Difference between revisions of "2015 AMC 12B Problems/Problem 4"
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===Note=== | ===Note=== | ||
− | This problem is also problem number 5 on the 2015 AMC 10B. | + | This problem is also problem number 5 on the 2015 AMC 10B, just with different names. |
[https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_5 This is the link.] | [https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_5 This is the link.] | ||
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==Solution 2== | ==Solution 2== | ||
− | We | + | We list the people out vertically for clarity, starting with Marzuq and working from there. |
<cmath>1 - </cmath> | <cmath>1 - </cmath> | ||
− | <cmath>2 Nabeel </cmath> | + | <cmath>2 \text{Nabeel} </cmath> |
− | <cmath>3 Rahul </cmath> | + | <cmath>3 \text{Rahul} </cmath> |
<cmath>4 - </cmath> | <cmath>4 - </cmath> | ||
− | <cmath>5 Rafsan </cmath> | + | <cmath>5 \text{Rafsan} </cmath> |
− | <cmath>6 Arabi </cmath> | + | <cmath>6 \text{Arabi} </cmath> |
<cmath>7 - </cmath> | <cmath>7 - </cmath> | ||
− | <cmath>8 Marzuq </cmath> | + | <cmath>8 \text{Marzuq} </cmath> |
Thus our answer is <math>\boxed{\textbf{(B)}\; \text{Marzuq}}</math>. | Thus our answer is <math>\boxed{\textbf{(B)}\; \text{Marzuq}}</math>. |
Latest revision as of 13:57, 23 November 2023
Contents
Note
This problem is also problem number 5 on the 2015 AMC 10B, just with different names. This is the link.
Problem
Lian, Marzuq, Rafsan, Arabi, Nabeel, and Rahul were in a 12-person race with 6 other people. Nabeel finished 6 places ahead of Marzuq. Arabi finished 1 place behind Rafsan. Lian finished 2 places behind Marzuq. Rafsan finished 2 places behind Rahul. Rahul finished 1 place behind Nabeel. Arabi finished in 6th place. Who finished in 8th place?
Solution 1
Let --- denote any of the 6 racers not named. Then the correct order looks like this:
Thus the 8th place runner is .
Solution 2
We list the people out vertically for clarity, starting with Marzuq and working from there.
Thus our answer is .
See Also
2015 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 3 |
Followed by Problem 5 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.