All practice questions

Question 4041Question →

You are serving as the Incident Commander following a major toxic chemical vapor release near a residential township. According to standard Disaster Management protocols and the Incident Command System (ICS), arrange the following emergency response actions in the correct chronological sequence from first to last:

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct chronological sequence is: (1) Establish the Incident Command Post (ICP) at a safe upwind location and formalize operational command authority; (2) Enforce immediate hazard perimeter isolation and issue shelter-in-place alerts for downwind civilian areas; (3) Deploy specialized hazardous materials (HazMat) mitigation units to seal the leak under continuous atmospheric monitoring; (4) Conduct post-mitigation environmental safety audits and authorize public re-entry into evacuated zones.

Under standard crisis management protocols, emergency response must follow a strict logical progression: first establishing unified command at a safe upwind base, second protecting public life safety through perimeter isolation, third executing tactical hazard containment, and fourth validating environmental safety before authorizing public re-entry.

Step-by-Step Solution

1

Determine the primary operational setup required by statutory emergency protocols.

Setting up the Incident Command Post upwind and formalizing command is step 1.

Without establishing central command and a safe vantage point, initial emergency actions risk being uncoordinated and endangering responders.

2

Identify immediate public safety and perimeter control measures.

Enforcing hazard isolation and notifying downwind residents to shelter in place is step 2.

Life safety of civilian populations takes immediate precedence over tactical operations once command is active.

3

Sequence specialized tactical hazard stabilization.

Deploying HazMat containment teams to stop the chemical leak under continuous air quality monitoring is step 3.

Tactical containment addresses the root threat while real-time data ensures responder safety.

4

Establish final recovery and de-escalation procedures.

Performing environmental clearance testing and issuing safe re-entry orders is step 4.

Re-occupancy of impacted zones can only occur after empirical testing confirms zero toxic residue.

Key Concept

Standard Operating Procedure (SOP) Sequence in Incident Command Systems

Estimated Time:1m 30s

Question 4042Question →

In a family, $X$ is the sister of $Y$ . $Y$ is married to $Z$ , who is the only son of $W$ . $V$ is the mother-in-law of $Z$ . How is $V$ related to $X$ ?

Show answer & explanation

Answer: Mother; mother

Answer

Mother

V

is the mother-in-law of

Z

, meaning

V

is the mother of

Z

's wife,

Y

. Since

X

is the sister of

Y

,

V

is also the mother of

X

.

Step-by-Step Solution

1

Identify the relationship between

Z

,

Y

, and

V

V

is the mother-in-law of

Z

. Since

Z

is married to

Y

,

V

is the mother of

Y

.

A person's mother-in-law is the mother of their spouse.

2

Connect

Y

's mother to

X

X

is the sister of

Y

, which means

X

and

Y

share the same mother.

Siblings share the same parents.

3

Deduce the relation of

V

to

X

Since

V

is the mother of

Y

and

X

is

Y

's sister,

V

is the mother of

X

.

The mother of a person's sibling is that person's mother.

Key Concept

Kinship and In-Law Relationship Deduction

Question 4043Question →

In a survey of $100$ residents in a locality, $60$ residents read Newspaper X, $50$ read Newspaper Y, and $20$ read both Newspaper X and Newspaper Y. How many residents read neither Newspaper X nor Newspaper Y?

10

20

30

40

Show answer & explanation

Answer: $10$

Answer

The number of residents who read neither Newspaper X nor Newspaper Y is 10.

The total number of residents who read at least one newspaper is given by

N(X \cup Y) = N(X) + N(Y) - N(X \cap Y) = 60 + 50 - 20 = 90

. Thus, the number of residents who read neither newspaper is

100 - 90 = 10

.

Step-by-Step Solution

1

Calculate the number of residents who read at least one newspaper using the Inclusion-Exclusion principle.

|X \cup Y| = 60 + 50 - 20 = 90

Simply adding total readers of X and Y double-counts the residents who read both newspapers.

2

Subtract the number of residents reading at least one newspaper from the total surveyed population.

Neither =

100 - 90 = 10

The universe of residents consists of those reading at least one newspaper and those reading neither.

Key Concept

Principle of Inclusion-Exclusion for two sets

Estimated Time:1m 0s

Question 4044Question →

What is the value of the following mathematical expression?

25 - [18 - \{12 - (6 - \overline{4 - 1})\}]

16

18

22

12

Show answer & explanation

Answer: 16

Answer

16

Following the BODMAS priority (Vinculum

\rightarrow

Round Brackets

\rightarrow

Curly Braces

\rightarrow

Square Brackets

\rightarrow

Outermost Subtraction):
1. Vinculum:

\overline{4 - 1} = 3

2. Round brackets:

6 - 3 = 3

3. Curly braces:

12 - 3 = 9

4. Square brackets:

18 - 9 = 9

5. Outer subtraction:

25 - 9 = 16

.
Thus, 16 is the mathematically correct answer.

