations on the text.

Question 8 |

Find the missing term:

60, 40, 55, 45, 50, 50,?

60, 40, 55, 45, 50, 50,?

45 | |

50 | |

55 | |

60 |

Question 8 Explanation:

Question 9 |

Find the missing alphabet:

T, r, O, m, J,?

T, r, O, m, J,?

h | |

i | |

I | |

g |

Question 9 Explanation:

In between r and T there is letter is s

Likewise m and O letter is n

Same h is the next letter to fit in that sequence

Likewise m and O letter is n

Same h is the next letter to fit in that sequence

Question 10 |

Here are some words translated from an artificial language.

Qmelaqali means fruitcake

Qalitiimmeo means cakewalk

Useguamao means buttercup

Which word could mean “cupcake”?

Qmelaqali means fruitcake

Qalitiimmeo means cakewalk

Useguamao means buttercup

Which word could mean “cupcake”?

qalitiiqali | |

amaotiimmeo | |

pakitreft | |

amaoqali |

Question 10 Explanation:

Qmelaqali→ fruitcake

We can divide artificial language into parts.

Qmela→ fruit

qali→ cake

Qalitiimmeo→ cakewalk

We can divide artificial language into parts.

Qali→ cake

tiimmeo→ walk

Useguamao → buttercup

We can divide artificial language into parts.

Usegu→ butter

amao→ cup

cupcake→ ?

amao→ cup

qali→ cake

So, option-D is correct answer.

We can divide artificial language into parts.

Qmela→ fruit

qali→ cake

Qalitiimmeo→ cakewalk

We can divide artificial language into parts.

Qali→ cake

tiimmeo→ walk

Useguamao → buttercup

We can divide artificial language into parts.

Usegu→ butter

amao→ cup

cupcake→ ?

amao→ cup

qali→ cake

So, option-D is correct answer.

Question 11 |

A man walks 5 Km towards south and then turns to the right. After walking 3 Km he turns to the left and walks 5 Km. Now in which direction is he from the starting place?

West | |

South | |

North-East | |

South- West |

Question 11 Explanation:

Question 12 |

If the consonants in the word ‘DROVE’ are first arranged alphabetically and the vowels are put in between two pairs of consonants in the alphabetical order, which of the following will be the fourth from the right end after the rearrangement?

D | |

E | |

R | |

O |

Question 12 Explanation:

In the word DROVE we have two vowels

Those are O and E

After consonants are arranged in alphabetically we got the word like DRV

But we have two place vowels in between two pairs of consonants we got finally DEROV

So right from fourth letter is E

Those are O and E

After consonants are arranged in alphabetically we got the word like DRV

But we have two place vowels in between two pairs of consonants we got finally DEROV

So right from fourth letter is E

Question 13 |

There is a queue in a ticketing office. Amanda is 10 th from the front while Murthy is 25 th from behind and Marta is just in the middle of the two. If there be 50 persons in the queue. What position does Marta occupy from the front?

16 | |

18 | |

15 | |

17 |

Question 13 Explanation:

Question 14 |

There are five janitors. Pali, Qureshi, Rohan, Sant and Timber. They all have a different height, Qureshi is shorter than only Timber and Sant is shorter than Pali and Qureshi. Who among them is the shortest?

Rohan | |

Sant | |

Pali | |

Data inadequate |

Question 14 Explanation:

From that question we can decide Pali Quereshi sant and timber but Rohan is not mentioned there. So Data inadequate is answer.

Question 15 |

A $ B means A is the father of B; A # B means A is the sister of B; A * B means A is the daughter of B and A @ B means A is the brother of B. Which of the following indicates that M is the wife of Q?

Q $ R # T @ M | |

Q $ R @ T # M | |

Q $ R * T # M | |

Q $ R @ T * M |

Question 15 Explanation:

Q $ R means R’s father is Q

R @ T means T’s brother is R

T * M means M’s Daughter T

Finally M’s Daughter is T and T’s Brother is R and R’s father is Q so Q’s wife is M.

