## STQC-NIELIT SC-B 2021

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*STQC-NIELIT SC-B 2021*. You scored %%SCORE%% out of %%TOTAL%%. Your performance has been rated as %%RATING%%
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Question 1 |

Q) Cloud, River, Mountain:

II | |

I | |

IV | |

III |

Question 1 Explanation:

Explanation:
cloud,river and mountain are three different things

Question 2 |

Q) Oxygen, Atmosphere, Nitrogen:

II | |

I | |

IV | |

III |

Question 2 Explanation:

Atmosphere consists of various gases. Oxygen and nitrogen are some of the gases present in the atmosphere.

Question 3 |

In each of the following questions, one number is wrong in the series. Find out the wrong number –
3, 5, 12, 39, 154, 772, 4634:

5 | |

3 | |

39 | |

154 |

Question 3 Explanation:

II number is 3*1+2 =5
III number is 5*2+2=12
IV number is 12*3+2=38
V number is 38*4+2=154
VI number is 154*5+2-772
So wrong number is 39

Question 4 |

701, 348, 173, 85, 41, 19, 8:

173 | |

41 | |

19 | |

348 |

Question 4 Explanation:

The wrong number is 348.
The correct sequence is 701,349,173,85,41,19,8
second number is 701- 352 = 349
third number is 349 - 176 = 173
fourth number is 173 - 88 = 85
fifth number is 85 - 44 = 41
sixth number is 41 - 22 = 19
and the seventh number is 19 - 11 = 8.
to get the next number in the sequence we are subtracting the sequence is 352,176,88,44,22,11. (which is every time we are dividing the number with 2)

Question 5 |

1, 9, 25, 49, 86, 121:

25 | |

121 | |

166 | |

86 |

Question 5 Explanation:

the correct sequence is 1,9,25,49,81,121= 12, 32 , 52, 72, 92, 112 ( which is squares of odd numbers)

Question 6 |

Rakesh is standing at a point. He walks 20 m towards the East and further 10m towards the South, then he walks 35m towards the West and further 5m towards the North, then he walks 15m towards the East. What is the straight distance in meters between his starting point and the point where he reached last?

0 | |

5 | |

10 | |

15 |

Question 6 Explanation:

https://solutionsadda.in/wp-content/uploads/2021/12/Zinc-1.1.jpg

Question 7 |

Directions (7 – 8): In the given questions below, a statement is given followed by two conclusions numbered I and II. You have to take the statement to be true. Read both the conclusions and decide which of the two or both following from the given statement. Give Answer:

(a)If only conclusion I follows (b) If only conclusion II follows

(b)If either I or II follows (d) If neither I nor II follows.

Q)A study of the planning commission reveals a boom in revenues. However, this has been of little avail owing to soaring expenditure. In the event, there has been a high dose of deficit financing, leading to marked rise in picket. Large financial outlays year after year had little impact on level of living.

I) A boom in revenues leads to rise in prices II) Large financial outlays should be avoided

(a)If only conclusion I follows (b) If only conclusion II follows

(b)If either I or II follows (d) If neither I nor II follows.

Q)A study of the planning commission reveals a boom in revenues. However, this has been of little avail owing to soaring expenditure. In the event, there has been a high dose of deficit financing, leading to marked rise in picket. Large financial outlays year after year had little impact on level of living.

I) A boom in revenues leads to rise in prices II) Large financial outlays should be avoided

If only conclusion I follows | |

If only conclusion II follows | |

If either I or II follows | |

If neither I nor II follows. |

Question 8 |

The top management has asked the four managers either to resign by tomorrow or face the order of service termination. Three of them have resigned till this very evening.

(I)The manager who did not resign yesterday will resign tomorrow (II)The management will terminate the service of one manager.

(I)The manager who did not resign yesterday will resign tomorrow (II)The management will terminate the service of one manager.

If only conclusion I follows | |

If only conclusion II follows | |

If either I or II follows | |

If neither I nor II follows |

Question 9 |

Who is the hockey player?

T | |

S | |

R | |

Q |

Question 9 Explanation:

P and S are unmarried ladies and do not participate in any game, so S is not a hockey player.
Q is the brother of R and is neither a chess player nor a hockey player. so Q is not a hockey player.
There is a married couple in which T is husband so R is wife. None of the ladies plays chess and football so R is a hockey player.

Question 10 |

Who is the wife of T?

Q | |

R | |

S | |

None of these |

Question 11 |

Which of the following is the correct group of ladies?

P, Q and R | |

Q, R and S | |

P. Q and S | |

P, R and S |

Question 11 Explanation:

P and S are unmarried ladies and R is the wife of T. So the correct group of ladies is P, R, and S.

Question 12 |

Who is the football player?

Q | |

R | |

S | |

T |

Question 12 Explanation:

R is a hockey player, S wont play any game, T is chess player, so Q is football player.

Question 13 |

Who is the chess player?

Q | |

R | |

S | |

T |

Question 14 |

Directions (1&18): Read the information given below and on the basis of the information, select the correct alternative for each question given after the information.

