Coal India 2020 Previous Year Questions (PYQs) with Solutions

Rohit Sharma is the first Indian cricketer and second in world cricket to complete 100 T-20 Internationals.

The Doctrine of Lapse was introduced by Lord Dalhousie during his tenure as Governor-General of India. Under this policy, if a ruler of a princely state died without a natural male heir, the state could be annexed by the British East India Company, and adopted heirs were not recognized for succession.

Anant Goenka is the correct answer. He was elected President of the Federation of Indian Chambers of Commerce and Industry (FICCI) for the 2025–26 term. As Vice Chairman of the RPG Group, he succeeded Sandeep Somany and assumed leadership of one of India's leading industry bodies.

Phulkari is a traditional embroidery style originating from the Punjab region. It is characterized by bright colors and floral patterns, making it a famous handicraft of the state of Punjab.

Nabakalebara is a major festival associated with the Jagannath Temple in Odisha. It is a special ritual in which the wooden idols of Lord Jagannath, Balabhadra, Subhadra, and Sudarshan are ceremonially replaced with new ones. The festival is observed at irregular intervals, usually every 12 to 19 years.

The Amazon rainforest is spread across nine countries in South America. These include Brazil, Peru, Colombia, Venezuela, Ecuador, Bolivia, Guyana, Suriname, and French Guiana.

Liquid sodium is used as a coolant in certain nuclear reactors, especially Fast Breeder Reactors (FBRs). Its high thermal conductivity, high boiling point, and low neutron absorption make it highly effective for transferring heat from the reactor core.

The correct answer is Option B because the International Hockey Federation (FIH) officially awarded the hosting rights for the 2026 Men's Hockey World Cup to Belgium and the Netherlands. This joint bid was selected as part of a strategic partnership between two leading hockey nations in Europe to showcase the sport's growth. Option A is incorrect because India and Sri Lanka have not been announced as hosts for this specific tournament, although they are active participants in the sport. Option C is also incorrect as New Zealand and Spain were not chosen for this event; while Spain has hosted previous editions, the 2026 slot belongs to the Benelux region. Option D is incorrect because China and Russia were not selected as co-hosts for the 2026 Men's Hockey World Cup. Therefore, based on official FIH announcements, Belgium and the Netherlands are confirmed as the hosts.

Sesame is one of the earliest domesticated oilseed crops in the world and has been cultivated in India for thousands of years. It is valued for its high-quality edible oil and nutritional content.

The Prime Minister of India is the only individual who receives SPG (Special Protection Group) protection under the current legal framework. Following amendments to the SPG Act in 2019, SPG cover is restricted to the serving Prime Minister and immediate family members residing at the official residence.

An anemometer is a scientific instrument used to measure wind speed. It typically consists of cups or propellers that rotate in response to air movement, allowing for accurate wind velocity readings.

The correct answer is Vipin Kumar. As of 2026, Vipin Kumar serves as the Chairman and Managing Director of the Airport Authority of India (AAI). He assumed this high-ranking position following his appointment by the Government of India, bringing extensive experience in aviation management and infrastructure development to lead the nation's premier civil aviation authority. His leadership focuses on expanding airport capacity, enhancing passenger experience, and modernizing air traffic management systems across the country. Option B (Rajnish Kumar) is incorrect as he is associated with other major Indian corporations, not AAI. Option C (Amit Khare) and Option D (Arvind Singh) are also incorrect as they do not hold the position of Chairman at AAI in 2026. Students should note that key positions in public sector organizations like AAI are often appointed by the government and can change over time, making awareness of current appointments important for competitive exams.

France is the correct answer. The 5th "No Money For Terror" Ministerial Conference was held in Paris, France, on 19 May 2026. The conference focused on combating terrorist financing, financial crime, misuse of financial innovations, and strengthening international cooperation.

IRNSS-1A was the first satellite of the Indian Regional Navigation Satellite System (IRNSS), now known as NavIC (Navigation with Indian Constellation). It was launched on 1 July 2013 and marked the beginning of India's indigenous regional navigation system.

India is mainly dependent on imports for mercury as domestic production is negligible. While lead, mica, and limestone have significant domestic availability, mercury imports are essential for various industrial applications.

The correct answer is Option D (Coconut). Coconut is the most cultivated crop in Kerala and occupies the largest area under cultivation among the state's crops. Kerala's warm and humid tropical climate, abundant rainfall, and coastal geography provide ideal conditions for coconut cultivation. The crop is an important source of income and employment and is widely used in food, oil production, coir manufacturing, and various traditional industries. Therefore, coconut is rightly regarded as the most cultivated crop of Kerala.

India's two operational Satellite Launch Pads—the First Launch Pad (FLP) and Second Launch Pad (SLP)—are located at the Satish Dhawan Space Centre in Sriharikota, Andhra Pradesh. These launch pads are used by Indian Space Research Organisation for launching PSLV, GSLV, LVM3, and other missions.

World Science Day for Peace and Development is officially celebrated on November 10th every year. This date was established by UNESCO in 2001 to promote the importance of science for peace and development. The correct answer is Option C (10th November) because it aligns with the globally recognized observance date set by international scientific organizations. Option A (30th August) is incorrect as it does not correspond to any major international science-related observance. Option B (4th November) is also wrong, as this date is not designated for World Science Day. Option D (11th October) is incorrect because it falls on the day after the actual celebration. Students should remember that November 10th is the fixed annual date for this important global event, making Option C the only accurate choice.

The Vice President of India is not directly appointed by the President, Prime Minister, or Supreme Court. Instead, Article 66 of the Indian Constitution mandates that the Vice President is elected by an electoral college. This body comprises all members of both Houses of Parliament, including the Lok Sabha and Rajya Sabha, using proportional representation through a single transferable vote system. Although the question uses the term 'appoints', in the context of constitutional processes, this refers to the selection mechanism. Option C correctly identifies 'The Electoral College of India' as the body responsible for this election. Options A, B, and D are incorrect because the President does not appoint them (though they administer the oath), the Supreme Court adjudicates disputes but does not select officials, and the Prime Minister has no constitutional role in this specific appointment process. Therefore, Option C is the accurate answer based on India's constitutional framework.

Apple Inc. developed the Mac Operating System, historically known as Mac OS. Microsoft develops Windows, while IBM and HP focus on other systems or hardware.

The Sayyid Dynasty was the fourth dynasty to rule the Delhi Sultanate, reigning from 1414 to 1451 CE. Khizar Khan established this dynasty in 1414 after defeating the Tughlaq ruler, effectively ending the Tughlaq rule in Delhi. He was originally a governor of Multan under Timur and later seized power, claiming descent from the Prophet Muhammad (Sayyid), which gave the dynasty its name. Therefore, Option C is correct.\nJalal-ud-din Khilji (Option A) founded the Khilji Dynasty in 1290, which preceded the Tughlaqs. Sikandar Lodhi (Option B) was a ruler of the Lodi Dynasty, which came much later in 1489. Nasir-ud-din Mahmud (Option D) was actually the last ruler of the Sayyid Dynasty, not its founder. Understanding the chronological sequence of Delhi Sultanate dynasties helps eliminate these distractors and confirms Khizar Khan as the correct answer.

Given:

• Average of 17 numbers = 48

• Average of first 9 numbers = 45

• Average of next 9 numbers = 52

Note:
The 9th number is common in both groups.
Step 1: Find total sum of 17 numbers
Total sum:
= 17 × 48
= 816
Step 2: Find sums of the two groups
Sum of first 9 numbers:
= 9 × 45
= 405
Sum of next 9 numbers:
= 9 × 52
= 468
Combined sum:
= 405 + 468
= 873
Step 3: Find the 9th number
The 9th number has been counted twice.
Therefore,
9th number:
= 873 − 816
= 57
Step 4: Find average after excluding 9th number
Remaining sum:
= 816 − 57
= 759
Remaining numbers:
= 16
Required average:
= 759/16
= 47.4375

Given:

Alloy A:
Copper : Zinc = 4 : 5

Copper % in A:
= 4/(4 + 5) × 100
= 4/9 × 100
= 44.44%

Alloy B:
Copper : Zinc = 7 : 2

Copper % in B:
= 7/(7 + 2) × 100
= 7/9 × 100
= 77.78%

A and B are mixed in the ratio 7 : 4.

Copper % in new alloy:
= (7 × 44.44 + 4 × 77.78)/(7 + 4)
= (311.08 + 311.12)/11
= 622.20/11
= 56.56%
≈ 56.57%

Concept

When quantities are divided in a ratio, write each quantity as its ratio part multiplied by one common value. A transfer subtracts the amount from the giver and adds the same amount to the receiver.

The new ratio then forms an equation in the common value. Solve that equation first, and only afterward calculate the requested initial total.

Application

1. Let the common value be x. The initial shares follow this mapping:

   Daughter	Initial share
   A	6x
   B	7x
   C	9x
   D	16x

2. After A transfers ₹600 to C, their shares become A = 6x − 600 and C = 9x + 600.

3. Use the resulting ratio A : C = 1 : 2: (6x − 600) / (9x + 600) = 1 / 2.

4. Cross-multiply and simplify: 2(6x − 600) = 9x + 600, so 12x − 1,200 = 9x + 600, hence 3x = 1,800 and x = 600.

5. The requested initial total is A + B + C = (6 + 7 + 9)x = 22 × 600 = ₹13,200.

Cross-check

With x = 600, the initial shares of A, B and C are ₹3,600, ₹4,200 and ₹5,400. After the transfer, A has ₹3,000 and C has ₹6,000, whose ratio is 1 : 2; their initial total is ₹13,200.

Given:
Sum of the first 10 terms = Sum of the first 12 terms
⇒ S₁₂ − S₁₀ = 0
⇒ 11th term + 12th term = 0
Let the first term = a and the common ratio = r. Then:
11th term = ar¹⁰ and 12th term = ar¹¹
So, ar¹⁰ + ar¹¹ = 0
⇒ ar¹⁰(1 + r) = 0
The common ratio is non-zero (given), so r ≠ 0, and a ≠ 0 (otherwise every sum would be 0, contradicting S₁₇ = 42). Hence ar¹⁰ ≠ 0, which forces 1 + r = 0, so r = −1.
Now, with r = −1 the GP becomes a, −a, a, −a, …
Since 17 is odd, the terms pair off and cancel, leaving:
S₁₇ = a = 42
Fourth term: T₄ = ar³ = 42 × (−1)³ = −42.

Given number:
3x9z8y
The number is divisible by 7, 11 and 13.
Since:
7 × 11 × 13 = 1001
Any number divisible by 7, 11 and 13 together must be divisible by 1001.
Now observe:
A 6-digit number of the form abcdef is divisible by 1001 if:
(abc) = (def)
Given number:
3x9z8y
So,
3x9 = z8y
Comparing digits:
z = 3
8 = x
y = 9
Therefore:
x = 8
y = 9
z = 3
Average of x, y, z:
= (8 + 9 + 3)/3
= 20/3

Given:
25 litres of 40% acid solution
35 litres of 70% acid solution

Amount of acid in first solution:
= 40% of 25
= (40/100) × 25
= 10 litres

Amount of acid in second solution:
= 70% of 35
= (70/100) × 35
= 24.5 litres

Total acid:
= 10 + 24.5
= 34.5 litres

Total solution:
= 25 + 35
= 60 litres

Concentration of acid in new solution:
= (34.5/60) × 100
= 57.5%

Given:
Length of ladder = 18 m
The ladder makes an angle of 30° with the wall.
We need to find the height where the ladder touches the wall.
In the right triangle formed:
Hypotenuse = 18 m
Angle with wall = 30°
Required height is adjacent to the angle.
Using:
cos 30° = Height/18
√3/2 = Height/18
Height = 18 × √3/2
= 9√3 m

Given:
First investment = Rs. 9000 at r%
Second investment = Rs. 13500 at (r − 2)%
Time = 5 years
Total simple interest = Rs. 9337.50

Using SI formula:
SI = (P × R × T)/100

Interest from first investment:
= (9000 × r × 5)/100
= 450r

Interest from second investment:
= (13500 × (r − 2) × 5)/100
= 675(r − 2)

Total interest:
450r + 675(r − 2) = 9337.50
450r + 675r − 1350 = 9337.50
1125r = 10687.50
r = 10687.50/1125
r = 9.5%

Given:
Height of pole house A = 21 m
Angle of depression from top of A to foot of B = 30°
Angle of elevation from top of A to top of B = 60°
Let the distance between the houses be x m.
From the angle of depression:
tan 30° = 21/x
1/√3 = 21/x
x = 21√3
Now, let the height of house B be h m.
Difference in heights:
= h − 21
Using angle of elevation:
tan 60° = (h − 21)/(21√3)
√3 = (h − 21)/(21√3)
Cross multiply:
h − 21 = 21 × 3
h − 21 = 63
h = 84 m
Since the obtained answer does not match the options, let us verify carefully.
Using:
Distance = 21√3 ≈ 36.37
Now,
tan 60° = (h − 21)/36.37
√3 = (h − 21)/36.37
h − 21 = 36.37 × 1.732
≈ 63
h ≈ 84 m
The options provided do not contain the correct value.