Step-by-Step Solution

1

Evaluate the expression under the vinculum bar

\overline{4 - 1} = 3

According to BODMAS, vinculum (bar bracket) has the highest priority.

2

Simplify the innermost round brackets

(6 - 3)

6 - 3 = 3

Evaluate terms inside round brackets next.

3

Simplify the curly braces

\{12 - 3\}

12 - 3 = 9

Evaluate terms inside curly braces next.

4

Simplify the square brackets

[18 - 9]

18 - 9 = 9

Evaluate terms inside square brackets next.

5

Perform final subtraction from 25

25 - 9 = 16

Complete the outermost operation.

Key Concept

BODMAS Rule with Vinculum (Bar Bracket)

Question 4045Question →

In a survey of $120$ civil service aspirants, $65$ read Newspaper A, $55$ read Newspaper B, and $45$ read Newspaper C. Additionally, $25$ read both A and B, $20$ read both B and C, $15$ read both A and C, and $8$ read all three newspapers. How many aspirants read exactly two of these newspapers?

36

60

52

44

Show answer & explanation

Answer: 36

Answer

36 aspirants read exactly two newspapers.

To find the number of aspirants who read exactly two newspapers, we must subtract the number of aspirants who read all three newspapers (8) from each of the two-newspaper intersection groups. The number of aspirants reading only A and B is

25 - 8 = 17

, only B and C is

20 - 8 = 12

, and only A and C is

15 - 8 = 7

. Summing these exclusive regions gives

17 + 12 + 7 = 36

.

Step-by-Step Solution

1

Identify the given set values and intersections

Total aspirants

N = 120

;

n(A \cap B) = 25

;

n(B \cap C) = 20

;

n(A \cap C) = 15

;

n(A \cap B \cap C) = 8

.

We need to extract overlapping region counts to isolate the 'exactly two' regions.

2

Calculate aspirants reading ONLY two newspapers for each pair

Only A and B =

25 - 8 = 17

; Only B and C =

20 - 8 = 12

; Only A and C =

15 - 8 = 7

.

The given pairwise intersections

n(A \cap B)

include those who read all three newspapers, so

n(A \cap B \cap C)

must be removed from each pair.

3

Sum the exclusive two-set regions

17 + 12 + 7 = 36

.

Adding these three mutually exclusive regions yields the total number of aspirants reading exactly two newspapers.

Key Concept

Venn Diagram set decomposition and region isolation

Question 4046Question →

The National Tiger Conservation Authority (NTCA) is a statutory body constituted under the provisions of which of the following Indian legislations?

Wildlife (Protection) Act, 1972

Environment (Protection) Act, 1986

Forest (Conservation) Act, 1980

Biological Diversity Act, 2002

Show answer & explanation

Answer: Wildlife (Protection) Act, 1972

Answer

The National Tiger Conservation Authority (NTCA) is constituted under the Wildlife (Protection) Act, 1972.

The National Tiger Conservation Authority (NTCA) is a statutory body under the Ministry of Environment, Forest and Climate Change. It was constituted under Section 38 IV-B of the Wildlife (Protection) Act, 1972, enacted through the Wildlife (Protection) Amendment Act of 2006.

Step-by-Step Solution

1

Identify the mandate of the National Tiger Conservation Authority (NTCA).

The NTCA is responsible for overseeing Project Tiger and standardizing tiger conservation guidelines across Indian states.

Understanding the primary mandate helps locate its governing enabling legislation.

2

Recall the legal framework establishing the NTCA.

The Wildlife (Protection) Act, 1972 was amended in 2006 to insert Section 38 IV-B, granting statutory status to the NTCA.

Statutory bodies derive their powers and structure directly from designated Acts of Parliament.

Key Concept

Statutory legal framework of national conservation authorities in India

Estimated Time:45s

Question 4047Question →

What is the unit digit of the expression $2^{44}$ ?

1

2

6

8

Show answer & explanation

Answer: 6

Answer

The unit digit of

2^{44}

is 6.

The unit digit of powers of 2 repeats in a pattern of 4 steps: 2, 4, 8, 6. Dividing the exponent 44 by 4 yields a remainder of 0. A remainder of 0 means the full 4th power in the cycle applies, so

2^4 = 16

yields a unit digit of 6.

Step-by-Step Solution

1

Identify the base and its cyclicity pattern.

The base is 2. The powers of 2 follow a cyclic pattern of unit digits:

2^1 = 2

,

2^2 = 4

,

2^3 = 8

,

2^4 = 16

(unit digit 6). The cyclicity of 2 is 4.

Unit digits repeat in a fixed pattern of length 4 for base 2.

2

Divide the exponent by the cyclicity period.

44 \div 4 = 11

with a remainder of

0

.

Determining the remainder tells us which position in the 4-step cycle applies.

3

Apply the cyclicity rule for remainder 0.

A remainder of 0 corresponds to the 4th power in the cycle (

2^4 = 16

), giving a unit digit of 6.

