Major Competitive Exams

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Major Competitive Exams

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Major Competitive Exams MCQ & Objective Questions

Major Competitive Exams play a crucial role in shaping the academic and professional futures of students in India. These exams not only assess knowledge but also test problem-solving skills and time management. Practicing MCQs and objective questions is essential for scoring better, as they help in familiarizing students with the exam format and identifying important questions that frequently appear in tests.

What You Will Practise Here

Key concepts and theories related to major subjects
Important formulas and their applications
Definitions of critical terms and terminologies
Diagrams and illustrations to enhance understanding
Practice questions that mirror actual exam patterns
Strategies for solving objective questions efficiently
Time management techniques for competitive exams

Exam Relevance

The topics covered under Major Competitive Exams are integral to various examinations such as CBSE, State Boards, NEET, and JEE. Students can expect to encounter a mix of conceptual and application-based questions that require a solid understanding of the subjects. Common question patterns include multiple-choice questions that test both knowledge and analytical skills, making it essential to be well-prepared with practice MCQs.

Common Mistakes Students Make

Rushing through questions without reading them carefully
Overlooking the negative marking scheme in MCQs
Confusing similar concepts or terms
Neglecting to review previous years’ question papers
Failing to manage time effectively during the exam

FAQs

Question: How can I improve my performance in Major Competitive Exams?
Answer: Regular practice of MCQs and understanding key concepts will significantly enhance your performance.

Question: What types of questions should I focus on for these exams?
Answer: Concentrate on important Major Competitive Exams questions that frequently appear in past papers and mock tests.

Question: Are there specific strategies for tackling objective questions?
Answer: Yes, practicing under timed conditions and reviewing mistakes can help develop effective strategies.

Start your journey towards success by solving practice MCQs today! Test your understanding and build confidence for your upcoming exams. Remember, consistent practice is the key to mastering Major Competitive Exams!

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Q. How many different ways can 4 people be seated at a round table?

A. 6
B. 12
C. 24
D. 30

Solution

The number of arrangements of n people at a round table is (n-1)!. For 4 people, it is 3! = 6.

Correct Answer: B — 12

Q. How many different ways can 4 prizes be distributed among 10 students? (2020)

A. 5040
B. 10000
C. 2100
D. 120

Solution

The number of ways to distribute 4 prizes among 10 students is 10P4 = 10! / (10-4)! = 5040.

Correct Answer: A — 5040

Q. How many different ways can 4 students be selected from a class of 10?

A. 210
B. 120
C. 240
D. 300

Solution

The number of ways to choose 4 students from 10 is 10C4 = 210.

Correct Answer: A — 210

Q. How many different ways can 4 students be selected from a group of 10?

A. 210
B. 120
C. 100
D. 90

Solution

The number of ways to choose 4 from 10 is given by 10C4 = 210.

Correct Answer: A — 210

Q. How many different ways can 6 people be seated at a round table?

A. 720
B. 120
C. 600
D. 480

Solution

The number of arrangements of 6 people at a round table is (6-1)! = 5! = 120.

Correct Answer: A — 720

Q. How many different ways can 6 people be seated in a row? (2020)

A. 720
B. 600
C. 360
D. 480

Solution

The number of arrangements of 6 people is 6! = 720.

Correct Answer: A — 720

Q. How many different ways can the letters of the word 'BOOK' be arranged? (2014)

A. 12
B. 24
C. 36
D. 48

Solution

The number of arrangements of the letters in 'BOOK' is 4! / 2! = 12.

Correct Answer: B — 24

Q. How many different ways can the letters of the word 'MATH' be arranged?

A. 12
B. 24
C. 36
D. 48

Solution

The number of arrangements of the letters in 'MATH' is 4! = 24.

Correct Answer: B — 24

Q. How many different ways can the letters of the word 'SCHOOL' be arranged?

A. 720
B. 360
C. 480
D. 600

Solution

The word 'SCHOOL' has 6 letters with 'O' repeated twice. The arrangements are 6! / 2! = 720 / 2 = 360.

Correct Answer: B — 360

Q. How many elements are in the power set of the empty set?

A. 0
B. 1
C. 2
D. Infinite

Solution

The power set of the empty set contains only one subset, which is the empty set itself. Thus, it has 1 element.

Correct Answer: B — 1

Q. How many essential amino acids are there for human adults?

A. 9
B. 10
C. 11
D. 12

Solution

There are 9 essential amino acids that must be obtained from the diet.

Correct Answer: A — 9

Q. How many essential amino acids are there that must be obtained from the diet?

A. 9
B. 10
C. 11
D. 12

Solution

There are 9 essential amino acids that must be obtained from the diet as the body cannot synthesize them.

Correct Answer: A — 9

Q. How many essential amino acids are there that the human body cannot synthesize?

A. 8
B. 9
C. 10
D. 11

Solution

There are 9 essential amino acids that the human body cannot synthesize and must be obtained from the diet.

Correct Answer: B — 9

Q. How many factors does the number 28 have?

A. 4
B. 6
C. 8
D. 10

Solution

The factors of 28 are 1, 2, 4, 7, 14, and 28, totaling 6 factors.

Correct Answer: B — 6

Q. How many factors does the number 36 have?

A. 6
B. 8
C. 9
D. 12

Solution

The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36, totaling 9 factors.