R @ T means T’s brother is R

T * M means M’s Daughter T

Finally M’s Daughter is T and T’s Brother is R and R’s father is Q so Q’s wife is M.

Question 16 |

Count the number of squares in the given figure.

32 | |

30 | |

29 | |

28 |

Question 16 Explanation:

For this type of questions there is one formula that is

[ n (n+1)(2n+1) ] / 6

Here n is 4 because it’s 4 x 4

[ 4 (4+1)(8+1) ] / 6

30

[ n (n+1)(2n+1) ] / 6

Here n is 4 because it’s 4 x 4

[ 4 (4+1)(8+1) ] / 6

30

Question 17 |

Who was the Viceroy of India, when Quit India Resolution was passed in 1942?

Lord Linlithgow | |

Lord Wavell | |

Lord Willingdon | |

Lord Mountbatten |

Question 18 |

When was the East India Association set up?

1866 | |

1857 | |

1836 | |

1885 |

Question 19 |

Who was the Spanish navigator who set out to discover India, but instead landed on the soil of America?

Christopher Columbus | |

Vasco Da Gama | |

James Cook | |

None of above |

Question 20 |

Which dynasty was ruling over north India when Alexander the great invaded India?

Gupta Dynasty | |

Maurya Dynasty | |

Sakya Dynasty | |

Nanda Dynasty |

Question 21 |

The roads of cities in the Indus Valley Civilization generally divided the city into

Rectangular Blocks | |

Circular Blocks | |

Triangular Blocks | |

None of above |

Question 22 |

In which year was Pulitzer Prize established?

1917 | |

1918 | |

1922 | |

1928 |

Question 23 |

‘Kanchipuram’ is in which of the following states?

Andhra Pradesh | |

Orissa | |

Kerala | |

Tamil Nadu |

Question 24 |

Which of the following is not a chief organ of the United Nations Organizations?

International Labour Organisation | |

Security Council | |

International Court of Justice | |

General Assembly |

Question 25 |

Which of the following is not a member of G-15?

Indonesia | |

Malaysia | |

Columbia | |

India |

Question 26 |

The group of metals Co, Ni, Fe may best called as

Transition metals | |

Main group metals | |

Alkali metals | |

Rare metals |

Question 27 |

Non stick cooking utensils are coated with

Black paint | |

PVC | |

Teflon | |

Polystyrene |

Question 28 |

The international township built near Pondicherry in India in coloration with UNEESCO is called

Elbaville | |

Auroville | |

Gayaville | |

Broadway |

Question 29 |

Irvin sold a book at a profit of 12%. If Irvin had sold it for Rs. 18 more, then 18% would have been gained. Find the cost price.

Rs. 600 | |

Rs. 300 | |

Rs. 400 | |

Rs. 200 |

Question 29 Explanation:

Explanation: Let CP = 100%

Sold on 12% profit means sold on 112%

now 18% would have been gained means 118% if Rs 18 was more it means 118% - 112% = 18 Rs

so 6%=18Rs

so 100% = (18/6)×100 = 3× 100 = 300 Rs

CP = 300

Sold on 12% profit means sold on 112%

now 18% would have been gained means 118% if Rs 18 was more it means 118% - 112% = 18 Rs

so 6%=18Rs

so 100% = (18/6)×100 = 3× 100 = 300 Rs

CP = 300

Question 30 |

In a group of 7 people, the average age is found to be 17 years. Two more people joined with an average age 19 years. One person left the group whose age was 25 years. What is the new average age of the group?

17.5 Years | |

16.5 Years | |

18 Years | |

16 Years |

Question 30 Explanation:

7 people, the average age is found to be 17 years means 17 * 7 = 119

Two more joined with an average age 19 years means 19 * 2 = 38 So total is 157

But one person left from that group whose age is 25 years 157 - 25 = 132

Remaining average of 8 people is 132 / 8

16.5 Year

Two more joined with an average age 19 years means 19 * 2 = 38 So total is 157

But one person left from that group whose age is 25 years 157 - 25 = 132

Remaining average of 8 people is 132 / 8

16.5 Year

Question 31 |

A 300 meter long metro train crosses a platform in a metro station in 39 seconds while it crosses a lamp post in 18 seconds. What is the length of the platform?

