2018 AMC 10B Problem 1

P1-10~1 minPrint

Basic Counting Quadrilaterals & Polygon Areas Set up the equation/formula and compute; no special trick needed

Kate bakes a $20$ -inch by $18$ -inch pan of cornbread. The cornbread is cut into pieces that measure $2$ inches by $2$ inches. How many pieces of cornbread does the pan contain?

Show solution without answering

Key insight

Count how many 2-inch pieces fit along each side of the pan, 10 along one side and 9 along the other, then multiply.

Solution

Picture the pan cut into a grid of little squares, then count rows and columns.

Along the $20$ -inch side, each piece takes up $2$ inches, so $20 \div 2 = 10$ pieces fit in a row.
Along the $18$ -inch side, $18 \div 2 = 9$ rows fit.
Both divisions come out even, so the grid fills the pan with nothing left over.
A grid with $10$ pieces in each of $9$ rows has $10 \times 9 = 90$ pieces.

The answer is $\boxed{\textbf{(A)}\ 90}$ .

Why this works

To count things laid out in a grid, multiply the number in one row by the number of rows. First check that the piece fits each side evenly. When it does, dividing the areas gives the same answer.

Alternative approach

Divide areas. The pan is $20 \times 18 = 360$ square inches. Each piece is $2 \times 2 = 4$ square inches. So there are $360 \div 4 = 90$ pieces.

Concise solution

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

Along the $20$ -inch side the pan is cut into $20/2 = 10$ pieces; along the $18$ -inch side into $18/2 = 9$ pieces. Since both dimensions divide evenly, the pieces form a $10 \times 9$ grid with nothing left over:
$10 \cdot 9 = 90.$

The answer is $\boxed{\textbf{(A)}\ 90}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Geometry, Chapter 8: Quadrilaterals
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques
Introduction to Geometry, Chapter 9: Polygons (§9.3)