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

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==Solution==
 
==Solution==
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We let Ian's speed and time equal <math>I_s</math> and <math>I_t</math>, respectively.
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We let Ian's speed and time equal <math>I_s</math> and <math>I_t</math>, respectively. Similarly, let Han's and Jan's speed and time be <math>H_s</math>, <math>H_t</math>, <math>J_s</math>, <math>J_t</math>. The problem gives us 5 [[equation]]s:
  
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We will represent Han's and Jan's speed and time similarly:
+
<cmath>\begin{align}
 +
H_s&=I_s+5 \\
 +
H_t&=I_t+1 \\
 +
J_t&=I_s+10 \\
 +
J_t&=I_t+2 \\
 +
H_s \cdot H_t & =I_s \cdot I_t+70 \end{align*}</cmath>
  
−
<math>H_s</math>, <math>H_t</math>, <math>J_s</math>, <math>J_t</math>
+
Substituting <math>(1)</math> and <math>(2)</math> equations into <math>(5)</math> gives:
  
−
The problem gives us 5 equations:
+
<cmath>(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 (*)</cmath>
−
 
 
−
<math>H_s=I_s+5</math>
 
−
 
 
−
<math>H_t=I_t+1</math>
 
−
 
 
−
<math>J_t=I_s+10</math>
 
−
 
 
−
<math>J_t=I_t+2</math>
 
−
 
 
−
<math>H_s H_t=I_s I_t+70</math>
 
−
 
 
−
Substituting the 1st and 2nd equations into the last gives:
 
−
 
 
−
<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> <math>(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>J_s J_t-I_s I_t=x</math>
+
<cmath>J_s J_t-I_s I_t=x,</cmath>
−
 
 
−
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:
+
Where <math>x</math> is the difference between Jan's and Ian's distances and the answer to the problem. Substituting <math>(3)</math> and <math>(4)</math> 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</math>
+
<cmath>(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</cmath>
  
−
<math>2I_s+10I_t+20=x \Longrightarrow 2(I_s+5I_t)+20=x</math>
+
<cmath>2I_s+10I_t+20=x \Longrightarrow 2(I_s+5I_t)+20=x</cmath>
  
−
Substituting <math>(I)</math> into this equation gives:
+
Substituting <math>(*)</math> into this equation gives:
  
−
<math>2(65)+20=x \Longrightarrow 130+20=x \Longrightarrow 150=x</math>
+
<cmath>2(65)+20=x \Longrightarrow 130+20=x \Longrightarrow 150=x</cmath>
  
−
Therefore, the answer is 150 miles or <math>\boxed{D}</math>
+
Therefore, the answer is <math>150</math> miles or <math>\boxed{\mathrm{(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}}
 +
 +
[[Category:Introductory Algebra Problems]]

Revision as of 17:36, 26 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. Similarly, let Han's and Jan's speed and time be $H_s$, $H_t$, $J_s$, $J_t$. The problem gives us 5 equations: 

\begin{align}
H_s&=I_s+5 \\
H_t&=I_t+1 \\
J_t&=I_s+10 \\
J_t&=I_t+2 \\
H_s \cdot H_t & =I_s \cdot I_t+70 \end{align*} (Error compiling LaTeX. Unknown error_msg)

Substituting $(1)$ and $(2)$ equations into $(5)$ gives:

\[(I_s+5)(I_t+1)=I_s I_t+70 \Longrightarrow I_s I_t+I_s+5I_t+5=I_s I_t+70 \Longrightarrow I_s+5I_t=65 \quad (*)\]

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 $(3)$ and $(4)$ 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 $(*)$ into this equation gives:

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

Therefore, the answer is $150$ miles or $\boxed{\mathrm{(D)}}$.

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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