Concept

Savings = Income − Expenditure, so a change in savings depends on how income and expenditure move together. Any ‘percent decrease’ is found as (decrease ÷ the ORIGINAL value) × 100, using the original savings as the base — never the new income, new expenditure, or new savings. When a problem gives only percentages (no rupee figures), assume a convenient base income, such as 100, to convert the percentages into workable numbers.

Application

1. Let the initial income = 100. Since she saves 18% of her income, original savings = 18% of 100 = 18, and original expenditure = 100 − 18 = 82.

2. Income rises by 15%, so new income = 100 × 1.15 = 115.

3. Expenditure rises by 21.6%, so new expenditure = 82 × 1.216 = 99.712.

4. New savings = new income − new expenditure = 115 − 99.712 = 15.288.

5. Decrease in savings = original savings − new savings = 18 − 15.288 = 2.712.

6. Percent decrease in savings = (decrease ÷ original savings) × 100 = (2.712 ÷ 18) × 100 ≈ 15.1%.

Cross-check

Cross-check with a different base to confirm the percentage does not depend on the assumed income: with income = 1000, original savings = 180, new income = 1150, new expenditure = 820 × 1.216 = 997.12, new savings = 152.88, decrease = 27.12, and (27.12 ÷ 180) × 100 ≈ 15.1% — the same result, matching 15.1.

Let C's income be 100 units. Since B is 40% less than C, B equals 60 units.

A is 60% more than B, so A equals 1.6 times 60, which is 96 units.

To find how much more C is than A, calculate (100 - 96) / 96 * 100. This equals approximately 4.2%.

Selling price of each article = Rs. 1836
For first article:
Gain = 20%
Let cost price be x.
120% of x = 1836
x = 1836 × 100/120
x = 1530
For second article:
Loss = 20%
Let cost price be y.
80% of y = 1836
y = 1836 × 100/80
y = 2295
Total cost price:
= 1530 + 2295
= 3825
Total selling price:
= 1836 + 1836
= 3672
Loss:
= 3825 − 3672
= 153
Loss percentage:
= (153/3825) × 100
= 4%

Let the original speed of the train be x km/hr.

Then:

• Reduced speed = (x − 15) km/hr

• Increased speed = (x + 60) km/hr

Let the total distance be D km.

Step 1: Form equations using time difference

According to the question:
When speed is reduced by 15 km/hr, the train becomes late by 45 hours.
So,
D/(x − 15) − D/x = 45
Taking LCM:
D[x − (x − 15)] / x(x − 15) = 45
15D / x(x − 15) = 45
D = 3x(x − 15) —— (1)
When speed is increased by 60 km/hr, the train arrives 72 hours earlier.
So,
D/x − D/(x + 60) = 72
Taking LCM:
D[(x + 60) − x] / x(x + 60) = 72
60D / x(x + 60) = 72
5D = 6x(x + 60) —— (2)

Step 2: Solve equations

From (1):
D = 3x(x − 15)
Substitute into (2):
5[3x(x − 15)] = 6x(x + 60)
15x(x − 15) = 6x(x + 60)
Divide by 3x:
5(x − 15) = 2(x + 60)
5x − 75 = 2x + 120
3x = 195
x = 65

Step 3: Find distance

Using equation (1):
D = 3 × 65 × (65 − 15)
= 3 × 65 × 50
= 9750 km

Given:

x = (√7 − √5)/(√5 + √7)

y is reciprocal of x.

So,

y = (√5 + √7)/(√7 − √5)

Now simplify x:

x = (√7 − √5)/(√5 + √7)

Multiply numerator and denominator by (√7 − √5):

x = (√7 − √5)²/(7 − 5)

= (7 + 5 − 2√35)/2

= (12 − 2√35)/2

= 6 − √35

Similarly,

y = 1/x = 6 + √35

Now,

x + y = 12

xy = 1

Using identity:

x³ + y³ = (x + y)³ − 3xy(x + y)

= 12³ − 3(1)(12)

= 1728 − 36

= 1692

Therefore,

√(x³ + y³)

= √1692

= √(36 × 47)

= 6√47

Given:

• Total students = 60

• Girls = 60% of the class

• Boys = the remaining students

• Average weight of boys = 62 kg

• Average weight of girls = 58 kg

Step 1: Find the number of boys and girls

Girls = 60% of 60
= (60/100) × 60
= 36
Boys = 60 − 36
= 24

Step 2: Find the total weight of boys and girls

Total weight of boys = 24 × 62 = 1488 kg
Total weight of girls = 36 × 58 = 2088 kg

Step 3: Find the average weight of the whole class

Total weight of the class = 1488 + 2088 = 3576 kg
Average weight = Total weight ÷ Total students
= 3576 ÷ 60
= 59.6 kg

Hence, the average weight of the whole class is 59.6 kg.

Given

• Each equal yearly instalment = Rs. 42135

• Rate of interest = 6% p.a., compounded annually

• Number of instalments = 2

Idea: The loan amount today equals the present value (PV) of all future instalments, discounted at the compound rate. Total interest = (sum of instalments paid) − (loan amount).

Step 1 — Present value of each instalment

First instalment (paid after 1 year):
PV₁ = 42135 ÷ 1.06 = Rs. 39750

Second instalment (paid after 2 years):
PV₂ = 42135 ÷ (1.06)² = 42135 ÷ 1.1236 = Rs. 37500

Step 2 — Loan amount (sum of present values)

Loan amount = PV₁ + PV₂ = 39750 + 37500 = Rs. 77250

Step 3 — Total amount repaid

Total repayment = 2 × 42135 = Rs. 84270

Step 4 — Total interest paid

Total interest = Total repayment − Loan amount = 84270 − 77250 = Rs. 7020

Hence, the total interest paid is Rs. 7020.

Given:
A and B together complete the work in 21 days.
Efficiency ratio:
A : B = 7 : 4
Let efficiencies be:
7x and 4x
Combined efficiency:
= 11x
Since total work is completed in 21 days:
11x = 1/21
x = 1/231
Efficiency of A:
= 7x
= 7/231
= 1/33
So, A alone completes 1 unit of work in 33 days.
Now,
Required work = 200/3 %
= 200/3 × 1/100
= 2/3 of the work
Time taken by A to complete 2/3 work:
= 33 × 2/3
= 22 days

Let a = 5.169 and b = 4 × 0.888 = 3.552.

The numerator becomes a³ + b³ and the denominator is a² - ab + b².

Using the identity a³ + b³ = (a+b)(a² - ab + b²), the expression simplifies to a + b.

Substituting values: 5.169 + 3.552 = 8.721.

Since the value is 8 + k, we have 8.721 = 8 + k, so k = 0.721.

Given:
√(12 − 2√35) + √(8 + 2√15) = √a + √b
We simplify each surd separately.
Step 1: Simplify √(12 − 2√35)
Assume:
√(12 − 2√35) = √m − √n
Squaring both sides:
12 − 2√35 = m + n − 2√mn
Comparing terms:
m + n = 12
mn = 35
The numbers satisfying these conditions are:
m = 7 and n = 5
Therefore,
√(12 − 2√35) = √7 − √5
Step 2: Simplify √(8 + 2√15)
Assume:
√(8 + 2√15) = √p + √q
Squaring both sides:
8 + 2√15 = p + q + 2√pq
Comparing terms:
p + q = 8
pq = 15
The numbers are:
p = 5 and q = 3
Therefore,
√(8 + 2√15) = √5 + √3
Step 3: Add the expressions
(√7 − √5) + (√5 + √3)
= √7 + √3
So,
a = 7 and b = 3
Therefore,
ab = 7 × 3 = 21

Given:
Third term of an Arithmetic Progression (AP) = 6
Let the AP be:
a − 2d, a − d, a, a + d, a + 2d
Here, the third term is:
a = 6
So the first five terms become:
6 − 2d, 6 − d, 6, 6 + d, 6 + 2d
Sum of first five terms:
= (6 − 2d) + (6 − d) + 6 + (6 + d) + (6 + 2d)
The d terms cancel out:
= 6 + 6 + 6 + 6 + 6
= 30

Given:

• A can complete the work in 12 days

• B can complete the work in 16 days

• Total payment = Rs. 8000

• With the help of C, the work is completed in 6 days

Step 1: Find efficiencies of A and B

Efficiency of A:
= 1/12 work per day
Efficiency of B:
= 1/16 work per day
Combined efficiency of A and B:
1/12 + 1/16
Taking LCM = 48
= 4/48 + 3/48
= 7/48

Step 2: Find total efficiency with C

Since the work is completed in 6 days:
Total efficiency of A + B + C
= 1/6
Efficiency of C:
= 1/6 − 7/48
Taking LCM = 48
= 8/48 − 7/48
= 1/48

Step 3: Find C’s share of money

Total work is divided according to efficiencies.
Ratio of efficiencies:
A : B : C
= 1/12 : 1/16 : 1/48
Multiplying by 48:
= 4 : 3 : 1
Total ratio:
4 + 3 + 1 = 8
C’s share:
= (1/8) × 8000
= 1000

First, factor the expression x⁴ + x²y² + y⁴ as (x² - xy + y²)(x² + xy + y²).

Substitute the given value 5 for (x² - xy + y²) to get 5(x² + xy + y²) = 29. Solve for x² + xy + y² to find it equals 5.8.

Subtract the original equation x² - xy + y² = 5 from this result to isolate 2xy. This gives 2xy = 0.8, so xy = 0.4.

Finally, multiply xy by 5 to find the value of 5xy is 2.

Assume total Cost Price (CP) of all goods = Rs. 100
Marked 40% above CP:
Marked Price (MP) = 140

Step 1: 30% goods sold at MP

Profit = 40%
Contribution to SP:
30% of 140
= 42

Step 2: 60% of remaining goods sold at 20% discount

Remaining goods after first sale:
100% − 30% = 70%
60% of remaining:
= 60% of 70%
= 42% of total goods
Selling price after 20% discount:
140 × 80/100 = 112
Contribution:
42% of 112
= 47.04

Step 3: Remaining goods sold at 40% discount

Remaining quantity:
100% − 30% − 42%
= 28%
Selling price after 40% discount:
140 × 60/100 = 84
Contribution:
28% of 84
= 23.52

Step 4: Total Selling Price

Total SP:
42 + 47.04 + 23.52
= 112.56
Total CP = 100
Profit:
112.56 − 100 = 12.56
Overall gain %:
12.56%

Let:

• Original distance = d

• Original speed = s

• Original time = t

Using the formula:
Speed = Distance / Time
So,
s = d/t
Now, according to the question:

• New distance = d/2

• New time = 3t

Let the new speed be s₁.
Then,
s₁ = (d/2)/(3t)
= d/6t
But,
s = d/t
Therefore,
s₁/s = (d/6t)/(d/t)
= 1/6

So,
s₁ : s = 1 : 6

Given:
x⁴ − 79x² + 1 = 0
We need to find the value of:
x³ + x⁻³
Step 1: Divide the equation by x²
x⁴/x² − 79x²/x² + 1/x² = 0
x² + 1/x² = 79
Step 2: Use identity
(x + 1/x)² = x² + 1/x² + 2
Substitute:
(x + 1/x)² = 79 + 2
= 81
x + 1/x = 9
Step 3: Use another identity
x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x)
Substitute:
= 9³ − 3 × 9
= 729 − 27
= 702

Concept: In blood-relation puzzles, a person who marries into a family takes an “in-law” relationship to their spouse’s parents: the husband of a couple’s daughter is their son-in-law. To find how any two people relate, first turn every sentence in the stem into a parent-child or marriage link, build one family tree from those links, then read the relationship straight off the tree.

