Competition · AMC preparation · step 4 of 4
AMC 8 · 2015 · #17
Grade 8 rate-ratioalgebraPick an answer.
AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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 · t₁ and d = (v+18) · t₂), 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 · t product comes out in miles.
Convert minutes to hours
Convert both times to hours to match the mph units: 20 min = hr and 12 min = hr.
Converting minutes into hours within the same time system is the Grade 5 "convert standard measurement units" move.
5.MD.A.1Analyze The UnitsName the unknowns
Name the unknowns: let d be the distance and v the rush-hour speed, so the clear-day speed is v + 18.
Using letters to stand for the two unknown quantities is Grade 6 "use variables to represent numbers."
6.EE.B.6Look For A PatternWrite a distance equation each day
Use distance = speed × time for each day: rush hour gives d = , clear day gives d = , same d both times.
Distance = rate × time is the Grade 6 rate-reasoning template.
6.RP.A.3Look For A PatternSet the two equal and solve
Set the two equal: from v = 3d and 5d = v + 18, substituting gives 5d = 3d + 18, so 2d = 18.
Solving a 2 × 2 linear system by substitution is the Grade 8 simultaneous-equations standard.
Because the drive covers the same route both days, the rush-hour distance expression and the clear-day distance expression must be equal, and solving that single equation pins down how far it is to school.
▸ Why?
Each expression really does equal the distance to school, because the miles covered are the steady speed multiplied by the time spent driving.
▸ Why?
Driving a fixed number of miles each hour for the trip's time stacks up that many equal hourly stretches, and totaling equal groups is exactly what multiplying speed by time does.
▸ Why?
Since both expressions equal that one distance, they equal each other, leaving a single equation whose only unknown is the rush-hour speed.
▸ Why?
The rush-hour expression equals the distance and the distance equals the clear-day expression, so the two expressions equal each other through the distance they share.
▸ Why?
Undoing the operations wrapped around the unknown with their opposites strips it down to one definite value for the distance.
▸ Why?
Subtracting the same amount from both sides reverses an addition and dividing both sides reverses a multiplication, so the unknown can be freed one step at a time.
Read off the distance
Read off the answer: the distance to school is d = 9 miles, which is choice (D).
Interpreting the solution of the system as the answer to the original word problem completes Tool #5.
8.EE.C.8Look For A PatternName 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.
- Convert minutes to hours
- Name the unknowns
- Write a distance equation each day
- Set the two equal and solve
- Read off the distance
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