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Difference between revisions of "2022 AMC 8 Problems/Problem 5"

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~MRENTHUSIASM
 
~MRENTHUSIASM
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==Video Solution  by Math-X (First understand the problem!!!)==
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https://youtu.be/oUEa7AjMF2A?si=AeqPj9sQgiBU3_lL&t=585
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~Math-X
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==Video Solution  (CREATIVE THINKING!!!)==
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https://youtu.be/NuRFgRUpV7o
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~Education, the Study of Everything
  
 
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==Video Solution==
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~STEMbreezy
 
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==Video Solution==
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https://youtu.be/Cb5as-fLejk
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~harungurcan
  
 
==See Also==  
 
==See Also==  
 
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{{AMC8 box|year=2022|num-b=4|num-a=6}}
 
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Latest revision as of 02:35, 25 September 2024

Problem

Anna and Bella are celebrating their birthdays together. Five years ago, when Bella turned $6$ years old, she received a newborn kitten as a birthday present. Today the sum of the ages of the two children and the kitten is $30$ years. How many years older than Bella is Anna?

$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } ~5$

Solution

Five years ago, Bella was $6$ years old, and the kitten was $0$ years old.

Today, Bella is $11$ years old, and the kitten is $5$ years old. It follows that Anna is $30-11-5=14$ years old.

Therefore, Anna is $14-11=\boxed{\textbf{(C) } 3}$ years older than Bella.

~MRENTHUSIASM

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/oUEa7AjMF2A?si=AeqPj9sQgiBU3_lL&t=585

~Math-X

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/NuRFgRUpV7o

~Education, the Study of Everything

Video Solution

https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=323

~Interstigation

Video Solution

https://youtu.be/WY0m0jAj__Y

~savannahsolver

Video Solution

https://youtu.be/Q0R6dnIO95Y?t=212

~STEMbreezy

Video Solution

https://youtu.be/Cb5as-fLejk

~harungurcan

See Also

2022 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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

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

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