Application:

1. Sukanya is Ayushman's wife, so Ayushman and Sukanya are married.

2. Lekha is Sukanya's sister, so Lekha and Sukanya share the same mother and father.

3. Nisha is Lekha's mother, and since Sukanya is Lekha's sister, Nisha is Sukanya's mother too.

4. Mukesh is Nisha's husband, which makes Mukesh Sukanya's father.

5. Piyush is Nisha's son, so Piyush is Sukanya's brother — Mukesh and Nisha's three children are Piyush, Lekha, and Sukanya.

6. Ayushman is married to Mukesh's daughter Sukanya, so Ayushman is Mukesh's son-in-law.

Cross-check: Viewed from Mukesh's side: his son Piyush's wife would be his daughter-in-law, while the husbands of his daughters Lekha and Sukanya are his sons-in-law. Ayushman, as Sukanya's husband, falls in that second group, confirming the relationship independently of the step-by-step trace.

Ayushman is Mukesh's son-in-law.

Identify the relationship in the given pair, Shark : Fish.
A shark is biologically classified as a type of fish (specifically, a cartilaginous fish). So the relationship is "member : its biological class".
Now test each option against this same "member : its biological class" pattern:
Bat : Bird → A bat is a mammal, not a bird. (Wrong class)
Dolphin : Fish → A dolphin is a mammal, not a fish. (Wrong class)
Whale : Mammal → A whale is biologically classified as a mammal. (Correct class)
Iguana : Amphibian → An iguana is a reptile, not an amphibian. (Wrong class)
Hence Whale : Mammal is the only pair that mirrors the logic of Shark : Fish.

- Write down the positional values of letters in PERSIST: P=16, E=5, R=18, S=19, I=9, S=19, T=20.
- Find the single-digit sum of the positional values (digital root) for each letter:
P -> 1 + 6 = 7
E -> 5
R -> 1 + 8 = 9
S -> 1 + 9 = 10 -> 1 + 0 = 1
I -> 9
S -> 1 + 9 = 10 -> 1 + 0 = 1
T -> 2 + 0 = 2
- Combining these digits gives 75911912, but the question states it is written as 7591912. Notice that the consecutive 'S' and 'I' positions are condensed, or the absolute values are written by grouping S(1), I(9), S(1) as 191. Let's apply the single-digit logic to ANCESTOR.
- Write down the positional values of letters in ANCESTOR: A=1, N=14, C=3, E=5, S=19, T=20, O=15, R=18.
- Convert each positional value to a single digit by adding its digits: A -> 1 N -> 1 + 4 = 5 C -> 3 E -> 5 S -> 1 + 9 = 10 -> 1 + 0 = 1 T -> 2 + 0 = 2 O -> 1 + 5 = 6 R -> 1 + 8 = 9
- Combining the resulting digits sequentially gives: 15351269.

Eligibility criteria (a candidate must satisfy all three):

1. Minimum 85% marks in the final-year examination.

2. At least 90% marks in the University's selection test.

3. Age between 21 and 25 years as on 01.01.2020.

Special relaxations:

• Fails (i) but has won at least one major Debate / Sports / Cultural prize -> refer the case to the Director.

• Fails (ii) and is from the reserved category -> a 2-year age relaxation is allowed.

Checking Deepinder Bhalla:

• Final-year marks = 79% -> below the required 85% (fails (i))

• Selection-test marks = 92% -> at least 90% (meets (ii))

• Born 17 Dec 1997 -> age on 01.01.2020 = 22 years, within 21-25 (meets (iii))

• Won 3 inter-university debate competitions -> a major Debate prize (satisfied)

She satisfies (ii) and (iii) and falls short only on (i). The reserved-category age relaxation does not apply, because she clears (ii) and no reserved-category status is given. However, having won a major Debate award, she fits the first special relaxation, so her case must be referred to the Director.

Correct answer: The case is to be referred to the Director of the Institute.

PLEAP → APPLE (Fruit)
SEHCWA → CASHEW (Nut)
RPOATIC → APRICOT (Fruit)
EOGARN → ORANGE (Fruit)

Only CASHEW is not a fruit.

Concept: When a large painted cube is cut into smaller unit cubes, only the cubes lying strictly inside (not touching any outer face) stay unpainted. These inner cubes form a smaller cube called the core.

Formula: Number of unpainted cubes = (n - 2)^3, where n is the number of unit cubes along one edge (Edge of large cube / Edge of small cube).

Given: Edge of the large cube = 4 cm, edge of each small cube = 1 cm.

Step 1 - Find n: n = 4 cm / 1 cm = 4.

Step 2 - Apply the formula: Unpainted cubes = (4 - 2)^3 = 2^3 = 2 x 2 x 2 = 8.

Therefore, exactly 8 of the smaller cubes have no paint on any face. The answer is 8.

Concept. A vertical plane mirror held to the right of a horizontal string produces a reflection in which two changes happen together: (1) the order of the characters is reversed end-for-end, and (2) every individual character is inverted left-to-right (laterally flipped). A figure is the exact mirror image only if BOTH changes are present.

Application to pR4GTF7Dq.

1. Reverse the order. Reading pR4GTF7Dq from right to left gives the character sequence q, D, 7, F, T, G, 4, R, p.

2. Laterally flip each glyph. q becomes p; D becomes a back-to-front D (bowl on the left); 7 becomes a reversed 7; F becomes a reversed F; T stays T (it is symmetric about a vertical axis); G becomes a reversed G; 4 becomes a reversed 4; R becomes a reversed R; and the final p becomes q.

3. Read off the result. The exact mirror image therefore begins with an upright p, ends with an upright q, and in between shows a reversed-D, reversed-7, reversed-F, upright T, reversed-G, reversed-4 and reversed-R. This is the figure whose first character is p and last character is q with every interior glyph turned back-to-front.

Cross-check. Confirm both conditions on the end characters, which are the easiest to verify: the original starts with p, so the reflection must END with q (a true reflection swaps p and q). A figure that ends in an upright D and d, or that begins with q and ends with p, has reversed the order but failed to flip one or more glyphs, so it is not the exact mirror image.

Concept

A triangle is any closed three-sided region whose three sides each run entirely along lines that are actually drawn. To count every triangle without missing or repeating any, work by SIZE: first count the smallest atomic triangles, then count the larger triangles that are built by joining two or more adjacent smaller ones along a shared line.

Applying it to this figure

The figure is a central square split into a 2×2 grid (one horizontal and one vertical midline) with a diamond inscribed by joining the four side-midpoints, plus one triangular wing attached on the left and one on the right. Count level by level:

1. Smallest triangles: inside the square, each of the 4 grid cells is cut by one diamond edge into 2 triangles, giving 4 × 2 = 8 smallest triangles.

2. Half-diamond triangles: joining two adjacent smallest triangles that share a diamond diagonal gives the four half-diamond triangles — top half, bottom half, left half and right half of the diamond. That is 4 triangles.

3. Wing triangles: the left wing is one triangle and the right wing is one triangle, giving 2 more.

Total = 8 + 4 + 2 = 14 triangles.

Cross-check

No larger composite triangle is created where a wing meets the square: a wing's slanted edge and the square's side do not continue into one straight line, so they cannot serve as a single side of a bigger triangle. Inside the square the only slanted lines are the four diamond edges, and the square has no corner-to-corner diagonals, so there are no half-square triangles. Re-counting by size confirms 8 + 4 + 2 with nothing left over, so the figure contains exactly 14 triangles.

- Let the expenditure of a single student in sections A, B, C, and D be a, b, c, and d respectively.
- Step 1: Establish the relational equations from the problem statement: 5a = 4b -> a = (4/5)b 12b = 9c -> c = (12/9)b = (4/3)b 6c = 8d -> d = (6/8)c = (3/4)c
- Step 2: Convert all variables into terms of 'b' to create a unified ratio: a = 0.8b c = 1.333b d = (3/4) (4/3)b = b
- Step 3: Set up the total expenditure equation using the student counts (A=15, B=24, C=15, D=20):
Total Expense = 15a + 24b + 15c + 20d = 23,370
Substitute the values: 15(4/5)b + 24b + 15(4/3)b + 20(b) = 23,370
Simplify: 12b + 24b + 20b + 20b = 23,370 76b = 23,370 b = 23,370 ÷ 76 = 307.50
- Step 4: Find the expenditure for Section D: * Since d = b, a student of Section D spent exactly Rs 307.50.

Concept

In a figure-series, treat the 8 marked cells of the box — the four corners and the four edge-midpoints — as points on a ring. Each shape advances by its own fixed number of these cells, in its own fixed direction, every frame; once that step size is confirmed from two consecutive frames, applying it once more predicts the next frame exactly.

Application

Track each shape across the three given boxes (positions named by where they sit inside the box):

• Star: middle-left → top-centre → middle-right. It advances two cells clockwise every frame, so next it lands at bottom-centre.

• Oval: top-centre → middle-left → bottom-centre. It moves two cells anticlockwise every frame, so next it lands at middle-right.

• Diamond: bottom-centre → top-right → middle-left. It advances a fixed step every frame (three cells anticlockwise), so next it lands at the bottom-right corner.

• Down-arrow: middle-right → bottom-right → bottom-centre (sharing the bottom edge with the oval in this frame). It advances one cell clockwise every frame, so next it lands at bottom-left.

Cross-check

The first two boxes alone fix every shape's step unambiguously, since no two shapes share a cell there: the star's middle-left-to-top-centre move and the down-arrow's middle-right-to-bottom-right move each cover exactly one named cell, and the oval's and diamond's moves are equally clear. Applying each shape's confirmed step a second time reproduces the third box exactly — including the down-arrow and oval both arriving at bottom-centre together, which is why that box shows them side by side. Applying the same steps once more, independently for each shape, is what fixes the fourth figure.

Result

Combining every shape's own fixed step gives one figure that satisfies all four simultaneously: the down-arrow at bottom-left, the oval at middle-right, the star at bottom-centre, and the diamond at the bottom-right corner. Every other figure breaks exactly one of these — either the diamond fails to advance at all, the diamond reverses to an earlier cell, or the down-arrow advances one cell too many.

Concept

These are class-relationship (syllogism) diagrams. The link between any two named groups is one of three kinds: one group is wholly a sub-type of the other (one circle drawn entirely inside the other), the groups partly share members (two circles that overlap), or the groups have nothing in common (two circles drawn fully apart). The correct diagram is the one whose circle arrangement matches the real-world relationship of every pair.

Application

Test each pair among Pen, Chair and Stationery:

• Pen and Stationery: a pen is a kind of stationery, so the whole Pen group sits within the Stationery group — the Pen circle must lie completely inside the Stationery circle.

• Chair and Stationery: a chair is furniture, not stationery, so they share no members — the Chair circle must stay completely apart from the Stationery circle.

• Pen and Chair: a pen is not a chair and a chair is not a pen, so these two also share no members — their circles must stay apart.

The arrangement that satisfies all three at once is: one smaller circle fully enclosed by a larger circle, plus a third circle standing entirely on its own, separate from both.

Cross-check

Reject any diagram with overlap (overlap would wrongly claim a chair is partly a pen or partly stationery) and reject any diagram that places two circles inside one outer circle (that would wrongly make a chair a type of stationery). Only the enclosed-circle-plus-one-detached-circle layout keeps Pen inside Stationery while leaving Chair unrelated to both.

