AMC 10 Step by Step

2017 AMC 10A Problem 3

P1-10~2 minPrint

Tamara has three rows of two 66 -feet by 22 -feet flower beds in her garden. The beds are separated and also surrounded by 11 -foot-wide walkways, as shown on the diagram. What is the total area of the walkways, in square feet?

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Key insight

Find the size of the whole garden, including every walkway strip, then subtract the six flower beds. What is left over is the walkway.

Solution

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 feet.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 feet.1 + 2 + 1 + 2 + 1 + 2 + 1 = 10 \text{ feet}.
3. Area of the garden: 15×10=15015 \times 10 = 150 square feet.
4. Area of the beds. Each bed is 6×2=126 \times 2 = 12 square feet, and there are 66 beds, so the beds cover 6×12=726 \times 12 = 72 square feet.
5. Walkways are everything that is not a bed:
150−72=78 square feet.150 - 72 = 78 \text{ square feet}.

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

Why this works

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.

Alternative approach

Add up the strips directly. There are 44 horizontal walkways, each running the full 1515-foot width: 4×15=604 \times 15 = 60 square feet. Between them sit 33 short vertical strips in each of the 33 bed rows, each 1×21 \times 2: that is 9×2=189 \times 2 = 18 square feet. Total 60+18=7860 + 18 = 78.

Concise solution

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

Measure the whole garden. Left to right there are two beds of width 66 and three walkways of width 11 (left edge, middle, right edge): 1+6+1+6+1=151 + 6 + 1 + 6 + 1 = 15 feet. Top to bottom there are three beds of height 22 and four walkways: 1+2+1+2+1+2+1=101 + 2 + 1 + 2 + 1 + 2 + 1 = 10 feet.

Total garden area: 15⋅10=15015 \cdot 10 = 150 square feet.

The six beds each measure 6×2=126 \times 2 = 12 square feet, for 6⋅12=726 \cdot 12 = 72 square feet of beds.

Walkways are everything else: 150−72=78150 - 72 = 78.

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

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.