AMC 8 · 2005 · #10
Grade 4 rate-ratioProblem
Joe had walked half way from home to school when he realized he was late. He ran the rest of the way to school. He ran 3 times as fast as he walked. Joe took 6 minutes to walk half way to school. How many minutes did it take Joe to get from home to school?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: Joe walks the first half of the way to school, then runs the second half. He runs $3$ times as fast as he walks, and walking the first half took $6$ minutes. How many minutes is the whole trip?
Givens: Walking the first half takes $6$ minutes; Running speed $= 3 \times$ walking speed; Both halves cover the same distance; Answer choices: (A) $7$, (B) $7.3$, (C) $7.7$, (D) $8$, (E) $8.3$
Unknowns: The total time, in minutes, from home to school
Understand
Restated: Joe walks the first half of the way to school, then runs the second half. He runs $3$ times as fast as he walks, and walking the first half took $6$ minutes. How many minutes is the whole trip?
Givens: Walking the first half takes $6$ minutes; Running speed $= 3 \times$ walking speed; Both halves cover the same distance; Answer choices: (A) $7$, (B) $7.3$, (C) $7.7$, (D) $8$, (E) $8.3$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #5 Look for a Pattern
Tool #7 (Identify Subproblems) splits the trip into two clean pieces: (a) the walking half, whose time is given, and (b) the running half, whose time we need. Each piece is one small calculation, and the total is just their sum. For the running piece, Tool #5 (Look for a Pattern) catches the inverse-proportion pattern: the running half covers the same distance as the walking half, but $3$ times as fast — so it takes $\tfrac{1}{3}$ as much time. Picking Tool #7 plus Tool #5 over Tool #13 (Convert to Algebra) keeps the whole solution at Grade 4-5 arithmetic with no equation-solving.
Execute — Answer: D
3.OA.A.3 Step 1 - Subproblem 1: write down the walking time.
- The problem gives it directly — Joe walked the first half in $6$ minutes.
💡 Half the problem is already solved for us. Name it and move on.
4.OA.A.2 Step 2 - Subproblem 2: find the running time.
- The running half is the same distance as the walking half, but Joe runs $3$ times as fast.
- Same distance and $3 \times$ the speed means $\tfrac{1}{3}$ the time — that is the inverse-proportion pattern between speed and time when distance is fixed.
💡 "$3$ times as fast" on the same trip is a multiplicative comparison: it shrinks the time by the same factor of $3$.
4.OA.A.3 Step 3 Add the two subproblem times to get the total trip time, and match it to a choice.
💡 Combining the two subproblem answers is the last move in any "split-it-up" plan.
3.OA.A.3 Subproblem 1: write down the walking time. The problem gives it directly — Joe w 4.OA.A.2 Subproblem 2: find the running time. The running half is the same distance as th 4.OA.A.3 Add the two subproblem times to get the total trip time, and match it to a choic Review
Reasonableness: Sanity check with concrete numbers. Suppose the half-distance is $6$ blocks. Walking $6$ blocks in $6$ minutes is $1$ block per minute. Running $3$ times as fast is $3$ blocks per minute, so the same $6$ blocks take $6 \div 3 = 2$ minutes. Total: $6 + 2 = 8$ minutes. Magnitude check: the answer must be more than $6$ (Joe still has half the trip to go after walking) and less than $12$ (he would only need $12$ if he walked the whole way), so $8$ sits in the right range. The trap answers $7.3$, $7.7$, and $8.3$ look like "$6 + \tfrac{6}{3}$" miscomputed with decimals, and $7$ would be the answer if Joe ran the second half $6$ times as fast.
Alternative: Tool #8 (Analyze the Units): walking $6$ minutes for half the trip gives a walking pace of $\tfrac{1 \text{ half-trip}}{6 \text{ min}}$. Running pace is $3 \times$ that, so $\tfrac{3 \text{ half-trips}}{6 \text{ min}} = \tfrac{1 \text{ half-trip}}{2 \text{ min}}$. Covering the second half therefore takes $2$ minutes, and the total is again $6 + 2 = 8$ minutes, choice (D).
CCSS standards used (min grade 4)
3.OA.A.3Use multiplication and division within $100$ to solve word problems (Reading the walking time directly from the problem ($6$ minutes) and using basic operations to set up the two subproblem times.)4.OA.A.2Multiply or divide to solve word problems involving multiplicative comparison (Turning "runs $3$ times as fast" into "running time is $\tfrac{1}{3}$ of the walking time" so $t_{\text{run}} = \tfrac{1}{3} \times 6 = 2$ minutes.)4.OA.A.3Solve multistep word problems using the four operations (Adding the walking time and the running time, $6 + 2 = 8$ minutes, to get the total trip time.)
⭐ Split the trip into the walking half and the running half. The running half is the same distance covered $3$ times as fast, so it takes $\tfrac{1}{3}$ the time — that one inverse-proportion idea turns this AMC 8 problem into a Grade 4 multiplicative-comparison exercise.
⭐ Split the trip into the walking half and the running half. The running half is the same distance covered $3$ times as fast, so it takes $\tfrac{1}{3}$ the time — that one inverse-proportion idea turns this AMC 8 problem into a Grade 4 multiplicative-comparison exercise.
More like this
Same archetype — closest grade level first.