Step 1 — Build one row. "X is second to the right of Z" places Z and X at the two ends of the same row, with one seat between them. Since "V is a neighbour of Z", V takes that middle seat. So this row reads: Z – V – X.

Step 2 — Build the other row. "Y is not at the end of any row", so Y sits in the middle of the opposite row. "W is the neighbour of Y" and W is diagonally opposite X, so W takes the end of that row that lies diagonally across from X. The only person left, U, takes the remaining end. So this row reads: U – Y – W (or W – Y – U), with U opposite Z.

Step 3 — Apply the interchange. V and U swap seats. U moves into V's old middle seat, which sits between Z and X.

Therefore, after the swap, U's neighbours are Z and X, which is option 4.

- Given relation string: R - V × T ÷ W
- Breakdown of relationships from left to right:
- R - V: 'R is the wife of V' (R is female, V is male, and they are a married couple: R(-) ⇔ V(+))
- V × T: 'V is brother of T' (V and T are siblings, sharing the same generation: V(+) — T)
- T ÷ W: 'T is the daughter of W' (T is female, and W is the parent of T: W is one generation above T)
- Generation and family tree tracking: Since V and T are siblings, W is the parent of both V and T. R is married to V, which means R is the wife of W's son.
- Therefore, R is the daughter-in-law of W.

Convert each statement into a height comparison (taller > shorter) and merge them into a single chain.

Statement 1: "Ravi is shorter than Kartik but taller than Gagan and Tarun" → Kartik > Ravi > Gagan, Tarun.

Statement 1 (continued): "…Gagan and Tarun, who are taller than Saksham" → Gagan, Tarun > Saksham.

Statement 2: "Sunil is taller than Aarav but shorter than Saksham" → Saksham > Sunil > Aarav.

Combined chain (tallest → shortest): Kartik > Ravi > Gagan/Tarun > Saksham > Sunil > Aarav.

Kartik sits at the top of the chain, so Kartik is the tallest of all.

The relationship between the words is based on intensity/degree.
- A 'Quarrel' is a minor argument, which escalates in intensity to a 'War'.
- Similarly, an 'Error' is a minor mistake, which escalates in intensity to a 'Blunder' (a massive, serious mistake).

The Logic: Each letter of the word is shifted by an alternating add/subtract pattern of consecutive natural numbers: -1, +2, -3, +4, -5, +6, -7.

Verifying the given pair REQUEST -> QGNYZYM:

• R - 1 = Q

• E + 2 = G

• Q - 3 = N

• U + 4 = Y

• E - 5 = Z

• S + 6 = Y

• T - 7 = M

So REQUEST is coded as QGNYZYM. The fourth term OTBHDIM is therefore the code of the missing word, so we reverse the shifts (+1, -2, +3, -4, +5, -6, +7) on OTBHDIM:

• O + 1 = P

• T - 2 = R

• B + 3 = E

• H - 4 = D

• D + 5 = I

• I - 6 = C

• M + 7 = T

The missing word is PREDICT. (Check: applying -1, +2, -3, +4, -5, +6, -7 to PREDICT gives back OTBHDIM.)

Given word: DOCUMENTARY — its letters are D, O, C, U, M, E, N, T, A, R, Y, each appearing exactly once.

Rule: a word can be formed only if every letter it needs is available in DOCUMENTARY, and no letter is used more times than it appears there.

• AERODUCT (A, E, R, O, D, U, C, T): every letter is available — can be formed.

• EDUCATOR (E, D, U, C, A, T, O, R): every letter is available — can be formed.

• DEMOCRACY (D, E, M, O, C, R, A, C, Y): it needs the letter 'C' twice, but DOCUMENTARY contains 'C' only once — cannot be formed.

• MONETARY (M, O, N, E, T, A, R, Y): every letter is available — can be formed.

Hence DEMOCRACY is the word that cannot be formed from the letters of DOCUMENTARY.

Concept: In this non-verbal "classification / odd-one-out" item, each figure pairs one outer polygon with one smaller inner polygon. The figures that belong together follow a consistent rule linking the pair — here, the outer and inner shapes always have the same number of sides. The one figure that breaks this side-count match is the odd one out.

Application — checking each figure:

Cross-check: Figures A, B and D each keep the outer and inner shape's side count identical (3-3, round-round, 5-5), confirming they follow the pairing rule. Figure C pairs a 6-sided hexagon with a 4-sided diamond — the side counts do not match, so Figure C alone violates the pattern the other three follow.

Hence, Figure C is the figure that is different from the other three alternatives.

- The Logic: Track the movement of the two symbols (Diamond '✧' and Plus '+') across the boxes dynamically.
- Movement of Diamond (✧):
- Box 1: Left-middle position
- Box 2: Bottom-middle position
- Box 3: Right-middle position
- Pattern: It moves clockwise along the sides (Left -> Bottom -> Right). Therefore, in Box 4, it must move to the Top-middle position.

- Movement of Plus (+):
- Box 1: Bottom-middle position
- Box 2: Top-middle position
- Box 3: Bottom-middle position
- Pattern: It alternates back and forth between Bottom and Top positions. Therefore, in Box 4, it must move to the Top-middle position.
- Final Combined Figure: Both the Diamond (✧) and the Plus (+) will sit at the top-middle area together, with the '+' symbol positioned slightly above/beside the diamond. This matches Option 3.

- The Logic: The statements describe nested relationships (concentric circles): Beetroots inside Carrots, Carrots inside Radishes, and Radishes inside White.
- Checking Conclusion I: Since all beetroots are inside carrots, all carrots are inside radishes, and all radishes are inside white, all beetroots are automatically inside white. (Follows)
- Checking Conclusion II: "All white are carrots" moves from the outermost circle to an inner circle, which is incorrect. Only "some white are carrots" is true. (Does not follow)
- Checking Conclusion III: "All radishes are beetroots" moves from an outer circle to an inner circle, which is incorrect. Only "some radishes are beetroots" is true. (Does not follow)
- Checking Conclusion IV: Since all carrots are inside radishes, and all radishes are inside white, all carrots are automatically inside white. (Follows)
- Therefore, only conclusions I and IV follow.

Concept

In a Statement–Courses of Action item, a proposed course of action follows only if it is a practical, feasible administrative step that a responsible authority would reasonably take in response to the stated situation, and if it does not work against the underlying purpose implied by the statement. A course of action that is impractical, merely restates the problem, or conflicts with that purpose does not follow, even if it appears superficially related to the situation.

Application to This Item

Statement:
It is estimated that 25 lakh people will visit the International Trade Fair at Pragati Maidan, Delhi.

We need to determine which course of action logically follows.

Analysis of Course of Action I:

“The Delhi Police should be on high alert and discourage people from visiting the fair.”

• Being on high alert may be reasonable for crowd management.

• However, discouraging people from visiting the fair is not logical because trade fairs are organized to attract visitors, not prevent them.

Therefore, Course of Action I does not logically follow.

Analysis of Course of Action II:

“The management of the fair should have enough ticket booths and shuttle services to avoid long queues.”

• A very large crowd is expected.

• Additional ticket booths and shuttle services would help manage the crowd efficiently.

Therefore, Course of Action II logically follows.

Why the Other Options Don't Hold

• Both I and II follow — this fails because course of action I's instruction to discourage attendance conflicts with the very purpose of holding a trade fair; grouping it with the genuinely sound course of action II is not justified.

• Neither I nor II follows — this fails because the ticket-booth and shuttle-service measure genuinely responds to the anticipated 25-lakh crowd and does not conflict with the fair's purpose, so ruling out every course of action is too sweeping a conclusion.

• Only I follows — this fails because approving the whole of course of action I overlooks that its discourage-attendance clause conflicts with the fair's purpose, even though the high-alert part alone would be reasonable.

Correct Option:
Only II follows

Step 1 — Spot the pattern. Count the letters in each word and compare with its code.

ARDENT has 6 letters and is coded 13.
CANOE has 5 letters and is coded 11.

Step 2 — Find the rule. In both cases, (number of letters × 2) + 1 gives the code:

ARDENT: (6 × 2) + 1 = 12 + 1 = 13 ✓
CANOE: (5 × 2) + 1 = 10 + 1 = 11 ✓

Step 3 — Apply to REPRESENT. REPRESENT has 9 letters.

Code = (9 × 2) + 1 = 18 + 1 = 19.

Hence, the code for REPRESENT is 19.

Pattern: in each group, divide the first number by 2 and square the result to get the second number, i.e. (First ÷ 2)² = Second.

18 → (18 ÷ 2)² = 9² = 81 ✓

22 → (22 ÷ 2)² = 11² = 121 ✓

38 → (38 ÷ 2)² = 19² = 361 ✓

28 → (28 ÷ 2)² = 14² = 196, but the group shows 195, so this group does not follow the rule.

Hence the group 28 - 195 is differently related from the other three.

The pattern works on the four digits of the two numbers: multiply the two digits of the first number, multiply the two digits of the second number, then add the two products.
Rule: (tens digit of 1st number × units digit of 1st number) + (tens digit of 2nd number × units digit of 2nd number) = Result

Verify with the given equations:
94 × 18 = 44 → (9 × 4) + (1 × 8) = 36 + 8 = 44 ✓
89 × 23 = 78 → (8 × 9) + (2 × 3) = 72 + 6 = 78 ✓

Apply the same rule to the unsolved equation:
75 × 36 = ? → (7 × 5) + (3 × 6) = 35 + 18 = 53

So 75 × 36 = 53.

Find the differences between consecutive terms of the series:

7 − 2 = 5
14 − 7 = 7
23 − 14 = 9
? − 23 = 11 → ? = 23 + 11 = 34
47 − 34 = 13

The differences are consecutive odd numbers: +5, +7, +9, +11, +13. Adding 11 to 23 gives the missing term, 34.

Correct answer: the segment 'many large number of'.

The sentence compares the number of educated workers in two groups (India and Sri Lanka vs Brazil) using 'than'. A comparison with 'than' must be supported by a comparative form. The phrase 'many large number of' provides no comparative and is itself ill-formed: it wrongly stacks 'many' (which goes straight before a plural noun, e.g. 'many workers') on top of the quantifier 'a large number of'. English uses one or the other, never both together.

The corrected sentence reads: 'India and Sri Lanka have a larger number of educated workers than Brazil' (or 'India and Sri Lanka have more educated workers than Brazil').

The remaining segments are fine: 'India and Sri Lanka' is a valid compound subject, the plural verb 'have' agrees with it, and 'educated workers than Brazil' is well-formed once the comparative is restored. Hence the error lies only in 'many large number of'.

The sentence is:
“The prize money was shared ______ all the members ______ the football team.”

• We use “among” when something is distributed within a group of more than two people.

• We use “of” with “members of the football team.”

So the correct sentence is:
“The prize money was shared among all the members of the football team.”
Correct Answer:
(c) among; of

The word that best replaces 'explaining something clearly' is 'exposition'. This noun refers to a comprehensive explanation or presentation of information, aligning with the teacher's skill in communication.

• Astrology → The study or belief that the movements and positions of stars and planets influence human life and events.
  Example: She believes in astrology and reads her horoscope daily.

• Imagination → The ability to form ideas, pictures, or concepts in the mind.
  Example: Children have a very active imagination.

• Prediction → A statement or forecast about what may happen in the future.
  Example: The weather prediction says it will rain tomorrow.

• Foresight → The ability to anticipate or think wisely about future events and consequences.
  Example: Her foresight helped the company avoid losses

The idiom 'watered down' means made weaker, milder, or less effective — just as a drink loses its strength when water is added to it.

In the sentence, the amended law is described as a 'watered down' version of the earlier law. This means the new law is weaker and less strict or forceful than the original — its strength has been diluted, not that it is hard to apply, made stronger, or literally placed in water.

Hence, the correct meaning of 'watered down' here is 'weaker or diluted'.

“Cautioned” means warned or advised carefully, which fits the context best.
The sentence requires a word indicating warning about safety. 'Cautioned' fits context of advising safe riding, while other options lack this meaning.