250 meter | |

350 meter | |

520 meter | |

300 meter |

Question 31 Explanation:

Speed = Distance/time = 300/18 = 50/3 m/sec

Let the length of the platform be x meters

Distance = Speed * Time

x+300 = (50/3) * 39

x = 350

Let the length of the platform be x meters

Distance = Speed * Time

x+300 = (50/3) * 39

x = 350

Question 32 |

Assume that a sum of money is divided equally among n girls. Each girl will receive $ 60. If another girl is added to the group and the sum is divided equally among all the girls, each child girl a $ 50 share. What is the sum of money?

$ 3000 | |

$ 300 | |

$ 110 | |

$ 10 |

Question 32 Explanation:

From that question 60 * n = 50 (n + 1)

n = 5

so sum of money is 60 * 5 = 300

n = 5

so sum of money is 60 * 5 = 300

Question 33 |

A tank can be filled by one tap in 10 minutes and by another in 30 minutes. Both the taps are kept open for 5 minutes and then the first one is shut off. In how many minutes more is the tank completely filled?

5 | |

7.5 | |

10 | |

12 |

Question 33 Explanation:

A tank can be filled in 10 mins

another one is in 30 mins and those taps are opened for 5 mins

so in one min one tap is 1/10 another one is 1/30

both can tank open for 5 mins so (1/10 + 1/30 ) * 5

= (2/15) * 5

= 10/15

= 2⁄3

Remaining work is 1- (2⁄3)

= 1⁄3

The left work has to be done by first Tap because first is shut off.

Work will be completed by Second Tap in= (1⁄3) * 30 = 10

another one is in 30 mins and those taps are opened for 5 mins

so in one min one tap is 1/10 another one is 1/30

both can tank open for 5 mins so (1/10 + 1/30 ) * 5

= (2/15) * 5

= 10/15

= 2⁄3

Remaining work is 1- (2⁄3)

= 1⁄3

The left work has to be done by first Tap because first is shut off.

Work will be completed by Second Tap in= (1⁄3) * 30 = 10

Question 34 |

From the four choices provided, choose the analogy that is most similar to the one in the question.

Wealth : Poverty

Wealth : Poverty

part: whole | |

good: excellent | |

prodigal : chary | |

wicked : sinful |

Question 34 Explanation:

Wealth is opposite to Poverty and Prodigal opposite word is chary.

Wealth: an abundance of valuable possessions or money

Poverty: The state of being extremely poor.

Prodigal: spending money or using resources freely and recklessly

chary: Cautiously or suspiciously reluctant to do something.

Wealth: an abundance of valuable possessions or money

Poverty: The state of being extremely poor.

Prodigal: spending money or using resources freely and recklessly

chary: Cautiously or suspiciously reluctant to do something.

Question 35 |

From the four choices provided, choose the analogy that is most similar to the one in the question.

Misfortune : Catastrophe

Misfortune : Catastrophe

miniature : big | |

limited : infinite | |

knowledge : learning | |

generosity : parsimony |

Question 35 Explanation:

Misfortune is related to Catastrophe and knowledge is related to learning.

Misfortune: An unfortunate condition or event.

Catastrophe: An event causing great and usually sudden damage or suffering.

Misfortune: An unfortunate condition or event.

Catastrophe: An event causing great and usually sudden damage or suffering.

Question 36 |

From the four choices provided, choose the analogy that is most similar to the one in the question.

Molecule : Atoms

Molecule : Atoms

family : sisters | |

light : bulb | |

tissue : cells | |

body : limb |

Question 36 Explanation:

Molecule is a group of atoms bonded together.