M and N are good at hockey and volleyball. O and M are good at hockey and baseball. P and N are good at cricket and volleyball. O, P and Q are good at football and basketball.

Q) Who is good at the largest number of games?

M and N are good at hockey and volleyball. O and M are good at hockey and baseball. P and N are good at cricket and volleyball. O, P and Q are good at football and basketball.

Q) Who is good at the largest number of games?

Q | |

P | |

O | |

N |

Question 15 |

M and N are good at hockey and volleyball. O and M are good at hockey and baseball. P and N are good at cricket and volleyball. O, P and Q are good at football and basketball.

Q) Who is good at cricket, baseball and volleyball?

Q) Who is good at cricket, baseball and volleyball?

Q | |

P | |

O | |

N |

Question 15 Explanation:

N is good at Volleyball, Cricket whereas O is only good at baseball according to the mentioned games in the question.

Question 16 |

Who is good at cricket, volleyball and hockey?

Q | |

P | |

O | |

N |

Question 17 |

Who among the following is good at four games?

Q | |

P | |

O | |

M |

Question 18 |

Who is good at baseball, hockey and volleyball?

Q | |

P | |

O | |

M |

Question 19 |

Read the information given below and on the basis of the information, select the correct alternative for each question given after the information.

(I) Six flats on a floor in two rows facing North and South are allotted to P, Q, R, S, T and U. (II) Q gets a north facing flat and is not next to S. (III) S and U get diagonally opposite flats. (IV) R next to U, gets a south facing flat and T gets a north facing flat. Whose flat is between Q and S?

(I) Six flats on a floor in two rows facing North and South are allotted to P, Q, R, S, T and U. (II) Q gets a north facing flat and is not next to S. (III) S and U get diagonally opposite flats. (IV) R next to U, gets a south facing flat and T gets a north facing flat. Whose flat is between Q and S?

T | |

U | |

R | |

P |

Question 20 |

The flats of which of the other pairs than SU, are diagonally opposite to each other?

PT | |

PQ | |

QR |

Question 20 Explanation:

Question 21 |

If the flats of T and P are interchanged, who’s flat will be next to that of U?

Q | |

T | |

P | |

R |

Question 21 Explanation:

https://solutionsadda.in/wp-content/uploads/2021/12/z1.2.jpg

Question 22 |

In a parallelogram ABCD, AP and BP are the angle bisectors of ∟DAB znc ∟ABC. Find ∟APB:

85 ^{0} | |

90 ^{0} | |

94 ^{0} | |

86 ^{0} |

Question 22 Explanation:

we know that, ∟DAB+ ∟ ABC=180
dividing by 2,
½ ∟ DAB+½ ∟ ABC= 180/2
∟ PAB + ∟PBA=90. ………(1 say)
hence, in triangle APB,
∟PAB +∟PBA + ∟APB=180. (Angle sum prop)
from 1,
angAPB+90= 180
ang APB=90

Question 23 |

A cuboid of dimension 24 cm × 9 cm × 8 cm is melted and smaller cubes of side 3 cm are formed. Find how many such cubes can be formed:

27 | |

64 | |

54 | |

32 |

Question 23 Explanation:

Volume of the cuboid : L*B*H = 24*9*8 = 1728 cm^3
Volume of the cube = (side)^3 = 3^3 = 27cm^3
Number of cubes possible = 1728 cm^3 / 27cm^3 = 64

Question 24 |

Instead of walking along two adjacent sides of a rectangular field, a boy took a short cut along the diagonal and saved a distance equal to half the longer side. Then, the ratio of the shorter side to the longer side is:

½ | |

2/3 | |

¼ | |

¾ |

Question 24 Explanation:

Let, Length = l say and breadth = b say.
According to given question we can write,
(l+b) - l2 + b2 = l/2
(l+b) -l/2 = l2 + b2
l + 2b = 2 l2 + b2 => l2+ 4b2 + 41lb = 4(l2 + b2)
b/l = 3/4

Question 25 |

In a fraction, the numerator is increased by 25% and the denominator is diminished by 10%. The fraction obtained is 5/9. The original fraction is:

2/5 | |

5/9 | |

3/5 | |

None of these |

Question 25 Explanation:

Let fraction is x/y
numerator increased by 25% then x will become 1.25x
denominator diminished by 10% then y will become 0.9x
From the given question we can write 1.25x0.9y= 59
xy= 25

Question 26 |

A factory employees skilled workers, unskilled, workers and clerks in the proportion 8 : 5 : 1 and the wages of a skilled worker, an unskilled worker and a clerk are in the ratio 5 : 2 : 3. When 20 unskilled workers are employed, the total daily wages of all (skilled workers, unskilled workers and clerks) amount to Rs. 318. The wages paid to each category of workers are:

Rs. 240, Rs. 60, Rs. 80 | |

Rs. 200, Rs. 90, Rs. 28 | |

Rs. 150, Rs. 108, Rs. 60 | |

Rs. 250, Rs. 50, Rs. 18 |

Question 26 Explanation:

