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2021 AMC 10B Problems/Problem 17: Difference between revisions

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==Problem==
==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 <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 <math>2</math> cards out of a set of <math>10</math> 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--<math>11,</math> Oscar--<math>4,</math> Aditi--<math>7,</math> Tyrone--<math>16,</math> Kim--<math>17.</math> 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>
<math>\textbf{(A) }\text{Ravon was given card 3.}</math>


==Solution==
<math>\textbf{(B) }\text{Aditi was given card 3.}</math>


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>
<math>\textbf{(C) }\text{Ravon was given card 4.}</math>


~smarty101 and smartypantsno_3
<math>\textbf{(D) }\text{Aditi was given card 4.}</math>
 
<math>\textbf{(E) }\text{Tyrone was given card 7.}</math>
 
==Solution 1 (Logical Deduction)==
By logical deduction, we consider the scores from lowest to highest:
<cmath>\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*}</cmath>
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>(\bigstar)</math> and pick <math>\textbf{(C)}.</math>
 
~smarty101 ~smartypantsno_3 ~SmileKat32 ~MRENTHUSIASM
 
==Solution 2 (Elimination)==
 
 
Rather than figuring out each person's cards individually, we can use elimination. Let's eliminate answer choices <math>\boxed{\textbf{(B)}}</math> and <math>\boxed{\textbf{(D)}}</math> since if one is true, then the other also must (Aditi's cards add up to <math>7</math>). However, since there is only one solution, it holds true that neither of these can be true. We also know that Oscar's cards added up to 4, and since he can't have two of the same, he must have a <math>1</math> and a <math>3</math>, eliminating <math>\boxed{\textbf{(A)}}</math>. Furthermore, if Tyrone got a <math>9</math>, his other card must be a <math>7</math>, which makes it impossible for Kim to have 17 points with the remaining cards. Thus, the only option remaining is <math>\boxed{\textbf{(C) }\text{Ravon was given card 4.}}</math>
 
~sigmacuber632
 
== 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==
{{AMC10 box|year=2021|ab=B|num-b=16|num-a=18}}
{{MAA Notice}}

Latest revision as of 18:19, 25 October 2025

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 (Logical Deduction)

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

Solution 2 (Elimination)

Rather than figuring out each person's cards individually, we can use elimination. Let's eliminate answer choices $\boxed{\textbf{(B)}}$ and $\boxed{\textbf{(D)}}$ since if one is true, then the other also must (Aditi's cards add up to $7$). However, since there is only one solution, it holds true that neither of these can be true. We also know that Oscar's cards added up to 4, and since he can't have two of the same, he must have a $1$ and a $3$, eliminating $\boxed{\textbf{(A)}}$. Furthermore, if Tyrone got a $9$, his other card must be a $7$, which makes it impossible for Kim to have 17 points with the remaining cards. Thus, the only option remaining is $\boxed{\textbf{(C) }\text{Ravon was given card 4.}}$

~sigmacuber632

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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