The correct spelling is accommodate, which has a double “c” and a double “m”: a-cc-o-mm-o-d-a-t-e.

Meaning: to provide space or lodging for someone, or to adjust or adapt to someone’s needs.

1. To provide space or lodging for someone
   Example: The hotel can accommodate 200 guests.

2. To adjust or adapt to someone’s needs or requests
   Example: The teacher accommodated the students by extending the deadline.

3. To fit in or make room for something
   Example: This bag can accommodate all your books.

Tip: Like “committee” and “occurrence”, this word doubles its consonants — keep both the double “c” and the double “m”.

The correct answer is c. fell
Correct sentence:
“Mohan Lal fell asleep while watching TV last night.”
Reason:
The phrase “last night” indicates a completed action in the past, so the simple past tense is required.

The highlighted word is "humility", which means modesty and a low or unassuming view of one's own importance.

The antonym we need is a word expressing the opposite quality – excessive self-importance. "Pride" fits exactly: it denotes a high or inflated opinion of one's own worth, the direct opposite of humility. ("Arrogance" and "conceit" are close synonyms of pride.)

"Submission" sits on the humble side of the scale rather than opposing it, while "concern" and "compassion" relate to care for others and are unrelated to the pride–humility contrast. Hence "pride" is the correct answer.

The word 'priceless' in the sentence describes treasures that are so valuable they cannot be assigned a monetary cost.
Therefore, we need to find the option with the closest meaning.\nOption D 'valuable' is the correct choice because it directly relates to high worth or importance, capturing the essence of 'priceless' in this context.
While 'memorable' (Option A) means worth remembering, it does not necessarily imply high monetary or intrinsic value.
'Cheap' (Option B) is an antonym, meaning low in cost.
'Prized' (Option C) means highly valued or cherished, which is close, but 'valuable' is the more direct synonym for something that has immense worth or price.
Thus, 'valuable' best replaces the highlighted word while maintaining the sentence's meaning.

The correct conjunction is 'so that' because it introduces a purpose clause explaining why you want to tell the story. This structure shows the relationship between telling something and achieving understanding.

The correct answer is 'ignore' (Option 2) as it is the synonym for 'neglect'.

Sentences with ignore

1. You should not ignore your teacher’s advice.

2. He ignored the warning signs and continued driving.
   
   Sentences with neglect

1. Parents should never neglect their children’s education.

2. The garden dried up because it was neglected for many weeks.

The correct sentence order is: 'If you can, you should try to take a walk every day.'
This arrangement places the subject first, followed by the modal verb and main action, with the conditional clause at the end.

Antonym questions test whether you know a word's core sense precisely enough to state its exact opposite — not just any word that looks or sounds related. Many English words cluster in families sharing a root (here, “-ceed” and “-cede”), so distractors are often built from that root while expressing a different relationship (such as sequence/order) instead of the one actually being tested (direction of movement).

The highlighted word “proceed” means to move forward or carry on in a course of action. Reversing this specific sense of forward movement gives “to move backward, withdraw, or go away” — exactly what “receding” expresses.

• proceeding — carries the same forward-moving sense as the highlighted word, so it is a near-synonym, not an opposite.

• succeeding — means coming after in sequence or achieving a goal; this concerns order, not direction of movement.

• preceding — means coming before in sequence; again an order relationship, not the movement-direction opposite required here.

Because “receding” is the only option that reverses the specific sense of forward movement carried by “proceed,” it is the correct antonym.

The sentence expresses a third conditional situation (an unreal past condition).
Correct structure:
Had + subject + past participle, ... would have ...
So the sentence becomes:
“Had he not resigned on his own, they would have been forced to dismiss him from service.”
Correct Answer:
(a) Had he not resigned

We need to identify the correctly spelled word.

• enaugarate → incorrect

• innagarate → incorrect

• inaugurate → correct

• inagurate → incorrect

The correct spelling is “inaugurate.”
Correct Answer:
(c) inaugurate

The error lies in the phrase “one kind.”

The intended expression is the idiom “one of a kind,” which means unique or the only one of its type. Omitting “of a” leaves the phrase incomplete and changes its meaning, so “one kind” is grammatically wrong here.

Corrected sentence: The coin collector was willing to pay a huge sum for the rare coin because it was one of a kind.

The sentence is:
“It is claimed that ______ in the telecom industry will bring the charges down for people.”
The word that best fits the context is “competition.”
More competition among telecom companies usually leads to lower prices for consumers.
Correct Answer:
(c) competition

The expression “whip up” means to excite, provoke, or stir up emotions in people.
In the sentence, the politician is trying to emotionally excite the crowd through his speech.
Correct Answer:
(b) stir emotionally

The correct sequence is Option C.
The paragraph begins by stating the observable phenomenon: 'Every day we read and hear about the crimes committed by younger people.' This sets the context.
Next, it introduces contributing factors, starting with media influence: 'Movies depict cruelty and violence as heroic rather than vicious.'
The third sentence adds another cause using the transition 'In addition,' linking parental neglect to the previous point.
Finally, the sentence starting with 'The result is that...' logically concludes the paragraph by summarizing the outcome of these combined factors.
Options A, B, and D fail because they either place the conclusion ('The result...') too early or disrupt the logical flow between causes and effects.
Specifically, Option A starts with movies but jumps to crime statistics before explaining the parental factor, creating a disjointed narrative.
Option B places the conclusion immediately after mentioning parents, which is premature since the media influence hasn't been introduced yet.
Therefore, Option C provides the most coherent cause-and-effect structure.

The correct answer is c. When Ratan asked him to take her with him, he laughed at the idea.

This incident does not provide comfort to the Postmaster. In fact, it relates to Ratan’s expectations and emotional hurt. The Postmaster’s laughter is directed at Ratan’s assumption, not something that relieved his homesickness. The comforting elements in the passage are his writing verses and talking to Ratan about his family, which eased his loneliness.

The correct answer is b. sometimes

The phrase “at times” means that something happens on some occasions but not regularly. In the sentence, it indicates that the Postmaster wrote verses occasionally when he felt like it, not frequently or habitually.

The correct answer is c. Human relationships have no monetary value.

The passage highlights emotional bonds between the Postmaster and Ratan, showing attachment, care, separation, and regret. The final lines especially emphasize that relationships cannot be measured or replaced materially (he gives money, but it has no emotional value for Ratan). The central idea is the emotional depth and non-material nature of human relationships.

The correct answer is a. sentimental

The passage focuses on emotional experiences—loneliness, attachment, care, separation, regret, and hope. Both the Postmaster and Ratan go through deep emotional conflict, and the ending highlights longing and unresolved feelings. This strongly reflects a sentimental tone.

The correct answer is b. Ratan was very much attached to the postmaster.

Ratan’s behaviour at the end of the passage shows that she continues to wander around the post office and still “fondly hopes” that the Postmaster will return. This clearly reflects her emotional attachment and deep affection for him, even after his departure and rejection of her expectations.

A schema diagram is a pictorial or graphical representation of a database structure. It visually maps out the entire organization of the database by displaying the relation names (tables), their attributes (columns), primary keys, and the foreign key relationships that link different tables together.

To solve this Preemptive Shortest Job First (SRTF) problem, we track the execution step-by-step using a Gantt chart based on the processes' remaining burst times:

• At t = 0: Only P0 arrives (Remaining Burst Time, RBT = 9). It executes.

• At t = 1: P1 arrives (RBT = 4). Since P1's burst time (4) is shorter than P0's remaining time (8), P0 is preempted. P1 runs until completion at t = 5 (since no other shorter job arrives at t=2, where P2 arrives with RBT = 9).

• At t = 5: Ready processes are P0 (RBT = 8) and P2 (RBT = 9). P0 executes next from t = 5 to t = 13.

• At t = 13: Finally, P2 executes from t = 13 to t = 22.

Waiting Time (WT = Turnaround Time - Burst Time):

• P0: (13 - 0) - 9 = 4 seconds

• P1: (5 - 1) - 4 = 0 seconds

• P2: (22 - 2) - 9 = 11 seconds

Average Waiting Time ={4 + 0 + 11}/3 = 15/3 = 5 seconds

Thus, the correct option is D.

The given code implements Euclid's algorithm. It repeatedly calls function(y, x % y) until y <= 0. When the remainder becomes 0, the current x is the greatest common divisor of the two numbers. Therefore, the function computes the GCD, also called HCF.

Correct option: A

The working set model is explicitly based on the principle of locality. This principle states that as a process executes, it moves from one locality (a set of pages that are actively used together) to another. By tracking and keeping the active pages (the working set) of a process in main memory, the operating system effectively minimizes page faults and prevents thrashing.

Thus, the correct option is A.

The correct answer is A: Stack. Depth First Search (DFS) explores a graph by going as deep as possible along each branch before backtracking. This Last-In, First-Out (LIFO) behavior is inherently managed using a Stack data structure (either explicitly or implicitly via recursion).

The encryption key e is selected such that 1 < e < φ(n) and gcd(e, φ(n)) = 1. This ensures e is coprime to φ(n), allowing a valid decryption key d to exist.

The State-Entry Problem X asks if a Turing Machine M ever visits a specific state q on input w. It is undecidable because solving it would easily solve the Halting Problem (by making the halting state q). However, it is partially decidable (recursively enumerable) because we can simulate M on w; if it ever enters state q, we can halt and answer "yes". Thus, Option A is correct.

The Commutative Law states that changing the sequence of variables in a logical operation does not affect the final output. In Boolean algebra, this applies to both OR (addition) and AND (multiplication) gates:

• OR operation: A + B = B + A

• AND operation: A . B = B . A

Thus, the correct expression matching the law is given in Option C.

To determine if the grammar S → m | mn | mno is LL(k), we find the minimum lookahead k to distinguish productions.

• FIRST3(m) = {m}

• FIRST3(mn) = {mn}

• FIRST3(mno) = {mno}

Since all three lookahead sets of length 3 are completely disjoint, a parser needs exactly 3 symbols of lookahead to make a unique decision. Thus, it is LL(3).

Correct Answer: The grammar is LL(3) (Option B).

To solve this, we must identify the Class of the IP address based on its first octet.

1. Identify the Class: The IP address is 128.252.144.84.

• Class A: 1 – 126

• Class B: 128 – 191

• Class C: 192 – 223

Since the first octet is 128, this is a Class B address.

2. Determine Network and Host ID: In Classful addressing, Class B uses a specific format for the 32-bit address:

• Network ID: The first 2 octets (16 bits).

• Host (Node) ID: The remaining 2 octets (16 bits).

3. Apply to the given IP:

• Network ID: 128.252.0.0 (The first two octets are kept, and the rest are set to zero).

• Node ID: While the "Node ID" can refer to the host portion (0.0.144.84), in many examination contexts, it is represented by showing the full IP to denote which specific node belongs to that network. Option D follows the standard convention of identifying the Network portion clearly.

Statement I is false because Packet Filter Firewalls primarily operate at the Network Layer (Layer 3), inspecting IP addresses and ports. Statement II is true; Circuit-Level Gateways operate at the Session Layer (Layer 5) to monitor TCP handshakes and session validity.

The correct option is C. AES (Advanced Encryption Standard) is a symmetric key symmetric cipher that uses the same secret key for both encryption and decryption. In contrast, El Gamal, Rabin, and RSA are all asymmetric (public key) cryptosystems that utilize a mathematically related public-private key pair.

The correct option is A. Let the initial state be q0 and the final state be q1. To reach the final state q1 and stay there, any sequence of inputs can be processed as long as the last symbol processed is 'a'. Every path ending at q1 must end with an 'a' transition from q0 or a self-loop 'a' on q1. Thus, the language consists of all strings over {a, b} ending with 'a', represented by (b + a + b)*a or simply (a+b)*a.

Intermediate Code Generation (ICG) acts as a machine-independent bridge between the front-end (source language) and back-end (target architecture) of a compiler. By targeting a standard abstract machine model, the front-end remains unchanged when porting the compiler to a new architecture. Only a new back-end needs to be written, significantly increasing compiler portability.