Let us assume the common ratio of workers x and common ratio of wages is y.
So, 5x=20, or x= 4
So, skilled worker=8x4=32, unskilled worker= 20, clerks=4
Hence total wages=32x5y+20x2y+4x3y=212y
So, 212 y=318
or, y=1.5
So, daily wages of skilled workers = 5*1.5 = 7.5 Daily wages of unskilled worker=2*1.5 = 3
And daily wages of clerks =3*1.5= 4.5
Therefore, Wages paid to skilled workers = 7.5 * 32 = 240
wages paid to unskilled worker = 3*20 = 60
wages paid to clerks = 4.5*4 = 18

Question 27 |

½ log10 25 – 2 log10 3 + log10 18 equals:

18 | |

1 | |

log _{10}3 | |

None of these |

Question 27 Explanation:

Given that, ½ log10 25 – 2 log10 3 + log10 18 = ½ log10 5^2 – 2 log10 3 + log10 (9*2)
= log10 5 – log10 9 + log10 9 + log10 2
= log10 5 + log10 2
= log10 10
= 1

Question 28 |

In a mixture of 60L, the ratio of milk and water is 2 : 1. If the ratio of milk and water is to be 1 : 2, then the amount of water to be further added must be:

40 L | |

30 L | |

20 L | |

60 L |

Question 28 Explanation:

In a mixture of 60L, the ratio of milk and water is 2: 1, then milk is 40L and water is 20L.
if the ratio becomes 1:2, we need to add 60L of water then water becomes 80L.

Question 29 |

If x = p, y = q, then which of the following are p and q respectively for pair of equations 3x – 7y + 10 = 0 and y – 2x – 3 = 0:

– 1, 1 | |

1, 1 | |

1, 0 | |

0, 1 |

Question 29 Explanation:

Explanation: (-1,1) satisfies both the given equations.

Question 30 |

The students in a class are seated according to their marks in the previous examination. Once it so happens that four of these students get equal marks and therefore the same rank. To decide their seating arrangement, the teacher wants to write down all possible arrangements, one in each separate bit of paper, in order to choose one of these by lots. How many bits of paper are required?

9 | |

12 | |

15 | |

24 |

Question 30 Explanation:

We are given that four students got equal marks.
On one bit of paper, one arrangement of rank is to be written.
Let the students be named as A,B,C and D.
Now, A can be treated as having rank I in 4 ways.
B can be treated as having rank II in 3 ways.
C can be treated as having rank III in 2 ways.
D can be treated as having rank IV in 1 way.
Therefore, total number of bits of paper required for all arrangement
=4×3×2×1=24

Question 31 |

A cylindrical vessel 60 cm in diameter is partially filled with water. A sphere 30 cm partially filled with water. A sphere 30 cm in a diameter is dropped into it. The increase in the level of water in the vessel is:

2 cm | |

3 cm | |

4 cm | |

5 cm |

Question 31 Explanation:

Volume of the sphere is 43r3 = 43(302)3= 4500
Then, the volume of the water displaced = 4500 cm3
Let, h be the increased level, then (602)2h = 4500
h = 5cm.

Question 32 |

A, B and C rented a pasture by paying Rs. 2,160 per month. They put 60, 60 and 20 sheep respectively. A sells 1/3 of his sheep to B after 6 months and after 3 months more C sells 2/5 of his sheep to A. find the rent paid C at the end of the year:

Rs. 4,355 | |

Rs. 3,888 | |

Rs. 2,464 | |

Rs. 6224 |

Question 33 |

In which of the following years was the percentage passed to appeared candidates from semi-urban area the least?

2015 | |

2016 | |

2017 | |

2018 |

Question 33 Explanation:

In 2015, percentage = (2513/7894) *100 = 31.83
In 2016, percentage = (2933/8562) *100 = 34.25
In 2017, percentage = (2468/8139) *100 = 30.32
In 2018, percentage = (3528/9432) *100 = 37.40
In 2017, the percentage was least in semi-urban area.

Question 34 |

What approximate value was the percentage drop in the number of sem-urban candidates appearing from 2016 to 2017?

15 | |

10 | |

5 | |

8 |

Question 34 Explanation:

Percentage drop = (8562-8139)/8562 = 423/8562 = 4.9% = 5%

Question 35 |

He was not only accused of theft __________ of conspiracy

Rather | |

But also | |

But even | |

Rather than |

Question 36 |

“The dress _____ him so well that she immediately ________ him on his appearance”. The words that best fill the blanks in the above sentence are:

Complemented, Complemented | |

Complimented, Complemented | |

Complimented, Complimented | |

Complemented, Complimented |

Question 37 |

Each of these questions (37 – 38) has an idiomatic expression followed by four options. Choose the one closest to its meaning.

Q) Stick to once guns:

Q) Stick to once guns:

Remain faithful to the cause | |

Suspect something | |

Make something fail | |

Be satisfied |

Question 38 |

Each of these questions (37 – 38) has an idiomatic expression followed by four options. Choose the one closest to its meaning.
Q) Talk shop:

Talk about once profession | |

Talk about shopping | |

Ridicule | |

Treat Lightly |

Question 38 Explanation:

Talk shop means discussing one's work even when not working.

