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

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Ravon, Oscar, Aditi, Tyrone, and Kim play a card game. Each person is given 2 cards out of a set of 10 cards numbered <math>1,2,3, \dots,10.</math> 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?
 
Ravon, Oscar, Aditi, Tyrone, and Kim play a card game. Each person is given 2 cards out of a set of 10 cards numbered <math>1,2,3, \dots,10.</math> 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?
  
<math> \textbf{(A) \ }\text{Ravon was given card 3.}  \qquad \textbf{(B) \ }\text{Aditi was given card 3.} \qquad \textbf{(C) \ }\text{Ravon was given card 4.} \qquad \textbf{(D) \ }\text{Aditi was given card 4} \qquad </math> <math> \textbf{(E) \ }\text{Tyrone was givencard 7}  </math>
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<math> \textbf{(A) \ }\text{Ravon was given card 3.}  \qquad \textbf{(B) \ }\text{Aditi was given card 3.} \qquad \textbf{(C) \ }\text{Ravon was given card 4.} \qquad \textbf{(D) \ }\text{Aditi was given card 4} \qquad </math> <math> \textbf{(E) \ }\text{Tyrone was given card 7}  </math>
  
 
==Solution==
 
==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{\qquad \textbf{(C) \ }\text{Ravon was given card 4.}}</math>
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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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== Video Solution by OmegaLearn (Using logical deduction) ==
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https://youtu.be/zO0EuKPXuT0
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~ pi_is_3.14
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==Video Solution by TheBeautyofMath==
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https://youtu.be/FV9AnyERgJQ?t=284
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~IceMatrix
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{{AMC10 box|year=2021|ab=B|num-b=16|num-a=18}}

Revision as of 06:48, 19 February 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?

$\textbf{(A) \ }\text{Ravon was given card 3.}  \qquad \textbf{(B) \ }\text{Aditi was given card 3.} \qquad \textbf{(C) \ }\text{Ravon was given card 4.} \qquad \textbf{(D) \ }\text{Aditi was given card 4} \qquad$ $\textbf{(E) \ }\text{Tyrone was given card 7}$

Solution

Oscar must be given 3 and 1, so we rule out $\textbf{(A) \ }$ and $\textbf{(B) \ }$. If Tyrone had card 7, then he would also have card 9, and then Kim must have 10 and 7 so we rule out $\textbf{(E) \ }$. If Aditi was given card 4, then she would have card 3, which Oscar already had. So the answer is $\boxed{ \textbf{(C) \ }\text{Ravon was given card 4.}}$

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

-SmileKat32

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


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