AMC 8 · 2020 · #24

Grade 7 geometry-2dalgebra

Archetype Arithmetic Expression Decomposition └ Sub-archetype Multi-Segment Path or Count Sum

area-rectanglespercentageratio-proportionlinear-equations-one-var convert-to-algebraidentify-subproblems ↑ Prerequisites: area-rectanglespercentagelinear-equations-one-var

📏 Long solution 💡 4 insights 📊 Diagram

Problem

A big square is tiled with n × n gray squares, each of side s inches, separated and bordered by strips of width d inches. The picture in the problem shows the layout for n=3. For n=24 the 576 gray tiles together cover 64% of the big square's area. Find the ratio

\frac{d}{s}

Pick an answer.

(A)

$frac{6}{25}$

(B)

$frac{1}{4}$

(C)

$frac{9}{25}$

(D)

$frac{7}{16}$

(E)

$frac{9}{16}$

AMC 8 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

n = 24 is too big to count borders by hand, so start with Tool #9 and use the n = 3 picture the problem already gives. Counting tiles vs. borders in that small case exposes the pattern "n tiles, n+1 borders" along one side; Tool #1 (the diagram is already supplied) makes the n+1 borders obvious — one between each pair, plus one at each end. Once the side length is ns + (n+1)d, Tool #13 turns the percent condition into a single linear equation in d and s that we can solve for the ratio.

1STEP 1

In the n=3 picture one side has 3 tiles and 4 border strips, so its length is 3s + 4d.

side(n=3) = 3s + 4d

2STEP 2

For general n a side has n tiles and n+1 borders, so its length is ns + (n+1)d; at n=24 this becomes 24s + 25d.

side(n) = n s + (n+1) d → side(24) = 24s + 25d

3STEP 3

The 576 tiles form a 24s-wide gray square of area (24s)², sitting in a big square of side 24s + 25d and area (24s + 25d)².

A_gray = (24s)², A_big = (24s + 25d)²

4STEP 4

For similar squares, area ratio = (side ratio)²; since 64% = ( $\frac{4}{5}$ )², the gray side is $\frac{4}{5}$ of the big side: $\frac{24s}{24s + 25d}$ = $\frac{4}{5}$ .

\frac{A_gray}{A_big}

\frac{64}{100}

= (

\frac{4}{5}

)² →

\frac{24s}{24s + 25d}

\frac{4}{5}

5STEP 5

Cross-multiply: 120s = 96s + 100d, so 24s = 100d and $\frac{d}{s}$ = $\frac{6}{25}$ , choice (A).

5 · 24s = 4(24s + 25d) → 120s = 96s + 100d → 24s = 100d →

\frac{d}{s}

\frac{24}{100}

\frac{6}{25}

→ (A)

Answer

\frac{6}{25}

Sanity-check the side ratio directly: the gray block is 24s wide and the big square is 24s + 25d wide, so the gray fraction (per side) should be

\frac{4}{5}

= 0.8. With

\frac{d}{s}

\frac{6}{25}

, one side is 24s + 25 ·

\frac{6}{25}

s = 24s + 6s = 30s, and

\frac{24s}{30s}

\frac{4}{5}

. Squaring gives

\frac{16}{25}

= 0.64 = 64% — exactly the given coverage, so

\frac{6}{25}

is consistent.

💡Key takeaway

This AMC 8 problem only needs Grade 7 scale-drawing reasoning — areas of similar squares scale by the square of the side ratio — that you already know!

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