2008 AMC 10A Problems/Problem 15

Problem

Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?

$\mathrm{(A)}\ 120\qquad\mathrm{(B)}\ 130\qquad\mathrm{(C)}\ 140\qquad\mathrm{(D)}\ 150\qquad\mathrm{(E)}\ 160$

Solution

We let Ian's speed and time equal $I_s$ and $I_t$ , respectively. Similarly, let Han's and Jan's speed and time be $H_s$ , $H_t$ , $J_s$ , $J_t$ . The problem gives us 5 equations:

\begin{align}
H_s&=I_s+5 \\
H_t&=I_t+1 \\
J_s&=I_s+10 \\
J_t&=I_t+2 \\
H_s \cdot H_t & =I_s \cdot I_t+70 \end{align*} (Error compiling LaTeX. Unknown error_msg)

Substituting $(1)$ and $(2)$ equations into $(5)$ gives:

$(I_s+5)(I_t+1)=I_s I_t+70 \Longrightarrow I_s I_t+I_s+5I_t+5=I_s I_t+70 \Longrightarrow I_s+5I_t=65 \quad (*)$

We are asked the difference between Jan's and Ian's distances, or

$J_s J_t-I_s I_t=x,$

Where $x$ is the difference between Jan's and Ian's distances and the answer to the problem. Substituting $(3)$ and $(4)$ equations into this equation gives:

$(I_s+10)(I_t+2)-I_s I_t=x \Longrightarrow I_s I_t+2I_s+10I_t+20-I_s I_t=x \Longrightarrow$

$2I_s+10I_t+20=x \Longrightarrow 2(I_s+5I_t)+20=x$

Substituting $(*)$ into this equation gives:

$2(65)+20=x \Longrightarrow 130+20=x \Longrightarrow 150=x$

Therefore, the answer is $150$ miles or $\boxed{\mathrm{(D)}}$ .

2008 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2008 AMC 10A Problems/Problem 15

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