# Difference between revisions of "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}

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 $\boxed{\mathrm{(D)}}$.

## Solution 2

Let Ian drive $D$ miles, at a speed of $R$, for some time(in hours) $T$. Hence, we have $D=RT$. We can find a similar equation for Han, who drove $D + 70$ miles, at a rate of $R+5$, for $T+1$ hours, giving us $D + 70 = (R + 5)(T + 1)$. We can do the same for Jan, giving us $D + x = (R + 10)(T + 2)$, where $x$ is how much further Jan traveled than Ian. We now have three equations: $$D= RT$$ $$D + 70 = (R+5)(T+1) = RT + R + 5T + 5$$ $$D + x = (R + 10)(T + 2) = RT + 10 T + 2R + 20.$$ Substituting $RT$ for $D$ in the second and third equations and cancelling gives us: $$70 = 5T + R + 5 \Longrightarrow 5T + R = 65$$ $$x = 10T + 2R + 20 \Longrightarrow x = 2(5T + R ) + 20 \Longrightarrow x= 2(65) + 20 = 150.$$ Since $x = 150$, our answer is $\boxed{\mathrm{D}}$.

## See also

 2008 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 14 Followed byProblem 16 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

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