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Difference between revisions of "2015 AMC 12B Problems/Problem 4"

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Let <math>-</math> denote any of the 6 racers not named. Then the correct order following all the logic looks like:
 
Let <math>-</math> denote any of the 6 racers not named. Then the correct order following all the logic looks like:
  
−
<cmath>-, \text{Rand}, \text{Todd}, -, \text{Jack}, \text{Marta}, -, \text{Hikmet}, -, \text{David}, -, -</cmath>
+
<cmath>-, \text{Nabil}, \text{Rahul}, -, \text{Rafsan}, \text{Arabi}, -, \text{Marzuq}, -, \text{Lian}, -, -</cmath>
  
−
Clearly the 8th place runner is <math>\fbox{\textbf{(B)}\; \text{Hikmet}}</math>.
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Clearly the 8th place runner is <math>\fbox{\textbf{(B)}\; \text{Marzuq}}</math>.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 03:58, 15 February 2021

Problem

Lian, Marzuq, Rafsan, Arabi, Nabil, and Rahul were in a 12-person race with 6 other people. Nabil finished 6 places ahead of Marzuq. Arabi finished 1 place behind Rafsab. Lian finished 2 places behind Marzuq. Rafsan finished 2 places behind Rahul. Rahul finished 1 place behind Nabil. Arabi finished in 6th place. Who finished in 8th place?

$\textbf{(A)}\; \text{Lian} \qquad\textbf{(B)}\; \text{Marzuq} \qquad\textbf{(C)}\; \text{Rafsan} \qquad\textbf{(D)}\; \text{Nabil} \qquad\textbf{(E)}\; \text{Rahul}$

Solution 1

Let $-$ denote any of the 6 racers not named. Then the correct order following all the logic looks like:

\[-, \text{Nabil}, \text{Rahul}, -, \text{Rafsan}, \text{Arabi}, -, \text{Marzuq}, -, \text{Lian}, -, -\]

Clearly the 8th place runner is $\fbox{\textbf{(B)}\; \text{Marzuq}}$.

Solution 2

We can list these out vertically to ensure clarity, starting with Marta and working from there.


\[1 -\] \[2 R\] \[3 T\] \[4 -\] \[5 J\] \[6 M\] \[7 -\] \[8 H\]

Thus our answer is $\fbox{\textbf{(B)}\; \text{Hikmet}}$.

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

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