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

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(Solution 5)
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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.  
 
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.  
 
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.  
 
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 would then be <math>9</math> miles (<math>(D)9</math>). ----LarryFlora
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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 14:20, 7 July 2021

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?

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

Solutions

Solution 1

Somehow 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 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 $2x/3 = 18$ 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 ($(D)9$).

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. Comparing these two situations, it gives us $5d-3d = 18$. So the distance to the school would be $9$ miles ($(D)9$). ----LarryFlora

See Also

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

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