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

(Solution 3)
(Solution 3)
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===Solution 3===
 
===Solution 3===
We set up an equation in terms of <math>d</math> the distance and <math>x</math> the speed In miles per hour. We have <math>\left (\dfrac{1}{3}\right )(x)=\left (\dfrac{1}{5}\right) (x+18)</math>
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We set up an equation in terms of <math>d</math> the distance and <math>x</math> the speed In miles per hour. We have <math>d=\left (\dfrac{1}{3}\right )(x)=\left (\dfrac{1}{5}\right) (x+18)</math>
 
<cmath>d=(5)(x)=(3)(x+18)</cmath>
 
<cmath>d=(5)(x)=(3)(x+18)</cmath>
 
<cmath>5x=3x+54</cmath>
 
<cmath>5x=3x+54</cmath>

Revision as of 18:08, 25 November 2015

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?

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

Solution 1

So $\frac{d}{v}=\frac{1}{3}$ 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

$d = rt$, $d$ is obviously constant

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

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

$\frac{2r}{15} = \frac{18}{5}$

$10r = 270$ 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 $d=\left (\dfrac{1}{3}\right )(x)=\left (\dfrac{1}{5}\right) (x+18)$ \[d=(5)(x)=(3)(x+18)\] \[5x=3x+54\] \[2x=54\] \[x=27\]

So $d=\dfrac{27}{3}=\boxed{\textbf{(D)},~9}$

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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