AMC 8 · 2019 · #16

Grade 6 rate-ratioalgebra

Archetype Arithmetic Expression Decomposition

ratelinear-equations-one-varfraction-arithmetic convert-to-algebraidentify-subproblems ↑ Prerequisites: ratelinear-equations-one-var

📏 Medium solution 💡 3 insights

Problem

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?

$\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135$

Pick an answer.

(A)

(B)

(C)

(D)

110

(E)

135

View mode:

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

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

#8 Analyze the Units 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.

$$t_1 = \dfrac{15 \text{ mi}}{30 \text{ mph}} = 0.5 \text{ hr}, \quad \text{avg speed} = \dfrac{15 + d}{0.5 + d/55}$$

💡 Tracking units (mi $\div$ mph $=$ hr) tells us exactly which numbers to add and divide — Grade 6 rate reasoning.

#6 Guess and Check 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.

$$\dfrac{105}{2.136} \approx 49.2 \text{ mph} < 50$$

💡 Bigger $d$ means more time at the fast $55$ mph leg, which pulls the overall average up toward $55$.

#3 Eliminate Possibilities 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.

$d = 45, 62, 90$ all give avg $< 50 \text{ mph} \;\Rightarrow\;$ ruled out

💡 Because average speed grows with $d$, anything below the failing $d = 90$ also fails — we save work.

#6 Guess and Check 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!

$$\dfrac{125}{2.5} = 50 \text{ mph} \;\checkmark$$

💡 $125 \div 2.5 = 50$ is a clean Grade 5 decimal division — no algebra needed.

#3 Eliminate Possibilities 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.

$$\dfrac{150}{2.955} \approx 50.8 \text{ mph} > 50 \;\Rightarrow\; \textbf{(D)}$$

💡 Average speed is monotonic in $d$, so once $110$ hits exactly $50$, no other choice can also work.

[1] #8 6.RP.A.3 First write down the fixed pieces. The first leg always takes the same time, no

[2] #6 6.RP.A.3 Try the middle choice (C) $d = 90$ first. Total distance $= 15 + 90 = 105$ mi. T

[3] #3 6.RP.A.3 Eliminate (A) $45$ and (B) $62$ — both are smaller than $90$, so they would give

[4] #6 5.NBT.B.7 Test (D) $d = 110$. Total distance $= 15 + 110 = 125$ mi. Second-leg time $= \tf

[5] #3 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.7 Add, 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.3 Use 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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