Difference between revisions of "2024 AMC 8 Problems/Problem 3"

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     <math>\textbf{(A)}\ 42 \qquad \textbf{(B)}\ 45\qquad \textbf{(C)}\ 49\qquad \textbf{(D)}\ 50\qquad \textbf{(E)}\ 52</math>
 
     <math>\textbf{(A)}\ 42 \qquad \textbf{(B)}\ 45\qquad \textbf{(C)}\ 49\qquad \textbf{(D)}\ 50\qquad \textbf{(E)}\ 52</math>
 
[[2024 AMC 8 Problems/Problem 3|Solution]]
 
  
 
==Solution 1==
 
==Solution 1==

Revision as of 12:27, 26 January 2024

Problem 3

Four squares of side length $4, 7, 9,$ and $10$ are arranged in increasing size order so that their left edges and bottom edges align. The squares alternate in color white-gray-white-gray, respectively, as shown in the figure. What is the area of the visible gray region in square units?

       [asy]         size(150);         filldraw((0,0)--(10,0)--(10,10)--(0,10)--cycle,gray(0.7),linewidth(1));                  filldraw((0,0)--(9,0)--(9,9)--(0,9)--cycle,white,linewidth(1));          filldraw((0,0)--(7,0)--(7,7)--(0,7)--cycle,gray(0.7),linewidth(1));          filldraw((0,0)--(4,0)--(4,4)--(0,4)--cycle,white,linewidth(1));          draw((11,0)--(11,4),linewidth(1));         draw((11,6)--(11,10),linewidth(1));         label("$10$",(11,5),fontsize(14pt));         draw((10.75,0)--(11.25,0),linewidth(1));         draw((10.75,10)--(11.25,10),linewidth(1));          draw((0,11)--(4,11),linewidth(1));         draw((6,11)--(10,11),linewidth(1));         draw((0,11.25)--(0,10.75),linewidth(1));         draw((10,11.25)--(10,10.75),linewidth(1));         label("$9$",(5,11),fontsize(14pt));          draw((-1,0)--(-1,1),linewidth(1));         draw((-1,3)--(-1,7),linewidth(1));         draw((-1.25,0)--(-0.75,0),linewidth(1));         draw((-1.25,7)--(-0.75,7),linewidth(1));         label("$7$",(-1,2),fontsize(14pt));          draw((0,-1)--(1,-1),linewidth(1));         draw((3,-1)--(4,-1),linewidth(1));         draw((0,-1.25)--(0,-.75),linewidth(1));         draw((4,-1.25)--(4,-.75),linewidth(1));         label("$4$",(2,-1),fontsize(14pt));         [/asy]
   $\textbf{(A)}\ 42 \qquad \textbf{(B)}\ 45\qquad \textbf{(C)}\ 49\qquad \textbf{(D)}\ 50\qquad \textbf{(E)}\ 52$

Solution 1

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 2

We convert each of the fractions to $4+2.5+0.04$. After adding up the values, we get $\boxed{6.54}$.

-ILoveMath31415926535

Video Solution 1 (easy to digest) by Power Solve

https://youtu.be/HE7JjZQ6xCk?si=39xd5CKI9nx-7lyV&t=118