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Difference between revisions of "2024 AMC 8 Problems/Problem 2"
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~Minor Formatting by GreenPlanet2050
 
~Minor Formatting by GreenPlanet2050
  
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==Video Solution by Central Valley Math Circle(Goes Through Full Thought Process)==
 +
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https://youtu.be/LoRrSfYvM9g
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~mr_mathman
 
==Video Solution by Math-X (MATH-X)==
 
==Video Solution by Math-X (MATH-X)==
 
https://youtu.be/BaE00H2SHQM?si=noTBPTosCtH31CpW&t=287
 
https://youtu.be/BaE00H2SHQM?si=noTBPTosCtH31CpW&t=287

Latest revision as of 01:32, 7 December 2024

Problem 2

What is the value of this expression in decimal form? \[\frac{44}{11} + \frac{110}{44} + \frac{44}{1100}\] $\textbf{(A) } 6.4\qquad\textbf{(B) } 6.504\qquad\textbf{(C) } 6.54\qquad\textbf{(D) } 6.9\qquad\textbf{(E) } 6.94$

Solution 1

We see that $\frac{44}{11}$ is $4$; $\frac{110}{44}$ simplifies to $\frac{5}{2}$, which is $2.5$;

and $\frac{44}{1100}$ simplifies to $\frac{1}{25}$, which is $0.04$;

$4+2.5+0.04$ reveals \[\frac{44}{11} + \frac{110}{44} + \frac{44}{1100}\] is $\boxed{\text{(C)\ 6.54}}$. ~

~Minor Formatting by GreenPlanet2050

Video Solution by Central Valley Math Circle(Goes Through Full Thought Process)

https://youtu.be/LoRrSfYvM9g

~mr_mathman

Video Solution by Math-X (MATH-X)

https://youtu.be/BaE00H2SHQM?si=noTBPTosCtH31CpW&t=287

~Math-X

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/5ZIFnqymdDQ?si=-FCGnA5WXQNp-JMF&t=175

~hsnacademy

Video Solution 1 (Quick and Easy!)

https://youtu.be/nzPT89ymlKk

~Education, the Study of Everything

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

https://youtu.be/BaE00H2SHQM?si=noTBPTosCtH31CpW&t=287

~Math-X

Video Solution by Daily Dose of Math (Understandable, Speedy, and Easy)

https://youtu.be/bSPWqeNO11M?si=HIzlxPjMfvGM5lxR

~Thesmartgreekmathdude

Video Solution by WhyMath

https://youtu.be/MasK0Z6wxR4

See Also

2024 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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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