Combined solids, frustum, conversions MCQs

Q1. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? What is the surface area of the resulting solid (use π = 22/7)?

• A.Diameter 7 cm, surface 332.5 cm^2
• B.Diameter 7 cm, surface 294 cm^2
• C.Diameter 14 cm, surface 449 cm^2
• D.Diameter 3.5 cm, surface 277.5 cm^2

Answer: A. Diameter 7 cm, surface 332.5 cm^2

Given 7 cm, asked for What is the greatest diameter the hemisphere can have?. By Greatest diameter = side = 7, so r = 3.5. Greatest diameter = side = 7, so r = 3.5 → Surface = 6(7)^2 − πr^2 + 2πr^2 = 294 + πr^2 = 294 + (22/7)(12.25) = 294 + 38.5 = 332.5 cm^2. Hence option A) Diameter 7 cm, surface 332.5 cm^2.

Q2. If the radius of a sphere is increased by 50%, by what percent does its volume increase?

• A.237.5%
• B.125%
• C.150%
• D.350%

Answer: A. 237.5%

Given 50, asked for the unknown. By New volume factor = (1.5)^3 = 3.375. New volume factor = (1.5)^3 = 3.375 → Increase = 2.375 = 237.5%. Hence option A) 237.5%.

Q3. Water in a canal of width 6 m, depth 1.5 m, flows at 10 km/h. How much area will it irrigate in 30 minutes if 8 cm of standing water is needed for irrigation?

• A.562500 m^2
• B.281250 m^2
• C.375000 m^2
• D.450000 m^2

Answer: A. 562500 m^2

Diagnose the question type - a typical math numerical. The data on the table: 6 m, 1.5 m, 10 km, 8 cm. We are after How much area will it irrigate in 30 minutes if 8 cm of standing water is needed for irrigation?. Tool of choice - Volume flowing in 30 min = 6 × 1.5 × (10000/2) = 45000 m^3. Rearrange it for the unknown before substituting. Numbers in: Volume flowing in 30 min = 6 × 1.5 × (10000/2) = 45000 m^3 → Area = V / 0.08 = 45000/0.08 = 562500 m^2. Lock in option A) 562500 m^2.