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
201
A spherical ball of lead 3 cm in radius is melted and recast into three spherical balls. The radii of two of these are 1.5 cm and 2 cm. Find the radius of the third ball.
Answer:
2.5 cm
Step 1: Volume is conserved. (4/3)π(3)³ = (4/3)π(1.5)³ + (4/3)π(2)³ + (4/3)π(r)³. Step 2: 27 = 3.375 + 8 + r³. Step 3: r³ = 27 - 11.375 = 15.625. r = ³√15.625 = 2.5 cm.
202
If the capacity of a cylindrical tank is 1848 m³ and the diameter of its base is 14 m, the depth of the tank is:
Answer:
12 m
Step 1: Radius = 7 m. Volume = πr²h. Step 2: 1848 = (22/7) × 49 × h = 154h. Step 3: h = 1848 / 154 = 12 m.
203
A copper sphere of radius 3 cm is beaten and drawn into a wire of diameter 0.2 cm. The length of the wire is:
Answer:
36 m
Step 1: Volume of sphere = (4/3)π(3)³ = 36π cm³. Step 2: Wire is a cylinder with r = 0.1 cm. Volume = π(0.1)²h = 0.01πh. Step 3: 36π = 0.01πh -> h = 3600 cm = 36 m.
204
The dimensions of a rectangular box are in the ratio 2:3:4 and the difference between the cost of covering it with sheet of paper at the rate of Rs. 4 and Rs. 4.50 per sq m is Rs. 416. Find the dimensions.
Answer:
8 m, 12 m, 16 m
Step 1: Let dimensions be 2x, 3x, 4x. TSA = 2(6x² + 12x² + 8x²) = 52x². Step 2: Difference in cost = Area × Difference in rate = 52x² × (4.50 - 4) = 52x² × 0.5 = 26x². Step 3: 26x² = 416 -> x² = 16 -> x = 4. Dimensions are 8, 12, 16.
205
A cylindrical tank is filled with water. A solid iron sphere of radius equal to half the radius of the cylinder is dropped into the tank. What fraction of water overflows?
Answer:
Depends on the height
Step 1: The amount of water that overflows is equal to the volume of the sphere. Step 2: Volume of sphere = (4/3)π(R/2)³ = (4/3)π(R³/8) = πR³/6. Step 3: Volume of cylinder = πR²H. The fraction is (πR³/6) / (πR²H) = R/(6H). Therefore, it depends on the height H of the cylinder.
206
What is the ratio of the surface area of a sphere to the total surface area of a hemisphere of the same radius?
Answer:
4:3
Step 1: Surface area of a sphere = 4πr². Step 2: Total surface area of a hemisphere = 3πr². Step 3: Ratio = 4πr² / 3πr² = 4:3.
207
A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. Find the height of the cone.
Answer:
14 cm
Step 1: External radius R = 4, Internal r = 2. Volume of hollow sphere = (4/3)π(R³ - r³) = (4/3)π(64 - 8) = (4/3)π(56). Step 2: Cone radius = 4. Volume = (1/3)π(4)²h = (16/3)πh. Step 3: Equating volumes: (4/3)π(56) = (16/3)πh -> 224 = 16h -> h = 14 cm.
208
Find the length of canvas 2 m in width required to make a conical tent 12 m in diameter and 8 m in slant height.
Answer:
75.4 m
Step 1: Radius = 12/2 = 6 m, Slant height = 8 m. Step 2: Curved surface area (amount of canvas) = πrl = (22/7) × 6 × 8 = 150.857 m². Step 3: Length of canvas = Area / width = 150.857 / 2 = 75.428 m. Approximated to 75.4 m.
209
A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Answer:
2.744 cm
Step 1: Volume of sphere = Volume of cylinder. (4/3)π(4.2)³ = π(6)²h. Step 2: (4/3) × 74.088 = 36 × h. Step 3: 98.784 = 36h -> h = 98.784 / 36 = 2.744 cm.
210
A cylindrical vessel 32 cm high and 18 cm as the radius of the base, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, the radius of its base is:
Answer:
36 cm
Step 1: Volume of cylindrical vessel = Volume of conical heap. π × r1² × h1 = (1/3) × π × r2² × h2. Step 2: π × 18² × 32 = (1/3) × π × r2² × 24. Step 3: 324 × 32 = 8 × r2² -> r2² = (324 × 32) / 8 = 324 × 4 = 1296. r2 = √1296 = 36 cm.