When the exponent is exactly divisible by the cyclicity length, the unit digit is determined by the last element of the cycle (

2^4

), not

2^0

.

Key Concept

Unit Digit Cyclicity for Base 2

Question 4048Question →

What is the simplified value of the following mathematical expression?

\frac{7}{2} \div \left[ 1 \frac{1}{4} + \left\{ 2 \frac{1}{2} - \frac{1}{2} \times \left( 2.5 - \overline{\frac{1}{4} + \frac{1}{6}} \right) \right\} \right] \times \frac{5}{12}

\frac{7}{13}

\frac{35}{61}

\frac{1008}{325}

\frac{5}{13}

Show answer & explanation

Answer: $\frac{7}{13}$

Answer

The simplified value of the expression is

\frac{7}{13}

.

Following strict BODMAS precedence: First, simplify the bar term

\overline{\frac{1}{4} + \frac{1}{6}} = \frac{5}{12}

. Next, resolve the round bracket

\frac{5}{2} - \frac{5}{12} = \frac{25}{12}

. Multiplying by

\frac{1}{2}

gives

\frac{25}{24}

. The curly bracket simplifies to

\frac{5}{2} - \frac{25}{24} = \frac{35}{24}

. The square bracket yields

\frac{5}{4} + \frac{35}{24} = \frac{65}{24}

. Finally, evaluating left to right gives

\frac{7}{2} \times \frac{24}{65} \times \frac{5}{12} = \frac{7}{13}

.

Step-by-Step Solution

1

Evaluate the expression under the vinculum (bar bracket)

\overline{\frac{1}{4} + \frac{1}{6}} = \frac{3 + 2}{12} = \frac{5}{12}

According to BODMAS, the vinculum takes precedence over round brackets.

2

Evaluate the terms inside the round brackets

2.5 - \frac{5}{12} = \frac{5}{2} - \frac{5}{12} = \frac{30 - 5}{12} = \frac{25}{12}

Convert decimal

2.5

to fraction

\frac{5}{2}

and perform subtraction.

3

Perform multiplication inside the curly brackets

\frac{1}{2} \times \frac{25}{12} = \frac{25}{24}

Multiplication inside brackets precedes subtraction.

4

Evaluate the terms inside the curly brackets

2 \frac{1}{2} - \frac{25}{24} = \frac{5}{2} - \frac{25}{24} = \frac{60 - 25}{24} = \frac{35}{24}

Complete the subtraction inside the curly brackets.

5

Evaluate the terms inside the square brackets

1 \frac{1}{4} + \frac{35}{24} = \frac{5}{4} + \frac{35}{24} = \frac{30 + 35}{24} = \frac{65}{24}

Perform addition within the square brackets.

6

Perform division and multiplication from left to right

\frac{7}{2} \div \frac{65}{24} \times \frac{5}{12} = \frac{7}{2} \times \frac{24}{65} \times \frac{5}{12} = \frac{84}{65} \times \frac{5}{12} = \frac{7}{13}

Division and multiplication have equal precedence and are executed strictly from left to right.

Key Concept

BODMAS Rule with Vinculum and Nested Brackets

Question 4049Question →

Let $N$ be the smallest positive integer that leaves remainders of $2$ , $4$ , and $6$ when divided by $5$ , $7$ , and $9$ , respectively. What is the remainder when $N^{2025}$ is divided by $17$ ?

11

6

2

5

Show answer & explanation

Answer: 11

Answer

The remainder when

N^{2025}

is divided by

17

is 11.

The number

N

satisfies

N \equiv -3 \pmod{\text{lcm}(5,7,9)}

, giving

N = 315 - 3 = 312

. Reducing

312 \pmod{17}

gives

6

. By Fermat's Little Theorem,

6^{16} \equiv 1 \pmod{17}

, so

6^{2025} \equiv 6^9 \pmod{17}

. Calculating

6^9 \pmod{17}

yields

-6 \pmod{17}

, which converts to positive remainder

17 - 6 = 11

.

Step-by-Step Solution

1

Determine the value of the smallest positive integer

N

N = 312

From the problem statement:

N \equiv 2 \equiv -3 \pmod 5

,

N \equiv 4 \equiv -3 \pmod 7

, and

N \equiv 6 \equiv -3 \pmod 9

. Therefore,

N + 3

must be divisible by

\text{lcm}(5, 7, 9) = 315

. The smallest positive integer is

N = 315 - 3 = 312

.

2

Reduce

N

modulo

17

N \equiv 6 \pmod{17}

Dividing

312

by

17

gives

312 = 17 \times 18 + 6

, so

312 \equiv 6 \pmod{17}

.

3

Apply Fermat's Little Theorem to reduce the power

6^{2025} \equiv 6^9 \pmod{17}

Since

17

is prime and

\gcd(6, 17) = 1

,

6^{16} \equiv 1 \pmod{17}

. Expressing the exponent as

2025 = 16 \times 126 + 9

yields

6^{2025} \equiv (6^{16})^{126} \times 6^9 \equiv 1^{126} \times 6^9 \equiv 6^9 \pmod{17}

.

