2020 AMC 10B Problem 2

P1-10~1 minPrint

Solid Geometry Set up the equation/formula and compute; no special trick needed

Carl has $5$ cubes each having side length $1$ , and Kate has $5$ cubes each having side length $2$ . What is the total volume of these $10$ cubes?

Show solution without answering

Key insight

A cube with side 2 has volume 2 times 2 times 2 = 8, not 2 or 4. So the total is 5 times 1 plus 5 times 8.

Solution

The volume of a cube is side $\times$ side $\times$ side.

Carl's cubes have side $1$ . Each has volume $1 \times 1 \times 1 = 1$ . Five of them give $5 \times 1 = 5$ .
Kate's cubes have side $2$ . Each has volume $2 \times 2 \times 2 = 8$ . Five of them give $5 \times 8 = 40$ .
Add the two totals: $5 + 40 = 45$ .

The answer is $\boxed{\textbf{(E)}\ 45}$ .

Why this works

When the side of a cube doubles, the volume does not double. It gets multiplied by $2 \times 2 \times 2 = 8$ , because the cube gets twice as long, twice as wide and twice as tall. Whenever a problem gives side lengths but asks about volume, turn each object into its volume first, then add.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

A cube of side $1$ has volume $1^3 = 1$ , so Carl's five cubes contribute $5 \cdot 1 = 5$ .

A cube of side $2$ has volume $2^3 = 8$ , so Kate's five cubes contribute $5 \cdot 8 = 40$ .

The total volume is $5 + 40 = 45$ .

The answer is $\boxed{\textbf{(E)}\ 45}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 14: Three-Dimensional Geometry
Introduction to Geometry, Chapter 15: Curved Surfaces