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2016 AMC 10B Problem 11

2016 AMC 10B Problem 11AMC 10 Step by Step

Name: Date: October 28, 2026

Linear Equations & Word Problems · difficulty 2 of 5 · about 3 min · position P11-15

Sola decided to fence in his rectangular garden. He bought $20$ fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly $4$ yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners. What is the area, in square yards, of Sola’s garden?

A) $256$
B) $336$
C) $384$
D) $448$
E) $512$

Answer

Key insight

Corners are shared, so n posts per short side and 2n per long side give 6n - 4 = 20. Each side is (posts - 1) gaps of 4 yards.

Solution

1. Say a short side has $n$ posts, counting its two corners. Then a long side has $2n$ posts.
2. Add up all four sides: $n + n + 2n + 2n = 6n$ . But each of the $4$ corner posts sits on two sides, so it was counted twice. Take away the extra $4$ :
$6n - 4 = 20, \qquad \text{so } n = 4.$
3. A short side has $4$ posts. Between $4$ posts in a row there are only $3$ gaps (picture posts at $0, 4, 8, 12$ yards). So the short side is $3 \cdot 4 = 12$ yards.
4. A long side has $2 \cdot 4 = 8$ posts, so $7$ gaps, and it is $7 \cdot 4 = 28$ yards.
5. Check: going all the way around a closed loop, the number of gaps equals the number of posts, $20$ . So the distance around is $20 \cdot 4 = 80$ yards, and indeed $12 + 28 + 12 + 28 = 80$ .
6. The area is $12 \cdot 28 = 336$ square yards.

The answer is $\boxed{\textbf{(B)}\ 336}$ .

Why this works

This is a "fence post" problem, and it has two separate traps. Corner posts belong to two sides, so they get counted twice if you add side by side. And along one straight side there is one fewer gap than posts, while around a closed loop there are exactly as many gaps as posts.

Alternative approach

Start from the loop. There are $20$ gaps all the way around, so the perimeter is $80$ yards and one short side plus one long side is $40$ yards. A short side with $n$ posts has $n - 1$ gaps and a long side has $2n - 1$ gaps, which is $3n - 2$ gaps together. So $4(3n - 2) = 40$ , giving $n = 4$ and sides $12$ and $28$ .

The trap

Multiplying the number of posts by 4 yards instead of the number of gaps (one less than the posts). That gives a 16 by 32 garden.

Concise solution

Let a short side carry $n$ posts, corners included; then a long side carries $2n$ . Adding up the four sides counts each corner post twice, so the total is
$2n + 2(2n) - 4 = 6n - 4 = 20,$
giving $n = 4$ .

A side with $4$ posts has $3$ gaps between them, so the short side is $3 \cdot 4 = 12$ yards. A side with $8$ posts has $7$ gaps, so the long side is $7 \cdot 4 = 28$ yards.

Quick check: the perimeter is $2(12 + 28) = 80$ yards, which is $20$ gaps of $4$ yards, one gap per post around a closed loop.

The area is $12 \cdot 28 = 336$ square yards.

The answer is $\boxed{\textbf{(B)}\ 336}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Geometry, Chapter 8: Quadrilaterals
Introduction to Algebra, Chapter 7: Proportion (§7.4)
Introduction to Geometry, Chapter 9: Polygons (§9.3)

Techniques

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