Question 39 |

Each of these questions (37 – 38) has an idiomatic expression followed by four options. Choose the one closest to its meaning.

Q) Identify the correct spelling out of the given options:

Q) Identify the correct spelling out of the given options:

Managable | |

Manageable | |

Managaeble | |

Managible |

Question 40 |

“Going by the ________ that many hands make light work, the school ________ involved all the students in the task”. The words that best fill the blanks in the above sentence are:

Principle, principal | |

Principal, principle | |

Principle, principle | |

principal, principal |

Question 40 Explanation:

Principle: a fundamental truth or proposition that serves as the foundation for a system of belief or behavior or for a chain of reasoning.
Principal: Head of the school.

Question 41 |

The fisherman, _________ the flood victims owed their lives, were rewarded by the Government.

Whom | |

To which | |

To whom | |

To that |

Question 42 |

Which of the following options is the closest in meaning to the word below? DELETERIOUS

Delaying | |

Glorious | |

Harmful | |

Graduating |

Question 42 Explanation:

DELETERIOUS: Harmful often in a subtle or unexpected way.

Question 43 |

If all transactions obey the __________, then all possible interleaved schedules (non – serial schedules) are serializable.

Lock based protocol | |

Two phase Locking protocol | |

Read – write lock based protocol | |

Timestamp based protocol |

Question 43 Explanation:

If all transactions obey the Two phase Locking protocol, then all possible interleaved schedules (non – serial schedules) are serializable.

Question 44 |

For Which one of the following sequences CAN NOT be a degree sequence of a graph of order 5?

3, 3, 2, 2, 2 | |

3, 3, 3, 3, 2 | |

3, 3, 3, 2, 2 | |

4, 3, 3, 2, 2 |

Question 45 |

A protected variable which can be accessed and changed by particular set of operation is called:

Interrupt | |

Monitor | |

Semaphore | |

IPC |

Question 45 Explanation:

Monitor is the collection of condition variables and procedures combined together in a special kind of module or a package. A protected variable which can be accessed and changed by a particular set of operations.

Question 46 |

0 | |

(p – q) (q – r) (r – p) | |

PQR | |

3pqr |

Question 47 |

a | |

b | |

c | |

d |

Question 48 |

What is the access point (AP) in a wireless LAN?

Device that allows wireless devices to connect to a wired network. | |

Wireless devices itself | |

Both device that allows wireless devices to connect to a wired network and wireless devices itself | |

All the nodes in the network. |

Question 48 Explanation:

A wireless access point (wireless AP) is a network device that transmits and receives data over a wireless local area network (WLAN).
Access point in a wireless network is any device that will allow the wireless devices to a wired network. A router is the best example of an Access Point.

Question 49 |

z | |

2z | |

4z | |

0 |

Question 50 |

The channel capacity of a noise free channel having M symbols is given by:

M | |

2 ^{M} | |

log M | |

None of these |

Question 50 Explanation:

The channel capacity of a noise free channel having M symbols is given by log M base 2.

Question 51 |

A language L is recognizable by a Turing machine M if and only if L is a _____ language.

Type 0 | |

Type 1 | |

Type 2 | |

Type 3 |

Question 51 Explanation:

Type 0 grammar generates recursively enumerable languages which are recognized by Turing Machine.

Question 52 |

An analog signal having 3 kHz bandwidth is sampled at 1.5 times the Nyquist rate. The successive samples are statistically independent. Each sample is quantized into one of 256 equally likely levels. The information rate of the source is:

3 kbps | |

72 kbps | |

256 kbps | |

9 kbps |

Question 53 |

A bottom-up parser generates __________ .

Rightmost derivation | |

Rightmost derivation in reverse | |

Leftmost derivation | |

Leftmost derivation in reverse |

Question 53 Explanation:

A bottom up parser generates rightmost derivation in reverse.

Question 54 |

In a paging scheme if page size is of 2048 bytes, then while accommodating a process of 72, 766 bytes, how much internal fragmentation occurs?

962 bytes | |

2048 bytes | |

1024 bytes | |

1086 bytes |

Question 55 |

The following type definition is for a ____________.
Type pointer = ↑node
Node = record
Data: integer
Link: pointer
End;

Structure | |

Link List | |

Stack | |

Doubly link list |

Question 56 |

For a hamming code of parity bit m = 5, what is the total bits and data bits?

(255, 247) | |

(127, 119) | |

(31, 26) | |

(0, 8) |

Question 56 Explanation:

Let m is the number of parity bits and n is the number of message bits.
So, m=5 and n=26.
Then, 2

^{m}≥ n+m+1 should be satisfied. 25 ≥ 26+5+1 32 ≥ 32Question 57 |

Which of the following sorting algorithms has the lowest worst-case complexity?

Merge sort | |

Bubble sort | |

Quick sort | |

Insertion sort |

Question 57 Explanation:

Merge sort → Worst case time complexity is O(nlogn)
Bubble, Quick and insertion sort → Worst case time complexity is O(n^2)

Question 58 |

What is WPA?