4

Evaluate

6^9 \pmod{17}

and convert to positive remainder

Remainder is

11

Computing successive powers modulo

17

:

6^2 = 36 \equiv 2 \pmod{17}

,

6^4 \equiv 2^2 = 4 \pmod{17}

, and

6^8 \equiv 4^2 = 16 \equiv -1 \pmod{17}

. Thus,

6^9 = 6^8 \times 6 \equiv (-1) \times 6 = -6 \pmod{17}

. Converting to a positive remainder gives

-6 + 17 = 11

.

Key Concept

Chinese Remainder Theorem (Constant Difference Method), Fermat's Little Theorem, and Negative Remainder Conversion

Question 4050Question →

In a survey of $500$ civil service aspirants preparing for State PSC examinations regarding their daily newspaper reading habits:
- $260$ aspirants read Newspaper A
- $220$ aspirants read Newspaper B
- $180$ aspirants read Newspaper C
- $90$ aspirants read both Newspaper A and Newspaper B
- $70$ aspirants read both Newspaper B and Newspaper C
- $80$ aspirants read both Newspaper A and Newspaper C
- $30$ aspirants read all three newspapers

How many aspirants read exactly one of these three newspapers?

180

240

270

420

Show answer & explanation

Answer: $270$

Answer

The number of aspirants who read exactly one newspaper is

270

.

The correct answer is derived by determining the number of readers exclusive to each single newspaper set. Subtracting all overlapping regions (both the exclusive two-paper readers and three-paper readers) from each newspaper's total gives

120

for Newspaper A only,

90

for Newspaper B only, and

60

for Newspaper C only. Adding these exclusive values yields

270

.

Step-by-Step Solution

1

Calculate the number of aspirants who read ONLY Newspaper A, ONLY Newspaper B, and ONLY Newspaper C by isolating two-set and three-set intersections.

Disjoint 2-set intersection counts (excluding all 3 newspapers):
- Reading A and B only:

90 - 30 = 60

- Reading B and C only:

70 - 30 = 40

- Reading A and C only:

80 - 30 = 50

The given two-newspaper intersection counts include the

30

aspirants who read all three newspapers.

2

Subtract the exclusive two-set and three-set intersection counts from each total newspaper count to find single-newspaper readers.

- Only Newspaper A:

260 - (60 + 50 + 30) = 120

- Only Newspaper B:

220 - (60 + 40 + 30) = 90

- Only Newspaper C:

180 - (50 + 40 + 30) = 60

To find readers of 'only' one paper, all overlaps must be subtracted from the total set count.

3

Sum the exclusive counts for Newspaper A, Newspaper B, and Newspaper C.

Total reading exactly one newspaper =

120 + 90 + 60 = 270

These three categories are mutually exclusive, so their sum gives the total count for 'exactly one'.

Key Concept

Principle of Inclusion-Exclusion for Three Sets

Question 4051Question →

What is the unit digit of the expression $N = 2^{40} + 4^{31} - 7^{38}$ ?

1

6

9

3

Show answer & explanation

Answer: 1

Answer

The unit digit of the given expression is 1.

The unit digits of the individual terms are 6, 4, and 9 respectively. Adding the first two unit digits gives

6 + 4 = 10

(unit digit 0). Subtracting 9 from 10 yields

10 - 9 = 1

. Therefore, the overall unit digit of the expression is 1.

Step-by-Step Solution

1

Find the unit digit of

2^{40}

using cyclicity.

The cyclicity of 2 is 4 (2, 4, 8, 6). Divide the exponent 40 by 4:

40 \pmod 4 = 0

. When the remainder is 0, take the 4th power:

2^4 = 16

, so the unit digit is 6.

Powers of 2 repeat their unit digits in cycles of 4.

2

Find the unit digit of

4^{31}

.

The cyclicity of 4 is 2 (4 for odd exponents, 6 for even exponents). Since 31 is odd, the unit digit is 4.

Odd powers of 4 always end in 4.

3

Find the unit digit of

7^{38}

.

The cyclicity of 7 is 4 (7, 9, 3, 1). Divide the exponent 38 by 4:

38 \pmod 4 = 2

.

7^2 = 49

, so the unit digit is 9.

A remainder of 2 corresponds to the second term in the cyclicity sequence of 7.

4

Combine the unit digits according to the expression

N = 2^{40} + 4^{31} - 7^{38}

.

Unit digit =

(6 + 4) - 9 = 10 - 9 = 1

.

Perform modular arithmetic modulo 10 to find the final unit digit.

