AMC 8 · 2000 · #16

Easy mode Grade 4

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Problem

Mateen has a rectangular backyard. He wants to walk exactly $1$ kilometer (which is $1000$ meters) in it.

He notices two different ways to do this:

If he walks along the long side back and forth, $25$ trips down the length adds up to $1000$ meters.
If he walks all the way around the backyard, $10$ laps around the perimeter adds up to $1000$ meters.

What is the area of Mateen's backyard, in square meters?

Pick an answer.

(A)

(B)

200

(C)

400

(D)

500

(E)

1000

View mode:

Toolkit + CCSS Solution

Understand

Restated: Mateen's rectangular backyard has the property that walking its length $25$ times covers $1000$ m, and walking its perimeter $10$ times also covers $1000$ m. Find the area of the backyard in square meters.

Givens: The backyard is a rectangle with length $L$ and width $W$ (in meters); $25$ trips along the length total $1000$ m; $10$ trips around the perimeter total $1000$ m; Answer choices: (A) $40$, (B) $200$, (C) $400$, (D) $500$, (E) $1000$

Unknowns: The area $A = L \times W$ of the backyard, in square meters

Understand

Plan

Primary tool: #13 Convert to Algebra

Secondary: #7 Identify Subproblems

Each English sentence in the problem turns directly into an equation, so Tool #13 (Convert to Algebra) is the natural fit: "25 lengths make 1000 m" becomes $25L = 1000$, and "10 perimeters make 1000 m" becomes $10P = 1000$. Tool #7 (Identify Subproblems) keeps the work orderly — solve for $L$ first, then for $P$, then back out $W$ from the perimeter formula, and finally compute the area. Each subproblem is a one-step Grade 4 division or substitution.

Execute — Answer: C

#13 Convert to Algebra 4.OA.A.3 Step 1

Translate the two sentences into equations.
Walking the length $25$ times covers $1000$ m, so $25L = 1000$.
Walking the perimeter $10$ times covers $1000$ m, so $10P = 1000$, where $P$ is the perimeter.

$$25L = 1000,\qquad 10P = 1000$$

💡 "$n$ trips of distance $d$ cover $D$" always means $n \cdot d = D$ — the Grade 4 multiplication-as-equal-groups idea.

#7 Identify Subproblems 4.OA.A.3 Step 2

Subproblem 1: solve for the length $L$.
Divide both sides of $25L = 1000$ by $25$.

$$L = \dfrac{1000}{25} = 40 \text{ m}$$

💡 $25 \times 40 = 1000$ is a familiar fact — four quarters make a dollar, so forty quarters make ten dollars.

#7 Identify Subproblems 4.OA.A.3 Step 3

Subproblem 2: solve for the perimeter $P$.
Divide both sides of $10P = 1000$ by $10$.

$$P = \dfrac{1000}{10} = 100 \text{ m}$$

💡 Dividing by $10$ shifts the digits one place — $1000 \to 100$.

#13 Convert to Algebra 4.MD.A.3 Step 4

Subproblem 3: use the rectangle perimeter formula $P = 2L + 2W$ to recover $W$.
Substitute $P = 100$ and $L = 40$, then solve.

$$100 = 2(40) + 2W \;\Rightarrow\; 100 - 80 = 2W \;\Rightarrow\; W = \dfrac{20}{2} = 10 \text{ m}$$

💡 Once two sides ($L = 40$) of the four-sided trip are accounted for, the other two sides must add to $100 - 80 = 20$, so each is $10$.

#7 Identify Subproblems 4.MD.A.3 Step 5

Subproblem 4: compute the area $A = L \times W$ with $L = 40$ and $W = 10$.

$$A = 40 \times 10 = 400 \text{ m}^2 \;\Rightarrow\; \textbf{(C)}$$

💡 Area of a rectangle is just length times width — the final Grade 4 step.

[1] #13 4.OA.A.3 Translate the two sentences into equations. Walking the length $25$ times covers

[2] #7 4.OA.A.3 Subproblem 1: solve for the length $L$. Divide both sides of $25L = 1000$ by $25

[3] #7 4.OA.A.3 Subproblem 2: solve for the perimeter $P$. Divide both sides of $10P = 1000$ by

[4] #13 4.MD.A.3 Subproblem 3: use the rectangle perimeter formula $P = 2L + 2W$ to recover $W$.

[5] #7 4.MD.A.3 Subproblem 4: compute the area $A = L \times W$ with $L = 40$ and $W = 10$.

Review

Reasonableness: Check the dimensions back against the original sentences. Length $L = 40$ m, so $25$ lengths give $25 \times 40 = 1000$ m — matches. Perimeter $P = 2(40) + 2(10) = 80 + 20 = 100$ m, so $10$ perimeters give $10 \times 100 = 1000$ m — also matches. The area $400$ m$^2$ falls between trap (A) $40$ (the length alone, forgetting to multiply by width) and (D) $500$ or (E) $1000$ (which mix area with perimeter or total distance). Answer (C) $400$ is the only choice consistent with $40 \times 10$.

Alternative: Tool #5 (Look for a Pattern) on the two trip counts: $25$ length-trips equal $10$ perimeter-trips, both equal $1000$ m. So $25L = 10P$, which simplifies to $P = 2.5 L$. With the rectangle formula $P = 2L + 2W$, this forces $2W = 0.5L$, i.e. $W = L/4$. Plug into $25L = 1000$ to get $L = 40$, then $W = 10$ and $A = 400$. Same answer (C), reached by spotting the ratio of trip counts first.

CCSS standards used (min grade 4)

4.OA.A.3 Solve multi-step word problems using the four operations with whole numbers (Turning the two trip-distance sentences into the equations $25L = 1000$ and $10P = 1000$, then dividing to recover $L = 40$ and $P = 100$.)
4.MD.A.3 Apply the area and perimeter formulas for rectangles in real-world and mathematical problems (Using $P = 2L + 2W$ to back out $W = 10$ from $P = 100$ and $L = 40$, then using $A = L \times W$ to compute the area $A = 40 \times 10 = 400$ m$^2$.)

⭐ Each sentence in a word problem can become one short equation. Solve them in order — length, then perimeter, then width, then area — and the answer is $40 \times 10 = 400$ m$^2$, choice (C).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 4)

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