Wi-fi protected access | |

Wired protected access | |

Wired process access | |

Wifi process access |

Question 58 Explanation:

WPA → Wi-fi protected access

Question 59 |

In Software Modeling ‘IS A’ represents ___________ relationship.

Aggregation | |

Overloading | |

Inheritance | |

Design Patterns |

Question 59 Explanation:

In Software Modeling ‘IS A’ represents Inheritance relationship.

Question 60 |

The following grammar is an example of __________.
A → A B C
CB → B c
A → a b c
bB → b b
B → c b a

Type 0 Grammar | |

Type 1 Grammar | |

Type 2 Grammer | |

Type 3 Grammar |

Question 60 Explanation:

Every production |LHS| <= |RHS| and LHS contains more than one symbol in second and fourth production, thus it is a Type 1 grammar.

Question 61 |

Delivery of the software product on time is considered as __________.

SDLC | |

User Satisfaction | |

Planning | |

UI/UX design for software |

Question 61 Explanation:

Delivery of the software product on time is considered as Planning

Question 62 |

(a + b)2 corresponds to the language:

{a + b, a + b} | |

{aa, ab, ba, bb} | |

{abab, baba} | |

{a + b, (a + b) ^{2}} |

Question 62 Explanation:

(a + b)

^{2}=(a+b)(a+b) which is concatenation of (a+b) with (a+b) (a+b)(a+b)= {aa, ab, ba, bb}Question 63 |

What is the value of f(4) using the following C code:
int f(int k) {
if k( < 3)
return k;
else
return f(k – 1) * f(k – 2) + f(k – 3);
}

5 | |

6 | |

7 | |

8 |

Question 63 Explanation:

https://solutionsadda.in/wp-content/uploads/2021/12/z1.7.jpg

Question 64 |

A | |

B | |

C | |

D |

Question 65 |

Who developed standards for the Only Statement reference model?

ANSI – American National Standards Institute | |

ISO – International Standards Organization | |

IEEE – Institute of Electrical and Electronics Engineering | |

ACM – Association for Computing Machinery |

Question 66 |

TCP / IP model does not have __________ layer but OSI model have this layer

Session layer | |

Transport layer | |

Application layer | |

Network layer |

Question 66 Explanation:

Both presentation and session layers are not presented in the TCP/IP model.

Question 67 |

A variable p is said to be live at point m if and only if:

P is assigned at point m | |

P is not assigned at pint m | |

Value of p at m would be used along some path in the flow graph starting at point m | |

Value of p at m would be used along some path in the flow graph ending at point m |

Question 67 Explanation:

A variable is live at some point if it holds a value that may be needed in the future, or equivalently if its value may be read before the next time the variable is written to. Thus, Value of p at m would be used along some path in the flow graph starting at point m then variable p is said to be live at point m.

Question 68 |

Integration testing, Unit Testing & System Testing are __________.

Fundamental logic of Testing | |

Level Testing | |

Core Testing | |

Testing Suites |

Question 69 |

Each layer of the OSI model receives services or data from a _________ layer.

Below Layer | |

Above Layer | |

Both (a) and (b) | |

None of the above |

Question 69 Explanation:

Each layer of the OSI model receives services or data from a above and below layer

Question 70 |

9/2 | |

6 | |

9 | |

5/2 | |

None Of the above |

Question 71 |

If a developer wants to transform a model into source code is also known as ________.

Backward Testing / Engineering | |

Forward Engineering | |

Forward Testing | |

Reverse Engineering |

Question 71 Explanation:

If a developer wants to transform a model into source code is also known as Reverse Engineering

Question 72 |

The maximum number of times the decrease key operation performed in Dijkstra’s algorithm will be equal to _________.

Total number of vertices | |

Total number of edges | |

Number of vertices – 1 | |

Number of edges – 1 |

Question 72 Explanation:

If the total number of edges in all adjacency lists is E, then there will be a total of E number of iterations, hence there will be a total of at most E decrease key operations.

Question 73 |

One root of x3 – x – 4 – 4 = 0 lies on (1, 2). In the bisection method, after the first iteration the root lies in the interval ______.

(1, 1.5) | |

(1.5, 2) | |

(1.25, 1.75) | |

( 1.75, 2) |

Question 73 Explanation:

One root of x<3>TEXT – x – 4 – 4 = 0 lies on (1, 2). In the bisection method, after the first iteration the root lies in the interval (1.5, 2)

Question 74 |

A | |

B | |

C | |

D |

Question 75 |

A wireless network interface controller can work in ___________.

