2024 AMC 10A Problem 2

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 $\boxed{\textbf{(B)}\ 246}$.

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.

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 $42x + 40y = 42 \cdot 3 + 40 \cdot 3 = 82 \cdot 3 = 246$.