Difference between revisions of "2022 AMC 8 Problems/Problem 6"

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~MRENTHUSIASM
 
~MRENTHUSIASM
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==Video Solution==
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https://youtu.be/Ij9pAy6tQSg?t=409
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~Interstigation
  
 
==See Also==  
 
==See Also==  
 
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{{AMC8 box|year=2022|num-b=5|num-a=7}}
 
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Revision as of 23:36, 18 February 2022

Problem

Three positive integers are equally spaced on a number line. The middle number is $15,$ and the largest number is $4$ times the smallest number. What is the smallest of these three numbers?

$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$

Solution

Let the smallest number be $x.$ It follows that the largest number is $4x.$

Since $x,15,$ and $4x$ are equally spaced on a number line, we have \begin{align*} 4x-15 &= 15-x \\ 5x &= 30 \\ x &= \boxed{\textbf{(C) } 6}. \end{align*} ~MRENTHUSIASM

Video Solution

https://youtu.be/Ij9pAy6tQSg?t=409

~Interstigation

See Also

2022 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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 AJHSME/AMC 8 Problems and Solutions

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