Correct Answer: It enhances the portability of the compiler system program (Option D).

The correct answer depends on the height definition used. If height is the number of edges on the longest path, the maximum height of a skewed binary tree with n nodes is (n - 1). If height is measured by nodes (levels), it is n. Based on Option D being marked correct, this platform defines height by edges, making Option D the intended correct choice, despite the conflicting feedback text in Option C.

Hash-table collision resolution splits into two families based on WHERE a colliding key ends up being stored. Open addressing (also called closed hashing) keeps every colliding key inside the hash table's own array, relocating it to another slot within that same array via a probe sequence (linear probing, quadratic probing, double hashing) — no extra memory is created. Close addressing (also called closed addressing, or chaining / open hashing) instead creates a separate auxiliary structure for each slot specifically to hold colliding keys — most commonly a linked list — so a colliding key is stored outside the table's own array, in memory allocated only when a collision happens.

Applying that test to the two named techniques: the technique described as contiguous storage places a colliding key into another position within the hash table's own array — no separate collision-handling memory is created for it, matching the open-addressing pattern. The technique described as linked storage creates a new node in a structure attached to the slot specifically to hold the colliding key, outside the table's own array — matching the close-addressing pattern exactly.

This lines up with how the two terms are traditionally split: close addressing (also called closed addressing) names the array-external, per-slot chaining approach, while open addressing (occasionally also called closed hashing — a separate point of terminology overlap) names the in-place, same-array approach. A technique that keeps every key inside the table's own array — however that array is organised — sits in the open-addressing family rather than close addressing.

So the linked-storage technique (Chains) is the close-addressing technique among the two listed here; the contiguous-storage technique (Buckets) belongs to the open-addressing family instead.

In set-associative mapping, a cache is partitioned into equal-size sets and each main-memory block is assigned to exactly one set. The governing rule is: set index = (block number) mod (number of cache sets). The modulus is the number of sets, not the number of cache blocks.

Applying the rule to this item:

1. Notation check: 2c and 2cm in the source are ordinary products, 2 × c and 2 × c × m; they are not powers of 2.

2. Cache size: the cache contains 2c blocks.

3. Associativity: two blocks are stored in each set.

4. Number of sets: S = (2c cache blocks) / (2 blocks per set) = c.

5. Mapping: block k is assigned to set k mod c.

Therefore, block k of main memory maps to set (k mod c) of the cache.

Contrast with the other values:

• (k mod m) uses the main-memory multiplier m rather than the cache-set count.

• (k mod 2c) uses the total number of cache blocks before grouping them two per set.

• (k mod 2cm) uses the total number of main-memory blocks rather than the cache-set count.

Cross-check: 2cm main-memory blocks distributed across c sets give 2cm / c = 2m candidate blocks per set, which is consistent with c cache sets.

Both statements are correct. A Full Binary Tree strictly restricts internal nodes to exactly two children (0 or 2 children total per node). A Complete Binary Tree ensures all levels are packed full sequentially from top-to-bottom, and the final level fills strictly from left-to-right without gaps.

• Disk Block Size (DBS): 1 KB = 210 bytes

• Disk Block Address (DBA) size: 32 bits = 4 bytes

• Number of entries per index block (N): DBS / DBA size = 1 KB / 4 bytes = 1024/4 = 256 = 28

Maximum File Size Formula:

Max File Size =(Direct DBAs + 1 x N + 1 x N2 + 1 x N3 x DBS)

The highest-order term (N3) dominant value determines the maximum approximate size:

Max File Size = N3 x DBS = (28)3 x 1 KB} = 224 x 210 bytes = 234 bytes = 16 GB

Thus, the exact maximum size is just slightly over 16 GB. The correct option is C.

A trigger follows the Event-Condition-Action framework:

1. Event: The change (like INSERT or UPDATE) that triggers the process.

2. Condition: A test or query to check if the action should proceed.

3. Action: The actual procedure executed. The correct option is D.

The Session Layer (Layer 5) establishes, manages, maintains, and terminates connections (sessions) between local and remote applications. This directly includes authentication and authorization steps like logging in and logging out.

The correct option is A. X.509 is a standard ITU-T format used to define the structure of digital certificates. It binds a public key to an identity using a digital signature from a trusted Certificate Authority (CA) and is widely used in TLS/SSL security protocols.

The ALTER TABLE command is used to change the structure of an existing table. It allows you to add, delete, or modify columns and constraints (like UNIQUE, CHECK, or FOREIGN KEY) without recreating the entire table. The correct option is D.

A symbol is useful if it is both generating and reachable.

1. Generating Symbols: S and P derive terminals (S → p, P → p). Q has no production rule, so it is non-generating.

2. Elimination: Since Q is useless, the production S → PQ is eliminated.

This leaves S → p and P → p. Now, P cannot be reached from the start symbol S, making P unreachable. Removing P leaves only S→ p.

Correct Answer: S → p (Option B).

The correct option is C (Leftmost derivation).

A top-down parser constructs the parse tree from the root (start symbol) down to the leaves (input tokens). When scanning the input string from left to right, it always expands the leftmost non-terminal symbol first at each step. This process corresponds exactly to finding a Leftmost Derivation (LMD) for the input string.

A Linked List is the most efficient data structure for polynomial addition. Each node represents a term containing a coefficient, an exponent, and a pointer to the next term. It dynamically handles any number of non-zero terms sorted by exponent order without wasting contiguous memory space. The correct option is C.

A super key is any set of one or more attributes whose values uniquely identify each tuple in a relation. A candidate key is only a minimal super key, and a primary key is a chosen candidate key. Since the question does not require minimality, the correct term is super key. Therefore, Option B is correct.

When a function calls itself recursively, the system needs to remember the execution state of the previous call. It uses a Stack data structure (called the system call stack) operating on a LIFO (Last-In, First-Out) basis to push active activation records and pop them when returning.

The capability of LR parsers to handle languages follows a strict hierarchy. A Canonical LR (CLR) parser is the most powerful bottom-up parser because it maintains distinct lookaheads for every state. LALR reduces the states by merging those with identical core items, which can introduce reduce-reduce conflicts, making CLR strictly more powerful than LALR.

Correct Answer: Canonical LR parser is more powerful than LALR parser (Option D).

A page fault is a hardware interrupt triggered when a running program references a virtual memory page that is not currently loaded into the physical main memory (RAM). The operating system must then fetch the required page from secondary storage (hard disk/SSD).

Thus, a page fault occurs when the page is not in main memory. Correct option is C.

A Queue follows the FIFO (First-In, First-Out) principle. Breadth First Search (BFS) explores a graph level-by-level, requiring nodes of the current level to be processed completely before moving to the next. A queue is uniquely suited to keep track of these nodes in the exact order they are discovered.

An Arithmetic Logic Unit (ALU) is a fundamental component of the CPU that handles calculations and logical decision-making.

• Statement 1 is correct: The ALU performs arithmetic operations such as addition (+), subtraction (-), multiplication (X), and division (/).

• Statement 2 is correct: It performs logical comparisons like "greater than" (>), "less than" (<), and "equal to" (=).

• Statement 3 is incorrect: Communicating with I/O devices is the responsibility of the Control Unit (CU) and the Bus system, not the ALU.

• Statement 4 is incorrect: Monitoring or managing the system is handled by the Control Unit and the Operating System.

Cycle Stealing is a method used in Direct Memory Access (DMA) data transfer. In this mode, the DMA controller "steals" or borrows a bus cycle from the CPU to transfer a single word of data.

• Since the CPU and DMA controller share the same system bus, the DMA controller requests the CPU to pause its bus activity for one clock cycle.

• The CPU suspends its operation for that specific cycle, allowing the DMA to transfer data directly between the I/O device and memory.

• This is highly efficient because the CPU doesn't stop entirely; it just slows down slightly while data is transferred in the background.

To find the Excess-3 code of a decimal number, add 3 to the given decimal number and then convert the resulting sum into its 4-bit binary equivalent:

1. Add 3 to the decimal digit:

   2 + 3 = 5

2. Convert the sum (5) to 4-bit binary:

   (5)10 = 01012

Thus, the Excess-3 code for decimal 2 is 0101. The correct option is A.

Concept

Under demand paging, a memory reference either finds its page already resident in main memory or it misses and must wait for the page to be brought in from secondary storage. The page-fault rate P is the probability of a miss. The Effective Memory Access Time (EMAT) is the probability-weighted average of the fast in-memory access and the slow fault-handling path:

EMAT = (1 - P) x M + P x S

Here M is the time for a normal main-memory access and S is the full page-fault service time (the complete cost of fetching the page and restarting the reference). The weights (1 - P) and P sum to 1, so EMAT always lies between M and S, moving toward S as the fault rate rises.

Application

1. Identify the given values: page-fault rate P = 0.65, main-memory access M = 20 ns, page-fault service time S = 100 ns.

2. Compute the no-fault contribution: (1 - P) x M = (1 - 0.65) x 20 = 0.35 x 20 = 7 ns.

3. Compute the fault contribution: P x S = 0.65 x 100 = 65 ns.

4. Add the two weighted terms: EMAT = 7 + 65 = 72 ns.

Cross-check

Sanity bound: EMAT must fall between M = 20 ns and S = 100 ns, and 72 ns does. Because the fault rate is high (0.65), the result should sit well above the midpoint of M and S and lean toward S, which 72 ns does, confirming the weighting is applied in the right direction.

The balance factor of a node is the difference between left subtree and right subtree.

The processing power of a CPU is not just about how fast it "thinks," but also how much data it can handle at once and how much memory it can access.

1. Data Bus Width: This determines how many bits of data the CPU can move in a single cycle. A 64-bit data bus can move twice as much data as a 32-bit bus, directly increasing throughput.

2. Clock Speed: This is the frequency at which a processor executes instructions. A higher clock speed (measured in GHz) means the CPU can perform more operations per second.

3. Address Bus Width: This determines the maximum amount of RAM the CPU can address (use). If the address bus is too narrow, the CPU is limited in the amount of memory it can utilize, which can create a bottleneck for data-heavy processing tasks.

A D flip-flop (Data or Delay flip-flop) transfers its input directly to the output after a clock trigger. The characteristic equation is Qnext = D. This means whatever logic level is present at the D input will appear exactly at the output state Q on the next active clock edge.

Thus, the correct option is B.

Data-manipulation language (DML) provides the functional capability to query data and perform modifications like inserting, deleting, or updating records (tuples) within the table structure. Unlike DDL, which defines the schema, DML interacts directly with the data itself. The correct option is B.

An Operator Grammar must satisfy two strict conditions:

1. No production can have an ε (null) on the right-hand side.

2. No two non-terminals (variables) can appear adjacently on the right-hand side.

• Rule i (A → BC): Violates condition 2 because B and C are adjacent variables.

• Rule ii (A → CcBb): Valid. Variables are separated by terminal c.

• Rule iii (A → BaC): Valid. Variables are separated by terminal a.

• Rule iv (A → ε): Violates condition 1 due to the null production.

Thus, i and iv violate the requirements.

Correct Answer: i and iv (Option D).

The Principle of Locality (or Locality of Reference) is the tendency of a processor to access the same set of memory locations repetitively over a short period of time. This principle is the fundamental reason why Cache Memory works effectively.

There are two main types:

1. Temporal Locality: If a memory location is accessed once, it is likely to be accessed again soon (e.g., loops).

2. Spatial Locality: If a memory location is accessed, nearby memory locations are likely to be accessed soon (e.g., arrays).

The correct option is D. In challenge-response authentication, the verifier sends a challenge (like a random nonce). The claimant processes it using their secret key and returns the response. This proves possession of the secret without ever sending the secret itself over the network.

In a Binary Search Tree, the first inserted element becomes the root. Here the root is 45.

LST (elements < 45): 35, 40, 33 -> 3 nodes
RST (elements > 45): 65, 70, 60, 75, 69 -> 5 nodes

Thus, LST = 3 and RST = 5. The correct option is C.

