2008 AMC 10A Problem 15

1. Name Ian's numbers. Say Ian drove for $t$ hours at $v$ miles per hour, so he drove $vt$ miles.
2. Han drove $1$ hour longer at $5$ mph faster, so Han's distance is $(v + 5)(t + 1)$. Multiply it out:
$$(v + 5)(t + 1) = vt + v + 5t + 5.$$
3. Han drove $70$ miles more than Ian's $vt$. The $vt$ parts match, so the extra part is $70$:
$$v + 5t + 5 = 70, \qquad\text{so}\qquad v + 5t = 65.$$
4. We cannot find $v$ and $t$ separately from one equation. But maybe we do not need to, so look at Jan. Jan's distance is
$$(v + 10)(t + 2) = vt + 2v + 10t + 20.$$
5. Jan's extra distance over Ian is $2v + 10t + 20$. The $2v + 10t$ part is exactly twice $v + 5t$:
$$2v + 10t + 20 = 2 \times 65 + 20 = 150.$$

The answer is $\boxed{\textbf{(D)}\ 150}$.

The problem has two unknowns but only one fact, which feels like not enough. The way out is to notice that the question asks for an amount built from the same combination, $v + 5t$, that the fact gives you. When a problem seems to be missing information, check whether you need the separate values at all.

Since any $v$ and $t$ with $v + 5t = 65$ will do, pick easy ones: $t = 1$ hour and $v = 60$ mph.

• Ian: $60 \times 1 = 60$ miles.
• Han: $65 \times 2 = 130$ miles, which is $70$ more than Ian. Good.
• Jan: $70 \times 3 = 210$ miles, which is $150$ more than Ian.

Let Ian drive for $t$ hours at $v$ miles per hour, covering $vt$ miles.

Han drove $(v + 5)(t + 1) = vt + v + 5t + 5$ miles, which is $70$ more than Ian's $vt$:
$$v + 5t + 5 = 70 \quad\Longrightarrow\quad v + 5t = 65.$$

Jan drove $(v + 10)(t + 2) = vt + 2v + 10t + 20$ miles, so Jan's excess over Ian is
$$2v + 10t + 20 = 2(v + 5t) + 20 = 2 \cdot 65 + 20 = 150.$$

The answer is $\boxed{\textbf{(D)}\ 150}$.