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!
Q. A person is standing 30 meters away from a flagpole. If the angle of elevation to the top of the flagpole is 30 degrees, what is the height of the flagpole?
Q. A person is standing 30 meters away from the foot of a tree. If the angle of elevation to the top of the tree is 60 degrees, what is the height of the tree?
A.
15√3 m
B.
30 m
C.
30√3 m
D.
45 m
Solution
Using tan(60°) = height/distance, we have height = distance * tan(60°) = 30√3 m.
Q. A person is standing 40 m away from a building and observes the top of the building at an angle of elevation of 60 degrees. What is the height of the building? (2023)
A.
20 m
B.
30 m
C.
40 m
D.
50 m
Solution
Height = distance * tan(60) = 40 * √3 ≈ 69.28 m, which rounds to 50 m.
Q. A person is standing 40 m away from a building and sees the top of the building at an angle of elevation of 45 degrees. What is the height of the building? (2020)
Q. A person is standing 40 m away from a building. If the angle of elevation to the top of the building is 60 degrees, what is the height of the building?
A.
20√3 m
B.
40 m
C.
30 m
D.
50 m
Solution
Using tan(60°) = height/40, we have √3 = height/40. Therefore, height = 40√3 m.
Q. A person is standing 40 m away from a building. If the angle of elevation to the top of the building is 30 degrees, what is the height of the building?
A.
20 m
B.
40 m
C.
30 m
D.
10 m
Solution
Using tan(30°) = height/distance, we have height = distance * tan(30°) = 40 * (1/√3) ≈ 20 m.
Q. A person is standing 40 meters away from a building and sees the top of the building at an angle of elevation of 45 degrees. What is the height of the building?
Q. A person is standing 40 meters away from a building and sees the top of the building at an angle of elevation of 60 degrees. What is the height of the building?
Q. A person is standing 40 meters away from a statue and measures the angle of elevation to the top of the statue as 53.13 degrees. What is the height of the statue?
A.
30 meters
B.
20 meters
C.
25 meters
D.
15 meters
Solution
Let h be the height of the statue. tan(53.13°) = h/40. Therefore, h = 40 * tan(53.13°) = 40 * 1.6 = 64 meters.
Q. A person is standing 40 meters away from a tower and sees the top of the tower at an angle of elevation of 60 degrees. What is the height of the tower?
Q. A person is standing 50 m away from a building. If the angle of elevation to the top of the building is 30 degrees, what is the height of the building?
A.
25√3 m
B.
50 m
C.
30 m
D.
40 m
Solution
Using tan(30°) = height/50, we have 1/√3 = height/50. Therefore, height = 50/√3 m.
Q. A person is standing 50 meters away from a vertical pole. If the angle of elevation of the top of the pole is 60 degrees, what is the height of the pole?
A.
25 m
B.
30 m
C.
35 m
D.
40 m
Solution
Using tan(60°) = height/50, we have √3 = height/50. Therefore, height = 50√3 ≈ 86.6 m.
Q. A person is standing 50 meters away from a vertical pole. If the angle of elevation to the top of the pole is 60 degrees, what is the height of the pole?
A.
25 m
B.
30 m
C.
35 m
D.
40 m
Solution
Using tan(60°) = height/50, we have √3 = height/50. Therefore, height = 50√3 ≈ 86.6 m.
Q. A person is standing at a distance of 20 m from a vertical pole. If the angle of elevation to the top of the pole is 45 degrees, what is the height of the pole?
A.
20 m
B.
10 m
C.
30 m
D.
15 m
Solution
Using tan(45°) = height/distance, we have height = distance * tan(45°) = 20 * 1 = 20 m.
Q. A person is standing at a distance of 40 m from a tree. If the angle of elevation to the top of the tree is 60 degrees, what is the height of the tree?
A.
20√3 m
B.
40 m
C.
30 m
D.
10√3 m
Solution
Using tan(60°) = height/distance, we have height = distance * tan(60°) = 40√3 m.
Q. A person is standing at a distance of 40 meters from the base of a building. If the angle of elevation to the top of the building is 60 degrees, what is the height of the building?
A.
20 m
B.
30 m
C.
40 m
D.
50 m
Solution
Using tan(60°) = height/40, we have √3 = height/40. Therefore, height = 40√3 ≈ 69.28 m.
Q. A person is standing on a hill 100 meters high. If he looks at a point on the ground at an angle of depression of 30 degrees, how far is the point from the base of the hill?
Q. A person is standing on a hill 80 m high. The angle of depression to a car on the ground is 60 degrees. How far is the car from the base of the hill?
A.
40 m
B.
80 m
C.
20√3 m
D.
40√3 m
Solution
Using tan(60°) = height/distance, we have distance = height/tan(60°) = 80/√3 = 40√3 m.
Q. A person is standing on a hill that is 80 meters high. If the angle of depression to a point on the ground is 45 degrees, how far is the point from the base of the hill?
A.
80 m
B.
40 m
C.
80√2 m
D.
40√2 m
Solution
Using tan(45°) = height/distance, we have distance = height/tan(45°) = 80/1 = 80 m.
Q. A person is standing on the ground and looking at the top of a building. If the angle of elevation is 45 degrees and the person is 20 meters away from the building, what is the height of the building?
Q. A person is standing on the ground and looking at the top of a tree. If the angle of elevation is 60 degrees and the person is 20 meters away from the tree, what is the height of the tree?
Q. A person needs 2000 calories a day and wants to get 15% of their calories from fats. How many grams of fat should they consume (1 gram of fat = 9 calories)? (2000)
A.
33 grams
B.
40 grams
C.
50 grams
D.
60 grams
Solution
15% of 2000 calories = 0.15 * 2000 = 300 calories from fats. 300 calories / 9 calories per gram = 33.33 grams, rounded to 33 grams.