Infrastructure mode | |

ad-hoc mode
| |

Both infrastructure mode and ad-hoc mode | |

WDS mode |

Question 75 Explanation:

A wireless network interface controller (WNIC) is a network interface controller which connects to a wireless network, such as Wi-Fi or Bluetooth, rather than a wired network, such as a Token Ring or Ethernet.
An 802.11 WNIC can operate in two modes known as infrastructure mode and ad hoc mode
Infrastructure mode:
In an infrastructure mode network the WNIC needs a wireless access point: all data is transferred using the access point as the central hub. All wireless nodes in an infrastructure mode network connect to an access point. All nodes connecting to the access point must have the same service set identifier (SSID) as the access point, and if a kind of wireless security is enabled on the access point (such as WEP or WPA), they must share the same keys or other authentication parameters.
Ad hoc mode:
In an ad hoc mode network the WNIC does not require an access point, but rather can interface with all other wireless nodes directly. All the nodes in an ad hoc network must have the same channel and SSID.

Question 76 |

If the Value of Register A = 98 h & Carry = 1. What will be the value of Register A after executing the RORC instruction 1 time?

AB h | |

CD h | |

EF h | |

AC h |

Question 77 |

Dijkstra’s Algorithm cannot be applied on ___________.

Directed and weighted graphs | |

Graphs having negative weight function | |

Unweighted graph | |

Undirected and unweighted graphs |

Question 77 Explanation:

Dijkstra’s Algorithm cannot be applied on Graphs having negative weights and negative weight cycle.

Question 78 |

A high resolution B/w TV picture contains 3 × 104 picture elements and 16 different brightness levels. Pictures are repeated at a rate of 24 per second. All levels have equal likelihood of occurrence and all picture elements are assumed to be independent. What will be the average rate of information carried by this TV picture source ?

288 Mbps | |

24 Mbps | |

132 Mbps | |

3 Mbps |

Question 78 Explanation:

https://solutionsadda.in/wp-content/uploads/2021/12/z1.15.jpg

Question 79 |

When a cache is 10 times faster than main memory, and the cache can be used 70% of the time, how much speed can be gained using cache?

≈ 10 | |

≈ 0.3 | |

≈ 0.7 | |

≈ 2.7 |

Question 80 |

100 elements can be sorted in 100 sec using bubble sort. In 400 sec, approximately _______ elements can be sorted.

100 | |

200 | |

300 | |

400 |

Question 80 Explanation:

Bubble sort worst case time complexity is O(n^2)
Worst case comparisons are (n(n-1))/2
For 100 elements = (100(100-1))/2
= 19,800 comparisons.
19,800*c= 100 sec
(n(n-1))/2 = 400 sec
(n(n-1))/2 = 79,200 sec
n=200

Question 81 |

The ____________ enables the software engineer to develop models of the information domain and functional domain at the same time.

Data flow diagram | |

State transition diagram | |

Control specification | |

Activity diagram |

Question 81 Explanation:

The Data flow diagram enables the software engineer to develop models of the information domain and functional domain at the same time.

Question 82 |

The ___________ of a relationship is 0 if there is no explicit reason for the relationship to occur or the relationship is optional.

Modality | |

Cardinality | |

Entity | |

Structured analysis |

Question 82 Explanation:

Cardinality and Modality are the indicators of the business rules around a relationship.
Cardinality refers to the maximum number of times an instance in one entity can be associated with instances in the related entity.
Modality refers to the minimum number of times an instance in one entity can be associated with an instance in the related entity.
Cardinality can be 1 or Many and the symbol is placed on the outside ends of the relationship line, closest to the entity, Modality can be 1 or 0 and the symbol is placed on the inside, next to the cardinality symbol

Question 83 |

What will be the Eulerian tour of the following graph G?

1 – 2 – 3 – 4 – 5 – 6 – 7 – 8 – 6 – 4 – 2 – 8 – 1 | |

1 – 2 – 3 – 4 – 5 – 6 – 7 – 8 | |

1 – 2 – 3 – 4 – 5 – 6 – 7 – 8 – 1 | |

8 – 7 – 6 – 5 – 4 – 3 – 2 – 1 |

Question 83 Explanation:

An Eulerian trail is a trail in a finite graph that visits every edge exactly once.
An Eulerian circuit or Eulerian cycle is an Eulerian trail that starts and ends on the same vertex.

Question 84 |

A | |

B | |

C | |

D |

Question 84 Explanation:

AB’+AB => A(B’+B) =A (B’+B is 1)

Question 85 |

The two numbers represented in Signed 2’s complement form are:
A = 11101101 and B = 11100110. If B is subtracted from A, the value obtained in signed 2’s complements is :

111000101 | |

00000111 | |

11111000 | |

10000011 |

Question 86 |

The number of un-labeled non-isomorphic graphs with four vertices is;

12 | |

11 | |

10 | |

9 |

Question 86 Explanation:

By examining the possibilities, we find 1 graph with 0 edges, 1 graph with 1 edge, 2 non-isomorphic graphs with 2 edges, 3 non-isomorphic graphs with 3 edges, 2 non isomorphic graphs with 4 edges, 1 graph with 5 edges and 1 graph with 6 edges. Altogether, we have 11 non-isomorphic graphs on 4 vertice

Question 87 |

DELETE [FROM] table [WHERE condition]; from the syntax if you omit the WHERE clause.