The core of this problem lies in the expression: x = (x = x + y) - (y = x - y);. This is an example of an expression that technically results in undefined behavior in standard C because it modifies the same variable multiple times without a sequence point. However, most modern compilers (and competitive exam environments) evaluate it using a specific mathematical sequence.

1. Initial Values: x = 5, y = 9.

2. Evaluating the Parentheses:

   • First sub-expression (x = x + y): x becomes 5 + 9 = 14. The value of this sub-expression is 14.

   • Second sub-expression (y = x - y): Now, x is 14. So, y becomes 14 - 9 = 5. The value of this sub-expression is 5.

3. Final Assignment for x:

   • x = (14) - (5) = 9.

4. Final Result: x = 9, y = 5.

A candidate key is any minimal attribute or set of attributes that can uniquely identify a row.

A primary key is the candidate key selected as the main unique identifier for the table. It must be unique and cannot contain NULL values.

Since the question asks for the selected main unique identifier, the correct answer is primary key.

Correct option: D

The correct option is B (FTTT). Statement A is false because the symbol table is accessed by all phases of a compiler. Statements B, C, and D are true: pretty printers format code structurally, YACC generates LALR(1) parsing tables, and the LALR(1) class of grammars is strictly larger than SLR(1).

The language requires strings to have a length >= 2 ending in 00 or 11. The minimal DFA needs 5 states:

1. Initial state (empty string)

2. Read a 0 (ends in 0)

3. Read a 1 (ends in 1)

4. Read 00 (Final state for 00)

5. Read 11 (Final state for 11)

A Stack follows the LIFO (Last In First Out) or FILO (First In Last Out) principle. The element inserted first is placed at the bottom, so it can only be removed after all subsequent elements above it are popped out. Thus, Option A is correct.

A sorting algorithm is defined as stable if it preserves the relative order of records with equal keys (identical values) as they appeared in the original input. If two items have identical keys, their initial positional sequence remains completely unchanged after the sorting process finishes.

Solution:

Given,

• Inorder: D B E F A G H C

• Preorder: A B D E F C G H

From the preorder traversal, the first node is the root.

Root = A

Splitting the inorder traversal about A:

• Left subtree: D B E F

• Right subtree: G H C

Left Subtree

Preorder: B D E F

• Root = B

• Inorder left of B = D

• Inorder right of B = E F

Thus:

• Left child of B = D

• Right child of B = E

• Right child of E = F

Right Subtree

Preorder: C G H

• Root = C

• Inorder left of C = G H

Thus:

• Left child of C = G

• Right child of G = H

The tree is:

        A
      /   \
     B     C
    / \   /
   D   E G
        \  \
         F  H

Now perform Postorder Traversal (Left → Right → Root):

• Left subtree of A: D F E B

• Right subtree of A: H G C

• Root: A

Therefore, the Postorder Traversal is:

D F E B H G C A

Answer: D F E B H G C A

The correct option is A. In SQL, functions like avg, min, max, sum, and count are called aggregate functions. They process a collection of multiple values across rows within a column and collapse them to return a single, summarized mathematical value.

Concept: In C, a sequence point marks a place where all side effects of prior evaluations are guaranteed complete before evaluation continues. Between two sequence points, modifying the same scalar object more than once, or modifying it while also reading its value for something other than computing the new stored value, is undefined behavior. The C standard also leaves the order in which a function call’s arguments are evaluated unspecified — a compiler may evaluate them left-to-right, right-to-left, or in any other order, and may schedule the associated side effects at different points.

1. The call is printf("%d %d %d", i++, i, ++i); with i initialized to 5.

2. All three arguments — i++, i, and ++i — are evaluated as part of setting up the same function call; the only sequence point here occurs right before the actual call, after every argument has been evaluated.

3. i++ modifies i (post-increment), ++i also modifies i (pre-increment), and the middle argument reads i directly — that is two modifications and one plain read of the same object across the three argument evaluations, with no sequence point separating them from each other.

4. Because no sequence point separates these evaluations, the compiler is free to choose any order among the three arguments and to apply the two increments at any point relative to the plain read — the standard places no obligation on what value the plain i sees or in which order the increments occur.

Cross-check: Different compilers, and even the same compiler under different optimization settings, are free to print different values for this exact program without violating the standard, because the standard simply does not define an outcome here. Assuming one specific evaluation order — say, strict left-to-right with each side effect applied immediately — only reproduces ONE of the many results a conforming compiler could legally produce; it does not establish what "the" output is.

Therefore, the only defensible classification for this program is undefined behavior — not any specific triplet of printed numbers.

Concept

LRU (Least Recently Used) replacement: on a page reference, if the page is already resident it is a hit; otherwise it is a page fault. When all frames are full and a fault occurs, the page that was used farthest back in time (the least recently used) is evicted to make room. The key discipline is to track the recency of each resident page, because that alone decides who is evicted — and an eviction can force a later re-reference of that same page to fault again.

Application — trace with 4 frames

Reference string = 2, 0, 1, 2, 4, 0, 5, 1, 4, 6, 4, 2, 1, 3, 0 (frames = 4). The bracket shows the resident set after each access.

1. 2: [2] → Fault (1)

2. 0: [2, 0] → Fault (2)

3. 1: [2, 0, 1] → Fault (3)

4. 2: [2, 0, 1] → Hit

5. 4: [2, 0, 1, 4] → Fault (4)

6. 0: [2, 0, 1, 4] → Hit

7. 5: [2, 0, 4, 5] — evict 1 (least recently used) → Fault (5)

8. 1: [0, 4, 5, 1] — evict 2 (least recently used) → Fault (6)

9. 4: [0, 4, 5, 1] → Hit

10. 6: [4, 5, 1, 6] — evict 0 → Fault (7)

11. 4: [4, 5, 1, 6] → Hit

12. 2: [4, 1, 6, 2] — evict 5 → Fault (8)

13. 1: [4, 1, 6, 2] → Hit

14. 3: [4, 1, 2, 3] — evict 6 → Fault (9)

15. 0: [1, 2, 3, 0] — evict 4 → Fault (10)

Why the count is 10, not 9: at the access to 5, the four resident pages are {2, 0, 1, 4} and their last-use times make 1 the least recently used, so 1 is evicted. The very next access is to 1, which is therefore no longer resident — it faults. Evicting any other page here (for example 2, which was used more recently) would be an LRU violation and is the usual source of the undercount.

Cross-check

Counting the fault markers above: faults occur at references 2, 0, 1, 4, 5, 1, 6, 2, 3, 0 — ten of the fifteen accesses miss, leaving five hits (at 2, 0, 4, 4, 1). Total page faults = 10.

The Rename operator is denoted by the Greek letter rho (ρ). It is used to rename either the relation (table) itself or the attributes (fields) within a relation. This is particularly useful in self-joins or to avoid ambiguity. The correct option is A.

Key idea: search time in a BST is proportional to the tree height, h.

• Balanced/average case: the height is O(log n), so search time is O(log n).

• Skewed worst case: the height can become O(n), so search can degrade to linear time. That is why the balanced-case qualifier is important.

Answer: O(log n)

ACID properties ensure reliable database transactions. They stand for:

1. Atomicity: All or nothing.

2. Consistency: Database remains in a valid state.

3. Isolation: Concurrent transactions don't interfere.

4. Durability: Changes are permanent after commit. The correct option is C.

The correct option is A. NAND gates (along with NOR gates) are functionally complete, universal logic gates. This means any fundamental boolean operation (AND, OR, NOT) and consequently any complex logic expression can be implemented entirely using only NAND gates.

The mutual exclusion problem arises when multiple concurrent processes access shared, non-shareable resources (like a shared variable, file, or printer) simultaneously. To prevent data inconsistency and race conditions, only one process must be allowed to execute its critical section at any given time.

Thus, the mutual exclusion problem occurs among processes that share resources. Correct option is C.

To search for an element in a singly or doubly linked list of length n, you must traverse it sequentially starting from the head node. In the worst case (if the element is at the end or not present), you have to check all n nodes. Therefore, the time complexity is O(n). Thus, Option C is correct.

Correct answer: Option B (0.0843 s and 1536 KBps)

Important unit note:
KBps means kilobytes per second. Kbps means kilobits per second. Since each sector size is given in bytes, the computed transfer rate is in KBps. If converted to bits, 1536 KBps = 1536 × 8 = 12288 Kbps.

Given:
Average seek time = 32 ns ≈ 0 s
Rotational speed = 360 rpm = 6 rotations/second
Sectors per track = 512
Sector size = 512 bytes

Track size:
512 sectors × 512 bytes = 262144 bytes = 256 KB

Average rotational latency:
One rotation takes 1/6 s.
Average rotational latency = (1/2) × (1/6) = 1/12 = 0.0833 s

Transfer time for 4 continuous sectors:
Data to read = 4 × 512 bytes = 2048 bytes = 2 KB
Transfer rate = 6 × 256 KB/s = 1536 KBps
Transfer time = 2/1536 s ≈ 0.00130 s

Total time:
Seek time + average rotational latency + transfer time
≈ 0 + 0.0833 + 0.00130
≈ 0.0846 s, represented by the closest option 0.0843 s.

Therefore, the intended answer is 0.0843 s and 1536 KBps.

The correct option is D. MSB stands for Most Significant Bit. In a binary number, it is the leftmost bit that carries the highest positional weight (2n-1). It determines the largest value component and often indicates the sign in signed number representations.

Shortest Job First (SJF) scheduling algorithm is provably optimal because it minimizes the average waiting time for a given set of processes. By scheduling the process with the shortest CPU burst time first, the waiting times of subsequent longer processes are reduced, moving the overall mathematical average to its minimum possible value.

Thus, the correct option is C.

1. Statement 1 is correct: Insertion Sort and Merge Sort preserve the relative order of equal elements, making them inherently stable.

2. Statement 2 is correct: Heap Sort uses a heap tree structure where non-adjacent elements are swapped, making it inherently unstable.

3. Statement 3 is correct: Standard Selection Sort is unstable due to long-distance swaps, but it can be made stable by changing the swapping mechanism to an insertion-like shift.

Therefore, all statements are correct. The correct option is B.

The correct option is C. Statement C is false because a subset of a recursively enumerable (RE) language is not necessarily recursive, or even RE. Since the set of all languages is uncountable but RE languages are countable, an RE language can have uncountably many subsets, most of which are non-RE.

In Tuple Relational Calculus (TRC), a query is expressed as { T | p(T) }. Here, T represents the tuple variable and p(T) is a formula (predicate) that describes the condition the tuple must satisfy. It non-procedurally describes "what" data is needed. The correct option is A.

Given Grammar:

S → AbBaCc | ε
A → aAb | ba
B → BBC | cb
C → cCa | ac

FIRST(A) = { a, b }

Since:
S → AbBaCc | ε

and A does not derive ε,

FIRST(S) = FIRST(A) ∪ { ε }

Therefore,

FIRST(S) = { a, b, ε }

Concept: In stop-and-wait flow control, the sender transmits one frame and then waits idle for its acknowledgement before sending the next. Link utilization (also called efficiency) is the fraction of time the link actually carries data, given by Tt / (Tt + 2Tp), where Tt is the transmission time and Tp is the one-way propagation time. The factor 2Tp accounts for the round trip the frame and its acknowledgement travel. This is a pure dimensionless ratio between 0 and 1 — it is not the throughput in bits per second.

Application: Substitute the given values step by step.

1. Transmission time Tt = frame size / bit rate = 4800 / 9600 = 0.5 s.

2. Propagation time Tp = distance / propagation speed = 2000 / 200000 = 0.01 s.

3. Efficiency = Tt / (Tt + 2Tp) = 0.5 / (0.5 + 2 × 0.01) = 0.5 / 0.52 = 0.9615.

Cross-check: Because the transmission time (0.5 s) is much larger than the round-trip propagation overhead (2 × 0.01 = 0.02 s), the sender stays busy almost the whole time, so the efficiency must be close to 1 — and 0.9615 ≈ 0.96 confirms this. (For reference, the actual data throughput would be 0.96 × 9600 ≈ 9231 bps; the utilization asked here is the ratio 0.96, not this bps value.)

