A solid cone is melted and the entire material is made into a sphere of volume 38,808 cm³. If the radius of the cone and the sphere are equal, find the height of the cone.

First get the radius from the sphere: ⅔πr³ = 38808, so r³ = 38808 × 3 × 7 / (4 × 22) = 9261, giving r = 21.
Melting conserves volume, so cone volume = sphere volume:
⅓πr²h = ⅔πr³ → h = 4r = 84 cm.
Worth memorising: whenever a cone is recast into a sphere of the same radius, h = 4r falls straight out of the formulas — you never need the numbers.

In a cylindrical vessel of height 14 cm and radius 5 cm, seven spheres of radius 3 cm are placed. How much water fills the vessel?

Cylinder volume = πr²h = (22/7) × 25 × 14 = 1100 cm³.
Seven spheres = 7 × ⅔π(3)³ = 7 × (4/3) × (22/7) × 27 = 792 cm³.
Water = 1100 − 792 = 308 cm³.
Note: the arrangement is not physically possible, but the question is purely a volume subtraction. Do not be thrown by that.

Two cubes, each of volume 4,096 cc, are joined to form a cuboid. Find the difference between the surface area of the cuboid and the total surface area of the two separate cubes.

Cube edge = ∛4096 = 16 cm.
Two separate cubes: 2 × 6 × 16² = 3,072 cm².
Joined cuboid is 32 × 16 × 16: 2(32·16 + 16·16 + 32·16) = 2,560 cm².
Difference = 512 cm².
Shortcut: joining hides exactly two faces, so the loss is 2 × 16² = 512 — no need to compute either surface area.

A rectangular plot of 210 m × 120 m is divided into 4 equal parts by two roads, each 12 m wide, running through the middle — one parallel to the length and one parallel to the breadth. Find the area of each part.

Total area = 210 × 120 = 25,200 m².
Road along the length = 210 × 12 = 2,520; road along the breadth = 120 × 12 = 1,440.
The two roads overlap at the crossing, so subtract it once: 12 × 12 = 144.
Roads = 2520 + 1440 − 144 = 3,816 m².
Remaining = 25,200 − 3,816 = 21,384, and each of the four parts = 5,346 m².
The trap: forgetting the overlap gives 3,960 for the roads and 5,310 per part. Crossing roads always share their intersection.

A wire of length 17 m is cut into two parts. One part forms a square and the other an equilateral triangle. If the square's area is A and its perimeter is B with B = 2A, find the ratio of the areas of the square and the triangle.

For the square: side = B/4, so A = B²/16.
Given B = 2A = B²/8, so B = 8 — the square has perimeter 8, side 2 and area 4.
The triangle takes the remaining 17 − 8 = 9 m, so its side is 3 and its area is (√3/4) × 9 = 9√3/4.
Ratio = 4 : 9√3/4 = 16 : 9√3, and multiplying both sides by √3 gives 16√3 : 27.
Note: the options are given in rationalised form, so an answer of 16 : 9√3 looks absent until you clear the surd from the second term.