Difference between revisions of "2024 AMC 8 Problems/Problem 2"
Countmath1 (talk | contribs) m (→See Also) |
|||
| Line 25: | Line 25: | ||
==Video Solution 1 (easy to digest) by Power Solve== | ==Video Solution 1 (easy to digest) by Power Solve== | ||
https://youtu.be/HE7JjZQ6xCk?si=4I0UO5oOVrC2vJep&t=29 | https://youtu.be/HE7JjZQ6xCk?si=4I0UO5oOVrC2vJep&t=29 | ||
| + | ==Video Solution 2 by SpreadTheMathLove== | ||
| + | https://www.youtube.com/watch?v=L83DxusGkSY | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2024|num-b=1|num-a=3}} | {{AMC8 box|year=2024|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 13:53, 26 January 2024
Contents
Problem
What is the value of the expression in decimal form?
![\[\frac{44}{11}+\frac{110}{44}+\frac{44}{1100}\]](http://latex.artofproblemsolving.com/a/0/2/a0239bc45b6cb760f0c8d1681b07160160efa56c.png)
Solution 1
We see 
, 
, and 
. Thus, 
~MrThinker
Solution 2
We can simplify this expression into 
. Now, taking the common denominator, we get ![\[\frac{200}{50}+\frac{125}{50}+\frac{2}{50}\]](http://latex.artofproblemsolving.com/f/a/f/fafe7fec668c75cd9b9f12bf444aec1e6c167719.png)
![\[= \frac{200+125+2}{50}\]](http://latex.artofproblemsolving.com/a/0/3/a03873103972cbf654a34dfd7bcf97b5db9413e1.png)
![\[= \frac{327}{50}\]](http://latex.artofproblemsolving.com/8/8/b/88b748fc048fd4ff8456c910e852f3e485a9832a.png)
![\[= \frac{654}{100}\]](http://latex.artofproblemsolving.com/b/e/f/bef73b25eb9085283b60990d080efe56451865ec.png)
![\[= \boxed{\textbf{(C) }6.54}\]](http://latex.artofproblemsolving.com/a/2/4/a2435d588ba21f0e8429dc29067aafe587a8ec58.png)
~Dreamer1297
Video Solution 1 (easy to digest) by Power Solve
https://youtu.be/HE7JjZQ6xCk?si=4I0UO5oOVrC2vJep&t=29
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=L83DxusGkSY
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. 
