Difference between revisions of "2015 AMC 8 Problems/Problem 17"
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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</math> is obviously constant | ||
| + | |||
| + | <math>\frac{1}{3} \times r = \frac{1}{5} \times (r + 18)</math> | ||
| + | |||
| + | <math>\frac{r}{3} = \frac{r}{5} + \frac{18}{5}</math> | ||
| + | |||
| + | <math>\frac{2r}{15} = \frac{18}{5}</math> | ||
| + | |||
| + | <math>10r = 270</math> so <math>r = 27</math>, plug into the first one and it's <math>\boxed{\textbf{B}~9}</math> miles to school | ||
| + | |||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2015|num-b=16|num-a=18}} | {{AMC8 box|year=2015|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:05, 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?
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
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. 
