Difference between revisions of "2015 AMC 8 Problems/Problem 17"
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| + | == Problem == | ||
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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? | 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? | ||
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</math> | </math> | ||
| − | + | ==Solution 1== | |
Somehow we get <math>\frac{d}{v}=\frac{1}{3}</math> and <math>\frac{d}{v+18}=\frac{1}{5}</math>. | 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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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>. | ||
| − | + | ==Solution 2== | |
<math>d = rt</math>, <math>d=\frac{1}{3} \times r = \frac{1}{5} \times (r + 18)</math> | <math>d = rt</math>, <math>d=\frac{1}{3} \times r = \frac{1}{5} \times (r + 18)</math> | ||
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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. | <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. | ||
| − | + | ==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>d=\left (\dfrac{1}{3}\right )(x)=\left (\dfrac{1}{5}\right) (x+18)</math>, giving | 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 | ||
<cmath>(5)(x)=(3)(x+18)</cmath> | <cmath>(5)(x)=(3)(x+18)</cmath> | ||
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Hence, <math>d=\dfrac{27}{3}=\boxed{\textbf{(D)}~9}</math>. | Hence, <math>d=\dfrac{27}{3}=\boxed{\textbf{(D)}~9}</math>. | ||
| − | + | ==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>). | 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>). | ||
Revision as of 02:14, 16 January 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?
Solution 1
Somehow we get 
and 
.
This gives 
, which gives 
, which then gives 
.
Solution 2
, 
so 
, plug into the first one and it's 
miles to school.
Solution 3
We set up an equation in terms of 
the distance and 
the speed In miles per hour. We have 
, giving
![\[(5)(x)=(3)(x+18)\]](http://latex.artofproblemsolving.com/d/1/4/d1462fd905bb7ab31d0490c0c5dd2fd85afb7028.png)
![\[5x=3x+54\]](http://latex.artofproblemsolving.com/0/f/5/0f50aceae271dd099c5020078db14e29f9105ae7.png)
![\[2x=54\]](http://latex.artofproblemsolving.com/8/3/a/83aea5404764b8ba4c4d66c2faa44946a0bebf46.png)
![\[x=27\]](http://latex.artofproblemsolving.com/d/a/a/daac91f4974ad61351248574f30790c620247e94.png)
Hence, 
.
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 
miles per hour. Solving for x gives us 27 miles per hour. Because 
minutes is a third of an hour, the distance would then be 
miles (
).
See Also
| 2015 AMC 8 (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
