AMC 10 · 2007 · #13
Grade 7 arithmeticYan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the ratio of Yan's distance from his home to his distance from the stadium?
Pick an answer.
AMC 10 2007 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: Yan stands somewhere on the straight road between his home and the stadium. He can either walk straight to the stadium, or walk back home first and then ride his bike to the stadium. His bike is $7$ times as fast as his walking. The two routes take the same amount of time. Find the ratio of his distance from home to his distance from the stadium.
Givens: Yan is at a point between his home and the stadium, on the straight road joining them; Route 1: walk directly from his spot to the stadium; Route 2: walk from his spot back home, then bike from home to the stadium; His biking speed is $7$ times his walking speed; Both routes take exactly the same amount of time; Answer choices: (A) $\frac{2}{3}$, (B) $\frac{3}{4}$, (C) $\frac{4}{5}$, (D) $\frac{5}{6}$, (E) $\frac{7}{8}$
Unknowns: The ratio of Yan's distance from home to his distance from the stadium
Understand
Restated: Yan stands somewhere on the straight road between his home and the stadium. He can either walk straight to the stadium, or walk back home first and then ride his bike to the stadium. His bike is $7$ times as fast as his walking. The two routes take the same amount of time. Find the ratio of his distance from home to his distance from the stadium.
Givens: Yan is at a point between his home and the stadium, on the straight road joining them; Route 1: walk directly from his spot to the stadium; Route 2: walk from his spot back home, then bike from home to the stadium; His biking speed is $7$ times his walking speed; Both routes take exactly the same amount of time; Answer choices: (A) $\frac{2}{3}$, (B) $\frac{3}{4}$, (C) $\frac{4}{5}$, (D) $\frac{5}{6}$, (E) $\frac{7}{8}$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #1 Draw a Diagram, #8 Analyze the Units, #13 Convert to Algebra
The problem gives no actual numbers, only relationships, so Tool #4 (Introduce a Variable) is the way in: name Yan's distance from home $h$, his distance from the stadium $s$, and his walking speed $w$. Tool #1 (Draw a Diagram) fixes the order home-Yan-stadium on a line, which is what makes the home-to-stadium distance equal $h+s$. Tool #8 (Analyze the Units) supplies the one physics fact needed: time $=$ distance $\div$ speed, so each route's time is a length over a speed. Tool #13 (Convert to Algebra) then turns 'both routes take the same time' into a single equation, and because every term carries the same $w$, the unknown speed cancels and only the ratio survives.
Execute — Answer: B
6.EE.B.6 Step 1 Draw the line and name the pieces
- Place home, Yan, and the stadium in that order on a straight line.
- Let $h$ be Yan's distance from home and $s$ his distance from the stadium.
- Then the full distance from home to the stadium is $h + s$.
- Let $w$ be his walking speed, so his biking speed is $7w$.
💡 Giving each unknown length and speed a letter lets both routes be written as expressions instead of mystery numbers.
6.RP.A.3 Step 2 Write each route's time
- Time is distance divided by speed.
- Route 1 walks the distance $s$ at speed $w$, taking $\frac{s}{w}$.
- Route 2 first walks home, a distance $h$ at speed $w$, taking $\frac{h}{w}$, then bikes the whole road home-to-stadium, a distance $h+s$ at speed $7w$, taking $\frac{h+s}{7w}$.
💡 Every leg of a trip contributes its own length-over-speed, and biking's larger speed sits in the denominator so that leg costs less time.
7.EE.B.4 Step 3 Set the two times equal
- The problem says both routes take the same time, so $t_1 = t_2$.
- That gives one equation linking $h$, $s$, and $w$.
💡 Equal travel times is the single fact the whole problem hangs on, so writing it as an equation captures everything at once.
7.EE.A.1 Step 4 Clear the speed and simplify
- Multiply every term by $7w$.
- The $w$ appears in every denominator, so it cancels completely and the walking speed disappears.
- This leaves $7s = 7h + (h+s)$.
- Combine the right side to get $7s = 8h + s$, and subtract $s$ from both sides to reach $6s = 8h$.
💡 Because the same speed multiplies every leg, it factors out entirely, which is why the answer never depends on how fast Yan actually walks.
6.RP.A.3 Step 5 Read off the ratio
- From $6s = 8h$, divide both sides by $s$ and by $8$ to isolate the ratio $\frac{h}{s} = \frac{6}{8} = \frac{3}{4}$.
- So Yan's distance from home is $\frac{3}{4}$ of his distance from the stadium, which is choice (B).
💡 An equation between two multiples of $h$ and $s$ is really a statement about their ratio, so dividing turns it straight into the answer.
6.EE.B.6 Place home, Yan, and the stadium in that order on a straight line. Let $h$ be Ya 6.RP.A.3 Time is distance divided by speed. Route 1 walks the distance $s$ at speed $w$, 7.EE.B.4 The problem says both routes take the same time, so $t_1 = t_2$. That gives one 7.EE.A.1 Multiply every term by $7w$. The $w$ appears in every denominator, so it cancels 6.RP.A.3 From $6s = 8h$, divide both sides by $s$ and by $8$ to isolate the ratio $\frac{ Review
Reasonableness: Test the ratio with concrete numbers. Take $h = 3$ and $s = 4$ (which give $\frac{h}{s} = \frac{3}{4}$) and let walking speed $w = 1$. Route 1 takes $\frac{s}{w} = 4$. Route 2 walks home in $\frac{h}{w} = 3$, then bikes the full $h+s = 7$ at speed $7$, taking $\frac{7}{7} = 1$, for a total of $3 + 1 = 4$. Both routes take $4$, exactly matching, so $\frac{3}{4}$ is correct. It also makes sense that Yan is closer to home than to the stadium: the backtrack home only pays off because the long ride afterward is very fast.
Alternative: Skip the walking-speed letter by measuring time in 'walking units,' where walking a distance $d$ costs $d$ and biking it costs $\frac{d}{7}$. Then route 1 costs $s$ and route 2 costs $h + \frac{h+s}{7}$. Setting $s = h + \frac{h+s}{7}$ and multiplying by $7$ gives $7s = 7h + h + s$, so $6s = 8h$ and $\frac{h}{s} = \frac{3}{4}$ again. Choosing units that make walking speed $1$ reaches the same equation with less writing.
CCSS standards used (min grade 7)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Naming Yan's distance from home $h$, his distance from the stadium $s$, and his walking speed $w$, and writing the home-to-stadium distance as $h+s$.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Turning each route into a time with time $=$ distance $\div$ speed, and reading the final ratio $\frac{h}{s}=\frac{3}{4}$ off the equation $6s=8h$.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Writing the equal-time condition as the equation $\frac{s}{w} = \frac{h}{w} + \frac{h+s}{7w}$.)7.EE.A.1Apply properties of operations to add, subtract, factor, and expand linear expressions (Multiplying through by $7w$ to cancel the speed and simplifying $7s = 7h + (h+s)$ down to $6s = 8h$.)
⭐ When a trip has no real numbers, name the distances and the speed with letters, write each time as distance over speed, and the unknown speed will cancel to leave just the ratio you want.
⭐ When a trip has no real numbers, name the distances and the speed with letters, write each time as distance over speed, and the unknown speed will cancel to leave just the ratio you want.
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