Tissue is any of the distinct types of material of which animals or plants are made, consisting of specialized cells and their products.

Tissue is any of the distinct types of material of which animals or plants are made, consisting of specialized cells and their products.

Question 37 |

From the four choices provided, choose the analogy that is most similar to the one in the question.

Limp : Walk

Limp : Walk

flap : fly | |

run : race | |

stutter : talk | |

chew : digest |

Question 37 Explanation:

Limp is related to walk and shutter is related to talk.

Limp: Walk with difficulty, typically because of a damaged or stiff leg or foot.

Limp: Walk with difficulty, typically because of a damaged or stiff leg or foot.

Question 38 |

From the four choices provided, choose the analogy that is most similar to the one in the question.

Riddle : Solve

Riddle : Solve

mirage : illusion | |

joke : amuse | |

tangle : unravel | |

target : aim |

Question 38 Explanation:

Riddle is related to Solve and tangle is related to unravel

Riddle: A question or statement intentionally phrased so as to require ingenuity in ascertaining its answer or meaning.

Tangle: Twist together into a confused mass.

unravel: undo (twisted, knitted, or woven threads).

Riddle: A question or statement intentionally phrased so as to require ingenuity in ascertaining its answer or meaning.

Tangle: Twist together into a confused mass.

unravel: undo (twisted, knitted, or woven threads).

Question 39 |

Fact 1 : All chickens are birds.

Fact 2 : Some chickens are hens.

Fact 3 : Female birds lay eggs.

If the first three statements are facts, which of the following statements must also be a fact? i. All birds lay eggs.

ii. Hens are birds.

iii. Some chickens are not hens.

Fact 2 : Some chickens are hens.

Fact 3 : Female birds lay eggs.

If the first three statements are facts, which of the following statements must also be a fact? i. All birds lay eggs.

ii. Hens are birds.

iii. Some chickens are not hens.

ii only | |

ii and iii only | |

I, ii and iii | |

None of the statements is a known fact |

Question 39 Explanation:

According to venn diagram, (i) is FALSE because male birds never lay eggs. (ii) and (iii) are TRUE

Question 40 |

What is the most essential thing for election?

President | |

Voter | |

November | |

Nation |

Question 41 |

What is the most essential thing for ovation?

Outburst | |

bravo | |

applause | |

encore |

Question 42 |

Introducing a man to her husband, a woman said, “His brother’s father is the only son of my grandfather.”How is the woman related to this man?

Mother | |

Aunt | |

Sister | |

Daughter |

Question 42 Explanation:

Only son of her grandfather -- Her father;

Man's brother's father -- man's father.

So, man's father is her father i.e., She is the man's sister.

Man's brother's father -- man's father.

So, man's father is her father i.e., She is the man's sister.

Question 43 |

Pointing to a photograph, a man said, “I have no brother or sister but that man’s father is my father’s son.” Whose photograph was it?

His Own | |

His Son | |

His Father | |

His Grandfather |

Question 43 Explanation:

Since the person who is telling has no brother or sister, so his father son is he himself.

So the man in the photograph is his son.

So the man in the photograph is his son.

Question 44 |

Find the number of quadrilaterals in the given figure.

6 | |

7 | |

9 | |

11 |

Question 44 Explanation:

→ The word "quadrilateral" is derived from the Latin words quadri, a variant of four, and latus, meaning "side".

→ A quadrilateral is a polygon with four edges (or sides) and four vertices or corners.

→ According to figure, we can label first and get total quadrilaterals.

Quadrilaterals are 1234, 1245, 1246, 1248, 3481, 3451, 3461, 4517, 4561, 6178 and 1748.

→ A quadrilateral is a polygon with four edges (or sides) and four vertices or corners.

→ According to figure, we can label first and get total quadrilaterals.

Quadrilaterals are 1234, 1245, 1246, 1248, 3481, 3451, 3461, 4517, 4561, 6178 and 1748.

Question 45 |

Count the number of convex pentagons in the adjoining figure.

