Competition · AMC preparation · step 4 of 4
AMC 8 · 2019 · #16
Grade 6 rate-ratioalgebraPick an answer.
AMC 8 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Since this is multiple choice and each candidate d gives a clean overall-average-speed check, Tool #6 (Guess and Check) on the five choices is the fastest honest path — far simpler than setting up and solving the rational equation (15+d)/(0.5 + d/55) = 50 with algebra. Tool #3 (Eliminate Possibilities) is the natural companion for any AMC multiple-choice question — we keep ruling out choices until one survives. Tool #8 (Analyze the Units) is the bookkeeping that keeps miles, mph, and hours consistent so the comparison to 50 mph is meaningful.
Write down the fixed times
The first leg is fixed: 15 mi ÷ 30 mph = 0.5 hr, so any candidate makes total distance 15 + d over total time 0.5 + hr.
Tracking units (mi ÷ mph = hr) tells us exactly which numbers to add and divide — Grade 6 rate reasoning.
For any additional distance d, the whole trip's average speed equals the total distance 15 + d divided by the total time 0.5 + d/55 hours.
▸ Why?
Average speed for the whole trip means the single steady speed that would cover the same total distance in the same total time, so it equals total distance divided by total time — not the plain average of 30 and 55.
▸ Why?
Going at a steady speed, the distance covered is that speed added once for every hour, so distance = speed × time.
▸ Why?
To recover the steady speed from distance = speed × time, you divide the distance by the time, because division undoes multiplication.
▸ Why?
The total distance is 15 + d because the first leg of 15 miles and the second leg of d miles join end to end with no gap or overlap.
▸ Why?
The total time is 0.5 + d/55 hours because the first leg's time and the second leg's time add up to the whole trip's time, with the first leg taking 15 ÷ 30 = 0.5 hour.
▸ Why?
The two legs happen one after the other with no gap or overlap, so their times add up to the whole trip's time.
▸ Why?
Each leg's time comes from dividing its distance by its speed, since division undoes the multiplication distance = speed × time.
Test the middle choice
Test the middle choice d = 90: total 105 mi over 0.5 + ≈ 2.14 hr gives avg ≈ 49.2 mph — just under 50, so we need a bigger d.
Bigger d means more time at the fast 55 mph leg, which pulls the overall average up toward 55.
6.RP.A.3Guess And CheckRule out the smaller choices
Since average speed grows with d, choices 45 and 62 (both below 90) fail even harder — eliminate them, leaving only 110 and 135.
Because average speed grows with d, anything below the failing d = 90 also fails — we save work.
6.RP.A.3Eliminate PossibilitiesTest 110 miles
Test d = 110: total 125 mi over 0.5 + = 2.5 hr gives 125 ÷ 2.5 = exactly 50 mph — a perfect hit.
125 ÷ 2.5 = 50 is a clean Grade 5 decimal division — no algebra needed.
5.NBT.B.7Guess And CheckRule out the largest choice
Check d = 135: it overshoots to avg ≈ 50.8 mph above 50, so 110 is the unique answer.
Average speed is monotonic in d, so once 110 hits exactly 50, no other choice can also work.
6.RP.A.3Eliminate PossibilitiesThis AMC 8 problem only needs Grade 6 rate reasoning — total distance divided by total time — that you already know!
- Write down the fixed times
- Test the middle choice
- Rule out the smaller choices
- Test 110 miles
- Rule out the largest choice
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