Difference between revisions of "2015 AMC 10B Problems/Problem 5"
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==Solution== | ==Solution== | ||
Because Marta was <math>6</math>th, Jack was <math>5</math>th, so Todd was <math>3</math>rd. Thus, Rand was <math>2</math>nd and the 8th place finisher was <math>\boxed{\mathbf{(B)}\ \mathrm{Hikmet}}</math> | Because Marta was <math>6</math>th, Jack was <math>5</math>th, so Todd was <math>3</math>rd. Thus, Rand was <math>2</math>nd and the 8th place finisher was <math>\boxed{\mathbf{(B)}\ \mathrm{Hikmet}}</math> | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/61oSoeT0ULA | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=B|num-b=4|num-a=6}} | {{AMC10 box|year=2015|ab=B|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 14:07, 25 June 2020
Contents
[hide]Problem
David, Hikmet, Jack, Marta, Rand, and Todd were in a 
-person race with 
other people. Rand finished places ahead of Hikmet. Marta finished 
place behind Jack. David finished 
places behind Hikmet. Jack finished places behind Todd. Todd finished place behind Rand. Marta finished in th place. Who finished in 
th place?
Solution
Because Marta was th, Jack was 
th, so Todd was 
rd. Thus, Rand was nd and the 8th place finisher was 
Video Solution
~savannahsolver
See Also
| 2015 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