16 | |

12 | |

8 | |

4 |

Question 45 Explanation:

A convex polygon is a simple polygon (not self-intersecting) in which no line segment between two points on the boundary ever goes outside the polygon.

The pentagons are 12468, 34682, 56824, 78246, 13467, 35681, 57823, 71245, 24578, 46712, 68134 and 82356.

The pentagons are 12468, 34682, 56824, 78246, 13467, 35681, 57823, 71245, 24578, 46712, 68134 and 82356.

Question 46 |

There is a family of six people whose nicknames are Pat, Qat, Rat, Sat, Tat and Uat. Their professionals are Engineer, Doctor, Teacher, Salesman, Manager and Lawyer. There are two married couples in the family. The Manager is the grandfather of Uat, who is an Engineer Rat, the Salesman, is married to the lady Teacher Qat is the mother of Uat and Tat. The Doctor, Sat is married to the
Manager.

How many male members are there in the family?

How many male members are there in the family?

Two | |

Three | |

Four | |

Data Inadequate |

Question 46 Explanation:

Question 47 |

There is a family of six people whose nicknames are Pat, Qat, Rat, Sat, Tat and Uat. Their professionals are Engineer, Doctor, Teacher, Salesman, Manager and Lawyer. There are two married couples in the family. The Manager is the grandfather of Uat, who is an Engineer Rat, the Salesman, is married to the lady Teacher Qat is the mother of Uat and Tat. The Doctor, Sat is married to the
Manager.

What is the professional of Pat?

What is the professional of Pat?

Lawyer | |

Lawyer or Teacher | |

Manager | |

None of these |

Question 47 Explanation:

Question 48 |

There is a family of six people whose nicknames are Pat, Qat, Rat, Sat, Tat and Uat. Their professionals are Engineer, Doctor, Teacher, Salesman, Manager and Lawyer. There are two married couples in the family. The Manager is the grandfather of Uat, who is an Engineer Rat, the Salesman, is married to the lady Teacher Qat is the mother of Uat and Tat. The Doctor, Sat is married to the
Manager.

Who are the married couples in the family?

Who are the married couples in the family?

Pat-Qat and Sat-Rat | |

Rat-Uat and Sat-Tat | |

Pat-Tat and Sat-Rat | |

Pat-Sat and Rat-Qat |

Question 48 Explanation:

Question 49 |

There is a family of six people whose nicknames are Pat, Qat, Rat, Sat, Tat and Uat. Their professionals are Engineer, Doctor, Teacher, Salesman, Manager and Lawyer. There are two married couples in the family. The Manager is the grandfather of Uat, who is an Engineer Rat, the Salesman, is married to the lady Teacher Qat is the mother of Uat and Tat. The Doctor, Sat is married to the Manager.

How Pat is related to Tat?

How Pat is related to Tat?

Father | |

Grandfather | |

Mother | |

Grandmother |

Question 49 Explanation:

Question 50 |

Shirin went to a bakery and bought items worth Rs. 25, out of which 30 paise went on sales tax on taxable purchases. If the tax rate was 6%, then what was the cost of the tax free items?

Rs. 12 | |

Rs. 19.70 | |

Rs. 19.10 | |

Rs. 18.80 |

Question 50 Explanation:

Total cost of the items he purchased = Rs.25

Given that out of this Rs.25, 30 Paise is given as tax => Total tax incurred = 30 Paise = Rs.30/100

Let the cost of the tax free items = x

Given that tax rate = 6%

( 25 - 30/100 - x ) * (6/100) = 30 / 100

6(25−0.3−x)=30

x=25−0.3−5=19.7

Given that out of this Rs.25, 30 Paise is given as tax => Total tax incurred = 30 Paise = Rs.30/100

Let the cost of the tax free items = x

Given that tax rate = 6%

( 25 - 30/100 - x ) * (6/100) = 30 / 100

6(25−0.3−x)=30

x=25−0.3−5=19.7

Question 51 |

The rate of simple interest on a sum of money is 6 percent per annum for the first 3 years, 8 per cent per annum for the next 5 years and 10 per cent per annum for the period beyond 8 years. If the simple interest accrued by the sum for a total period of 10 years is Rs. 1,560, what is the sum?

