Difference between revisions of "2008 AMC 10A Problems/Problem 15"
Stevenmeow (talk | contribs) (→Solution) |
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Substituting the 3rd and 4th equations into this equation gives: | Substituting the 3rd and 4th equations into this equation gives: | ||
− | <math>(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</math> | + | <math>(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</math> |
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+ | <math>\Longrightarrow 2(I_s+5I_t)+20=x</math> | ||
Substituting (I) into this equation gives: | Substituting (I) into this equation gives: |
Revision as of 17:40, 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?
Solution
We let Ian's speed and time equal and , respectively.
We will represent Han's and Jan's speed and time similarly:
, , ,
The problem gives us 5 equations:
Substituting the 1st and 2nd equations into the last gives:
(I)
We are asked the difference between Jan's and Ian's distances, or
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:
Substituting (I) into this equation gives:
Therefore, the answer is 150 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 |