2017 AMC 10A Problem 3

The walkways have an awkward shape, but the whole garden is a simple rectangle. So find the whole garden's area and take away the beds.

1. Width of the garden. Going left to right you cross a walkway, a bed, a walkway, a bed, a walkway:
$$1 + 6 + 1 + 6 + 1 = 15 \text{ feet}.$$
2. Height of the garden. Going bottom to top you cross four walkways and three beds:
$$1 + 2 + 1 + 2 + 1 + 2 + 1 = 10 \text{ feet}.$$
3. Area of the garden: $15 \times 10 = 150$ square feet.
4. Area of the beds. Each bed is $6 \times 2 = 12$ square feet, and there are $6$ beds, so the beds cover $6 \times 12 = 72$ square feet.
5. Walkways are everything that is not a bed:
$$150 - 72 = 78 \text{ square feet}.$$

The answer is $\boxed{\textbf{(B)}\ 78}$.

When a region has a messy shape, look for a simple shape that contains it and subtract the pieces that do not belong. The only place to slip is counting the strips: because the walkways go around the outside too, each direction has one more walkway than it has beds.

Add up the strips directly. There are $4$ horizontal walkways, each running the full $15$-foot width: $4 \times 15 = 60$ square feet. Between them sit $3$ short vertical strips in each of the $3$ bed rows, each $1 \times 2$: that is $9 \times 2 = 18$ square feet. Total $60 + 18 = 78$.

Measure the whole garden. Left to right there are two beds of width $6$ and three walkways of width $1$ (left edge, middle, right edge): $1 + 6 + 1 + 6 + 1 = 15$ feet. Top to bottom there are three beds of height $2$ and four walkways: $1 + 2 + 1 + 2 + 1 + 2 + 1 = 10$ feet.

Total garden area: $15 \cdot 10 = 150$ square feet.

The six beds each measure $6 \times 2 = 12$ square feet, for $6 \cdot 12 = 72$ square feet of beds.

Walkways are everything else: $150 - 72 = 78$.

The answer is $\boxed{\textbf{(B)}\ 78}$.