All rows in the table are deleted | |

It will give you an error | |

No rows will be deleted | |

Only one row will be deleted |

Question 88 |

The number of 4 digit numbers which contain not more than two different digits is:

576 | |

567 | |

513 | |

504 |

Question 89 |

___________ is the class of decision problems that can be solved by non-deterministic polynomial algorithms.

NP | |

P | |

Hard | |

Complete |

Question 89 Explanation:

NP is a set of problems that can be solved by a Non-deterministic Turing Machine in Polynomial time.

Question 90 |

The spectral efficiency of QPSK null to null having a width 4 Hz will be:

½ | |

1 | |

¼ | |

4 |

Question 91 |

Two Important lexical categories are ____________.

White Space | |

Comments | |

White Space & Comments | |

None of these options |

Question 91 Explanation:

Whitespace and comments are ignored by lexical analysers, so these are not any categories of lexical analysis phase.

Question 92 |

Which address is used to identify a process on a host by the transport layer?

Physical address | |

Logical address | |

Port address | |

Specific address |

Question 92 Explanation:

A port is a communication endpoint. At the software level, within an operating system, a port is a logical construct that identifies a specific process or a type of network service.
A port is identified for each transport protocol and address combination by a 16-bit unsigned number, known as the port number. The most common transport protocols that use port numbers are the Transmission Control Protocol (TCP) and the User Datagram Protocol (UDP).

Question 93 |

Convert (503201)

_{6}into (?)_{4}.12122231 | |

11000011 | |

21222301 | |

22323301 |

Question 93 Explanation:

https://solutionsadda.in/wp-content/uploads/2021/12/z1.19.jpg

Question 94 |

Number of entity types participating in E-R diagrams is represented by:

Degree of relationship | |

Structure of relationship | |

Instance of relationship | |

Role of relationship |

Question 94 Explanation:

The number of different entity sets participating in a relationship set is called the degree of a relationship set.

Question 95 |

If data is stored in AC = 5F h and DR = C2 h what is the value of AC after AC^DR operation?

9D | |

42 | |

DF | |

DE |

Question 96 |

A sequence of statement of the form x = y op z is called a:

Three address code | |

Syntax tree | |

Postfix notation | |

Operator |

Question 96 Explanation:

Three-address code(TAC) is an intermediate code used by optimizing compilers to aid in the implementation of code-improving transformations.
Each TAC instruction has at most three operands and is typically a combination of assignment and a binary operator.
For example, t1 := t2 + t3. The name derives from the use of three operands in these statements even though instructions with fewer operands may occur.

Question 97 |

ABC* + is the postfix form of:

A * B + C | |

A* + BC | |

A + B * C | |

None of these |

Question 97 Explanation:

A+B*C ===> A+[BC*] => ABC*+

Question 98 |

y = 10 cos (1800 πt) + 20 cos (2000 πt) + 10 cos (220 πt). Find the modulation index (μ) of the given wave.

0.3 | |

0.5 | |

0.7 | |

1 |

Question 99 |

Write Recurrence of Quicksort in the worst case.

T(n) = T(n – 1) + 1 | |

T(n) = T(n – 1) + n | |

T(n) = 2T(n – 1) + n | |

T(n) = T(n/3) + T(2n/2) + n |

Question 99 Explanation:

Worst Case Analysis: T(N) = N + T(N-1)
Recurrence Relation: T(0) = T(1) = 0 (base case)
T(N) = N + T(N-1)
T(N-1) = (N-1) + T(N-2)
T(N-2) = (N-2) + T(N-3)
;;
T(3) = 3 + T(2)
T(2) = 2 + T(1)
T(1) = 0
Hence, T(N) = N + (N-1) + (N-2) ... + 3 + 2
≈ (N^2)/2
=O(N^2)

Question 100 |

Which of the following Boolean expressions is not logically equivalent to all of the rest?

ab + (cd)’ + cd + bd’ | |

a (b + c) + cd | |

ab + ac + (cd)’ | |

bd' + c’d’ + ab + cd |

Question 101 |

W – 2, X-3, Y-1, Z-4 | |

W – 3, X-4, Y-2, Z-1 | |

W – 3, X-1, Y-4, Z-2 | |

W – 3, X-1, Y-2, Z-4 |

Question 101 Explanation:

Condition coverage → White-box testing
Equivalence class partitioning → Black-box testing
Volume testing → Performance testing
Alpha testing → System testing

Question 102 |

Let B

_{n}denote the number of full binary trees with n vertices. Then a recurrence relation for B_{n}is:Bn = Bn – 1 + O(1) | |

Bn = 2Bn – 1 + O(1) | |

Bn = Bn – 1 + O(n) | |

Bn = 2Bn – 1 + O(n) |

Question 102 Explanation:

Bn full binary trees with b vertices
Recurrence relation is Bn = 2Bn – 1 + O(1)
Time complexity is O(2n)

Question 103 |

Print columns A & B from relation R when C = 10 | |

Print C = 10 from relation R | |

Print all data of relation R when C = 10 | |

Print A, B, C from relation B when C = 10 |

Question 104 |

XOR Gate | |

XOR Gate | |

OR Gate | |

NAND Gate |

Question 104 Explanation:

The equation for the output of the diagram is (A+B)(A’+B’).
(A+B)(A’+B’)= AA’+AB’+BA’+BB’
= 0+AB’+A’B+0
= AB’+A’B
= A ⊕ B

Question 105 |

If A and B are two sets. A binary relation from set A and set B is any subset of the _________.

