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Difference between revisions of "2021 AMC 10B Problems/Problem 17"

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(Undo revision 164714 by Puffer13 (talk) We wrote this solution COLLECTIVELY. So, credits should be given to all contributors.)
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<math>\textbf{(E) }\text{Tyrone was given card 7.}</math>
 
<math>\textbf{(E) }\text{Tyrone was given card 7.}</math>
  
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==Solution 1==
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==Solution==
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By logical deduction, we consider the scores from lowest to highest:
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Oscar must be given 3 and 1, so we rule out <math>\textbf{(A) \ }</math> and <math>\textbf{(B) \ }</math>. If Tyrone had card 7, then he would also have card 9, and then Kim must have 10 and 7 so we rule out <math>\textbf{(E) \ }</math>. If Aditi was given card 4, then she would have card 3, which Oscar already had. So the answer is <math>\boxed{ \textbf{(C) }\text{Ravon was given card 4.}}</math>
 
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~smarty101 and smartypantsno_3
 
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==Solution 2==
 
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Oscar must be given 3 and 1. Aditi cannot be given 3 or 1, so she must have 2 and 5. Similarly, Ravon cannot be given 1, 2, 3, or 5, so he must have 4 and 7, and the answer is <math>\boxed{ \textbf{(C) }\text{Ravon was given card 4.}}</math>.
 
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-SmileKat32
 
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==Solution 3 (Comprehensive but Unnecessary)==
 
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By observations, we consider the scores from lowest to highest. We make the following logical deduction:
 
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<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
\text{Oscar's score is 4.} &\implies \text{Oscar is given cards 1 and 3.} \\
 
\text{Oscar's score is 4.} &\implies \text{Oscar is given cards 1 and 3.} \\
 
&\implies \text{Aditi is given cards 2 and 5.} \\
 
&\implies \text{Aditi is given cards 2 and 5.} \\
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&\implies \text{Ravon is given cards 4 and 7.} \hspace{13mm} (*) \\
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&\implies \text{Ravon is given cards 4 and 7.} && (\bigstar) \\
 
&\implies \text{Tyrone is given cards 6 and 10.} \\
 
&\implies \text{Tyrone is given cards 6 and 10.} \\
 
&\implies \text{Kim is given cards 8 and 9.}
 
&\implies \text{Kim is given cards 8 and 9.}
 
\end{align*}</cmath>
 
\end{align*}</cmath>
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Therefore, the answer is <math>\boxed{\textbf{(C) }\text{Ravon was given card 4.}}</math>
 
Therefore, the answer is <math>\boxed{\textbf{(C) }\text{Ravon was given card 4.}}</math>
  
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Of course, if we read the answer choices sooner, then we can stop at <math>(*)</math> and pick <math>\textbf{(C)}.</math>
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Certainly, if we read the answer choices sooner, then we can stop at <math>(\bigstar)</math> and pick <math>\textbf{(C)}.</math>
  
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~MRENTHUSIASM
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~smarty101 ~smartypantsno_3 ~SmileKat32 ~MRENTHUSIASM
  
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== Video Solution by OmegaLearn (Using logical deduction) ==
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== Video Solution by OmegaLearn (Using Logical Deduction) ==
 
https://youtu.be/zO0EuKPXuT0
 
https://youtu.be/zO0EuKPXuT0
  
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~IceMatrix
 
~IceMatrix
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==Video Solution by Interstigation==
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https://youtu.be/8BPKs24eyes
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~Interstigation
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021|ab=B|num-b=16|num-a=18}}
 
{{AMC10 box|year=2021|ab=B|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 15:18, 6 November 2021

Problem

Ravon, Oscar, Aditi, Tyrone, and Kim play a card game. Each person is given $2$ cards out of a set of $10$ cards numbered $1,2,3, \dots,10.$ The score of a player is the sum of the numbers of their cards. The scores of the players are as follows: Ravon--$11,$ Oscar--$4,$ Aditi--$7,$ Tyrone--$16,$ Kim--$17.$ Which of the following statements is true?

$\textbf{(A) }\text{Ravon was given card 3.}$

$\textbf{(B) }\text{Aditi was given card 3.}$

$\textbf{(C) }\text{Ravon was given card 4.}$

$\textbf{(D) }\text{Aditi was given card 4.}$

$\textbf{(E) }\text{Tyrone was given card 7.}$

Solution

By logical deduction, we consider the scores from lowest to highest: \begin{align*} \text{Oscar's score is 4.} &\implies \text{Oscar is given cards 1 and 3.} \\ &\implies \text{Aditi is given cards 2 and 5.} \\ &\implies \text{Ravon is given cards 4 and 7.} && (\bigstar) \\ &\implies \text{Tyrone is given cards 6 and 10.} \\ &\implies \text{Kim is given cards 8 and 9.} \end{align*} Therefore, the answer is $\boxed{\textbf{(C) }\text{Ravon was given card 4.}}$

Certainly, if we read the answer choices sooner, then we can stop at $(\bigstar)$ and pick $\textbf{(C)}.$

~smarty101 ~smartypantsno_3 ~SmileKat32 ~MRENTHUSIASM

Video Solution by OmegaLearn (Using Logical Deduction)

https://youtu.be/zO0EuKPXuT0

~ pi_is_3.14

Video Solution by TheBeautyofMath

https://youtu.be/FV9AnyERgJQ?t=284

~IceMatrix

Video Solution by Interstigation

https://youtu.be/8BPKs24eyes

~Interstigation

See Also

2021 AMC 10B (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 AMC 10 Problems and Solutions

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