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Difference between revisions of "2008 AMC 10A Problems/Problem 15"

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Substituting the 1st and 2nd equations into the last gives:
 
Substituting the 1st and 2nd equations into the last gives:
  
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<math>(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</math> (I)
+
<math>(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 (I)</math>
  
 
We are asked the difference between Jan's and Ian's distances, or
 
We are asked the difference between Jan's and Ian's distances, or
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<math>2I_s+10I_t+20=x \Longrightarrow 2(I_s+5I_t)+20=x</math>
 
<math>2I_s+10I_t+20=x \Longrightarrow 2(I_s+5I_t)+20=x</math>
  
−
Substituting (I) into this equation gives:
+
Substituting <math>(I)</math> into this equation gives:
  
 
<math>2(65)+20=x \Longrightarrow 130+20=x \Longrightarrow 150=x</math>
 
<math>2(65)+20=x \Longrightarrow 130+20=x \Longrightarrow 150=x</math>

Revision as of 17:43, 25 June 2008

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:

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

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:

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

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

Substituting $(I)$ into this equation gives:

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

Therefore, the answer is 150 miles or $\boxed{D}$

See also

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
All AMC 10 Problems and Solutions
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