AMC 8 · 2015 · #17
Grade 8 rate-ratioalgebraProblem
Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?
Pick an answer.
AMC 8 2015 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: Jeremy's dad drives him to school in $20$ minutes during rush hour traffic. On a clear day his dad drives $18$ mph faster and the trip takes only $12$ minutes. The route is the same both days. How long is the route in miles?
Givens: Rush-hour trip time: $20$ minutes; Clear-day trip time: $12$ minutes; Clear-day speed is $18$ mph greater than rush-hour speed; Same distance both days; Answer choices: (A) $4$, (B) $6$, (C) $8$, (D) $9$, (E) $12$ (miles)
Unknowns: The one-way distance $d$ from home to school, in miles
Understand
Restated: Jeremy's dad drives him to school in $20$ minutes during rush hour traffic. On a clear day his dad drives $18$ mph faster and the trip takes only $12$ minutes. The route is the same both days. How long is the route in miles?
Givens: Rush-hour trip time: $20$ minutes; Clear-day trip time: $12$ minutes; Clear-day speed is $18$ mph greater than rush-hour speed; Same distance both days; Answer choices: (A) $4$, (B) $6$, (C) $8$, (D) $9$, (E) $12$ (miles)
Plan
Primary tool: #5 Find a Variable
Secondary: #8 Analyze the Units
Two unknowns are floating around — the distance $d$ and the rush-hour speed $v$ — but the problem only asks for $d$. Tool #5 (Find a Variable) says: name both, write what each scenario tells you ($d = v \cdot t_1$ and $d = (v+18) \cdot t_2$), then use the shared distance to eliminate $v$ and solve for $d$. Tool #8 (Analyze the Units) handles the trap that times are in minutes while speeds are in mph: convert $20$ min and $12$ min to hours first so every $r \cdot t$ product comes out in miles.
Execute — Answer: D
5.MD.A.1 Step 1 - Convert both travel times from minutes to hours so they match the mph units.
- $20$ min $= \tfrac{20}{60} = \tfrac{1}{3}$ hr and $12$ min $= \tfrac{12}{60} = \tfrac{1}{5}$ hr.
💡 Converting minutes into hours within the same time system is the Grade 5 "convert standard measurement units" move.
6.EE.B.6 Step 2 - Name the unknowns.
- Let $d$ be the distance to school in miles and $v$ be the rush-hour speed in mph.
- The clear-day speed is then $v + 18$ mph.
- This is the Tool #5 step — every quantity now has a label we can compute with.
💡 Using letters to stand for the two unknown quantities is Grade 6 "use variables to represent numbers."
6.RP.A.3 Step 3 - Write the distance equation for each day using $d = r \cdot t$.
- Rush hour: $d = v \cdot \tfrac{1}{3}$.
- Clear day: $d = (v+18) \cdot \tfrac{1}{5}$.
- Same $d$ on both sides because it is the same route.
💡 Distance $=$ rate $\times$ time is the Grade 6 rate-reasoning template.
8.EE.C.8 Step 4 - Set the two expressions for $d$ equal and solve the resulting system.
- Multiplying both equations to clear fractions gives $v = 3d$ from the first and $5d = v + 18$ from the second; substituting yields $5d = 3d + 18$, so $2d = 18$.
💡 Solving a $2 \times 2$ linear system by substitution is the Grade 8 simultaneous-equations standard.
8.EE.C.8 Step 5 - Read off the answer.
- The distance to school is $d = 9$ miles, matching choice (D).
- (As a side check, the rush-hour speed comes out to $v = 27$ mph and the clear-day speed to $45$ mph — both plausible.)
💡 Interpreting the solution of the system as the answer to the original word problem completes Tool #5.
5.MD.A.1 Convert both travel times from minutes to hours so they match the mph units. $20 6.EE.B.6 Name the unknowns. Let $d$ be the distance to school in miles and $v$ be the rus 6.RP.A.3 Write the distance equation for each day using $d = r \cdot t$. Rush hour: $d = 8.EE.C.8 Set the two expressions for $d$ equal and solve the resulting system. Multiplyin 8.EE.C.8 Read off the answer. The distance to school is $d = 9$ miles, matching choice (D Review
Reasonableness: Check both scenarios with $d = 9$ miles. Rush hour: $v = 27$ mph for $\tfrac{1}{3}$ hr gives $27 \times \tfrac{1}{3} = 9$ miles. Clear day: $v+18 = 45$ mph for $\tfrac{1}{5}$ hr gives $45 \times \tfrac{1}{5} = 9$ miles. Both match, and $27$ mph in traffic plus $45$ mph on a clear road are realistic driving speeds — not a sanity-check red flag.
Alternative: Tool #6 (Guess and Check) on the choices: pick $d$, then rush-hour speed is $3d$ and clear-day speed is $5d$, so the gap is $5d - 3d = 2d$, which must equal $18$. Thus $d = 9$, immediately landing on (D) without writing a system. The other choices give gaps of $8, 12, 16, 24$ — none equal $18$.
CCSS standards used (min grade 8)
5.MD.A.1Convert among different-sized standard measurement units within a given system (Converting $20$ min to $\tfrac{1}{3}$ hr and $12$ min to $\tfrac{1}{5}$ hr so the times match the mph speed unit.)6.EE.B.6Use variables to represent numbers and write expressions when solving real-world problems (Naming the distance $d$ and the rush-hour speed $v$, then writing the clear-day speed as $v + 18$.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Applying $\text{distance} = \text{rate} \times \text{time}$ to each scenario to write $d = \tfrac{v}{3}$ and $d = \tfrac{v+18}{5}$.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Solving the $2 \times 2$ system in $d$ and $v$ by substitution to get $v = 27$ and $d = 9$.)
⭐ Name what you don't know with a letter, write one equation per scenario, and let the matching distance do the work — that is the Grade 8 simultaneous-equations move.
⭐ Name what you don't know with a letter, write one equation per scenario, and let the matching distance do the work — that is the Grade 8 simultaneous-equations move.
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