Result: The link utilization is approximately 0.96.

The correct option is D (The 10's complement of a decimal number is equal to 9's complement of given number plus one).

Mathematically, for a base-r number system, the r's complement is calculated by adding 1 to the least significant digit of its (r-1)'s complement. For decimal numbers (base r=10), the 10's complement is obtained by finding the 9's complement first and then adding 1 to it (10's Complement = 9's Complement + 1).

The maximum throughput (S) of Slotted ALOHA is achieved when the load G = 1, given by S = G . e-G. For G = 1, S = 1/e = 0.368 or 36.8%. This is exactly double the efficiency of Pure ALOHA (18.4%).

The correct option is D. Not every regular grammar is LL(1) because left-recursion or common prefixes can cause conflicts. However, every regular set (language) can be represented by at least one deterministic, conflict-free LR(1) grammar because regular languages are a strict subset of deterministic context-free languages (DCFLs).

Given: Each RAM chip has capacity 1024 × 1, which means 1024 bits = 1K bit.

• Required memory: 16K bytes = 16K × 8 bits = 128K bits.

• Capacity of one chip: 1K bit.

• Number of chips: 128K bits ÷ 1K bit = 128.

Answer: 128 chips

The language L = {an b2m | n, m >= 1} requires at least one a, followed by an even, non-zero number of b's (minimum string is abb). Constructing the minimal DFA requires: an initial state, a state tracking the first a, a state for the first b, an accepting state for the second b, and a dead state for invalid structures. Total states = 5. Thus, Option C is correct.

In computer architecture, the access time refers to the time it takes for the CPU to retrieve data from a storage location. The memory hierarchy is designed such that the closer a storage unit is to the CPU's execution unit, the faster (lower access time) it is.

Registers are located directly inside the CPU. They are the smallest and fastest storage elements in a computer system, operating at the same speed as the processor itself.

The correct option is A: A + B. Using the redundancy law (A + ĀB = A + B), the first two terms simplify to A + B. Adding the third term gives (A + B) + AB̅. Applying the distributive law, A(1 + B̅) + B = A + B.

Correct Answer: C (Both I and II are true)

• Statement I is true: By definition, a language is context-free (CFL) if and only if there exists a Pushdown Automaton (PDA) that accepts it, establishing a perfect equivalence.

• Statement II is true: Structurally, a PDA is formally defined as a finite automaton enhanced with an external, infinite memory storage known as a pushdown stack.

A Deque (Double-Ended Queue) is a linear data structure that generalizes a queue. It permits the flexibility of inserting and deleting elements from both structural endpoints: the front and the rear. Unlike standard queues (FIFO) or stacks (LIFO), it does not restrict access to a single side. The correct option is B.

An edge-triggered flip-flop changes states only during specific clock transitions. A negative edge-triggered flip-flop responds when the clock pulse drops from a logic high state (1) to a logic low state (0). This falling edge initiates the data transfer from input to output.

Thus, the correct option is C.

The given relational algebra expression selects boats with color = 'red', performs a natural join (⋈) across Boats, Reserves, and Sailors, and projects (π) only sname. In SQL, a natural join is achieved via cross-product in the FROM clause combined with matching primary/foreign key equality predicates (S.sid = R.sid AND R.bid = B.bid) in the WHERE clause alongside the filter condition.

Merge Sort follows the Divide and Conquer paradigm. The recurrence relation is T(n) = 2T(n/2) + θ(n). Solving this using Master's Theorem or a recursion tree gives a consistent time complexity of O(n.log n) for best, average, and worst cases. The correct option is C.

The correct answer is Option D (Articulation point / Cut vertex). In graph theory, an articulation point (or cut vertex) is any vertex whose removal increases the number of connected components in the graph, thereby disconnecting a previously connected graph.

SOP stands for Sum of Products. A Canonical SOP form is a logic expression where every individual product term contains all the variables of the function (either in complemented or uncomplemented form). These individual product terms are called Minterms. Therefore, a canonical SOP expression consists entirely of a sum of minterms.

Thus, the correct option is D.

CONCEPT: A state in the canonical LR(0) automaton carries a shift-reduce or reduce-reduce conflict only when TWO DIFFERENT items in that SAME state compete for the same lookahead symbol. SLR(1) restricts each reduce action to the lookaheads in FOLLOW of the reduced nonterminal — one GLOBAL, context-blind set covering every place that nonterminal ever appears in the grammar. LALR(1) instead computes, per state, the precise lookaheads a valid derivation can actually deliver there, which is never broader than FOLLOW (FOLLOW is the union across every context, not just the ones reaching this one state) — so wherever SLR(1)'s coarser FOLLOW-based check already avoids a clash in a state, LALR(1)'s narrower, state-specific lookaheads cannot manufacture a new clash there. So if a state in this grammar's LR(0) automaton never holds more than one completed item, and never holds a completed item alongside a shift action on the same symbol, there is only ONE production's item present to begin with — no amount of lookahead splitting can introduce a second, competing item into that state, so both SLR(1) and LALR(1) succeed there.

APPLICATION:

1. FOLLOW(A): A occurs in S → Aa (followed by 'a') and S → bAc (followed by 'c'), so FOLLOW(A) = {a, c}.

2. Build the canonical LR(0) item sets. The start state contains S′→•S, S→•Aa, S→•bAc, S→•dc, S→•bda, A→•a.

3. On symbol 'a' from the start state, the automaton moves to the state {A→a•} — a single completed item with no other item sharing that state.

4. After prefix 'b', the state {S→b•Ac, S→b•da, A→•a} also transitions on 'a' to that SAME state {A→a•} (identical item set), so the two paths converge on one state instead of colliding inside it.

5. Every other position that completes a handle — {S→Aa•}, {S→dc•}, {S→bAc•}, {S→bda•} — is likewise reached as a state containing exactly that one completed item, with no shift action or second reduce competing for any symbol in that same state.

6. So no state in the LR(0) automaton ever holds two competing actions on the same input symbol — the construction has zero shift-reduce or reduce-reduce conflicts even before any lookahead restriction is applied.

CROSS-CHECK: Every reduce-bearing state found above — {A→a•}, {S→Aa•}, {S→dc•}, {S→bAc•}, {S→bda•} — carries exactly one item, so there is never a second, competing item for LALR(1)'s finer per-state lookaheads to clash with. The FOLLOW(A) = {a, c} overlap only matters when it forces two DIFFERENT items into the same state, which never happens here. So the grammar succeeds under SLR(1) (the coarser check), and therefore also under LALR(1), whose lookaheads are never broader than SLR(1)'s.

Result: the grammar is parsable by both SLR(1) and LALR(1).

The correct option is A (First normal).

By definition, a relation is in the First Normal Form (1NF) if and only if the domain of each attribute contains only atomic (indivisible) values. It strictly prohibits multi-valued attributes, composite attributes, or non-atomic collections like lists and sets within a single field.

A collision occurs when two or more distinct keys are mapped to the same hash value or "bucket" by a hash function. If that primary bucket is already full (occupied), the new record cannot be stored there, resulting in a collision. The correct option is B.

The correct option is A (ALTER VIEW).

In database dialects that support explicit view recompilation (such as Oracle SQL), the ALTER VIEW view_name COMPILE; statement is used to manually recompile an invalid or outdated view. This occurs when underlying base tables are modified, ensuring dependencies are verified without recreating the view definition.

Concept

Booth's algorithm multiplies two signed numbers held in two's-complement form. Each register is interpreted as a signed two's-complement value (a leading 1 marks a negative number), and the algorithm produces the ordinary signed product of those two values. So the first step is always to decode each register into its signed decimal value, then multiply.

Application

1. Decode the multiplicand register (11101). It has 5 bits and the leading bit is 1, so it is negative. Taking the two's complement: invert 11101 -> 00010, then add 1 -> 00011 = 3. Hence the multiplicand = -3.

2. Decode the multiplier register (1100). It has 4 bits and the leading bit is 1, so it is negative. Two's complement: invert 1100 -> 0011, then add 1 -> 0100 = 4. Hence the multiplier = -4.

3. Multiply the decoded signed values: (-3) x (-4). A negative times a negative is positive, with magnitude 3 x 4 = 12.

4. Therefore the product produced by Booth's algorithm is 1210.

Cross-check

Range sanity check: a 5-bit value lies in [-16, 15] and a 4-bit value in [-8, 7], so any product is bounded by 16 x 8 = 128 in magnitude. A magnitude of 12 sits comfortably inside this range, while values such as 812 are far outside it and cannot be a valid product of these registers - confirming that 1210 is the result.

The question describes operations on relations that parallel algebraic operations on numbers, making Option C (relational algebra) the correct concept. However, the platform has incorrectly marked Option A as correct. Relational algebra is a procedural query language containing formal operators like selection, projection, and join to manipulate relations.

The correct option is D (Intersection). The class of Context-Free Languages (CFLs) generated by Context-Free Grammars is closed under union, concatenation, and Kleene star operations. However, it is not closed under intersection (and complementation).

A unary operation is an operation that is performed on a single relation. In Relational Algebra, Projection (𝝅) and Selection (σ) are primary unary operations. Intersection, Join, and Cartesian Product are binary operations because they require two relations to operate. The correct option is B.

Concept: Prim's algorithm grows a Minimum Spanning Tree from a start vertex by repeatedly adding the cheapest edge that crosses from the tree to a vertex outside it (the cut property). To pick that cheapest crossing edge fast, every vertex keeps a key = the smallest edge weight connecting it to the current tree, and these keys are stored in a min-priority queue so that extract-min always returns the next vertex to attach.

Initialization: every vertex starts with key = infinity, then the start (root) vertex's key is set to 0 so it is the first one extracted. So the root's key is 0 while all other vertices begin at infinity and are lowered as cheaper edges are discovered.

Application — checking each statement against this method:

• "binomial max heap for the priority queue" — Prim's needs the MINIMUM-weight vertex each step, which a min-priority queue (min-heap) provides. A max-heap returns the largest key, so it cannot drive Prim's; this claim does not hold.

• "root key initialized to 0" — consistent with the initialization above: the start vertex's key is set to 0 (others are infinity), so this holds.

• "O(E log V) using a binary heap" — with a binary heap each of the O(E) decrease-key and O(V) extract-min operations costs O(log V), giving O(E log V); this holds.

• "O(E + V log V) using a Fibonacci heap" — a Fibonacci heap does decrease-key in O(1) amortized and extract-min in O(log V) amortized, giving O(E + V log V); this holds.

Cross-check: three statements match standard Prim's behaviour, so the single statement that is false about Prim's algorithm is the one claiming a binomial max heap can represent the priority queue.

Correct Answer: D (II and IV)

• II is Undecidable: Checking if a context-free language is regular is a well-known undecidable problem (proven via Rice's Theorem or reduction from the Post Correspondence Problem).

• IV is Undecidable: Type 0 languages are Recursively Enumerable (RE). The membership problem for $RE$ languages is equivalent to the Halting Problem, which is undecidable.

• I & III are Decidable: Membership for CFLs is decidable (CYK algorithm). A Finite State Automaton (DFA/NFA) always halts on all inputs because it processes exactly one input symbol per transition and has no infinite loops without consuming input; hence, its halting problem is trivially decidable.

DMA (Direct Memory Access) allows data transfer directly between I/O devices and memory without CPU intervention. This mechanism improves system efficiency by freeing the processor from managing data transfers manually.

Dr. Vinaya Shetty became the first Indian and Asian to be elected as the General Vice President (GVP) of the International Organisation Development Association (IODA) since its inception in 1986. She is a management consultant from Mumbai and has been associated with IODA's organizational development initiatives since 2009

The Ministry of Finance launched ICEDASH and ATITHI initiatives to improve monitoring, pace of Customs clearance, and facilitate arriving international passengers.

The solution must explain that Bajaj Allianz General Insurance launched the 'Farmitra' mobile app to assist farmers.

India decided not to join the Regional Comprehensive Economic Partnership agreement. Hence India is the correct answer.