Key Concept

Unit Digit Cyclicity and Modular Addition/Subtraction

Question 4052Question →

Match each complex administrative dilemma (Column I) with the primary ethical doctrine or administrative principle (Column II) that should strictly govern the officer's decision-making process in that specific context.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The environmental clearance dilemma strictly aligns with statutory due process; the election affidavit error aligns with administrative proportionality; the intelligence report release aligns with maintaining public order; and the tender evaluation aligns with institutional objectivity and recusal.

Each dilemma presents a specific tension between competing values. The correct matching identifies the overriding administrative principle that resolves the tension according to constitutional and civil service norms. Bypassing environmental laws tests statutory due process. Minor procedural errors invoke proportionality. Releasing riot-inducing information is governed by public order exceptions. Finally, any familial stake in a tender mandates recusal.

Step-by-Step Solution

1

Analyze the core ethical tension in the District Forest Officer scenario.

The tension is between utilitarian public health needs and environmental laws.

Identifying that civil servants cannot unilaterally bypass the law points directly to the 'Adherence to Statutory Due Process' principle.

2

Evaluate the Election Returning Officer's dilemma regarding the spelling error.

The tension is between extreme procedural strictness and substantive justice.

This necessitates the 'Principle of Administrative Proportionality' so the administrative remedy fits the minor nature of the defect.

3

Examine the Police Commissioner's dilemma regarding the intelligence report.

The tension is between the right to information and the immediate threat to life and peace.

Constitutional and administrative norms dictate that 'Maintenance of Public Order' acts as a valid constraint on absolute transparency.

4

Assess the Tender Evaluation Authority's conflict.

The tension is between bureaucratic efficiency (best bid) and perceived bias (sibling's stake).

The absolute necessity to avoid even the appearance of impropriety requires the 'Doctrine of Institutional Objectivity and Recusal'.

Key Concept

Application of Administrative Doctrines to Ethical Dilemmas

Question 4053Question →

What is the unit digit of $7^{82}$ ?

Show answer & explanation

Answer: 9

Answer

The unit digit of

7^{82}

is 9.

The unit digits of powers of 7 repeat in a pattern of 4 (7, 9, 3, 1). Dividing the exponent 82 by 4 yields a remainder of 2. The second number in the cyclic pattern is 9, making 9 the unit digit of

7^{82}

.

Step-by-Step Solution

1

Find the cyclicity of the base number 7

The unit digits follow a repeating 4-step sequence: 7, 9, 3, 1.

Powers of 7 cycle every 4 powers because

7^1=7

,

7^2=49

,

7^3=343

, and

7^4=2401

.

2

Divide the exponent by the cyclicity period

82 \div 4 = 20

with a remainder of 2.

The remainder determines the equivalent power position within the 4-step cycle.

3

Determine the unit digit from the remainder

The unit digit of

7^2

is 9.

A remainder of 2 corresponds to

7^2

, giving 9.

Key Concept

Unit Digit Cyclicity

Question 4054Question →

Let $N = 75x38y2$ be a seven-digit number that is completely divisible by $72$ , where $x$ and $y$ are single-digit natural numbers such that $x > y$ . What is the remainder when $2^{x \cdot y + 1}$ is divided by $13$ ?

2

11

3

6

Show answer & explanation

Answer: $2$

Answer

The remainder when

2^{x \cdot y + 1}

is divided by

13

is

2

.

The seven-digit number

75x38y2

is divisible by

72

, which means it must satisfy divisibility by both

8

and

9

. The last three digits

8y2

are divisible by

8

only when

y = 3

or

y = 7

. The sum of digits

25 + x + y

is divisible by

9

when

x + y = 2

or

x + y = 11

. Testing

y = 3

gives

x = 8

, which satisfies

x > y

. Testing

y = 7

gives

x = 4

, which violates

x > y

. Thus,

x = 8

and

y = 3

, giving

x \cdot y + 1 = 25

. By Fermat's Little Theorem,

2^{12} \equiv 1 \pmod{13}

, so

2^{25} = (2^{12})^2 \cdot 2^1 \equiv 2 \pmod{13}

.

Step-by-Step Solution

1

Apply divisibility rule by 8 to find possible values of y

y

can be either

3

or

7

A number is divisible by

72

if it is divisible by both

8

and

9

. For divisibility by

8

, the last three digits

8y2

must be divisible by

8

. Testing single digits gives

832 / 8 = 104

and

872 / 8 = 109

.

2

Apply divisibility rule by 9 to find corresponding x values and enforce x > y

x = 8

and

y = 3

For divisibility by

9

, the sum of digits

(7 + 5 + x + 3 + 8 + y + 2 = 25 + x + y)

must be a multiple of

9

. If

y = 3

,

28 + x = 36 \implies x = 8

, satisfying

x > y

. If

y = 7

,

32 + x = 36 \implies x = 4

, which violates

x > y

.

3

Evaluate the exponent x * y + 1

Exponent = 25

Substituting

x = 8

and

y = 3

gives

8 \cdot 3 + 1 = 25

.

