https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Dreambca&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T19:14:57ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_17&diff=1513222015 AMC 8 Problems/Problem 172021-04-11T02:10:31Z<p>Dreambca: /* Solution 2 */</p>
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<div>== Problem ==<br />
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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?<br />
<br />
<math><br />
\textbf{(A) } 4 \qquad<br />
\textbf{(B) } 6 \qquad<br />
\textbf{(C) } 8 \qquad<br />
\textbf{(D) } 9 \qquad<br />
\textbf{(E) } 12<br />
</math><br />
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==Solutions==<br />
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===Solution 1===<br />
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Somehow we get <math>\frac{d}{v}=\frac{1}{3}</math> and <math>\frac{d}{v+18}=\frac{1}{5}</math>.<br />
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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>.<br />
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===Solution 2===<br />
<math>d = rt</math>, <math>d=\frac{1}{3} \times r = \frac{1}{5} \times (r + 18)</math><br />
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<math>\frac{r}{3} = \frac{r}{5} + \frac{18}{5}</math><br />
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<math>10r = 270</math> so <math>r = 27</math>, plug into the first one and it's <math>\boxed{\textbf{(D)}~9}</math> miles to school.<br />
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===Solution 3===<br />
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>, giving<br />
<cmath>(5)(x)=(3)(x+18)</cmath><br />
<cmath>5x=3x+54</cmath><br />
<cmath>2x=54</cmath><br />
<cmath>x=27</cmath><br />
Hence, <math>d=\dfrac{27}{3}=\boxed{\textbf{(D)}~9}</math>.<br />
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===Solution 4===<br />
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>).<br />
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==See Also==<br />
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{{AMC8 box|year=2015|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Dreambca