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

m (Solution 3 (Comprehensive but Unnecessary))
(All solutions to this problem are incredibly similar (repetitive). So, I cleaned up this page by keeping a comprehensive solution. I will give credit to all users who ever contributed to the written solutions. Let me know if you are unhappy with this edit)
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<math>\textbf{(E) }\text{Tyrone was given card 7.}</math>
 
<math>\textbf{(E) }\text{Tyrone was given card 7.}</math>
  
==Solution 1==
+
==Solution==
 
 
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>
 
 
 
~smarty101 and smartypantsno_3
 
 
 
==Solution 2==
 
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>.
 
 
 
-SmileKat32
 
 
 
==Solution 3 (Comprehensive but Unnecessary)==
 
 
By observations, we consider the scores from lowest to highest. We make the following logical deduction:
 
By observations, we consider the scores from lowest to highest. We make the following logical deduction:
 
<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.} \
&\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.}
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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>
  
Certainly, 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>
  
~MRENTHUSIASM
+
~smarty101 ~smartypantsno_3 ~SmileKat32 ~MRENTHUSIASM
  
 
== Video Solution by OmegaLearn (Using logical deduction) ==
 
== Video Solution by OmegaLearn (Using logical deduction) ==

Revision as of 21:19, 7 September 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 observations, we consider the scores from lowest to highest. We make the following logical deduction: \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 (ProblemsAnswer KeyResources)
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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