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Difference between revisions of "2024 AMC 8 Problems/Problem 2"
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-thebanker88
 
-thebanker88
  
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==Video Solution by Math-X (First fully understand the problem!!!)==
 +
https://youtu.be/BaE00H2SHQM?si=7javrSzAbBZ0EKkD&t=287
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~Math-X
  
 
==Video Solution 1 (easy to digest) by Power Solve==
 
==Video Solution 1 (easy to digest) by Power Solve==

Revision as of 07:39, 31 January 2024

Problem

What is the value of the 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 $\frac{44}{11}=4$, $\frac{110}{44}=2.5$, and $\frac{44}{1100}=0.04$. Thus, $4+2.5+0.04=\boxed{\textbf{(C) }6.54}$


~MrThinker

Solution 2

We can simplify this expression into $4+\frac{5}{2}+\frac{1}{25}$. Now, taking the common denominator, we get \[\frac{200}{50}+\frac{125}{50}+\frac{2}{50}\] \[= \frac{200+125+2}{50}\] \[= \frac{327}{50}\] \[= \frac{654}{100}\] \[= \boxed{\textbf{(C) }6.54}\]

~Dreamer1297

Solution 3

Convert all of them into the same demoninator of $1100$. We have $\frac{4400}{1100} + \frac{2750}{1100} + \frac{44}{1100} = \frac{7194}{1100} = \boxed{\textbf{(C) }6.54}$ ~andliu766


Solution 4(fastest)

Use 4400 as the common denominator.

$\frac{17600}{4400} + \frac{11000}{4400} + \frac{176}{4400} = \frac{17600+11000+176}{4400} = \frac{28776}{4400} = \boxed{\textbf{(C) }6.54}$

-thebanker88

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

https://youtu.be/BaE00H2SHQM?si=7javrSzAbBZ0EKkD&t=287 ~Math-X

Video Solution 1 (easy to digest) by Power Solve

https://youtu.be/HE7JjZQ6xCk?si=4I0UO5oOVrC2vJep&t=29

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=Ylw-kJkSpq8

~NiuniuMaths

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=L83DxusGkSY

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=NLoyzNyvFKU

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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