AMC 10 · 2024 · #2
Grade 8 algebrarate-ratioProblem
A model used to estimate the time it will take to hike to the top of the mountain on a trail is of the form where and are constants, is the time in minutes, is the length of the trail in miles, and is the altitude gain in feet. The model estimates that it will take minutes to hike to the top if a trail is miles long and ascends feet, as well as if a trail is miles long and ascends feet. How many minutes does the model estimates it will take to hike to the top if the trail is miles long and ascends feet?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: A hiking-time model is $T = aL + bG$, where $L$ is trail length in miles, $G$ is altitude gain in feet, and $T$ is time in minutes. Two different trails both take $69$ minutes: one is $1.5$ mi with $800$ ft gain, the other is $1.2$ mi with $1100$ ft gain. Using the same model, find $T$ for a trail that is $4.2$ mi long with $4000$ ft gain.
Givens: Model: $T = aL + bG$ with unknown constants $a, b$; Trail 1: $L=1.5,\ G=800,\ T=69$; Trail 2: $L=1.2,\ G=1100,\ T=69$; Target trail: $L=4.2,\ G=4000$; Answer choices: (A) $240$, (B) $246$, (C) $252$, (D) $258$, (E) $264$
Unknowns: The estimated time $T$ for the $4.2$ mi, $4000$ ft trail
Understand
Restated: A hiking-time model is $T = aL + bG$, where $L$ is trail length in miles, $G$ is altitude gain in feet, and $T$ is time in minutes. Two different trails both take $69$ minutes: one is $1.5$ mi with $800$ ft gain, the other is $1.2$ mi with $1100$ ft gain. Using the same model, find $T$ for a trail that is $4.2$ mi long with $4000$ ft gain.
Givens: Model: $T = aL + bG$ with unknown constants $a, b$; Trail 1: $L=1.5,\ G=800,\ T=69$; Trail 2: $L=1.2,\ G=1100,\ T=69$; Target trail: $L=4.2,\ G=4000$; Answer choices: (A) $240$, (B) $246$, (C) $252$, (D) $258$, (E) $264$
Plan
Primary tool: #13 Convert to Algebra
Secondary: #7 Identify Subproblems
The problem hands us a formula with two unknown constants and two data points, which is the textbook setup for Tool #13 (Convert to Algebra) — translate each trail into one linear equation, then solve the $2\times 2$ system. Tool #7 (Identify Subproblems) splits the work into two clean parts: (i) recover the constants $a$ and $b$ from Trails 1 and 2, then (ii) plug them into $T = aL + bG$ for the target trail. The two scenarios share the same $T = 69$, so subtracting (or equating) the equations cancels $T$ immediately and gives a one-line relation between $a$ and $b$.
Execute — Answer: B
8.F.B.4 Step 1 - Subproblem 1: write a linear equation for each trail.
- Plug each $(L, G, T)$ triple into $T = aL + bG$.
💡 Grade 8 "construct a function from data": each trail is one $(L, G) \mapsto T$ data point, and we are pinning down the linear model that fits both.
8.EE.C.8 Step 2 - Both equations equal $69$, so the right-hand sides are equal.
- Subtract Trail 2 from Trail 1 to eliminate the constant and get a direct relation between $a$ and $b$.
💡 Grade 8 systems of equations: equal $T$ values let us subtract the two equations and collapse a 2-variable system into a single relation $a = 1000b$.
7.EE.B.4 Step 3 Substitute $a = 1000b$ back into Trail 1 to solve for $b$, then read off $a$.
💡 Grade 7 one-variable equation: once one unknown is written in terms of the other, the system reduces to $2300b = 69$, which divides cleanly.
8.F.B.4 Step 4 Subproblem 2: plug the recovered constants and the new trail's $(L, G) = (4.2,\ 4000)$ into the model.
💡 Now that the model is fully known, applying it to a new $(L, G)$ is just one substitution — the second subproblem is a single line of arithmetic.
8.F.B.4 Subproblem 1: write a linear equation for each trail. Plug each $(L, G, T)$ trip 8.EE.C.8 Both equations equal $69$, so the right-hand sides are equal. Subtract Trail 2 f 7.EE.B.4 Substitute $a = 1000b$ back into Trail 1 to solve for $b$, then read off $a$. 8.F.B.4 Subproblem 2: plug the recovered constants and the new trail's $(L, G) = (4.2,\ Review
Reasonableness: Sanity-check the constants on Trail 2: $1.2(30) + 1100(0.03) = 36 + 33 = 69$. Matches. The numbers also pass a units check: $a = 30$ min/mile and $b = 0.03$ min/ft, so a longer, steeper trail naturally takes more time. For the target, $4.2$ mi is about $3\times$ longer than Trail 1 and $4000$ ft is $5\times$ steeper, so a time of $246 \approx 3.5 \times 69$ is in the right ballpark.
Alternative: Tool #3 (Eliminate Possibilities): once $a = 30$ and $b = 3/100$ are known, you could plug them into each answer choice and verify $30(4.2) + 0.03(4000) = 246$ matches only (B). Or, without solving the system, notice $4.2 = 1.5 + 1.2 + 1.5$ and $4000 = 800 + 1100 + ?$ — but that linear-combination shortcut is messier than just solving the system, which is why Tool #13 is the cleaner primary choice.
CCSS standards used (min grade 8)
8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Setting up and solving the $2\times 2$ system $1.5a + 800b = 69$ and $1.2a + 1100b = 69$ for the constants $a$ and $b$.)8.F.B.4Construct a function to model a linear relationship between two quantities (Reading $T = aL + bG$ as a linear model in $L$ and $G$, fitting it to the two data points, then evaluating it at $(L, G) = (4.2,\ 4000)$.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Reducing the system to the single-variable equation $2300b = 69$ and solving for $b$, then back-substituting to find $a$.)
⭐ Two trails with the same time give two equations in the same two unknowns — solve the little system once, and the model is yours to use on any new trail.
⭐ Two trails with the same time give two equations in the same two unknowns — solve the little system once, and the model is yours to use on any new trail.