Difference between revisions of "2015 AMC 8 Problems/Problem 17"

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\textbf{(E) } 12
 
\textbf{(E) } 12
 
</math>
 
</math>
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==Video Solution==
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https://www.youtube.com/watch?v=TFm1jNgB4QM
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https://youtu.be/4wZ4ToIyrnw
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~savannahsolver
  
 
==Solutions==
 
==Solutions==
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===Solution 1===
 
===Solution 1===
  
Somehow we get <math>\frac{d}{v}=\frac{1}{3}</math> and <math>\frac{d}{v+18}=\frac{1}{5}</math>.
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Somehow(yes no maybe so?) we get <math>\frac{d}{v}=\frac{1}{3}</math> and <math>\frac{d}{v+18}=\frac{1}{5}</math>.
  
 
This gives <math>d=\frac{1}{5}v+3.6=\frac{1}{3}v</math>, which gives <math>v=27</math>, which then gives <math>d=\boxed{\textbf{(D)}~9}</math>.
 
This gives <math>d=\frac{1}{5}v+3.6=\frac{1}{3}v</math>, which gives <math>v=27</math>, which then gives <math>d=\boxed{\textbf{(D)}~9}</math>.
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===Solution 4===
 
===Solution 4===
Since it takes 3/5 of the original time for him to get to school when there is no traffic, the speed must be 5/3 of the speed in traffic or 2/3 more. Letting x be the distance he can drive with traffic in 1 hour, we have <math>2x/3 = 18</math> miles per hour. Solving for x gives us 27 miles per hour. Because <math>20</math> minutes is a third of an hour, the distance would then be <math>9</math> miles (<math>(D)9</math>).
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Since it takes <math>\frac{3}{5}</math> of the original time for him to get to school when there is no traffic, the speed must be <math>\frac{5}{3}</math> of the speed in no traffic or <math>\frac{2}{3}</math> more. Letting <math>x</math> be the rate and we know that <math>\frac{5}{3}x = x + 18</math>, so we have <math>\frac{2x}{3} = 18</math> miles per hour. Solving for <math>x</math> gives us <math>27</math> miles per hour. Because <math>20</math> minutes is a third of an hour, the distance would then be <math>9</math> miles <math>\boxed{\textbf{(D)}~9}</math>.
  
 
===Solution 5===
 
===Solution 5===
When driving in rush hour traffic, he drives 20 minutes for one distance (<math>1d</math>) to the school. It means he drives 60 minutes for 3 distances (<math>3d</math>) to the school.  
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When driving in rush hour traffic, he drives <math>20</math> minutes for one distance (<math>1d</math>) to the school. It means he drives <math>60</math> minutes for <math>3</math> distances (<math>3d</math>) to the school. When driving in no traffic hours, he drives <math>12</math> minutes for one distance (<math>1d</math>) to the school. It means he drives <math>60</math> minutes for <math>5</math> distances (<math>5d</math>) to the school. Subtracting these two situations, it gives us <math>5d-3d = 18 = 2d</math>, then <math>d=\frac{18}{2}=9</math>. So the distance to the school would be <math>\boxed{\textbf{(D)}~9}</math> miles. ----LarryFlora
When driving in no traffic hours, he drives 12 minutes for one distance (<math>1d</math>) to the school. It means he drives 60 minutes for 5 distances (<math>5d</math>) to the school.  
 
Comparing these two situations, it gives us <math>5d-3d = 18</math>. So the distance to the school would be <math>9</math> miles (<math>(D)9</math>). ----LarryFlora
 
  
 
==See Also==
 
==See Also==

Revision as of 20:12, 18 July 2022

Problem

Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?

$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$

Video Solution

https://www.youtube.com/watch?v=TFm1jNgB4QM

https://youtu.be/4wZ4ToIyrnw

~savannahsolver

Solutions

Solution 1

Somehow(yes no maybe so?) we get $\frac{d}{v}=\frac{1}{3}$ and $\frac{d}{v+18}=\frac{1}{5}$.

This gives $d=\frac{1}{5}v+3.6=\frac{1}{3}v$, which gives $v=27$, which then gives $d=\boxed{\textbf{(D)}~9}$.

Solution 2

$d = rt$, $d=\frac{1}{3} \times r = \frac{1}{5} \times (r + 18)$

$\frac{r}{3} = \frac{r}{5} + \frac{18}{5}$


$10r = 270$ so $r = 27$, plug into the first one and it's $\boxed{\textbf{(D)}~9}$ miles to school.

Solution 3

We set up an equation in terms of $d$ the distance and $x$ the speed In miles per hour. We have $d=\left (\dfrac{1}{3}\right )(x)=\left (\dfrac{1}{5}\right) (x+18)$, giving \[(5)(x)=(3)(x+18)\] \[5x=3x+54\] \[2x=54\] \[x=27\] Hence, $d=\dfrac{27}{3}=\boxed{\textbf{(D)}~9}$.

Solution 4

Since it takes $\frac{3}{5}$ of the original time for him to get to school when there is no traffic, the speed must be $\frac{5}{3}$ of the speed in no traffic or $\frac{2}{3}$ more. Letting $x$ be the rate and we know that $\frac{5}{3}x = x + 18$, so we have $\frac{2x}{3} = 18$ miles per hour. Solving for $x$ gives us $27$ miles per hour. Because $20$ minutes is a third of an hour, the distance would then be $9$ miles $\boxed{\textbf{(D)}~9}$.

Solution 5

When driving in rush hour traffic, he drives $20$ minutes for one distance ($1d$) to the school. It means he drives $60$ minutes for $3$ distances ($3d$) to the school. When driving in no traffic hours, he drives $12$ minutes for one distance ($1d$) to the school. It means he drives $60$ minutes for $5$ distances ($5d$) to the school. Subtracting these two situations, it gives us $5d-3d = 18 = 2d$, then $d=\frac{18}{2}=9$. So the distance to the school would be $\boxed{\textbf{(D)}~9}$ miles. ----LarryFlora

See Also

2015 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AJHSME/AMC 8 Problems and Solutions

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