4

Calculate 2^25 mod 13 using Fermat's Little Theorem

Remainder is 2

Since

13

is prime, Fermat's Little Theorem states

2^{12} \equiv 1 \pmod{13}

. Therefore,

2^{25} = (2^{12})^2 \cdot 2^1 \equiv 1^2 \cdot 2 \equiv 2 \pmod{13}

.

Key Concept

Combining composite divisibility rules (8 and 9) with Fermat's Little Theorem for large power remainder evaluation.

Question 4055Question →

A survey was conducted among $300$ State PSC aspirants regarding their preparation for three subjects: General Studies ( $GS$ ), General Aptitude Test ( $C$ ), and Optional Subject ( $O$ ). The data collected is as follows:
- $180$ candidates prepare for $GS$
- $140$ candidates prepare for $C$
- $120$ candidates prepare for $O$
- $80$ candidates prepare for both $GS$ and $C$
- $50$ candidates prepare for both $C$ and $O$
- $60$ candidates prepare for both $GS$ and $O$
- $30$ candidates prepare for all three subjects

Which of the following statements are correct? (Select all correct statements)

Select all that apply

The number of candidates preparing for exactly two subjects is

100

.

The number of candidates preparing for at least two subjects is

130

.

The number of candidates preparing for only General Studies (

GS

) is

80

.

The number of candidates who do not prepare for any of the three subjects is

20

.

Show answer & explanation

Answer: The number of candidates preparing for exactly two subjects is $100$ .; The number of candidates preparing for at least two subjects is $130$ .; The number of candidates who do not prepare for any of the three subjects is $20$ .

Answer

The correct statements are those asserting that the number of candidates preparing for exactly two subjects is 100, the number of candidates preparing for at least two subjects is 130, and the number of candidates preparing for none of the three subjects is 20.

The statements asserting that exactly two subjects equal 100, at least two subjects equal 130, and none of the subjects equal 20 are all mathematically accurate based on set region decomposition: exactly two subjects count is

50 + 20 + 30 = 100

; at least two subjects is

100 + 30 = 130

; and outside all sets is

300 - 280 = 20

.

Step-by-Step Solution

1

Identify the central region (all three subjects).

The number of candidates preparing for all three subjects

n(GS \cap C \cap O) = 30

.

This value serves as the base subtraction term for all pairwise intersections.

2

Calculate the counts for candidates preparing for exactly two subjects.

GS and C only =

80 - 30 = 50

; C and O only =

50 - 30 = 20

; GS and O only =

60 - 30 = 30

. Total exactly two subjects =

50 + 20 + 30 = 100

.

Subtracting the triple intersection from each dual intersection isolates regions with exactly two subjects.

3

Calculate candidates preparing for only one subject.

Only GS =

180 - (50 + 30 + 30) = 70

; Only C =

140 - (50 + 30 + 20) = 40

; Only O =

120 - (30 + 30 + 20) = 40

.

Subtracting all overlapping regions from total set cardinalities yields single-subject counts.

4

Calculate the total union and the remainder outside all sets.

Total in at least one subject =

70 + 40 + 40 + 50 + 20 + 30 + 30 = 280

. Neither subject =

300 - 280 = 20

.

Applying inclusion-exclusion principle determines the complete universe coverage.

Key Concept

Three-Set Principle of Inclusion-Exclusion

Estimated Time:2m 0s

Question 4056Question →

The following data grid presents the cargo volume handled (in thousands of Metric Tonnes, '000 MT) across five maritime ports from FY 2021–22 to FY 2024–25:

Port	FY 2021–22	FY 2022–23	FY 2023–24	FY 2024–25
Port Alpha	450	540	648	777.6
Port Beta	800	720	648	583.2
Port Gamma	300	375	450	562.5
Port Delta	600	660	792	871.2
Port Epsilon	500	550	605	726

Which of the following statements regarding the performance of these ports are correct? Select all that apply.

Select all that apply

Port Alpha registered a constant year-on-year cargo growth rate of 20% across all three consecutive periods.

The combined total cargo volume handled by all five ports increased by more than 30% in FY 2024–25 compared to FY 2021–22.

Port Beta accounted for more than 20% of the total cargo volume handled across all five ports in FY 2024–25.

Port Gamma achieved the largest absolute increase in cargo volume between FY 2021–22 and FY 2024–25.

Show answer & explanation

Answer: Port Alpha registered a constant year-on-year cargo growth rate of 20% across all three consecutive periods.; The combined total cargo volume handled by all five ports increased by more than 30% in FY 2024–25 compared to FY 2021–22.

Answer

The statements confirming that Port Alpha registered a constant year-on-year growth rate of 20% and that the combined total cargo volume grew by more than 30% are correct.

The statement regarding Port Alpha is correct because the year-on-year growth rate is uniformly

20\%

for every year (

90/450 = 20\%

,

108/540 = 20\%

, and

129.6/648 = 20\%

). The statement regarding combined total growth is also correct because the total cargo grew from

2650

thousand MT to

3520.5

thousand MT, which equals a

32.85\%

increase—clearly exceeding

30\%

.

