Difference between revisions of "2021 AMC 10B Problems/Problem 17"
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\text{Oscar's score is 4.} &\Longrightarrow \text{Oscar was given cards 1 and 3.} \\ | \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{Aditi was given cards 2 and 5.} \\ | ||
− | &\Longrightarrow \text{Ravon was given cards 4 and 7.} \hspace{ | + | &\Longrightarrow \text{Ravon was given cards 4 and 7.} \hspace{13mm} (*) \\ |
&\Longrightarrow \text{Tyrone was given cards 6 and 10.} \\ | &\Longrightarrow \text{Tyrone was given cards 6 and 10.} \\ | ||
&\Longrightarrow \text{Kim was given cards 8 and 9.} | &\Longrightarrow \text{Kim was given cards 8 and 9.} |
Revision as of 21:26, 1 April 2021
Contents
Problem
Ravon, Oscar, Aditi, Tyrone, and Kim play a card game. Each person is given cards out of a set of
cards numbered
The score of a player is the sum of the numbers of their cards. The scores of the players are as follows: Ravon--
Oscar--
Aditi--
Tyrone--
Kim--
Which of the following statements is true?
Solution 1
Oscar must be given 3 and 1, so we rule out and
. If Tyrone had card 7, then he would also have card 9, and then Kim must have 10 and 7 so we rule out
. If Aditi was given card 4, then she would have card 3, which Oscar already had. So the answer is
~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 .
-SmileKat32
Solution 3 (Comprehensive, but Unnecessary)
Using observations, we consider the scores from lowest to highest. We make the following logical deduction:
Therefore, the answer is
Of course, if we look at the answer choices earlier, then we can stop at
~MRENTHUSIASM
Video Solution by OmegaLearn (Using logical deduction)
~ pi_is_3.14
Video Solution by TheBeautyofMath
https://youtu.be/FV9AnyERgJQ?t=284
~IceMatrix
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.