Correct Answer: B — 8

Q. How many Fundamental Duties are enshrined in the Constitution of India?

A. 10
B. 11
C. 12
D. 9

Solution

There are 11 Fundamental Duties listed in Article 51A of the Indian Constitution.

Correct Answer: B — 11

Q. How many fundamental rights are guaranteed by the Indian Constitution?

A. 5
B. 6
C. 7
D. 4

Solution

The Indian Constitution originally provided for 7 fundamental rights, but currently there are 6 after the Right to Property was removed from the list.

Correct Answer: B — 6

Q. How many grams are in 0.25 moles of glucose (C6H12O6)?

A. 30 g
B. 45 g
C. 75 g
D. 90 g

Solution

Molar mass of C6H12O6 = 6*12 + 12*1 + 6*16 = 180 g/mol. Mass = 0.25 moles x 180 g/mol = 45 g.

Correct Answer: D — 90 g

Q. How many grams are in 0.25 moles of KCl?

A. 35 g
B. 70 g
C. 17.5 g
D. 140 g

Solution

Molar mass of KCl = 39 + 35.5 = 74.5 g/mol. Mass = moles x molar mass = 0.25 moles x 74.5 g/mol = 18.625 g.

Correct Answer: B — 70 g

Q. How many grams are in 0.25 moles of sulfuric acid (H2SO4)?

A. 49 g
B. 98 g
C. 24.5 g
D. 12.25 g

Solution

Molar mass of H2SO4 = 98 g/mol. Mass = moles x molar mass = 0.25 moles x 98 g/mol = 24.5 g.

Correct Answer: A — 49 g

Q. How many grams are in 0.5 moles of glucose (C6H12O6)? (2023)

A. 90 g
B. 180 g
C. 45 g
D. 60 g

Solution

Molar mass of C6H12O6 = 6*12 + 12*1 + 6*16 = 180 g/mol. Mass = 0.5 moles x 180 g/mol = 90 g.

Correct Answer: A — 90 g

Q. How many grams are in 0.5 moles of sodium (Na)?

A. 11 g
B. 22 g
C. 5.5 g
D. 44 g

Solution

The molar mass of sodium is approximately 23 g/mol. Therefore, 0.5 moles of Na weigh 0.5 x 23 g = 11 g.

Correct Answer: A — 11 g

Q. How many grams are in 2 moles of carbon dioxide (CO2)?

A. 44 g
B. 88 g
C. 22 g
D. 66 g

Solution

The molar mass of CO2 is 44 g/mol. Thus, 2 moles of CO2 = 2 x 44 g = 88 g.

Correct Answer: B — 88 g

Q. How many grams are in 3 moles of glucose (C6H12O6)?

A. 180 g
B. 360 g
C. 540 g
D. 90 g

Solution

Molar mass of glucose = 6*12 + 12*1 + 6*16 = 180 g/mol. Mass = moles x molar mass = 3 moles x 180 g/mol = 540 g.

Correct Answer: B — 360 g

Q. How many grams are in 4 moles of potassium chloride (KCl)? (2022)

A. 74 g
B. 148 g
C. 296 g
D. 37 g

Solution

Molar mass of KCl = 39 + 35.5 = 74.5 g/mol. Mass = 4 moles x 74.5 g/mol = 298 g.

Correct Answer: C — 296 g

Q. How many grams are in 4 moles of sulfuric acid (H2SO4)? (2023)

A. 196 g
B. 98 g
C. 392 g
D. 294 g

Solution

Molar mass of H2SO4 = 2*1 + 32 + 4*16 = 98 g/mol. Mass = 4 moles x 98 g/mol = 392 g.

Correct Answer: C — 392 g

Q. How many grams of CO2 are produced from the complete combustion of 1 mole of C3H8?

A. 44 g
B. 88 g
C. 132 g
D. 22 g

Solution

C3H8 + 5O2 → 3CO2 + 4H2O. 1 mole of C3H8 produces 3 moles of CO2. Mass = 3 * 44 g = 132 g.

Correct Answer: B — 88 g

Q. How many grams of KCl are needed to prepare 0.5 M solution in 500 mL of water? (2023)

A. 14.9 g
B. 7.45 g
C. 29.8 g
D. 3.73 g

Solution

Molarity = moles/volume(L). Moles = 0.5 mol/L * 0.5 L = 0.25 mol. Mass = moles * molar mass = 0.25 mol * 74.5 g/mol = 18.625 g.

Correct Answer: B — 7.45 g

Q. How many grams of KCl are needed to prepare 0.5 moles of KCl solution?

A. 37.5 g
B. 74.5 g
C. 50 g
D. 100 g

Solution

The molar mass of KCl is 39 g/mol (K) + 35.5 g/mol (Cl) = 74.5 g/mol. Therefore, 0.5 moles of KCl will weigh 0.5 x 74.5 g = 37.25 g.

Correct Answer: B — 74.5 g

Q. How many grams of KCl are needed to prepare 0.5 moles of KCl solution? (Molar mass of KCl = 74.5 g/mol)

A. 37.25 g
B. 74.5 g
C. 148.5 g
D. 18.5 g

Solution

To find the mass, use the formula: mass = moles x molar mass. Thus, 0.5 moles x 74.5 g/mol = 37.25 g.

Correct Answer: A — 37.25 g

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