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 We will represent Han's and Jan's speed and time similarly: $H_s$ , $H_t$ , $J_s$ , $J_t$ The problem gives us 5 equations: $H_s=I_s+5$ $H_t=I_t+1$ $J_t=I_s+10$ $J_t=I_t+2$ $H_s H_t=I_s I_t+70$ Substituting the 1st and 2nd equations into the last gives: $(I_s+5)(I_t+1)=I_s I_t+70$ $I_s I_t+I_s+5I_t+5=I_s I_t+70$ $I_s+5I_t=65$ (I) 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 the 3rd and 4th equations into this equation gives: $(I_s+10)(I_t+2)-I_s I_t=x$ $I_s I_t+2I_s+10I_t+20-I_s I_t=x$ $2I_s+10I_t+20=x$ 2(I_s+5I_t)+20=x Substituting (I) into this equation gives: $2(65)+20=x$ $130+20=x$ $150=x$ Therefore, the answer is 150 miles or $\boxed{D}$

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

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