Step-by-Step Solution

1

Calculate year-on-year percentage growth for Port Alpha.

FY 21–22 to 22–23:

(540 - 450) / 450 = 0.20

(

20\%

). FY 22–23 to 23–24:

(648 - 540) / 540 = 0.20

(

20\%

). FY 23–24 to 24–25:

(777.6 - 648) / 648 = 0.20

(

20\%

).

Verify if the annual growth rate remains constant across all three consecutive periods.

2

Calculate total cargo volume for FY 2021–22 and FY 2024–25 across all ports.

Total FY 2021–22

= 450 + 800 + 300 + 600 + 500 = 2650

thousand MT. Total FY 2024–25

= 777.6 + 583.2 + 562.5 + 871.2 + 726 = 3520.5

thousand MT.

Determine the combined aggregate performance for percentage growth evaluation.

3

Compute total overall growth percentage from FY 2021–22 to FY 2024–25.

Percentage Growth

= \frac{3520.5 - 2650}{2650} \times 100\% = \frac{870.5}{2650} \times 100\% \approx 32.85\%

.

Compare calculated growth (

32.85\%

) against the target threshold of

30\%

.

4

Calculate Port Beta's share of total cargo in FY 2024–25.

Share

= \frac{583.2}{3520.5} \times 100\% \approx 16.57\%

.

Test whether Port Beta's share exceeded

20\%

in FY 2024–25.

5

Compare absolute growth values across all ports between FY 2021–22 and FY 2024–25.

Alpha:

327.6

, Beta:

-216.8

, Gamma:

262.5

, Delta:

271.2

, Epsilon:

226.0

.

Identify the port with the maximum absolute increase.

Key Concept

Tabular Data Interpretation: Multi-Period Rate Analysis and Baseline Distinctions

Question 4057Question →

Calculate the exact numerical value of the following mathematical expression by applying the standard order of operations (BODMAS):

36.4 - \left[ 12.8 + \left\{ 6.5 \times \left( 4.4 - \overline{2.6 + 0.8} \right) \right\} \div 1.3 \right]

Show answer & explanation

Answer: 18.6

Answer

The simplified value of the mathematical expression is 18.6.

Strict application of BODMAS requires simplifying grouping symbols from the innermost vinculum outward, followed by resolving division prior to addition within brackets, yielding an exact answer of 18.6.

Step-by-Step Solution

1

Evaluate the expression underneath the vinculum (bar line)

\overline{2.6 + 0.8} = 3.4

The vinculum functions as an innermost bracket with highest evaluation priority.

2

Evaluate the terms inside the round brackets

4.4 - 3.4 = 1.0

Process operations inside round brackets next.

3

Perform multiplication within the curly braces

6.5 \times 1.0 = 6.5

Complete operations within the curly braces.

4

Execute division inside the square brackets

6.5 \div 1.3 = 5.0

Division takes precedence over addition according to BODMAS rules.

5

Perform addition within the square brackets

12.8 + 5.0 = 17.8

Finish evaluating all operations contained within the square brackets.

6

Perform final subtraction from left to right

36.4 - 17.8 = 18.6

Complete the outermost subtraction to arrive at the final simplified value.

Key Concept

BODMAS Rule with Vinculum and Decimals

Question 4058Question →

What is the simplified value of $\sqrt{7 + 4\sqrt{3}} - \sqrt{3}$ ?

Show answer & explanation

Answer: 2

Answer

The simplified value is 2.

Expressing

7 + 4\sqrt{3}

as

(2 + \sqrt{3})^2

allows the square root to simplify directly to

2 + \sqrt{3}

. Subtracting

\sqrt{3}

leaves the exact numerical answer 2.

Step-by-Step Solution

1

Rewrite the expression under the square root as a perfect square of a binomial.

7 + 4\sqrt{3} = 2^2 + (\sqrt{3})^2 + 2(2)(\sqrt{3}) = (2 + \sqrt{3})^2

Using the identity

(a+b)^2 = a^2 + b^2 + 2ab

, setting

a = 2

and

b = \sqrt{3}

yields

a^2 + b^2 = 4 + 3 = 7

and

2ab = 4\sqrt{3}

.

2

Evaluate the square root of the perfect square.

\sqrt{(2 + \sqrt{3})^2} = 2 + \sqrt{3}

The principal square root of a positive squared expression

\sqrt{x^2}

is

x

.

3

Perform the subtraction indicated in the stem.

(2 + \sqrt{3}) - \sqrt{3} = 2

The radical terms

\sqrt{3}

and

-\sqrt{3}

cancel out, leaving the integer 2.

