Difference between revisions of "2008 AMC 10A Problems/Problem 15"
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==Problem== | ==Problem== | ||
| − | Yesterday | + | Yesterday Darla drove 1 hour longer than Carla at an average speed 5 miles per hour faster than Carla. Marla drove 2 hours longer than Carla at an average speed 10 miles per hour faster than Carla. Darla drove 70 miles more than Carla. How many more miles did Marla drive than Carla? |
<math>\mathrm{(A)}\ 120\qquad\mathrm{(B)}\ 130\qquad\mathrm{(C)}\ 140\qquad\mathrm{(D)}\ 150\qquad\mathrm{(E)}\ 160</math> | <math>\mathrm{(A)}\ 120\qquad\mathrm{(B)}\ 130\qquad\mathrm{(C)}\ 140\qquad\mathrm{(D)}\ 150\qquad\mathrm{(E)}\ 160</math> | ||
Revision as of 22:52, 12 November 2019
Contents
Problem
Yesterday Darla drove 1 hour longer than Carla at an average speed 5 miles per hour faster than Carla. Marla drove 2 hours longer than Carla at an average speed 10 miles per hour faster than Carla. Darla drove 70 miles more than Carla. How many more miles did Marla drive than Carla?
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 
.
Solution 2
Let Ian drive 
miles, at a speed of 
, for some time(in hours) 
. Hence, we have 
. We can find a similar equation for Han, who drove 
miles, at a rate of 
, for 
hours, giving us 
. We can do the same for Jan, giving us 
, where is how much further Jan traveled than Ian. We now have three equations:
![\[D= RT\]](http://latex.artofproblemsolving.com/1/d/5/1d5cbfc18b3322e0788409ed138a8a1ac4b99155.png)
![\[D + 70 = (R+5)(T+1) = RT + R + 5T + 5\]](http://latex.artofproblemsolving.com/3/1/0/31027237132b64757f91712ff52e33fc2cbd3018.png)
![\[D + x = (R + 10)(T + 2) = RT + 10 T + 2R + 20.\]](http://latex.artofproblemsolving.com/a/9/d/a9d1f5b641067dd0d155883fa393b8c412522701.png)
Substituting 
for in the second and third equations and cancelling gives us:
![\[70 = 5T + R + 5 \Longrightarrow 5T + R = 65\]](http://latex.artofproblemsolving.com/c/6/9/c690bd68c5e5f0156d4ebb0a47ac6e80f960b39c.png)
![\[x = 10T + 2R + 20 \Longrightarrow x = 2(5T + R ) + 20 \Longrightarrow x= 2(65) + 20 = 150.\]](http://latex.artofproblemsolving.com/5/e/2/5e24637771f1bb2acebbb8fa1d079798bda10b0e.png)
Since 
, our answer is 
.
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
