AMC 8 · 2019 · #16
Grade 6 rate-ratioalgebraProblem
Qiang drives 15 miles at an average speed of 30 miles per hour. How many additional miles will he have to drive at 55 miles per hour to average 50 miles per hour for the entire trip?
Pick an answer.
AMC 8 2019 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: Qiang has already driven $15$ miles at $30$ mph. He now wants to keep going at $55$ mph for some extra distance $d$ so that, when you average over the whole trip, the overall speed comes out to exactly $50$ mph. How many extra miles $d$ does he need to drive?
Givens: First leg: $15$ miles at $30$ mph; Second leg: unknown distance $d$ at $55$ mph; Target average speed for the whole trip: $50$ mph; Average speed $=$ total distance $\div$ total time; Answer choices: (A) $45$, (B) $62$, (C) $90$, (D) $110$, (E) $135$ miles
Unknowns: The additional distance $d$ (in miles) Qiang must drive at $55$ mph
Understand
Restated: Qiang has already driven $15$ miles at $30$ mph. He now wants to keep going at $55$ mph for some extra distance $d$ so that, when you average over the whole trip, the overall speed comes out to exactly $50$ mph. How many extra miles $d$ does he need to drive?
Givens: First leg: $15$ miles at $30$ mph; Second leg: unknown distance $d$ at $55$ mph; Target average speed for the whole trip: $50$ mph; Average speed $=$ total distance $\div$ total time; Answer choices: (A) $45$, (B) $62$, (C) $90$, (D) $110$, (E) $135$ miles
Plan
Primary tool: #6 Guess and Check
Secondary: #3 Eliminate Possibilities, #8 Analyze the Units
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 $\tfrac{15+d}{0.5 + d/55} = 50$ 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.
Execute — Answer: D
6.RP.A.3 Step 1 - First write down the fixed pieces.
- The first leg always takes the same time, no matter what $d$ is: $15$ mi $\div$ $30$ mph $= 0.5$ hr.
- So for any candidate $d$, total distance is $15 + d$ and total time is $0.5 + \dfrac{d}{55}$ hr.
💡 Tracking units (mi $\div$ mph $=$ hr) tells us exactly which numbers to add and divide — Grade 6 rate reasoning.
6.RP.A.3 Step 2 - Try the middle choice (C) $d = 90$ first.
- Total distance $= 15 + 90 = 105$ mi.
- Time on the second leg $= \tfrac{90}{55} = \tfrac{18}{11} \approx 1.636$ hr, so total time $\approx 0.5 + 1.636 = 2.136$ hr.
- Average speed $\approx \tfrac{105}{2.136} \approx 49.2$ mph $< 50$.
- So $d = 90$ is slightly too small — we need a bigger $d$ to pull the average up.
💡 Bigger $d$ means more time at the fast $55$ mph leg, which pulls the overall average up toward $55$.
6.RP.A.3 Step 3 - Eliminate (A) $45$ and (B) $62$ — both are smaller than $90$, so they would give an even lower average speed than $49.2$ mph, which is already below $50$.
- Only (D) $110$ and (E) $135$ remain.
💡 Because average speed grows with $d$, anything below the failing $d = 90$ also fails — we save work.
5.NBT.B.7 Step 4 - Test (D) $d = 110$.
- Total distance $= 15 + 110 = 125$ mi.
- Second-leg time $= \tfrac{110}{55} = 2$ hr, so total time $= 0.5 + 2 = 2.5$ hr.
- Average speed $= \tfrac{125}{2.5} = 50$ mph.
- Exact hit!
💡 $125 \div 2.5 = 50$ is a clean Grade 5 decimal division — no algebra needed.
6.RP.A.3 Step 5 - Confirm by ruling out (E) $135$.
- Second-leg time $= \tfrac{135}{55} \approx 2.455$ hr, total time $\approx 2.955$ hr, average $\approx \tfrac{150}{2.955} \approx 50.8$ mph $> 50$.
- So $135$ overshoots, and $110$ 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.3 First write down the fixed pieces. The first leg always takes the same time, no 6.RP.A.3 Try the middle choice (C) $d = 90$ first. Total distance $= 15 + 90 = 105$ mi. T 6.RP.A.3 Eliminate (A) $45$ and (B) $62$ — both are smaller than $90$, so they would give 5.NBT.B.7 Test (D) $d = 110$. Total distance $= 15 + 110 = 125$ mi. Second-leg time $= \tf 6.RP.A.3 Confirm by ruling out (E) $135$. Second-leg time $= \tfrac{135}{55} \approx 2.45 Review
Reasonableness: A common wrong intuition is to average $30$ and $55$ to get $42.5$ and panic that no $d$ can reach $50$. But $d$ only changes the *second* leg, and Qiang spends much more *time* on the slow leg ($0.5$ hr) than the fast leg if $d$ is small. To pull the time-weighted average up to $50$, the fast leg has to dominate the trip — needing $2$ full hours at $55$ mph (i.e. $110$ miles) versus just $0.5$ hr at $30$ mph. The ratio of fast-time to slow-time being $4{:}1$ matches the way $50$ sits between $30$ and $55$ ($50$ is $20$ above $30$ but only $5$ below $55$, i.e. $4{:}1$ — the same ratio). That cross-check confirms (D) $110$.
Alternative: Tool #13 (Convert to Algebra) sets up the single equation $\dfrac{15+d}{0.5 + d/55} = 50$. Cross-multiplying gives $15 + d = 25 + \tfrac{10d}{11}$, so $\tfrac{d}{11} = 10$ and $d = 110$. Same answer, but it requires comfort with rational equations and Grade 7+ algebraic manipulation — heavier than the guess-and-check route the multiple-choice format allows.
CCSS standards used (min grade 6)
5.NBT.B.7Add, subtract, multiply, and divide decimals to hundredths (Computing $125 \div 2.5 = 50$ and the other decimal arithmetic that converts each candidate distance into an average speed.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Treating average speed as total distance $\div$ total time, comparing each candidate's average to the $50$ mph target, and using the monotonic relationship between $d$ and average speed to eliminate choices.)
⭐ This AMC 8 problem only needs Grade 6 rate reasoning — total distance divided by total time — that you already know!
⭐ This AMC 8 problem only needs Grade 6 rate reasoning — total distance divided by total time — that you already know!
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