Sixty-four cubes of sides 2 cm each are combined to form a cube of side 8 cm. If four of the smaller cubes along the diagonal of a surface are removed from the surface of the large cube, which one of the following statements about the surface area of this solid object is true?

Correct — C, it is more than the surface area of the large cube. Start with the intact solid: a cube of side 8 cm has area 6 × 8² = 384 cm². Each of its faces is a 4 × 4 array of 2 cm squares, so the diagonal of a face runs through exactly four small cubes — the two at the ends sit at corners of the large cube, the two in the middle sit inside the face. Now count what one removal does. A small cube that shows k of its own faces on the outside loses those k faces and opens up the 6 − k faces it had shared with its neighbours, a net change of (6 − 2k) little squares of 4 cm² each. A corner cube has k = 3, so 6 − 6 = 0 and nothing changes. A cube buried in the middle of a face has k = 1, so 6 − 2 = 4 extra squares, or +16 cm². The four diagonal cubes give 0 + 16 + 16 + 0 = +32 cm², taking the solid to 416 cm². Diagonally placed cubes meet only along an edge, never across a whole face, so the removals do not interfere with one another.

• (a)It is equal to the surface area of the large cube. — This is what you get if you assume every removed cube is a corner cube. Only the two at the ends of the diagonal are; the two in the middle of the face each open a pit with a floor and four walls where a single 4 cm² square used to be.
• (b)It is less than the surface area of the large cube. — Taking a small cube out can never shrink the area. Removing k outer faces exposes 6 − k inner ones, and 6 − k is at least as large as k for every k up to 3, which is the most any cube on this solid can show.
• (d)Insufficient data — Everything needed is printed. Sixty-four cubes of side 2 cm fix the large cube at 8 cm, a face is a 4 × 4 grid, and its diagonal picks out four cubes in a fixed pattern of two corners and two interiors. All six faces behave identically by symmetry, so it does not matter which one you choose.

Cutting a hole into a solid removes some outer skin and reveals fresh inner skin, and the surface area moves by the difference between the two. For a unit cube pulled out of a larger stack, the outer skin lost is the number of its faces that were already exposed, and the inner skin gained is the number that were touching neighbours. Those two numbers add up to six, so the net change is 6 minus twice the exposed count — a rule that settles the whole question without any arithmetic on 384.

The wording asks only whether the area goes up, down or stays put, which is a hint that classifying the four cubes matters more than adding numbers. A candidate's instinct is that removing material shrinks the object, and for volume that instinct is right. Surface area behaves the opposite way here, because a pit contributes five new walls in place of the one square it swallowed. The one genuine subtlety is spotting that a face diagonal of a 4 × 4 grid passes through two corner cells, where the loss and the gain cancel exactly.

• A cube of side a has surface area 6a², so the 8 cm cube measures 384 cm² and each 2 cm cube face is 4 cm².
• Sixty-four cubes of side 2 cm build an 8 cm cube, and every face of it is a 4 × 4 grid of 2 cm squares.
• Pull out a small cube showing k faces and the area changes by (6 − 2k) × 4 cm²: 0 for a corner (k = 3), +8 cm² for an edge cube (k = 2), +16 cm² for one inside a face (k = 1).
• A face diagonal of the 4 × 4 grid hits cells (1,1), (2,2), (3,3) and (4,4) — two corners and two face interiors.
• The four removals add 32 cm², so the pitted solid measures 416 cm² against the original 384 cm².

• Assuming that taking material away must reduce the surface area.
• Forgetting that two of the four cubes on a face diagonal sit at corners of the large cube, where nothing changes.
• Counting only the four walls of a pit and leaving out its floor.

Set as a qualitative comparison rather than a calculation, so the marks go to whoever classifies each removed cube by the number of faces it had on the outside.

• practice — not a real PYQ

  A cube of side 4 cm is built from 64 cubes of side 1 cm. One corner cube is taken out. What is the surface area of the solid that remains?

  • (a)96 cm²
  • (b)93 cm²
  • (c)99 cm²
  • (d)90 cm²

  Answer(a) 96 cm² — a corner cube shows 3 faces and hides 3, so the loss and the gain cancel and the area stays at 6 × 4² = 96 cm².

• practice — not a real PYQ

  From the middle of one face of a 6 cm cube built out of 27 cubes of side 2 cm, one small cube is removed. By how much does the surface area change?

  • (a)It falls by 4 cm²
  • (b)It rises by 16 cm²
  • (c)It is unchanged
  • (d)It rises by 24 cm²

  Answer(b) It rises by 16 cm² — one 4 cm² square is lost and a pit with a floor and four walls, five squares of 4 cm², is opened.