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2024 AMC 10A Problem 2

2024 AMC 10A Problem 2AMC 10 Step by Step

Name: Date: October 28, 2026

Linear Equations & Word Problems · difficulty 1 of 5 · about 2 min · position P1-10

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 $T=aL+bG,$ where $a$ and $b$ are constants, $T$ is the time in minutes, $L$ is the length of the trail in miles, and $G$ is the altitude gain in feet. The model estimates that it will take $69$ minutes to hike to the top if a trail is $1.5$ miles long and ascends $800$ feet, as well as if a trail is $1.2$ miles long and ascends $1100$ feet. How many minutes does the model estimates it will take to hike to the top if the trail is $4.2$ miles long and ascends $4000$ feet?

A) $240$
B) $246$
C) $252$
D) $258$
E) $264$

Answer

Key insight

Both trails take 69 minutes, so subtract one equation from the other. That gives a simple link between a and b: a is 1000 times b.

Solution

The formula $T = aL + bG$ has two unknown numbers, $a$ and $b$ . The problem gives two trails with known times, so each trail gives one equation. Two equations are enough to find two unknowns.

1. The first trail has $L = 1.5$ , $G = 800$ and $T = 69$ :
$1.5a + 800b = 69.$
2. The second trail has $L = 1.2$ , $G = 1100$ and $T = 69$ :
$1.2a + 1100b = 69.$
3. Both right sides are $69$ , so subtracting the second equation from the first makes the $69$ disappear:
$0.3a - 300b = 0.$
So $0.3a = 300b$ . Multiply both sides by $10$ : $3a = 3000b$ , so $a = 1000b$ .
4. Put $a = 1000b$ into the first equation:
$1.5 \cdot 1000b + 800b = 1500b + 800b = 2300b = 69.$
So $b = \dfrac{69}{2300} = 0.03$ , and $a = 1000 \cdot 0.03 = 30$ .
5. Check with the second trail: $1.2 \cdot 30 + 1100 \cdot 0.03 = 36 + 33 = 69$ . Good.
6. Now the new trail, with $L = 4.2$ and $G = 4000$ :
$T = 4.2 \cdot 30 + 4000 \cdot 0.03 = 126 + 120 = 246.$

The answer is $\boxed{\textbf{(B)}\ 246}$ .

Why this works

A formula with two unknown constants is pinned down by two examples. When both examples give the same result, subtracting one equation from the other is the fastest first move, because the equal right sides cancel. Always finish by answering the actual question, which here is the time for a third trail.

Alternative approach

The decimals and the big numbers can be made friendlier first. Write $a = 10x$ and $b = \dfrac{y}{100}$ . Then $1.5a = 15x$ and $800b = 8y$ , and the other terms change the same way. The two equations become
$15x + 8y = 69, \qquad 12x + 11y = 69.$
In each equation the two coefficients add to $23$ , and $69$ is three times $23$ . So $x = 3$ and $y = 3$ work in both. That means $a = 30$ and $b = 0.03$ , as before.

The new trail asks for $4.2a + 4000b$ , which is $42x + 40y = 42 \cdot 3 + 40 \cdot 3 = 82 \cdot 3 = 246$ .

The trap

Making a decimal slip when subtracting, or finding a and b and then forgetting to work out the time for the new trail.

Concise solution

The two hikes give
$1.5a + 800b = 69, \qquad 1.2a + 1100b = 69.$
Subtract the second from the first: $0.3a - 300b = 0$ , so $a = 1000b$ .

Put this into the first equation: $1500b + 800b = 2300b = 69$ , hence $b = 0.03$ and $a = 30$ .

For the new trail,
$T = 4.2\cdot 30 + 4000\cdot 0.03 = 126 + 120 = 246.$

The answer is $\boxed{\textbf{(B)}\ 246}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Algebra, Chapter 5: Multi-Variable Linear Equations (§5.2, §5.3, §5.6)
Introduction to Algebra, Chapter 7: Proportion (§7.4)
Intermediate Algebra, Chapter 1: Basic Techniques for Solving Equations (§1.4)

Techniques

Set up the equation/formula and compute; no special trick needed