Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(wrote solution)
(→‎Solution)
Line 25: Line 25:
 
Substituting the 1st and 2nd equations into the last gives:
 
Substituting the 1st and 2nd equations into the last gives:
  
−
<math>(I_s+5)(I_t+1)=I_s I_t+70</math>
+
<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 I_t+I_s+5I_t+5=I_s I_t+70</math>
 
−
 
 
−
<math>I_s+5I_t=65</math> (I)
 
  
 
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
Line 39: Line 35:
 
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</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 \Longrightarrow 2(I_s+5I_t)+20=x</math>
−
 
 
−
<math>I_s I_t+2I_s+10I_t+20-I_s I_t=x</math>
 
−
 
 
−
<math>2I_s+10I_t+20=x</math>
 
−
 
 
−
<math>2(I_s+5I_t)+20=x</math>
 
  
 
Substituting (I) into this equation gives:
 
Substituting (I) into this equation gives:
  
−
<math>2(65)+20=x</math>
+
<math>2(65)+20=x \Longrightarrow 130+20=x \Longrightarrow 150=x</math>
−
 
 
−
<math>130+20=x</math>
 
−
 
 
−
<math>150=x</math>
 
  
 
Therefore, the answer is 150 miles or <math>\boxed{D}</math>
 
Therefore, the answer is 150 miles or <math>\boxed{D}</math>
−
 
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2008|ab=A|num-b=14|num-a=16}}
 
{{AMC10 box|year=2008|ab=A|num-b=14|num-a=16}}

Revision as of 18:39, 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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10A_Problems/Problem_15&oldid=26776"