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2016 AMC 10B Problem 11

2016 AMC 10B Problem 11AMC 10 Step by Step

Name: Date: September 13, 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; a side spans (posts - 1) gaps of 4 yards.

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 $\boxed{\textbf{(B)}\ 336}$ .

Why this works

Two separate fencepost facts appear here. Around a closed loop the number of posts equals the number of gaps, but along a single side the number of gaps is one less than the number of posts. Counting posts side by side double-counts the corners, so subtract $4$ . Set up the count as a linear equation, then convert posts to lengths with care.

Alternative approach

Around the loop there are $20$ gaps, so the perimeter is $80$ and length plus width is $40$ . A short side of $n$ posts has length $4(n-1)$ and a long side $4(2n-1)$ ; $4(3n - 2) = 40$ gives $n = 4$ , so the sides are $12$ and $28$ .

The trap

Multiplying the number of posts by 4 yards instead of the number of gaps (posts minus 1), which gives 16 by 32 = 512.

Common mistakes

Multiplying the number of posts by $4$ yards instead of the number of gaps (posts minus $1$ ), which gives $16 \times 32 = 512$ .
Forgetting that the corner posts belong to two sides, so writing $2n + 4n = 20$ and getting a non-integer $n$ .

Techniques

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