Mathematics/General Ability MCQs
Topic Notes: Mathematics/General Ability
MCQs and preparation resources for competitive exams, covering important concepts, past papers, and detailed explanations.
Plato
- Biography: Ancient Greek philosopher (427–347 BCE), student of Socrates and teacher of Aristotle, founder of the Academy in Athens.
- Important Ideas:
- Theory of Forms
- Philosopher-King
- Ideal State
31
Find the distance between points (2, 5) and (6, 10).
Answer:
6.4
Step-by-step solution: 1. Use distance formula: sqrt[(x2 - x1)² + (y2 - y1)²]. 2. Compute squares: (4)² + (5)². 3. Distance = sqrt(41) = 6.4.
32
Find the distance between points (5, 5) and (9, 10).
Answer:
6.4
Step-by-step solution: 1. Use distance formula: sqrt[(x2 - x1)² + (y2 - y1)²]. 2. Compute squares: (4)² + (5)². 3. Distance = sqrt(41) = 6.4.
33
Find the distance between points (5, 3) and (9, 8).
Answer:
6.4
Step-by-step solution: 1. Use distance formula: sqrt[(x2 - x1)² + (y2 - y1)²]. 2. Compute squares: (4)² + (5)². 3. Distance = sqrt(41) = 6.4.
34
Find the distance between points (3, 6) and (7, 11).
Answer:
6.4
Step-by-step solution: 1. Use distance formula: sqrt[(x2 - x1)² + (y2 - y1)²]. 2. Compute squares: (4)² + (5)². 3. Distance = sqrt(41) = 6.4.
35
Find the distance between points (5, 7) and (9, 12).
Answer:
6.4
Step-by-step solution: 1. Use distance formula: sqrt[(x2 - x1)² + (y2 - y1)²]. 2. Compute squares: (4)² + (5)². 3. Distance = sqrt(41) = 6.4.
36
Find the distance between points (5, 8) and (9, 13).
Answer:
6.4
Step-by-step solution: 1. Use distance formula: sqrt[(x2 - x1)² + (y2 - y1)²]. 2. Compute squares: (4)² + (5)². 3. Distance = sqrt(41) = 6.4.
37
Find the equation of a line passing through the origin that is parallel to the line y = 5x + 2.
Answer:
y = 5x
Step 1: Parallel lines share the exact same slope. The given line y = 5x + 2 has a slope of 5. Step 2: The new line must also have a slope of 5, making its equation y = 5x + c. Step 3: Because it passes through the origin (0,0), its y-intercept c is 0. The equation is y = 5x.
38
What is the reflection of the point (3, -2) across the origin?
Answer:
(-3, 2)
Step 1: A reflection across the origin inverts both the x and y coordinates simultaneously. Step 2: The transformation rule is (x, y) becomes (-x, -y). Step 3: Applying this to (3, -2) results in (-3, 2).
39
What is the reflection of the point (-4, 5) across the y-axis?
Answer:
(4, 5)
Step 1: Reflecting a point across the y-axis flips its horizontal position but leaves its vertical position unchanged. Step 2: The rule for reflection across the y-axis changes (x, y) to (-x, y). Step 3: Applying this to (-4, 5) yields (-(-4), 5), which is (4, 5).
40
The circumcenter of a right-angled triangle is always located at:
Answer:
The midpoint of the hypotenuse
Step 1: A right-angled triangle can be inscribed in a semicircle. Step 2: According to Thales's theorem, the hypotenuse of the triangle acts as the diameter of that circumscribed circle. Step 3: The center of the circle (circumcenter) is precisely the midpoint of the diameter (the hypotenuse).