AMC 8 · 2000 · #16
Easy mode Grade 4Problem
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.
AMC 8 2000 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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
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) $1000$
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
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.
💡 "$n$ trips of distance $d$ cover $D$" always means $n \cdot d = D$ — the Grade 4 multiplication-as-equal-groups idea.
4.OA.A.3 Step 2 - Subproblem 1: solve for the length $L$.
- Divide both sides of $25L = 1000$ by $25$.
💡 $25 \times 40 = 1000$ is a familiar fact — four quarters make a dollar, so forty quarters make ten dollars.
4.OA.A.3 Step 3 - Subproblem 2: solve for the perimeter $P$.
- Divide both sides of $10P = 1000$ by $10$.
💡 Dividing by $10$ shifts the digits one place — $1000 \to 100$.
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.
💡 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 $10$.
4.MD.A.3 Step 5 Subproblem 4: compute the area $A = L \times W$ with $L = 40$ and $W = 10$.
💡 Area of a rectangle is just length times width — the final Grade 4 step.
4.OA.A.3 Translate the two sentences into equations. Walking the length $25$ times covers 4.OA.A.3 Subproblem 1: solve for the length $L$. Divide both sides of $25L = 1000$ by $25 4.OA.A.3 Subproblem 2: solve for the perimeter $P$. Divide both sides of $10P = 1000$ by 4.MD.A.3 Subproblem 3: use the rectangle perimeter formula $P = 2L + 2W$ to recover $W$. 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.3Solve 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.3Apply 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).
⭐ 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).
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