Rs. 1,500 | |

Rs. 3,000 | |

Rs. 2,000 | |

Data Inadequate |

Question 51 Explanation:

Sum = (1560 * 100 ) / ( 3 * 6 + 8 * 5 + 2 * 10 ) = 2000

Question 52 |

Adrian starts a start-up with a capital of Rs. 85,000. Brian joins in the start-up with Rs. 42,500 after sometime. For how much period does Brian join, if the profits at the end of the year are divided in the ratio of 3 : 1?

5 months | |

6 months | |

7 months | |

8 months |

Question 52 Explanation:

Let B joins for X months.

Then,

(Profit of A)/(Profit of B) = 3/1

(85000 * 12)/(42500 *X) = 3/1

85000 * 12 = 42500 * 3 * X

X = 8.

B Joined for 8 months

Then,

(Profit of A)/(Profit of B) = 3/1

(85000 * 12)/(42500 *X) = 3/1

85000 * 12 = 42500 * 3 * X

X = 8.

B Joined for 8 months

Question 53 |

A car travels at an average of 50 miles per hour for 2.5 hours and then travels at a speed of 70 miles per hour for 1.5 hours. How far did the car travel in the entire 4 hours?

210 miles | |

230 miles | |

250 miles | |

260 miles |

Question 53 Explanation:

Speed= 50 miles/ hour Time=2.5 hour

speed=70 miles / hour Time=1.5 hour

Speed= Distance/ time

Distance= speed x Time

=50×2.5+70×1.5

=125+105

=230 miles

speed=70 miles / hour Time=1.5 hour

Speed= Distance/ time

Distance= speed x Time

=50×2.5+70×1.5

=125+105

=230 miles

Question 54 |

By selling 45 limes for Rs. 40, a woman loses 20%. How many should she sell for Rs. 24 to gain 20% in the transaction?

16 | |

18 | |

20 | |

22 |

Question 54 Explanation:

SP of 45 limes is Rs.40

loss = 20%

So CP of 45 limes would be 40*100/80= Rs. 50

To gain 20% She should sell these 45 limes for =50*20/100=Rs. 60

Now If she sells for Rs. 60 = 45 limes

So for Rs. 24 he would sell = (24/60)*45 = 18 limes

loss = 20%

So CP of 45 limes would be 40*100/80= Rs. 50

To gain 20% She should sell these 45 limes for =50*20/100=Rs. 60

Now If she sells for Rs. 60 = 45 limes

So for Rs. 24 he would sell = (24/60)*45 = 18 limes

Question 55 |

The prestigious Ramon Magsaysay Award was conferred upon Ms. Kiran Bedi for her excellent contribution to which of the following fields?

Literature | |

Community Welfare | |

Government | |

Journalism |

Question 56 |

Who among the following is not a recipient of ‘Dada Saheb Phalke’ Award?

Ramanand Sagar | |

Raj Kapoor | |

V. Shantaram | |

Ashok Kumar |

Question 57 |

What is part of a database that holds only one type of information?

Report | |

Field | |

Record | |

File |

Question 57 Explanation:

Record holds only specific and one type of information.

Question 58 |

‘.JPG’ extension refers usually to what kind of file?

System file | |

Animation/movie file | |

MS Encarta document | |

Image file |

Question 58 Explanation:

‘.JPG’ extension used for image files.

Question 59 |

Which of the following is not written by Munshi Premchand?

Gaban | |

Godan | |

Guide | |

Manasorovar |

Question 60 |

The famous Dilwara Temples are situated in

Uttar Pradesh | |

Rajasthan | |

Maharashtra | |

Madhya Pradesh |

There are 60 questions to complete.