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

m (Added in the "See Also" and MAA Notice.)
(Added in Sol 3.)
Line 3: Line 3:
 
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 given card 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>
 +
 
 +
<math>\textbf{(D) }\text{Aditi was given card 4.}</math>
 +
 
 +
<math>\textbf{(E) }\text{Tyrone was given card 7.}</math>
 +
 
 +
==Solution 1==
 +
 
 +
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
 
~smarty101 and smartypantsno_3
  
 
==Solution 2==
 
==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>.
+
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
 
-SmileKat32
 +
 +
==Solution 3 (Comprehensive)==
 +
Using observations, we consider the scores from lowest to highest. We make the following logical deduction:
 +
 +
<cmath>\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.} \\
 +
&\Longrightarrow \text{Tyrone was given cards 6 and 10.} \\
 +
&\Longrightarrow \text{Kim was given cards 8 and 9.}
 +
\end{align*}</cmath>
 +
 +
Therefore, the answer is <math>\boxed{\textbf{(C) }\text{Ravon was given card 4.}}</math>
 +
 +
Of course, if we look at the answer choices earlier, then we can stop after line 3 of the block of logical statements.
 +
 +
~MRENTHUSIASM
  
 
== Video Solution by OmegaLearn (Using logical deduction) ==
 
== Video Solution by OmegaLearn (Using logical deduction) ==

Revision as of 16:08, 5 March 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 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)

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.} \\ &\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 after line 3 of the block of logical statements.

~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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png