Key Concept

Simplification of Nested Surds

Question 4059Question →

The table below provides operational details regarding procurement, storage loss, public distribution system (PDS) allocation, and unit storage costs for rice across four agricultural zones of a state during FY 2025–26:

Zone	Rice Procured (in '000 MT)	Storage & Transit Loss (% of Procured Rice)	Distribution to PDS (% of Net Retained Rice)	Unit Storage Cost per MT of Net Retained Rice (₹)
Zone Alpha	250	4.0%	80%	₹ 450
Zone Beta	180	5.0%	75%	₹ 500
Zone Gamma	200	2.5%	85%	₹ 400
Zone Delta	300	6.0%	70%	₹ 350

*Note: Net Retained Rice = Total Rice Procured $-$ Storage & Transit Loss.*

What is the total quantity of rice (in MT) retained as undistributed buffer stock across all four zones combined after completing the PDS distribution?

204,600\text{ MT}

215,000\text{ MT}

20,460\text{ MT}

162,600\text{ MT}

Show answer & explanation

Answer: $204,600\text{ MT}$

Answer

204,600\text{ MT}

The correct answer is derived by first subtracting the percentage storage loss from the gross procurement of each zone to establish the net retained rice base (

240,000\text{ MT}

for Alpha,

171,000\text{ MT}

for Beta,

195,000\text{ MT}

for Gamma, and

282,000\text{ MT}

for Delta). Multiplying these net figures by their respective remaining undistributed percentages (

20\%

,

25\%

,

15\%

, and

30\%

) gives buffer quantities of

48,000\text{ MT}

,

42,750\text{ MT}

,

29,250\text{ MT}

, and

84,600\text{ MT}

. The sum of these four quantities equals

204,600\text{ MT}

.

Step-by-Step Solution

1

Convert procured quantities into metric tonnes (MT) and compute net retained rice for each zone.

Zone Alpha:

250,000 \times (1 - 0.04) = 240,000\text{ MT}

.
Zone Beta:

180,000 \times (1 - 0.05) = 171,000\text{ MT}

.
Zone Gamma:

200,000 \times (1 - 0.025) = 195,000\text{ MT}

.
Zone Delta:

300,000 \times (1 - 0.06) = 282,000\text{ MT}

.

Storage loss must be deducted from gross procurement to get net retained rice.

2

Calculate the undistributed buffer stock percentage for each zone.

Zone Alpha:

100\% - 80\% = 20\%

.
Zone Beta:

100\% - 75\% = 25\%

.
Zone Gamma:

100\% - 85\% = 15\%

.
Zone Delta:

100\% - 70\% = 30\%

.

Buffer stock represents the remaining fraction of net retained rice after PDS distribution.

3

Calculate the volume of undistributed buffer stock for each zone.

Zone Alpha:

240,000 \times 0.20 = 48,000\text{ MT}

.
Zone Beta:

171,000 \times 0.25 = 42,750\text{ MT}

.
Zone Gamma:

195,000 \times 0.15 = 29,250\text{ MT}

.
Zone Delta:

282,000 \times 0.30 = 84,600\text{ MT}

.

Multiply net retained rice by the respective buffer stock percentage.

4

Sum the buffer stock volumes across all four zones.

48,000 + 42,750 + 29,250 + 84,600 = 204,600\text{ MT}

.

To find the combined total buffer stock for the state.

Key Concept

Multi-stage sequential percentage calculations on tabular net base values.

Question 4060Question →

Find the exact numerical value of the following mathematical expression evaluated strictly according to the VBODMAS rule:

75\% \text{ of } 160 - \left[ 3.5 \times 8 + \left\{ 48 \div \left( 14 - \overline{5 \times 2} \right) \right\} \right]

Show answer & explanation

Answer: 80

Answer

The simplified numerical value of the expression is

80

.

Evaluating the expression following strict VBODMAS hierarchy: bar expression

\overline{5 \times 2} = 10

, round brackets

14 - 10 = 4

, curly brackets

48 \div 4 = 12

, square brackets

3.5 \times 8 + 12 = 40

, and percentage 'of' term

75\% \text{ of } 160 = 120

. Finally,

120 - 40 = 80

.

Step-by-Step Solution

1

Evaluate the expression under the vinculum (bar)

\overline{5 \times 2} = 10

According to VBODMAS, operations grouped under a bar take highest priority.

2

Evaluate inside the innermost round brackets

(14 - 10)

14 - 10 = 4

Perform subtraction inside the parentheses.

3

Evaluate inside the curly brackets

\{48 \div 4\}

48 \div 4 = 12

Divide the number outside the round bracket by the result of the round bracket.

4

Evaluate inside the square brackets

[3.5 \times 8 + 12]

28 + 12 = 40

Perform multiplication (

3.5 \times 8 = 28

) before adding

12

.

5

Calculate the percentage term

75\% \text{ of } 160

\frac{75}{100} \times 160 = 120

Evaluate the 'of' operation before final subtraction.

6

Perform final subtraction

120 - 40

80

Subtract the total result of the bracketed expression from the percentage term.

Key Concept

Order of Operations (VBODMAS Rule)

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

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Key Concept

Common Mistakes

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Key Concept

Common Mistakes

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept