In order for Mateen to walk a kilometer (1000m) in his rectangular backyard, he must walk the length 25 times or walk its perimeter 10 times. What is the area of Mateen's backyard in square meters?
Try it yourself first — the explanation is most useful after you’ve attempted it.
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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
#13 Convert to Algebra 4.OA.A.3Step 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.3Step 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.3Step 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.3Step 4
Subproblem 3: use the rectangle perimeter formula $P = 2L + 2W$ to recover $W$.
💡 Area of a rectangle is just length times width — the final Grade 4 step.
[1]
#13 4.OA.A.3Translate the two sentences into equations. Walking the length $25$ times covers
[2]
#7 4.OA.A.3Subproblem 1: solve for the length $L$. Divide both sides of $25L = 1000$ by $25
[3]
#7 4.OA.A.3Subproblem 2: solve for the perimeter $P$. Divide both sides of $10P = 1000$ by
[4]
#13 4.MD.A.3Subproblem 3: use the rectangle perimeter formula $P = 2L + 2W$ to recover $W$.
[5]
#7 4.MD.A.3Subproblem 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).
⭐ 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).
