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

Revision as of 18:18, 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 ( $3$ + $4$ = $7$ ). However, since there is only one solution, it holds true that neither of these can be true. We also know that Kyle'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

@@ Line 30: / Line 30: @@
 ==Solution 2 (Elimination)==
-Notice that answer choices <math>\boxed{\textbf{(B)}}</math> and <math>\boxed{\textbf{(D)}}</math> must both be false because if one was true, the other would also be true, but two answer choices can't both be true, so they're out. If Tyrone was given a <math>7</math>, then he must've also gotten a <math>9</math>, but the only possible pairs of cards Kim could get are <math>7</math> and <math>10</math>, or <math>8</math> and <math>9</math>, and both wouldn't work if Tyrone got a <math>7</math> and <math>9</math>, so answer choice <math>\boxed{\textbf{(E)}}</math> is out. We now know that Tyrone got a <math>6</math> and <math>10</math>, and Kim got an <math>8</math> and <math>9</math>, so Ravon couldn't have gotten a <math>3</math> and <math>8</math>, ruling out <math>\boxed{\textbf{(A)}}</math>. Therefore, the only answer choice left is <math>\boxed{\textbf{(C) }\text{Ravon was given card 4.}}</math>
-~NXC
+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 (<math>3</math> + <math>4</math> = <math>7</math>). However, since there is only one solution, it holds true that neither of these can be true. We also know that Kyle'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) ==

2021 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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