AMC 10 · 2009 · #4

Grade 6 rate-ratio

Archetype Meeting Time with Piecewise Rates └ Sub-archetype Piecewise-Leg Time Balance

rateunit-conversionfraction-arithmetic identify-subproblemswork-backwards ↑ Prerequisites: rate

📏 Medium solution 💡 2 insights

Problem

Eric plans to compete in a triathlon. He can average $2$ miles per hour in the $\frac{1}{4}$ -mile swim and $6$ miles per hour in the $3$ -mile run. His goal is to finish the triathlon in $2$ hours. To accomplish his goal what must his average speed in miles per hour, be for the $15$ -mile bicycle ride?

Pick an answer.

(A)

$\ \frac{120}{11}$

(B)

$\ 11$

(C)

$\ \frac{56}{5}$

(D)

$\ \frac{45}{4}$

(E)

$\ 12$

AMC 10 2009 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.

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Toolkit + CCSS Solution

Understand

Restated: Eric does a triathlon with three legs: a $\frac{1}{4}$-mile swim at $2$ miles per hour, a $3$-mile run at $6$ miles per hour, and a $15$-mile bike ride. He wants the whole thing to take exactly $2$ hours. Find the average speed, in miles per hour, the bike ride must have.

Givens: Swim: $\frac{1}{4}$ mile at $2$ miles per hour; Run: $3$ miles at $6$ miles per hour; Bike: $15$ miles at an unknown speed; Total time for all three legs must equal $2$ hours; Answer choices: (A) $\frac{120}{11}$, (B) $11$, (C) $\frac{56}{5}$, (D) $\frac{45}{4}$, (E) $12$

Unknowns: The average speed, in miles per hour, needed for the $15$-mile bike ride

Understand

Plan

Primary tool: #8 Analyze the Units

Secondary: #7 Identify Subproblems, #11 Work Backwards

This is a distance-speed-time problem, so Tool #8 (Analyze the Units) is the spine: the relationship time $=$ distance $\div$ speed converts every leg into a number of hours and, at the end, turns a leftover time back into a speed. Tool #7 (Identify Subproblems) splits the race into its three legs so the two known legs can be measured first. Tool #11 (Work Backwards) is the finishing move: the finish time ($2$ hours) is fixed, so we subtract the known leg times to see how much time the bike is allowed, then read off the speed that fits that time.

Execute — Answer: A

#8 Analyze the Units 6.RP.A.3 Step 1

Set the time budget rule

Time, distance, and speed are tied together by time $=$ distance $\div$ speed, which comes out in hours when distance is in miles and speed is in miles per hour.
The three leg times must add up to the $2$-hour goal, so if we find the swim and run times, whatever is left is the time the bike ride gets.

$$\text{time}=\dfrac{\text{distance}}{\text{speed}},\qquad t_{\text{swim}}+t_{\text{run}}+t_{\text{bike}}=2\text{ hours}$$

💡 A fixed finish time is a budget: the three legs share $2$ hours, so time spent early is time taken from the bike.

#7 Identify Subproblems 6.RP.A.3 Step 2

Time the swim and the run

Handle the two legs whose speeds are known.
Swim: $\frac{1}{4}$ mile at $2$ miles per hour takes $\frac{1}{4}\div 2=\frac{1}{8}$ hour.
Run: $3$ miles at $6$ miles per hour takes $3\div 6=\frac{1}{2}$ hour.

$$t_{\text{swim}}=\dfrac{1/4}{2}=\dfrac{1}{8}\text{ h},\qquad t_{\text{run}}=\dfrac{3}{6}=\dfrac{1}{2}\text{ h}$$

💡 Each leg is its own tiny problem: divide its distance by its speed to get its hours.

#11 Work Backwards 5.NF.A.1 Step 3

Find the bike's time budget

Add the two known times, then subtract from the $2$-hour goal to see how long the bike ride may take.
First $\frac{1}{8}+\frac{1}{2}=\frac{1}{8}+\frac{4}{8}=\frac{5}{8}$ hour.
The bike then gets $2-\frac{5}{8}=\frac{16}{8}-\frac{5}{8}=\frac{11}{8}$ hour.

$$t_{\text{swim}}+t_{\text{run}}=\dfrac{1}{8}+\dfrac{4}{8}=\dfrac{5}{8}\text{ h},\qquad t_{\text{bike}}=2-\dfrac{5}{8}=\dfrac{11}{8}\text{ h}$$

💡 Whatever the swim and run do not use, the bike is free to use — so subtract from the $2$-hour budget.

#8 Analyze the Units 6.NS.A.1 Step 4

Turn the time back into a speed

The bike must cover $15$ miles in $\frac{11}{8}$ hour, so its speed is distance divided by time.
Dividing by $\frac{11}{8}$ means multiplying by $\frac{8}{11}$: $15\div\frac{11}{8}=15\times\frac{8}{11}=\frac{120}{11}$ miles per hour.
That is choice (A).

$$\text{speed}=\dfrac{15}{\,11/8\,}=15\times\dfrac{8}{11}=\dfrac{120}{11}\text{ mph}\ \textbf{(A)}$$

💡 Speed is just distance shared out over the time you have, so divide the miles by the hours.

[1] #8 6.RP.A.3 Time, distance, and speed are tied together by time $=$ distance $\div$ speed, w

[2] #7 6.RP.A.3 Handle the two legs whose speeds are known. Swim: $\frac{1}{4}$ mile at $2$ mile

[3] #11 5.NF.A.1 Add the two known times, then subtract from the $2$-hour goal to see how long th

[4] #8 6.NS.A.1 The bike must cover $15$ miles in $\frac{11}{8}$ hour, so its speed is distance

Review

Reasonableness: Check the total time with speed $\frac{120}{11}$: the bike takes $15\div\frac{120}{11}=\frac{15\times 11}{120}=\frac{165}{120}=\frac{11}{8}$ hour, and $\frac{1}{8}+\frac{1}{2}+\frac{11}{8}=\frac{1}{8}+\frac{4}{8}+\frac{11}{8}=\frac{16}{8}=2$ hours exactly, matching the goal. The value $\frac{120}{11}\approx 10.9$ mph is a believable bike speed — faster than the run but well short of the smaller choices being too large — so answer (A) holds.

Alternative: Work purely in fractions of the $2$-hour budget without a separate speed formula: the swim eats $\frac{1}{8}$ h and the run $\frac{4}{8}$ h, leaving $\frac{11}{8}$ h. Since $15$ miles must fit into $\frac{11}{8}$ h, scale up to one hour by multiplying by $\frac{8}{11}$, giving $\frac{120}{11}$ miles in one hour — the required speed.

CCSS standards used (min grade 6)

6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Using time $=$ distance $\div$ speed to turn each leg into hours and to convert the bike's leftover time back into a speed.)
5.NF.A.1 Add and subtract fractions with unlike denominators (Combining $\frac{1}{8}+\frac{1}{2}=\frac{5}{8}$ and subtracting $2-\frac{5}{8}=\frac{11}{8}$ to find the bike's time budget.)
6.NS.A.1 Interpret and compute quotients of fractions and solve word problems (Dividing $15$ miles by $\frac{11}{8}$ hour, i.e. $15\times\frac{8}{11}=\frac{120}{11}$, to get the required bike speed.)

⭐ A fixed finish time is a budget of hours: measure the legs you can, subtract to see what time is left, then divide the miles by that time to get the speed.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Set the time budget rule

Time the swim and the run

Find the bike's time budget

Turn the time back into a speed

Review

CCSS standards used (min grade 6)

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