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Difference between revisions of "2024 AMC 8 Problems/Problem 2"

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==Solution 1==  
 
==Solution 1==  
  
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We see that 44/11 is 4;  
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We see that <math>\frac{44}{11}</math> is <math>4</math>;  
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110/44 simplifies to 5/2, which is 2.5;  
+
<math>\frac{110}{44}</math> simplifies to <math>\frac{5}{2}</math>, which is <math>2.5</math>;  
  
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and 44/1100 simplifies to 1/25, which is 0.04;  
+
and <math>\frac{44}{1100}</math> simplifies to <math>\frac{1}{25}</math>, which is <math>0.04</math>;  
  
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4+2.5+0.04 reveals <cmath>\frac{44}{11} + \frac{110}{44} + \frac{44}{1100}</cmath> is <math>\boxed{\textbf{(C)\ 6.54}}</math>.  
+
<math>4+2.5+0.04</math> reveals <cmath>\frac{44}{11} + \frac{110}{44} + \frac{44}{1100}</cmath> is <math>\boxed{\textbf{(C)\ 6.54}}</math>.  
 
~ le_petit_chouetteur from TSMV
 
~ le_petit_chouetteur from TSMV
 +
 +
~Miner Formatting by GreenPlanet2050
  
 
==Video Solution 1 (Quick and Easy!)==
 
==Video Solution 1 (Quick and Easy!)==

Revision as of 19:46, 4 April 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 $\boxed{\textbf{(C)\ 6.54}}$. ~ le_petit_chouetteur from TSMV

~Miner Formatting by GreenPlanet2050

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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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