2021 AMC 10B Problems/Problem 17

Revision as of 20:26, 1 April 2021 by MRENTHUSIASM (talk | contribs) (Solution 3 (Comprehensive, but Unnecessary))

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 1

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

Solution 3 (Comprehensive, but Unnecessary)

Using observations, we consider the scores from lowest to highest. We make the following logical deduction:

\begin{align*} \text{Oscar's score is 4.} &\Longrightarrow \text{Oscar was given cards 1 and 3.} \\ &\Longrightarrow \text{Aditi was given cards 2 and 5.} \\ &\Longrightarrow \text{Ravon was given cards 4 and 7.} \hspace{13mm} (*) \\ &\Longrightarrow \text{Tyrone was given cards 6 and 10.} \\ &\Longrightarrow \text{Kim was given cards 8 and 9.} \end{align*}

Therefore, the answer is $\boxed{\textbf{(C) }\text{Ravon was given card 4.}}$

Of course, if we look at the answer choices earlier, then we can stop at $(*).$

~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

See Also

2021 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AMC 10 Problems and Solutions

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