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Difference between revisions of "2009 AMC 10A Problems/Problem 20"

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In all, Lauren will have to bike <math>20 - \frac{15}{4} = \frac{80}{4} - \frac{15}{4} = \frac{65}{4}km</math>. Because her speed is <math>\frac{1}{4} \textbf{km/min}</math>, the time elapsed will be <math>\frac{\frac{65}{4}}{\frac{1}{4}} = \boxed{\textbf{D) 65}}</math>
 
In all, Lauren will have to bike <math>20 - \frac{15}{4} = \frac{80}{4} - \frac{15}{4} = \frac{65}{4}km</math>. Because her speed is <math>\frac{1}{4} \textbf{km/min}</math>, the time elapsed will be <math>\frac{\frac{65}{4}}{\frac{1}{4}} = \boxed{\textbf{D) 65}}</math>
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==Solution 3==
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Since the distance between them decreases at a rate of <math>1</math> kilometer per minute when they are both biking, their combined speed is <math>1</math> kilometer per minute. Andrea travels three times as fast as Lauren, so they travel at speeds of <math>\frac{3}{4}</math> kilometers per minute and <math>\frac{1}{4}</math> kilometers per minute, respectively.
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After <math>5</math> minutes, the distance between them will have decreased by <math>5</math> kilometers, so they will be <math>20-5 = 15</math> kilometers apart when Andrea stops. Then, Lauren will take <math>\frac{15}{\frac{1}{4}} = 15*4=60</math> more minutes to reach Andrea.
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They started to bike <math>5</math> minutes before Andrea stopped, so the total time Lauren passed from the time they started biking to the time Lauren reached Andrea is <math>60+5=65</math> minutes. Hence, the answer is <math>\boxed{\textbf{(B) } 65}</math>. ~azc1027
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 14:40, 4 June 2023

Problem

Andrea and Lauren are $20$ kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of $1$ kilometer per minute. After $5$ minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from the time they started to bike does Lauren reach Andrea?

$\mathrm{(A)}\ 20 \qquad \mathrm{(B)}\ 30 \qquad \mathrm{(C)}\ 55 \qquad \mathrm{(D)}\ 65 \qquad \mathrm{(E)}\ 80$

Solution 1

Let their speeds in kilometers per hour be $v_A$ and $v_L$. We know that $v_A=3v_L$ and that $v_A+v_L=60$. (The second equation follows from the fact that $1\mathrm km/min = 60\mathrm km/h$.) This solves to $v_A=45$ and $v_L=15$.

As the distance decreases at a rate of $1$ kilometer per minute, after $5$ minutes the distance between them will be $20-5=15$ kilometers.

From this point on, only Lauren will be riding her bike. As there are $15$ kilometers remaining and $v_L=15$, she will need exactly an hour to get to Andrea. Therefore the total time in minutes is $5+60 = \boxed{65}$.

Solution 2

Because the speed of Andrea is 3 times as fast as Lauren and the distance between them is decreasing at a rate of 1 kilometer per minute, Andrea's speed is $\frac{3}{4} \textbf{km/min}$, and Lauren's $\frac{1}{4} \textbf{km/min}$. Therefore, after 5 minutes, Andrea will have biked $\frac{3}{4} \cdot 5 = \frac{15}{4}km$.

In all, Lauren will have to bike $20 - \frac{15}{4} = \frac{80}{4} - \frac{15}{4} = \frac{65}{4}km$. Because her speed is $\frac{1}{4} \textbf{km/min}$, the time elapsed will be $\frac{\frac{65}{4}}{\frac{1}{4}} = \boxed{\textbf{D) 65}}$

Solution 3

Since the distance between them decreases at a rate of $1$ kilometer per minute when they are both biking, their combined speed is $1$ kilometer per minute. Andrea travels three times as fast as Lauren, so they travel at speeds of $\frac{3}{4}$ kilometers per minute and $\frac{1}{4}$ kilometers per minute, respectively.

After $5$ minutes, the distance between them will have decreased by $5$ kilometers, so they will be $20-5 = 15$ kilometers apart when Andrea stops. Then, Lauren will take $\frac{15}{\frac{1}{4}} = 15*4=60$ more minutes to reach Andrea.

They started to bike $5$ minutes before Andrea stopped, so the total time Lauren passed from the time they started biking to the time Lauren reached Andrea is $60+5=65$ minutes. Hence, the answer is $\boxed{\textbf{(B) } 65}$. ~azc1027

Video Solution

https://youtu.be/YJigRBg0LIQ

~savannahsolver

See Also

2009 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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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