Cartesian Product A × B | |

Union A u B | |

Intersection A u B | |

Addition A + B |

Question 105 Explanation:

If A and B are sets, then a binary relation R from A to B is a subset of the Cartesian product of A and B (A x B)

Question 106 |

Which notation gives the lower bound of a function?

Ө – Notation | |

O – Notation | |

sigma - notation | |

None of these |

Question 106 Explanation:

BigOmega→ Lower bound [ Generally we call it as Best case)
BigTheta → Average bound [ Generally we call it as Average case)
BigOh → Upper bound [ Generally we call it as Worst case)

Question 107 |

The bit rate of a digital communication system is R kbit/s. The modulation used is 32-QAM. The minimum bandwidth required for ISI free transmission is:

R/10 Hz | |

R/10 kHz | |

R/5 Hz | |

R/5 kHz |

Question 108 |

Assume that a DBA issued the following create table command:
Create table A (Aid, ………..)
Storage (initial 20480, next 20480, maxextents 8, minextents 3, pctincrease 0),
How many bytes of disk space will be allocated to this file when it is first created?

163, 840 bytes | |

20480 bytes | |

61,440 bytes | |

8 bytes |

Question 109 |

A | |

B | |

C | |

D |

Question 109 Explanation:

One bit is used for signs . Remaining bits are 15
Maximum value stored in signed integer is 2

_{15}– 1Question 110 |

Which type of illustration lists the functionality of the whole project?

DFD – 0 | |

Class Diagram | |

Use case of Diagram | |

State Diagram |

Question 110 Explanation:

Class diagram is a static diagram. It represents the static view of an application. Class diagram is not only used for visualizing, describing, and documenting different aspects of a system but also for constructing executable code of the software application

Question 111 |

A binary tree of depth K is called a full binary tree of depth K, if it has exactly ____________nodes.

K | |

2 ^{k} | |

2 ^{k} – 1 | |

2 ^{k} + 1 |

Question 112 |

In _____________, machine is executing operating system instruction:

System Mode | |

User Mode | |

Normal Mode | |

Safe Mode |

Question 112 Explanation:

A processor in a computer running Windows has two different modes: user mode and kernel mode.
The processor switches between the two modes depending on what type of code is running on the processor. Applications run in user mode, and core operating system components run in kernel mode

Question 113 |

__________ is the elapsed time between the time a program or job is submitted and the time when it is completed.

Response time | |

Turnaround time | |

Waiting time | |

Throughput |

Question 113 Explanation:

Turnaround time (TAT) means the amount of time taken to complete a process or fulfill a request

Question 114 |

The operating system stores an _________ in order to decide to which user to grant which access rights to which file?

File allocation table | |

Process control block | |

Access control matrix | |

File control matrix |

Question 114 Explanation:

An Access Control Matrix or Access Matrix is an abstract, formal security model of protection state in computer systems, that characterizes the rights of each subject with respect to every object in the system.
An access matrix can be envisioned as a rectangular array of cells, with one row per subject and one column per object. The entry in a cell – that is, the entry for a particular subject-object pair – indicates the access mode that the subject is permitted to exercise on the object. Each column is equivalent to an access control list for the object; and each row is equivalent to an access profile for the subject

Question 115 |

If the size of the logical address space is 2m and the page size is 2n addressing units, then the high order m-n bits of a logical address designate the ________.

Offset | |

Page No | |

Frame No | |

Physical address |

Question 115 Explanation:

If the size of the logical address space is 2m and the page size is 2n addressing units, then the high order “m-n” bits of a logical address designate the page number and the “m” low order bits designate the page offset

Question 116 |

Following are implicitly provided in C programming language:

Output facility | |

Input facility | |

Both Input and Output facility | |

No input and Output facility |

Question 116 Explanation:

A Binary Tree is full if every node has 0 or 2 children. So, in such case, the binary tree with K leaves contains a total of 2*K - 1 node.

Question 117 |

A | |

B | |

C | |

D |

Question 118 |

Which is the layer that converts Packets to Frames and Frames to Packets in the OSI model?

Physical Layer | |

Data Link Layer | |

Network Layer | |

Transport Layer |

Question 118 Explanation:

Network layer converts Packets to Frames and Frames to Packets in the OSI model

Question 119 |

Leaves a and b unchanged | |

a doubles and stores in a | |

B doubles and stores in a | |

Exchanges a and b |

Question 119 Explanation:

The given code swaps(exchanges ) two variable values without using a third variable.

Question 120 |

The bit transmission rate in a pulse coded modulation system with number of quantisation levels 8 and maximum signal frequency 4000 Hz is given by:

16 kpbs | |

24 kpbs | |

32 kpbs | |

8 kpbs |

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