Difference between revisions of "2015 AMC 8 Problems/Problem 17"
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<cmath>x=27</cmath> | <cmath>x=27</cmath> | ||
| − | So <math>d=\dfrac{27}{3}=\boxed{\textbf{(D)} | + | So <math>d=\dfrac{27}{3}=\boxed{\textbf{(D)}~9}</math> |
==See Also== | ==See Also== | ||
Revision as of 11:41, 30 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?
Contents
Solution 1
So 
and 
.
This gives 
, which gives 
, which then gives 
Solution 2
, 
is obviously constant
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 
![\[d=(5)(x)=(3)(x+18)\]](http://latex.artofproblemsolving.com/7/e/c/7ecd2768b6ac29529ea25c082bceaa740c25df26.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)
So 
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 | ||
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