A solid sphere is placed inside a cube such that it touches all six faces. What percentage of the cube’s volume is not occupied by the sphere?

Correct — A.
Let the cube's side be a. The sphere touches all six faces, so its diameter = a and its radius = a⁄2.
Sphere volume = (4⁄3)π(a⁄2)³ = πa³⁄6
Share filled = (πa³⁄6) ÷ a³ = π⁄6

With π = 3.14: π⁄6 = 3.14 ÷ 6 ≈ 0.52333
Share empty = 1 − 0.52333 = 0.47667 ≈ 47.67% → option (a).

Why the others are wrong

• (b)
  48.56% — 48.56% empty would mean the sphere fills 51.44%, so π⁄6 = 0.5144 and π ≈ 3.086. Real π, as 3.14 or 22⁄7, leaves between 47.6% and 47.7% empty.
• (c)
  49.28% — 49.28% empty would mean the sphere fills 50.72%, forcing π ≈ 3.043. π is about 3.14, so the π⁄6 ratio cannot produce this value.
• (d)
  42.54% — 42.54% empty would mean the sphere fills 57.46%, forcing π ≈ 3.448. The sphere fills only a little over half the cube, about 52.4%.

A sphere that touches all six faces of a cube is inscribed: its diameter equals the cube's side. The side cancels, so the filled share is π⁄6 ≈ 52.4% whatever the cube's size.

The empty share is therefore 1 − π⁄6, and the value of π you use moves only the last decimal.

Key facts

• A sphere inscribed in a cube of side a has radius a⁄2.
• Sphere volume = (4⁄3)πr³, which becomes πa³⁄6 when r = a⁄2.
• The inscribed sphere fills π⁄6 of the cube and leaves 1 − π⁄6, about 47.6%, empty.
• In the reverse case, a cube inscribed in a sphere, the cube's space diagonal a√3 equals the sphere's diameter.

Common traps

• Taking the sphere's radius equal to the cube's side instead of half of it
• Reporting the filled share, about 52.4%, when the question asks for the empty share

No directly related past PYQ was found.