Difference between revisions of "2021 AMC 10B Problems/Problem 17"
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<math>\textbf{(E) }\text{Tyrone was given card 7.}</math> | <math>\textbf{(E) }\text{Tyrone was given card 7.}</math> | ||
− | ==Solution | + | ==Solution== |
− | + | By logical deduction, we consider the scores from lowest to highest: | |
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\text{Oscar's score is 4.} &\implies \text{Oscar is given cards 1 and 3.} \ | \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{Aditi is given cards 2 and 5.} \ | ||
− | &\implies \text{Ravon is given cards 4 and 7.} \ | + | &\implies \text{Ravon is given cards 4 and 7.} && (\bigstar) \ |
&\implies \text{Tyrone is given cards 6 and 10.} \ | &\implies \text{Tyrone is given cards 6 and 10.} \ | ||
&\implies \text{Kim is given cards 8 and 9.} | &\implies \text{Kim is given cards 8 and 9.} | ||
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Therefore, the answer is <math>\boxed{\textbf{(C) }\text{Ravon was given card 4.}}</math> | 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>( | + | Certainly, if we read the answer choices sooner, then we can stop at <math>(\bigstar)</math> and pick <math>\textbf{(C)}.</math> |
− | ~MRENTHUSIASM | + | ~smarty101 ~smartypantsno_3 ~SmileKat32 ~MRENTHUSIASM |
− | == Video Solution by OmegaLearn (Using | + | == Video Solution by OmegaLearn (Using Logical Deduction) == |
https://youtu.be/zO0EuKPXuT0 | https://youtu.be/zO0EuKPXuT0 | ||
Latest revision as of 15:18, 6 November 2021
Contents
[hide]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
By logical deduction, we consider the scores from lowest to highest: Therefore, the answer is
Certainly, if we read the answer choices sooner, then we can stop at and pick
~smarty101 ~smartypantsno_3 ~SmileKat32 ~MRENTHUSIASM
Video Solution by OmegaLearn (Using Logical Deduction)
~ pi_is_3.14
Video Solution by TheBeautyofMath
https://youtu.be/FV9AnyERgJQ?t=284
~IceMatrix
Video Solution by Interstigation
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.