Difference between revisions of "2008 AMC 10A Problems/Problem 15"
(→Solution) |
|||
| Line 12: | Line 12: | ||
J_s&=I_s+10 \\ | J_s&=I_s+10 \\ | ||
J_t&=I_t+2 \\ | J_t&=I_t+2 \\ | ||
| − | H_s \cdot H_t & =I_s \cdot I_t+70 \end{align | + | H_s \cdot H_t & =I_s \cdot I_t+70 \end{align}</cmath> |
Substituting <math>(1)</math> and <math>(2)</math> equations into <math>(5)</math> gives: | Substituting <math>(1)</math> and <math>(2)</math> equations into <math>(5)</math> gives: | ||
Revision as of 10:00, 20 September 2015
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?
Solution
We let Ian's speed and time equal 
and 
, respectively. Similarly, let Han's and Jan's speed and time be 
, 
, 
, 
. The problem gives us 5 equations:
Substituting 
and 
equations into 
gives:
![\[(I_s+5)(I_t+1)=I_s I_t+70 \Longrightarrow I_s I_t+I_s+5I_t+5=I_s I_t+70 \Longrightarrow I_s+5I_t=65 \quad (*)\]](http://latex.artofproblemsolving.com/9/5/9/959c4ab7dc335d539d5d0e1edd070aa6c29de010.png)
We are asked the difference between Jan's and Ian's distances, or
![\[J_s J_t-I_s I_t=x,\]](http://latex.artofproblemsolving.com/9/0/b/90bfcf79cfe47ece92e7c7c48a8d0a1536625f89.png)
Where 
is the difference between Jan's and Ian's distances and the answer to the problem. Substituting 
and 
equations into this equation gives:
![\[(I_s+10)(I_t+2)-I_s I_t=x \Longrightarrow I_s I_t+2I_s+10I_t+20-I_s I_t=x \Longrightarrow\]](http://latex.artofproblemsolving.com/d/c/2/dc251ae3c94cbac68f792c1a5844f68e12d781ea.png)
![\[2I_s+10I_t+20=x \Longrightarrow 2(I_s+5I_t)+20=x\]](http://latex.artofproblemsolving.com/0/0/a/00a7eed557bc9d5cd8132f3055c34976067dcf7b.png)
Substituting 
into this equation gives:
![\[2(65)+20=x \Longrightarrow 130+20=x \Longrightarrow 150=x\]](http://latex.artofproblemsolving.com/5/b/7/5b74cd0faa3b7a2d71e6c0589c59dc87925f188e.png)
Therefore, the answer is 
miles or 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
